[ { "id": 495, "tag": "OPTICS", "content": "In 1845, Faraday studied the influence of electricity and magnetism on polarized light and discovered that heavy glass, originally non-optically active, exhibited optical activity under the influence of a strong magnetic field, causing the plane of polarization of polarized light to rotate. This was the first time humanity recognized the relationship between electromagnetic phenomena and light phenomena. The magneto-optical rotation effect later became known as the Faraday Effect.\n\nIn the magneto-optical rotation effect, the rotation angle of the vibration plane \\(\\beta\\) is proportional to the distance \\(d\\) that light travels through the medium and is also proportional to the magnetic induction intensity \\(B\\) within the medium, expressed as:\n\\[\n\\beta = vBd\n\\]\nwhere \\(v\\) is the proportionality constant, known as the Verdet constant, which depends on the properties of the medium and is also related to the wavelength of the incident light. Assume that the medium can be modeled using the Lorentz model, where the medium is composed of atoms, with \\(N\\) atoms per unit volume, and each atom contains \\(Z\\) electrons in the outer shell. The outer electrons deviate from their equilibrium positions and are subject to linear restoring forces: \\(\\vec{f} = -k\\vec{r}\\), with the harmonic resonance frequency of the outer shell electrons being \\(\\omega_0\\). The interaction between the incident light and the medium exclusively involves the outer shell electrons.\n\nCircularly polarized light can be regarded as the result of the synthesis of two linearly polarized light waves with the same frequency and direction, mutually perpendicular vibration axes, and a stable phase relationship. The two linearly polarized light waves have the same amplitude and a phase difference of \\(\\pm\\pi/2\\), representing left-handed and right-handed circularly polarized light, respectively. Similarly, a linearly polarized light wave can be decomposed into left-handed and right-handed circularly polarized components.\n\nBased on these models, calculate the Verdet constant \\(\\beta\\) for incident light with an angular frequency of \\(\\omega\\). Assume that the medium does not absorb the incident light and its refractive index near the angular frequency \\(\\omega\\) is \\(n\\).", "solution": "Solution: The propagation of light waves in a medium under an applied magnetic field involves the motion of electrons:\n\\[m\\ddot{\\vec{r}} = -k\\vec{r} - g\\vec{v} - e(\\vec{E} + \\vec{v} \\times \\vec{B})\\]\n\\(\\vec{E}\\) is the electric field intensity of the incident light wave; \\(\\vec{B}\\) is the magnetic induction intensity applied to the medium; \\(g\\) is the damping term, and since the medium does not absorb the incident light, it exhibits weak damping and low loss, thus neglecting the damping term:\n\\[m\\ddot{\\vec{r}} = -k\\vec{r} - e(\\vec{E} + \\vec{v} \\times \\vec{B})\\]\nLet \\(\\vec{B}\\) be along the positive \\(z\\)-axis, which is also the propagation direction of the light wave, thus:\n\\[\n\\begin{cases}\nm\\ddot{x} + e\\dot{y}B + kx = -eE_x \\\\\nm\\ddot{y} - e\\dot{x}B + ky = -eE_y\n\\end{cases}\n\\]\nAssuming the incident wave is a steady-state wave, the relationship between the electric field and time: \\(e^{-i\\omega t}\\)\nThe relationship between the perturbation term of the forced vibration of the outer electrons and time: \\(e^{-i\\omega t}\\)\nThus: \\(\\frac{\\partial}{\\partial t} \\equiv -i\\omega\\), \\(\\frac{\\partial^2}{\\partial t^2} \\equiv -\\omega^2\\), therefore:\n\\[\n\\begin{cases}\n(\\omega_0^2 - \\omega^2)r_x - i\\omega\\Omega r_y = -\\frac{e}{m}E_x \\\\\n(\\omega_0^2 - \\omega^2)r_y + i\\omega\\Omega r_x = -\\frac{e}{m}E_y\n\\end{cases}\n\\]\nWhere:\n\\(\\Omega = \\frac{eB}{m}\\), \\(\\omega_0 = \\sqrt{\\frac{k}{m}}\\)\n\nFor left-hand circularly polarized light, \\(E_+ = E_x + iE_y\\), then: \\(r_+ = r_x + ir_y\\)\n\\[(\\omega_0^2 - \\omega^2 - \\omega\\Omega)r_+ = -\\frac{e}{m}E_+\\]\nSolving this, we get:\n\\[r_+ = -\\frac{e}{m(\\omega_0^2 - \\omega^2 - \\omega\\Omega)}E_+\\]\nThe polarization vector: The atomic number density of the medium is \\(N(1/m^3)\\), with each atom providing \\(Z\\) electrons weakly bound in the outer layer;\n\\[\\vec{P}_+ = NZ(-e)\\vec{r}_+ = \\frac{NZ\\frac{e^2}{m}}{\\omega_0^2 - \\omega^2 - \\omega\\Omega}E_+\\]\nSince \\(P = \\varepsilon_0\\chi E\\), thus:\n\\[\\chi_+ = \\frac{NZ\\frac{e^2}{\\varepsilon_0 m}}{\\omega_0^2 - \\omega^2 - \\omega\\Omega}\\]\nThe refractive index for left-hand circularly polarized light is:\n\\[n_+^2 = \\varepsilon_+ = (1 + \\chi_+) = 1 + \\frac{NZ\\frac{e^2}{\\varepsilon_0 m}}{\\omega_0^2 - \\omega^2 - \\omega\\Omega}\\]\n\nFor right-hand circularly polarized light \\(E_- = E_x - iE_y\\), then: \\(r_- = r_x - ir_y\\)\n\\[(\\omega_0^2 - \\omega^2 + \\omega\\Omega)r_- = -\\frac{e}{m}E_-\\]\nSolving this, we get:\n\\[r_- = -\\frac{e}{m(\\omega_0^2 - \\omega^2 + \\omega\\Omega)}E_-\\]\n\\[\\vec{P}_- = NZ(-e)\\vec{r}_- = \\frac{NZ\\frac{e^2}{m}}{\\omega_0^2 - \\omega^2 + \\omega\\Omega}E_-\\]\nThus:\n\\[\\chi_- = \\frac{NZ\\frac{e^2}{\\varepsilon_0 m}}{\\omega_0^2 - \\omega^2 + \\omega\\Omega}\\]\nThe refractive index for right-hand circularly polarized light is:\n\\[n_-^2 = \\varepsilon_- = (1 + \\chi_-) = 1 + \\frac{NZ\\frac{e^2}{\\varepsilon_0 m}}{\\omega_0^2 - \\omega^2 + \\omega\\Omega}\\]\nThat is, the refractive indices for left-hand and right-hand circularly polarized light are different:\n\\[n_{\\pm}^2 = \\varepsilon_{\\pm} = (1 + \\chi_{\\pm}) = 1 + \\frac{NZ\\frac{e^2}{\\varepsilon_0 m}}{\\omega_0^2 - \\omega^2 \\mp \\omega\\Omega}\\]\n\nAny linearly polarized light can be decomposed into two circularly polarized lights with the same frequency, equal amplitude, and a stable phase relationship. When linearly polarized light passes through a medium with an external magnetic field aligned with the direction of propagation, the rotation angle of the polarization plane is:\n\\[\\beta = \\frac{1}{2}(\\alpha_+ - \\alpha_-) = \\frac{\\pi}{\\lambda}(n_+ - n_-)d\\]\nwhere \\(d\\) is the propagation length of light in the medium.\n\nFind \\((n_+ - n_-)\\)\n\\[\n\\begin{align*}\nn_+^2 - n_-^2 &= (n_+ - n_-)(n_+ + n_-)\\\\\n&= NZ\\frac{e^2}{\\varepsilon_0 m} \\cdot \\frac{2\\omega\\Omega}{(\\omega_0^2 - \\omega^2)^2 - (\\omega\\Omega)^2}\n\\end{align*}\n\\]\nGenerally, the cyclotron frequency \\(\\Omega\\) is much smaller than the optical frequency \\(\\omega\\), and the optical frequency \\(\\omega\\) is not near the resonance frequency \\(\\omega_0\\), hence:\n\\[n_+ + n_- \\approx 2n\\]\n\\[n^2 \\approx 1 + NZ\\frac{e^2}{\\varepsilon_0 m} \\cdot \\frac{1}{\\omega_0^2 - \\omega^2}\\]\nTherefore:\n\\[n_+ - n_- = NZ\\frac{e^2}{\\varepsilon_0 mn} \\cdot \\frac{\\omega\\Omega}{(\\omega_0^2 - \\omega^2)^2}\\]\nThus:\n\\[\n\\begin{align*}\n\\beta &= \\frac{\\pi}{\\lambda}(n_+ - n_-)d = \\frac{\\omega}{2c}NZ\\frac{e^2}{\\varepsilon_0 mn} \\cdot \\frac{\\omega\\Omega}{(\\omega_0^2 - \\omega^2)^2}Bd = \\frac{NZ e^3}{2c\\varepsilon_0 m^2 n} \\cdot \\frac{\\omega^2}{(\\omega_0^2 - \\omega^2)^2} \\cdot Bd\\\\\n&= vBd\n\\end{align*}\n\\]\nThe Verdet constant is obtained as:\n\\[v = \\frac{NZ e^3}{2c\\varepsilon_0 m^2 n} \\cdot \\frac{\\omega^2}{(\\omega_0^2 - \\omega^2)^2}\\]", "answer": "\\[v = \\frac{NZ e^3}{2c\\varepsilon_0 m^2 n} \\cdot \\frac{\\omega^2}{(\\omega_0^2 - \\omega^2)^2}\\] " }, { "id": 600, "tag": "OPTICS", "content": "The Yang's double slit interference experiment consists of three parts: light source, double slit, and receiving screen. In the experiment, we generated a line light source by placing a point light source behind the slit $S_0$, and the light was then projected onto the receiving screen through the double slits $S_1$ and $S_2$, forming interference fringes. Now we will perform the following actions on this device:\nPlace polarizer $P_0$ next to slit $S_0$, and polarizers $P_1$ and $P_2$ next to slits $S_1$ and $S_2$, respectively, so that the transmission direction of $P_0$ is parallel to $P_1$ and forms a $\\theta=60\\degree$ angle with $P_2$.\nTry to determine the contrast of the stripes on the receiving screen at this time.", "solution": "The magnitude of electric field intensity is set to A after passing through polarizer P0.\n\nThe optical amplitude through P1 is A, and the optical amplitude through P2 is $\\ frac 12 $A.\nLet the direction parallel to P0 be the x direction and the direction perpendicular to it be the y direction.\nstandard\n$$\nE_{1x}=A,E_{2x}=\\frac{1}{4} A,E_{2y}=\\frac {\\sqrt{3}}{4} A\n$$\n$E2 {1x} $and $E2 {2x} $are coherently superimposed, with a total light intensity of\n$$\nI=I_{0}\\left(\\frac 54 +\\frac 12 \\cos\\delta\\right)\n$$\nThe lining ratio is\n$$\n\\gamma=\\frac 25\n$$", "answer": "$$\\gamma = \\frac{2}{5}$$" }, { "id": 285, "tag": "MECHANICS", "content": "A small bug with a mass of $m$ crawls on a disk with a radius of $2R$. Relative to the disk, its crawling trajectory is a circle of radius $R$ that passes through the center of the disk. The disk rotates with a constant angular velocity $\\omega$ about an axis passing through its center and perpendicular to the plane of the disk. The bug's angular crawling velocity relative to the disk is in the same direction as the disk's angular velocity and has the same magnitude. Solve for the maximum force $F_{max}$ between the bug and the disk required to maintain this motion (neglecting gravity).", "solution": "The problem can use the acceleration transformation formula for rotating reference frames, where the actual acceleration of the small insect \n\n$$\n\\overrightarrow{a}=\\overrightarrow{a}+2\\overrightarrow{\\omega}\\times\\overrightarrow{v}+\\dot{\\overrightarrow{\\omega}}\\times\\overrightarrow{r}-\\omega^{2}\\overrightarrow{r}\n$$\n\nHere, $\\vec{a}^{\\prime}$ is the acceleration relative to the rotating system, and since the rotating system is a uniform circular motion, \n\n$$\n\\dot{a_{\\mathrm{\\scriptsize~=~}}}\\omega^{2}R\n$$\n\nis the speed relative to the rotating system, with a magnitude of $\\omega R$. Thus, the magnitude of the Coriolis acceleration is: \n\n$$\n2\\omega v^{'}=2\\omega^{2}R\n$$\n\nThe last term is the centripetal acceleration, which is directly proportional to the distance from the insect to the center of the disk, \n\n$$\n\\omega^{2}r=2\\omega^{2}R\\cos\\left(\\frac{\\theta}{2}\\right)\n$$\n\n(Analysis of each acceleration, 7 points) \n\nThese three acceleration components are combined to derive the total acceleration magnitude as: \n\n$$\na=\\omega^{2}R{\\sqrt{\\left(3+2\\cos\\left({\\frac{\\theta}{2}}\\right)^{2}\\right)^{2}+4\\cos\\left({\\frac{\\theta}{2}}\\right)^{2}\\sin\\left({\\frac{\\theta}{2}}\\right)^{2}}}=\\omega^{2}R{\\sqrt{9+16\\cos\\left({\\frac{\\theta}{2}}\\right)^{2}}}\n$$\n\nwhich is a constant, so the maximum force is \n\n$$\nF_{m a x}=5m\\omega^{2}R\n$$", "answer": "$$F_{\\max} = 5m\\omega^2R$$" }, { "id": 321, "tag": "MECHANICS", "content": "The civilization of *Three-Body* changes the values of fundamental physical constants, such as the Planck constant, inside the \"water drop\" to alter the range of strong forces. \nWhat kind of effects would this produce? This problem provides a suitable discussion about this question. \n\nHideki Yukawa pointed out that the propagator of nuclear forces is the $\\pi$ meson. One can equivalently simplify the effect by considering that protons and protons, protons and neutrons, and neutrons and neutrons have an identical strong-force potential energy. For a force field with a propagator of mass, Yukawa potential and its range can be expressed as: \n\n$$\nU = -\\frac{A\\mathrm{e}^{-\\frac{2\\pi m c r}{h}}}{r}\\quad,\\quad\\lambda = \\frac{h}{2\\pi m c}\n$$ \n\nFor the deuteron (the strong-force binding between a proton and a neutron), consider two particles rapidly rotating around the center of mass. If the distance between the proton and the neutron is precisely $r$ and they do not collide, find the angle rotated around the center of mass during one complete cycle of small radial oscillation. Express the answer in terms of $\\lambda$ and $r$.", "solution": "$$\nF=-\\frac{A(1+\\frac{2\\pi m c r}{h})e^{-\\frac{2\\pi m c r}{h}}}{r^{2}}\n$$ \n\nNewton's Law \n\n$$\nm\\omega^{2}\\frac{r}{2}=\\frac{A(1+\\frac{2\\pi m c r}{h})e^{-\\frac{2\\pi m c r}{h}}}{r^{2}}\n$$ \n\nEffective Potential Energy \n\n$$\nV_{eff}=-\\frac{A e^{-\\frac{2\\pi m c r}{h}}}{r}+\\frac{L^{2}}{2\\cdot m/2\\cdot r^{2}}\n$$ \n\nSolving the Second Derivative \n\n$$\n\\frac{d^{2}V_{eff}}{d r^{2}}=\\frac{A e^{-\\frac{r}{\\lambda}}}{r^{3}}\\left(1+\\frac{r}{\\lambda}-\\frac{r^{2}}{\\lambda^{2}}\\right)\n$$ \n\nAngular Frequency \n\n$$\n\\omega_{0}=\\sqrt{\\frac{2\\frac{d^{2}V_{eff}}{d r^{2}}}{m}}\n$$ \n\nGeometric Relation \n\n$$\n\\varphi=2\\pi\\cdot\\frac{\\omega}{\\omega_{0}}\n$$ \n\nResulting \n$$\n\\varphi=2\\pi\\sqrt{\\frac{1+r/\\lambda}{1+r/\\lambda-r^2/\\lambda^2}}\n$$", "answer": "$$\\varphi = 2\\pi \\sqrt{\\frac{1 + \\frac{r}{\\lambda}}{1 + \\frac{r}{\\lambda} - \\frac{r^2}{\\lambda^2}}}$$" }, { "id": 674, "tag": "MECHANICS", "content": "A coin is placed at rest on the edge of a smooth tabletop, with only a small portion of its right side extending beyond the edge. The coin can be considered as a uniform disk with mass $m$, radius $r$, and gravitational acceleration $g$. A vertical impulse $I$ is applied to the right side of the coin. During its subsequent motion after the impact, the coin might fly off the table at some point. Determine the minimum initial velocity of the center of mass, $v_{0, \\min}$, required for the coin to leave the table, expressed in terms of $g$ and $r$.\",\n", "solution": "The moment of inertia of the coin about the axis is \\[J=\\frac14mr^2\\] According to the impulse and impulse-momentum theorem, about the center of mass we have \\[I+I_N=m v_0, Ir-I_Nr=J\\omega_0\\] There is also the velocity relation \\[v_0=\\omega_0 r\\] Therefore, we have \\[v_0=\\frac{8I}{5m}, \\omega_0=\\frac{8I}{5mr}\\] After being struck, assume the angle between the coin and the horizontal direction is \\(\\phi\\). Because the tabletop is smooth, the horizontal coordinate of the center of mass remains unchanged, so \\[\\vec r_c=(0,r\\sin\\phi)\\] Thus, we have \\[E_k=\\frac 12m\\dot y_c^2+\\frac12 J\\dot\\phi ^2=\\frac12mr^2\\left(\\cos^2\\phi+\\frac 14\\right)\\dot\\phi^2\\] \\[E_p=mgy_c=mgr\\sin\\phi\\] Thus, mechanical energy is conserved \\[\\frac 18mr^2(4\\cos^2\\phi+1)\\dot\\phi^2+mgr\\sin\\phi=\\frac{8I^2}{5m}\\] From this, we can obtain \\[\\dot\\phi^2=\\frac{64I^2-40m^2gr\\sin\\phi}{5m^2r^2(4\\cos^2\\phi+1)}\\] Since \\(\\frac{\\mathrm d\\dot\\phi^2}{\\mathrm d\\phi}=2\\ddot\\phi\\), we have \\[\\ddot\\phi = \\frac{4\\cos\\phi}{5r^2(4\\cos^2\\phi+1)^2}\\left(64\\frac{I^2}{m^2}\\sin\\phi-5gr(5+4\\sin^2\\phi)\\right)\\] Hence, we can obtain \\[\\begin{align*}\\ddot y_c&=r(\\ddot\\phi\\cos\\phi-\\dot\\phi^2\\sin\\phi)\\\\ &=\\frac{4\\cos^2\\phi}{5r(4\\cos^2\\phi+1)^2}\\left(64\\frac{I^2}{m^2}\\sin\\phi-5gr(5+4\\sin^2\\phi)\\right)-\\frac{(64 I^2-40 m^2gr\\sin\\phi)\\sin\\phi}{5m^2r(4\\cos^2\\phi+1)}\\end{align*}\\] When there is no support force from the tabletop, the coin leaves the table, at that moment \\[m\\ddot y_c=-mg\\] At this point \\(\\phi=\\theta\\), \\[20\\sin^2\\theta-64\\frac{I^2}{m^2gr}\\sin\\theta+25=0\\] It can be seen that the solution for \\(\\sin\\theta\\) can take all values in \\((0,1]\\), which means for it to jump, we have \\[I=m\\sqrt{gr}\\left(\\sqrt{\\frac{20\\sin^2\\theta+25}{64\\sin\\theta}}\\right)_{\\min}=\\frac{3\\sqrt5}{8}m\\sqrt{gr}\\] Thus we have \\[v_{0,\\min}=\\frac{3\\sqrt 5}{5}\\sqrt{gr}\\]\n", "answer": "$$\\frac{3\\sqrt{5}}{5}\\sqrt{gr}$$" }, { "id": 83, "tag": "ELECTRICITY", "content": "\n\n---\n\nA concentric spherical shell with inner and outer radii of $R_{1}=R$ and $R_{2}=2^{1/3}R$, and magnetic permeability $\\mu=3\\mu_{0}$, is placed in an external uniform magnetic field with magnetic flux density $B_{0}$ and the magnetic field generated by a fixed magnetic dipole located at the center of the sphere. The magnetic dipole has a magnetic moment $m$ that is aligned with the external magnetic field. The system reaches magnetostatic equilibrium. To solve this problem, we can use the method of magnetic scalar potential. The concept is to express the magnetic field intensity as ${\\pmb{H}}=-\\nabla\\varphi$, where $\\varphi$ is referred to as the magnetic scalar potential. This approach applies to all three regions: the space inside the shell, the space outside the shell, and the region within the shell itself. The relationship between magnetic flux density and magnetic field intensity is $B=\\mu H$.\n\nIf the magnetic moment is rotated by an angle $\\alpha$, calculate the torque it experiences. The answer should be expressed in terms of $m$, $B_0$, and $\\alpha$. Please verify the result before outputting.\n\n---", "solution": "(1) \n\nLet: \n\n$$\n\\varphi={\\left\\{\\begin{array}{l l}{A_{1}{\\frac{r}{R}}\\cos\\theta+B_{1}{\\frac{R^{2}}{r^{2}}}\\cos\\theta}&{(rk R)}\\end{array}\\right.}\n$$ \n(Tangential) continuity $(k=2^{1/3})$ \n$$\n\\begin{array}{c}{{A_{1}+B_{1}=A_{2}+B_{2}}}\\ {{\\qquad}}\\ {{k A_{2}+k^{-2}B_{2}=k A_{3}+k^{-2}B_{3}}}\\end{array}\n$$ \n\nMagnetic induction normal continuity: \n\n$$\n\\mu_{0}A_{1}-2\\mu_{0}B_{1}=\\mu A_{2}-2\\mu B_{2}\n$$ \n\n$$\nk\\mu A_{2}-2k^{-2}\\mu B_{2}=k\\mu_{0}A_{3}-2k^{-2}\\mu_{0}B_{3}\n$$ \n\nAccording to the boundary condition at infinity: \n\n$$\nA_{3}=-\\frac{B_{0}R}{\\mu_{0}}\n$$ \n\nAccording to the boundary condition at the magnetic dipole: \n\n$$\nB_{1}=\\frac{m}{4\\pi R^{2}}\n$$ \n\nThe solution is: \n\n$$\n\\begin{array}{c}{{A_{2}=\\displaystyle\\frac{21}{31}A_{3}+\\displaystyle\\frac{6}{31}B_{1}}}\\ {{{}}}\\ {{B_{2}=\\displaystyle\\frac{6}{31}A_{3}+\\displaystyle\\frac{15}{31}B_{1}}}\\ {{{}}}\\ {{A_{1}=\\displaystyle\\frac{27}{31}A_{3}-\\displaystyle\\frac{10}{31}B_{1}}}\\ {{{}}}\\ {{B_{3}=\\displaystyle-\\frac{14}{31}A_{3}+\\displaystyle\\frac{27}{31}B_{1}}}\\end{array}\n$$ \n\nSubstituting: \n\n$$\n\\varphi=\\left\\{\\begin{array}{l l}{\\left(-\\frac{27}{31}\\frac{B_{0}R}{\\mu_{0}}-\\frac{10}{31}\\frac{m}{4\\pi R^{2}}\\right)\\frac{r}{R}\\cos\\theta+\\frac{m}{4\\pi R^{2}}\\frac{R^{2}}{r^{2}}\\cos\\theta}&{(rk R)}\\end{array}\\right.\n$$ \n\n(2)\nMagnetic field induced by external magnetic field: \n\n$$\nB^{\\prime}=\\frac{27}{31}B_{0}\n$$ \n\nThus, torque: \n\n$$\n{\\cal M}=\\underline{{{m\\times B}}}\n$$ \n\nResult: \n\n$$\nM=-{\\frac{27}{31}}m B_{0}\\sin\\alpha\n$$", "answer": "$$ M = -\\frac{27}{31} m B_0 \\sin \\alpha $$" }, { "id": 54, "tag": "MECHANICS", "content": "Three small balls are connected in series with three light strings to form a line, and the end of one of the strings is hung from the ceiling. The strings are non-extensible, with a length of $l$, and the mass of each small ball is $m$.\n\nInitially, the system is stationary and vertical. A hammer strikes one of the small balls in a horizontal direction, causing the ball to acquire an instantaneous velocity of $v_0$. Determine the instantaneous tension in the middle string when the topmost ball is struck. (The gravitational acceleration is $g$.)", "solution": "Strike the top ball, the acceleration of the top ball is:\n$$a=\\frac{v_0^2}{l}$$\n\nSwitching to the reference frame of the top ball, let the tensions from top to bottom be $T_1, T_2, T_3$. Applying Newton's second law to the middle ball:\n$$T_2-T_3-mg-m\\frac{v_0^2}{l}=m\\frac{v_0^2}{l}$$\n\nThe acceleration of the bottom ball is: $$a=2\\frac{v_0^2}{l}$$\n\nApplying Newton's second law to the bottom ball: $$T_3-mg=2m\\frac{v_0^2}{l}$$\n\nThus, we have: $$T_2=2mg+4m\\frac{v_0^2}{l}$$", "answer": "$$T_2 = 2mg + 4m \\frac{v_0^2}{l}$$" }, { "id": 10, "tag": "ELECTRICITY", "content": "The region in space where $x > 0$ and $y > 0$ is a vacuum, while the remaining region is a conductor. The surfaces of the conductor are the $xOz$ plane and the $yOz$ plane. A point charge $q$ is fixed at the point $(a, b, c)$ in the vacuum, and the system has reached electrostatic equilibrium. Find the magnitude of electric field intensity on the surface of the conductor at the $xOz$ plane, ${E}(x, +0, z)$.", "solution": "The infinite conductor and the point at infinity are equipotential, $U{=}0$, with the boundary condition that the tangential electric field $E_{\\tau}{=}0$. By the uniqueness theorem of electrostatics, the induced charges on the conductor can be equivalently replaced by image charges to determine the contribution to the electric potential in the external space of the conductor. The top view of the image charges is shown below.\n\nTherefore, the answer is\n\\(\\vec{E}\\left(x,+0,z\\right)=\\left(\\frac{-2q}{4\\pi\\varepsilon_{0}}\\frac{b}{r_{1}^{3}}{+}\\frac{2q}{4\\pi\\varepsilon_{0}}\\frac{b}{r_{2}^{3}}\\right)\\hat{y}=\\frac{q b}{2\\pi\\varepsilon_{0}}\\biggl[\\bigl[(a+x)^{2}+(z-c)^{2}+b^{2}\\bigr]^{\\frac{-3}{2}}-\\bigl[(x-a)^{2}+(z-c)^{2}+b^{2}\\bigr]^{\\frac{-3}{2}}\\biggr]\\hat{y},\\) \\(\\vec{E}\\left(+0,y,z\\right)=(-\\frac{2q}{4\\pi\\varepsilon_{0}}\\frac{a}{r_{1}^{3}}+\\frac{2q}{4\\pi\\varepsilon_{0}}\\frac{a}{r_{2}^{3}})\\hat{x}=\\frac{q a}{2\\pi\\varepsilon_{0}}\\Bigg[\\Big[(y+b)^{2}+\\big(z-c\\big)^{2}+a^{2}\\Big]^{\\frac{-3}{2}}-\\Big[\\big(y-b\\big)^{2}+\\big(z-c\\big)^{2}+a^{2}\\Big]^{\\frac{-3}{2}}\\Bigg]\\hat{x}\\)", "answer": "$$\\frac{q b}{2\\pi \\varepsilon_0} \\left[ ((a+x)^2 + (z-c)^2 + b^2)^{-3/2} - ((x-a)^2 + (z-c)^2 + b^2)^{-3/2} \\right]$$" }, { "id": 33, "tag": "THERMODYNAMICS", "content": "Consider an infinite-length black body with inner and outer cylinders, which are in contact with heat sources at temperatures $T_{1}$ and $T_{2}$, respectively; assume that the temperature of the heat sources remains constant. Let the inner cylinder have a radius $r$, the outer cylinder have a radius $R$, and the distance between the axes of the inner and outer cylinders be $b$, with $r < b < R$ and $r + b < R$. Find the power $p(\\theta)$ absorbed per unit area from the heat source at angle $\\theta$ on the surface of the outer cylinder (i.e., the power density at $\\theta$), where $\\theta$ is the angle between the line connecting a point on the surface of the outer cylinder and the center of the outer cylinder, and the line connecting the centers of the inner and outer cylinders. The Stefan-Boltzmann constant is denoted as $\\sigma$.", "solution": "In the case of non-coaxial, we examine the long strip-shaped area element at the outer wall reaching $\\theta + \\mathrm{d} \\theta$:\n\nThe heat flux projected onto the inner axis is:\n\n$$\n\\frac{d S}{2}\\sigma T_{2}^{4}\\int_{\\mathrm{arcsin}\\left({\\frac{b \\sin\\theta}{D}}\\right)-\\mathrm{arcsin}({\\frac{r}{D}})}^{\\mathrm{arcsin}\\left({\\frac{b \\sin\\theta}{D}}\\right)+\\mathrm{arcsin}({\\frac{r}{D}})}\\cos\\theta \\, d\\theta\n$$\n\nThe above equation equals $\\sigma T_{2}^{4}\\frac{r(R-b \\cos \\theta)}{R^{2}+b^{2}-2R b \\cos \\theta}\\mathrm{d}S$, where dS is the area of the strip-shaped area element. According to the zeroth law of thermodynamics, the heat flux sent from the inner cylinder to the strip-shaped area element at $\\theta + \\mathrm{d} \\theta$ on the outer cylinder is $\\begin{array}{r}{\\sigma T_{1}^{4}\\frac{r(R-b \\cos\\theta)}{R^{2}+b^{2}-2R b \\cos\\theta}\\mathrm{d} S}\\end{array}$: Similarly, according to the zeroth law of thermodynamics,\n\nIt is known that the heat flux sent from other places on the outer cylinder to this area element is\n\n$$\n\\begin{array}{r}{(\\sigma T_{2}^{4}-\\sigma T_{2}^{4}\\frac{r(R-b}{R^{2}+b^{2}-2R b \\cos \\theta})\\mathrm{d} S}\\end{array}\n$$\n\nIt can be seen that the heat absorbed by the unit area element at $\\theta$ from the heat source with temperature $T_{2}$ is equal to\n\n$$\n\\begin{array}{r}{(\\sigma T_{2}^{4}-\\sigma T_{1}^{4})\\frac{r(R-b)}{R^{2}+b^{2}-2R b \\cos \\theta}}\\end{array}\n$$", "answer": "$$\np(\\theta) = (\\sigma T_2^4 - \\sigma T_1^4) \\frac{r(R - b \\cos \\theta)}{R^2 + b^2 - 2Rb \\cos \\theta}\n$$" }, { "id": 41, "tag": "MECHANICS", "content": "The rocket ascends vertically from the Earth's surface with a constant acceleration $a = k_1g_0$, where $g_0$ is the gravitational acceleration at the Earth's surface. Inside the rocket, there is a smooth groove containing a testing apparatus. When the rocket reaches a height $h$ above the ground, the pressure exerted by the instrument on the bottom of the groove is $k_2$ times what it was at the time of liftoff. Given that the radius of the Earth is $R$, find $h$.", "solution": "At the moment of takeoff, $N_1=(1+k_1)mg_0$, where $m$ is the mass of the tester. \nWhen ascending to a height $h$, the gravitational acceleration changes to $g=g_0(\\dfrac{R}{R+h})^2$ \nThe pressure at this time is $N_2=mg_0(\\dfrac{R}{R+h})^2+k_1mg_0$ \nSince $N_2=k_2N_1$, solving the equations together gives\n$h=R(\\dfrac{1}{\\sqrt{k_2(1+k_1)-k_1}}-1)$", "answer": "$$R\\left(\\frac{1}{\\sqrt{k_2(1+k_1)-k_1}}-1\\right)$$" }, { "id": 453, "tag": "ELECTRICITY", "content": "Most media exhibit absorption of light, where the intensity of light decreases as it penetrates deeper into the medium.\n\nLet monochromatic parallel light (with angular frequency $\\omega$) pass through a uniform medium with a refractive index of $n$. Experimental results show that, over a reasonably wide range of light intensities, after passing through a small thickness, the decrease in light intensity is proportional to both the light intensity itself and this small thickness. The proportionality coefficient is denoted as $\\alpha$.\n\nNext, we will estimate the coefficient $\\alpha$ based on the classical oscillator model of atoms. In this model, the atomic nucleus can be considered stationary, while the electron, when in motion, is bound by the nucleus. This binding force can be approximated as a linear restoring force: $-m\\omega_{0}^{2}x$. Here, $\\mathsf{m}$ is the mass of the electron, $\\omega_{0}$ is the natural angular frequency (describing the strength of the restoring force), and $x$ is the distance between the electron and the nucleus. Simultaneously, when the electron is in motion, there often exists a damping force: $-m\\gamma v$. However, since the damping is usually quite weak, it can be approximately assumed that the electron undergoes simple harmonic motion within any given period.\n\nAdditionally, it is important to know that the power of radiation emitted by a charged particle in accelerated motion is given by:\n\n$$\nP = {\\frac{e^{2}{\\overline{{{\\dot{v}}^{2}}}}}{6\\pi\\varepsilon_{0}c^{3}}}\n$$ \n\nBased on this model, derive the expression for $\\alpha$. \n\n**Hint**: The primary interaction between electromagnetic waves and the medium originates from the contribution of the electric field component. $e$ is the magnitude of the electron charge, $\\overline{{{\\dot{v}}^{2}}}$ represents the average value of the square of the acceleration, $\\gamma$ is a positive constant, $\\varepsilon_{0}$ is the vacuum permittivity, $v$ is the velocity of the electron, $c$ is the speed of light in vacuum, and $N$ is the number density of electrons.", "solution": "The dynamics equation for electrons:\n\n$$\nm\\ddot{x}=-m\\gamma\\dot{x}-m\\omega_{0}^2x-eE_0e^{i\\omega t}\n$$\n\nSubstituting $x=\\tilde{x}e^{i\\omega t}$ yields $\\tilde{x}=\\frac{-eE_0/m}{-\\omega^2+\\omega_0^2+i\\gamma \\omega}$\n\n$$\n\\tilde{v}=i\\omega \\tilde{x}=\\frac{-i\\omega eE_0/m}{-\\omega^2+\\omega_0^2+i\\gamma \\omega}\n$$\n\n$$\n\\dot{\\tilde{v}}=\\frac{-\\omega^2 eE_0/m}{-\\omega^2+\\omega_0^2+i\\gamma \\omega}\n$$\n\n$$\n\\overline{\\dot{v^2}}=\\frac{\\frac{1}{2}(\\omega^2 eE_0/m)^2}{(\\omega_0^2-\\omega^2)^2+\\gamma^2 \\omega^2}\n$$\n\n$$\nP=-\\frac{e^2\\overline{\\dot{v^2}}}{6\\pi\\epsilon_0 c^3}=-\\frac{1}{12\\pi\\epsilon_0 c^3}\\frac{(\\omega^2 e^2 E_0/m)^2}{(\\omega_0^2-\\omega^2)^2+\\gamma^2 \\omega^2}\n$$\n\nThe energy flux density of the electromagnetic wave:\n\n$$\nI=\\frac{1}{2}nc\\epsilon_0E_0^2\n$$\n\nAssuming the number density of electrons is \\(N\\), then the energy loss per unit area in a length \\(dx\\) is \\(NPdx=dI\\).\n\nAlso, from $-dI=\\alpha Idx$, we have\n\n$$\n\\alpha=-\\frac{dI}{Idx}=-\\frac{NP}{I}\n$$\n\nThus,\n\n$$\n\\alpha=\\frac{N}{6\\pi n\\epsilon_0^2c^4}\\frac{(\\frac{\\omega^2 e^2}{m})^2}{(\\omega_0^2-\\omega^2)^2+\\gamma^2 \\omega^2}\n$$", "answer": "$$\n\\alpha=\\frac{N}{6\\pi n\\epsilon_0^2c^4}\\frac{(\\frac{\\omega^2 e^2}{m})^2}{(\\omega_0^2-\\omega^2)^2+\\gamma^2 \\omega^2}\n$$\n" }, { "id": 281, "tag": "THERMODYNAMICS", "content": "Due to the imbalance of molecular forces on the surface layer of a liquid, tension will arise along any boundary of the surface layer. In this context, the so-called \"capillary phenomenon\" occurs, affecting the geometric shape of a free liquid surface. Assume a static large container exists in a uniform atmosphere. The container holds a liquid in static equilibrium, and its side walls are vertical planes. The contact angle between the liquid and the side wall is $\\theta$ $(0 < \\theta < \\frac{\\pi}{2})$, the density of the liquid is $\\rho$, its surface tension coefficient is $\\sigma$, atmospheric pressure is $p_{0}$, and the gravitational acceleration is $g$. \n\nFind the additional height $h$ of the liquid surface rise at the junction between the container wall and the liquid surface due to the capillary phenomenon, expressing the result using $\\theta, \\rho, \\sigma, g$. \nHint: The formula for the radius of curvature in a two-dimensional Cartesian coordinate system is: $R = {\\frac{(1+z^{\\prime}(x)^{2})^{\\frac{3}{2}}}{\\mid z^{\\prime\\prime}(x)\\mid}}$.", "solution": "Choose $x\\sim x+\\mathrm{d}x.$ , analyze the surface microelement of the liquid located at a height between $z\\sim z+\\mathrm{d}z$, considering the liquid pressure $p_{i}$ at that point, it is evident that: \n\n$$\np_{0}=p_{i}+\\rho g z\n$$ \n\nNext, consider the forces perpendicular to the liquid surface. From the Laplace's additional pressure formula: \n\n$$\np_{i}+\\frac{\\sigma}{R}=p_{0}\n$$ \n\n$R$ is the radius of curvature of the liquid surface equation at that point. By combining equations (5) and (6), we obtain: \n\n$$\nz=z(x)=\\cfrac{\\sigma}{\\rho g R}=\\cfrac{\\sigma z^{\\\"v}(x)}{\\rho g(1+z^{\\prime}(x)^{2})^{\\frac{3}{2}}}\n$$ \n\nNote that here, due to the concave shape of the liquid surface, $z\"(x)$ is positive. \n\nIntegrate equation (7): \n\n$$\n\\int_{h}^{0}z\\mathrm{d}z=\\frac{\\sigma}{\\rho g}\\int_{h}^{0}\\frac{z^{\\prime}(x)}{\\left(1+z^{\\prime}(x)^{2}\\right)^{\\frac{3}{2}}}\\mathrm{d}z=\\frac{\\sigma}{\\rho g}\\int_{z^{\\prime}(0)}^{z^{\\prime}(\\infty)}\\frac{\\mathrm{d}z^{\\prime}(x)}{\\left(1+z^{\\prime}(x)^{2}\\right)^{\\frac{3}{2}}}=\\frac{\\sigma}{\\rho g}\\int_{\\frac{1}{\\sqrt{1+z\\left(\\left(x\\right)^{2}\\right)}}}^{\\frac{1}{\\sqrt{1+z\\left(0\\right)^{2}}}}\\mathrm{d}\\left(\\frac{1}{\\sqrt{1+z^{\\prime}(x)^{2}}}\\right)\n$$ \n\nHere, the boundary conditions are specifically pointed out: $z(0)=h,z(\\infty)=0,z^{\\prime}(\\infty)=0$ (the liquid surface approaches horizontal infinitely far away). Thus, $h$ can be expressed as: \n\n$$\n\\frac{1}{2}h^{2}=\\frac{\\sigma}{\\rho g}[1-\\frac{1}{\\sqrt{1+z^{\\prime}(0)^{2}}}]\n$$ \n\nThe contact angle $\\theta$ satisfies: \n\n$$\nz^{\\prime}(0)=\\tan(\\frac{\\pi}{2}-\\theta)=-\\cot\\theta\n$$ \n\nSubstitute to calculate, and the final result is: \n\n$$\nh={\\sqrt{\\frac{2\\sigma}{\\rho g}(1-\\sin\\theta)}}\n$$", "answer": "$$\\sqrt{\\frac{2\\sigma}{\\rho g}(1-\\sin \\theta)}$$" }, { "id": 95, "tag": "MECHANICS", "content": "At the North Pole, the gravitational acceleration is considered a constant vector of magnitude $g$, and the Earth's rotation angular velocity $\\Omega$ is along the $z$-axis pointing upward (the $z$-axis direction is vertically upward). A mass $M$ carriage has a mass $m=0.1M$ particle suspended from its ceiling by a light string of length $l$. It is known that $\\frac{g}{l} = \\omega_{0}^{2} = 10\\Omega^{2}$. Initially, the particle is at rest, hanging vertically. The carriage slides frictionlessly along a track in the $x$-direction. Suddenly, the carriage is acted upon by a constant forward force $F$ in the $x$-direction. Determine the expression for the $x$-coordinate of the particle relative to the carriage's reference frame as a function of time $t$. Use the small-angle approximation. At $t=0$, $x=0$. Express the answer in terms of $M$ and $\\Omega$. Consider the effects of the Coriolis force caused by the Earth's rotation in this scenario.", "solution": "The two components of the force exerted on the small ball can be approximated as: \n\n$$\n(F_{x},F_{y})=\\left(-\\frac{m g}{l}x,-\\frac{m g}{l}y\\right)\n$$\n\nAssume the coordinate of the carriage is $x$. Then the Coriolis force acting on the small ball is: \n\n$$\n(C_{x},C_{y})=\\Big(2m\\Omega\\dot{y},-2m\\Omega(\\dot{X}+\\dot{x})\\Big)\n$$\n\nNewton's laws: \n\n$$\nF_{\\pi}+C_{x}=m\\left(\\ddot{X}+\\ddot{x}\\right)\n$$\n\n$$\nF_{y}+C_{y}=m{\\ddot{y}}\n$$\n\n$$\nF-F_{x}=M\\ddot{X}\n$$\n\nRearranging: \n\n$$\n\\ddot{X}-\\frac{m g}{M l}x=\\frac{F}{M}\n$$\n\n$$\n\\ddot{X}+\\ddot{x}+\\frac{g}{l}x-2\\Omega\\dot{y}=0\n$$\n\n$$\n\\ddot{y}+\\frac{g}{l}y+2\\Omega(\\dot{X}+\\dot{x})=0\n$$\n\nWe find the following particular solutions $\\scriptstyle X={\\frac{1}{2}}a t^{2}$ ${\\pmb y}=b t$ ${\\pmb x}={\\pmb c}$ \n\n$$\n\\begin{array}{c}{{a-\\displaystyle\\frac{m}{M}\\omega_{0}^{2}c=\\displaystyle\\frac{F}{M}}}\\ {{}}\\ {{a+\\omega_{0}^{2}c-2\\Omega b=0}}\\ {{}}\\ {{\\omega_{0}^{2}b+2\\Omega a=0}}\\end{array}\n$$\n\nFrom this, we obtain: \n\n$$\nX_{0}=\\frac{25F}{57M}t^{2}\\quad,\\quad y_{0}=-\\frac{10F}{57M\\Omega}t\\quad,\\quad x_{0}=-\\frac{7F}{57M\\Omega^{2}}\n$$\n\nThe final solution should be $X=X_{0}+X_{1}$, ${\\mathfrak{y}}={\\mathfrak{y}}_{0}+{\\mathfrak{y}}_{1}$ ${\\boldsymbol{x}}={\\boldsymbol{x}}_{0}+{\\boldsymbol{x}}_{1}$, where $X_{1}, y_{1}, x_{1}$ satisfy the corresponding homogeneous equations. Let:\n\n$$\nX_{1}, x_{1}, y_{1}=A, B, C\\mathrm{e}^{i\\omega t}\n$$\n\nSubstituting these into the equations results in the matrix equation: \n\n$$\n\\left[\\begin{array}{c c c}{{-\\omega^{2}}}&{{-m\\omega_{0}^{2}/M}}&{{0}}\\ {{-\\omega^{2}}}&{{-\\omega^{2}+\\omega_{0}^{2}}}&{{-2\\mathrm{i}\\Omega\\omega}}\\ {{2\\mathrm{i}\\Omega\\omega}}&{{2\\mathrm{i}\\Omega\\omega}}&{{-\\omega^{2}+\\omega_{0}^{2}}}\\end{array}\\right]\\left[\\begin{array}{c}{{A}}\\ {{B}}\\ {{C}}\\end{array}\\right]=\\left[\\begin{array}{l}{{0}}\\ {{0}}\\ {{0}}\\ {{0}}\\end{array}\\right]\n$$\n\nTaking the determinant of the coefficient matrix equal to zero allows it to be factored: \n\n$$\n\\omega^{2}\\left(\\omega^{2}-6\\Omega^{2}\\right)\\left(\\omega^{2}-19\\Omega^{2}\\right)=0\n$$\n\nThe case $\\omega^{2}=0$ corresponds to the trivial solution: \n\n$$\nX_{1}, x_{1}, y_{1}=C_{1}, 0, 0\n$$\n\nThe solution for $\\omega=\\sqrt{6}\\Omega$ is: \n\n$$\nX_{1}, x_{1}, y_{1}=C_{2}, -6C_{2}, \\frac{5\\sqrt{6}}{2}\\mathrm{i}C_{2}\\mathrm{e}^{\\mathrm{i}\\omega t}\n$$\n\nThe solution for $\\omega=19\\Omega$ is: \n\n$$\nX_{1}, x_{1}, y_{1}=C_{3}, -19C_{3}, -4\\sqrt{19}\\mathrm{i}C_{3}\\mathrm{e}^{\\mathrm{i}\\omega t}\n$$\n\nBy superimposing these three solutions and considering the initial conditions: \n\n$$\nC_{1}+C_{2}+C_{3}=0\n$$\n\n$$\n-6C_{2}-19C_{3}+c=0\n$$\n\n$$\n-\\frac{5\\sqrt{6}}{2}\\cdot\\sqrt{6}\\Omega\\cdot C_{2}+4\\sqrt{19}\\cdot\\sqrt{19}\\Omega\\cdot C_{3}+b=0\n$$\n\nWe obtain: \n\n$$\nC_{1}=\\frac{59F}{3249M\\Omega^{2}}\\quad,\\quad C_{2}=-\\frac{2F}{117M\\Omega^{2}}\\quad,\\quad C_{3}=-\\frac{5F}{4693M\\Omega^{2}}\n$$\n\nFinally, substituting these into the equations and taking the real part: \n\n$$\n\\boxed{x=-\\frac{7F}{57M\\Omega^{2}}+\\frac{4F}{39M\\Omega^{2}}\\cos{\\sqrt{6}}\\Omega t+\\frac{5F}{247M\\Omega^{2}}\\cos{\\sqrt{19}}\\Omega t}\n$$", "answer": "$$\n\\boxed{x=-\\frac{7F}{57M\\Omega^2}+\\frac{4F}{39M\\Omega^2}\\cos(\\sqrt{6}\\Omega t)+\\frac{5F}{247M\\Omega^2}\\cos(\\sqrt{19}\\Omega t)}\n$$" }, { "id": 101, "tag": "MECHANICS", "content": "The problem discusses a stick figure model. The stick figure's head is a uniform sphere with a radius of $r$ and a mass of $m$, while the rest of the body consists of uniform rods with negligible thickness and a mass per unit length of $\\lambda$. All parts are connected by hinges. Specifically, the torso and both arms have a length of $l$, while the legs have a length of $1.2l$. $\\theta$ represents the angle between the arms and the torso, and $\\varphi$ represents the angle between the legs and the extended torso line (not an obtuse angle). The stick figure is composed of a head, a torso, two arms, and two legs. Initially, the stick figure is in a state where $\\theta=\\textstyle{\\frac{\\pi}{2}}$ and $\\varphi={\\frac{\\pi}{4}}$. The head has a mass of $m=0.6\\lambda l$, with the result expressed in terms of $\\lambda$. \n\nThe stick figure is a simulation of a real biological body, so if necessary, the connections between the torso and the arms/legs, in addition to providing interaction forces, can also exert a couple (i.e., a torque), even if there is no interaction force between them. This net torque is independent of the choice of reference point. In the following discussion, this will be referred to simply as a \"couple.\"\n\nDuring a certain free-fall motion, the stick figure, overwhelmed with panic, forgets to bring its legs together and swings its arms and torso with an amplitude of $\\theta_{0}$ (small angle) and an angular frequency of $\\omega$. As a result, its legs passively oscillate near $\\varphi={\\frac{\\pi}{4}}$. Find the amplitude of resonance for the legs (seek the steady-state solution starting from the initial position). Assume that the stick figure can actively generate a couple for the arms but not for the legs.", "solution": "\n\n(1) The forces are already marked on the diagram. List Newton's laws and rotational equations, assuming the displacement of the driven object is $\\pmb{x}$, positive upwards.\n\nVertical direction of the leg\n\n$$\nT_{3}=1.2\\lambda l\\times\\left(0.6l\\times{\\frac{d^{2}}{d t^{2}}}\\left({\\mathrm{cos}}\\varphi\\right)-{\\ddot{x}}\\right)\n$$\n\nVertical direction of the arm\n\n$$\nT_{1}=\\lambda l\\left(\\ddot{x}-0.5l\\times\\frac{d^{2}}{d t^{2}}\\left(\\mathrm{cos}\\theta\\right)\\right)\n$$\n\nVertical direction of the torso (head)\n\n$$\n2\\left(T_{3}-T_{1}\\right)=(m+\\lambda l)\\ddot{x}\n$$\n\nHorizontal direction of the leg\n\n$$\n-T_{4}=1.2\\lambda l\\times0.6l\\times\\frac{d^{2}}{d t^{2}}\\left(\\mathrm{sin}\\varphi\\right)\n$$\n\nHorizontal direction of the arm\n\n$$\n-T_{2}=\\lambda l\\times0.5l\\times{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{sin}\\theta\\right)\n$$\n\nAngular momentum theorem for the leg\n\n$$\nT_{3}\\times0.6l\\mathrm{sin}\\varphi+T_{4}\\times0.6l\\mathrm{cos}\\varphi-M=\\frac{1}{12}\\times1.2\\lambda l\\times(1.2l)^{2}\\ddot{\\varphi}\n$$\n\nAngular momentum theorem for the arm\n\n$$\nT_{2}\\times0.5l\\mathrm{cos}\\theta-T_{1}\\times0.5l\\mathrm{sin}\\theta=\\frac{1}{12}\\times\\lambda l\\times l^{2}\\ddot{\\theta}\n$$\n\nSubstitute in, we have\n\n$$\n\\frac{1}{12}l^{2}{\\ddot{\\theta}}+0.25l^{2}{\\mathrm{cos}}\\theta\\frac{d^{2}}{d t^{2}}{\\mathrm{sin}}\\theta+\\frac{T_{1}}{\\lambda}\\times0.5{\\mathrm{sin}}\\theta=0\n$$\n\nBy combining the previous equations, we get\n\n$$\nT_{1}=\\left({\\frac{6}{25}}{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{cos}\\varphi\\right)-{\\frac{1}{3}}{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{cos}\\theta\\right)\\right)\\lambda l^{2}\n$$\n\nBy substituting, we obtain the final expression\n\n$$\n\\left(\\frac{1}{3}-\\frac{1}{12}\\mathrm{sin}^{2}\\theta\\right)\\ddot{\\theta}-\\frac{1}{12}\\mathrm{sin}\\theta\\mathrm{cos}\\theta\\dot{\\theta}^{2}=\\frac{3}{25}\\omega^{2}\\mathrm{sin}\\theta\\mathrm{cos}\\varphi\n$$\n\nTherefore,\n\n$$\n\\begin{array}{c}{{A(\\theta)=\\displaystyle\\frac{1}{3}-\\frac{1}{12}\\mathrm{sin}^{2}\\theta}}\\ {{{}}}\\ {{B(\\theta)=-\\displaystyle\\frac{1}{12}\\mathrm{sin}\\theta\\mathrm{cos}\\theta}}\\ {{{}}}\\ {{C(\\theta)=\\displaystyle\\frac{3}{25}\\omega^{2}\\mathrm{sin}\\theta}}\\end{array}\n$$\n\nSubstituting back into the previous angular momentum theorem, we have\n\n$$\n\\begin{array}{r l r}{M=\\frac{9}{25}\\mathrm{sin}\\varphi\\lambda l^{3}\\left(\\frac{12}{25}\\mathrm{cos}\\varphi\\omega^{2}+\\frac{1}{3}\\mathrm{cos}\\theta\\dot{\\theta}^{2}+\\frac{1}{3}\\mathrm{sin}\\theta\\ddot{\\theta}\\right)}&{{}=\\frac{12}{25}\\frac{\\mathrm{sin}\\varphi}{\\mathrm{sin}\\theta}\\ddot{\\theta}}\\end{array}\n$$\n\nThere still remain the first five equations from the first question and the modified angular momentum theorem\n\n$$\nT_{3}\\times0.6l\\mathrm{sin}\\varphi+T_{4}\\times0.6l\\mathrm{cos}\\varphi=\\frac{1}{12}\\times1.2\\lambda l\\times(1.2l)^{2}\\ddot{\\varphi}\n$$\n\nSubstitute and combine, we get\n\n$$\nT_{3}\\times0.6l\\mathrm{sin}\\varphi-1.2\\times0.6\\lambda l^{2}{\\frac{d^{2}}{d t^{2}}}(\\mathrm{sin}\\varphi)\\times0.6l\\mathrm{cos}\\varphi={\\frac{1}{12}}\\times1.2\\lambda l\\times(1.2l)^{2}\\ddot{\\varphi}\n$$\n\n$$\nT_{3}=\\lambda l\\left(\\ddot{x}-0.5l\\frac{d^{2}}{d t^{2}}\\left(\\mathrm{cos}\\theta\\right)\\right)+\\frac{1}{2}(m+\\lambda l)\\ddot{x}\n$$\n\nCombining, we get\n\n$$\nT_{3}=-\\frac{1}{5}\\lambda l^{2}\\frac{d^{2}}{d t^{2}}\\left(\\mathrm{cos}\\theta\\right)+\\frac{54}{125}\\lambda l^{2}\\frac{d^{2}}{d t^{2}}\\left(\\mathrm{cos}\\varphi\\right)\n$$\n\nSubstituting the above equation, we obtain the final expression\n\n$$\n\\left({\\frac{54}{125}}{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{cos}\\varphi\\right)-{\\frac{1}{5}}{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{cos}\\theta\\right)\\right)\\mathrm{sin}\\varphi-{\\frac{18}{25}}{\\frac{d^{2}}{d t^{2}}}\\left(\\mathrm{sin}\\varphi\\right)\\mathrm{cos}\\varphi={\\frac{6}{25}}{\\ddot{\\varphi}}\n$$\n\nLet = +, then retain terms up to the first order of and θ, we get\n\n$$\n\\ddot{\\delta}=-\\frac{25\\sqrt{2}}{204}\\omega^{2}\\theta_{0}\\mathrm{cos}\\omega t\n$$\n\nThat is\n\n$$\nA={\\frac{25{\\sqrt{2}}}{204}}\\theta_{0}\n$$", "answer": "$$\nA = \\frac{25 \\sqrt{2}}{204} \\theta_0\n$$" }, { "id": 630, "tag": "THERMODYNAMICS", "content": "In this problem, we study a simple \\\"gas-fueled rocket.\\\" The main body of the rocket is a plastic bottle, which can take off after adding a certain amount of fuel gas and igniting it. It is known that the external atmospheric pressure is $P_{0}$, and the initial pressure of the gas in the bottle is $P_{0}$, with a temperature of $T_{0}$. The proportion of the molar amount of fuel gas is $\\alpha$, and the rest is air. Alcohol is chosen as the fuel, with the reaction equation in air as: $C_{2}H_{5}OH(g) + 3O_{2}(g) \\rightarrow 3H_{2}O(g) + 2CO_{2}(g)$. Assume that the air contains only $N_{2}$ and $O_{2}$, all gases are considered as ideal gases, and none of the vibrational degrees of freedom are excited. It is known that under conditions of pressure $P_{0}$ and temperature $T_{0}$, the reaction heat is $-\\lambda$ (enthalpy change for $1\\mathrm{mol}$ of ethanol reacting under isothermal and isobaric conditions). Ignore heat loss, assume the reaction is complete, and that there is air remaining, with the gas volume remaining constant throughout. Find the pressure $P_{1}$ of the gas inside the bottle after the gas has completely reacted upon ignition.", "solution": "Let the initial moles of air be $n_{0}$, and the moles of alcohol be $$ n_{x}=\\frac{\\alpha n_{0}}{1-\\alpha} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\\left(n_{x}+n_{0}\\right)R T_{0}} \\\\ {P_{0}V_{1}=\\left(2n_{x}+n_{0}\\right)R T_{0}} $$ According to the definition of enthalpy change, we have $$ \\Delta{{H}_{0}}=\\Delta{{U}_{0}}+\\Delta\\left(P V\\right)=\\Delta{{U}_{0}}+{{P}_{0}}\\left({{V}_{1}}-{{V}_{0}}\\right) $$ It follows that $$ \\Delta U_{0}=-n_{x}\\lambda-n_{x}RT_{0} $$ Since heat loss is neglected and no gas escapes from the bottle, the reaction is adiabatic and the volume does not change, giving us $$ \\Delta U=\\Delta U_{\\mathrm{0}}+\\Big[\\frac{5}{2}R\\cdot\\big(n_{0}-n_{x}\\big)+3R\\cdot 3n_{x}\\Big]\\big(T_{1}-T_{0}\\big)=0 $$ We obtain $$ T_{1}=T_{0}+\\frac{2n_{x}\\left(\\lambda+R T_{0}\\right)}{R\\left(5n_{0}+13n_{x}\\right)}=T_{0}+\\frac{2\\alpha\\left(\\lambda+R T_{0}\\right)}{R\\left(5+8\\alpha\\right)} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\\left(n_{x}+n_{0}\\right)R T_{0}} \\\\ {P_{1}V_{0}=\\left(2n_{x}+n_{0}\\right)R T_{1}} $$ We find $$ P_{1}=P_{0}\\frac{(1+\\alpha)(5+10\\alpha+2\\alpha(\\frac{\\lambda}{R T_{0}}))}{5+8\\alpha} $$", "answer": "$$P_0 \\frac{(1+\\alpha)(5+10\\alpha+2\\alpha \\frac{\\lambda}{R T_0})}{5+8\\alpha}$$" }, { "id": 134, "tag": "MECHANICS", "content": "A fox is escaping along a straight line $AB$ at a constant speed $v_{1}$. A hound is pursuing it at a constant speed $v_{2}$, always aimed at the fox. At a certain moment, the fox is at $F$ and the hound is at $D$. $FD \\bot AB$ and $FD = L$. Find the magnitude of the hound's acceleration at this moment.", "solution": "To solve $\\textcircled{1}$, the hunting dog performs uniform circular motion. Thus, the tangential acceleration is zero, while the normal acceleration changes. Assume that during a very short period of time $\\mathrm{d}t$ after the given moment, the curvature radius of the trajectory of the hunting dog is $\\rho,$ then the normal acceleration is \n\n$$\na_{n}={\\frac{v_{2}^{2}}{\\rho}}.\n$$\n\nDuring the time interval $\\mathrm{d}t$, the fox and the hunting dog respectively reach positions $F^{\\prime}$ and $D^{\\prime}$ (see Figure 1.7(a) in the problem). The angle turned by the hunting dog's direction of motion is \n\n$$\n\\alpha={\\frac{\\widehat{D D^{\\prime}}}{\\rho}}={\\frac{v_{2}\\mathrm{d}t}{\\rho}}.\n$$\n\nHowever, the distance traveled by the fox is \n\n$$\n\\begin{array}{r}{\\overline{{F F^{\\prime}}}=v_{1}\\mathrm{d}t\\approx L\\tan\\alpha\\approx\\alpha L,}\\end{array}\n$$\n\nwhich implies \n\n$$\n\\alpha={\\frac{v_{1}\\mathrm{d}t}{L}}.\n$$\n\nThus, \n\n$$\n\\frac{\\upsilon_{2}\\mathrm{d}t}{\\rho}=\\frac{\\upsilon_{1}\\mathrm{d}t}{L},\n$$\n\nand from this, the curvature radius can be obtained as \n\n$$\n\\rho=\\frac{L v_{2}}{v_{1}}.\n$$\n\nTherefore, the magnitude of the acceleration is \n\n$$\na_{n}={\\frac{v_{2}^{2}}{\\rho}}={\\frac{v_{1}v_{2}}{L}}.\n$$\n\nIt should be noted that the acceleration obtained this way corresponds only to the specific moment mentioned in the problem, not to the acceleration at any arbitrary position during the chasing process.", "answer": "$$\\frac{v_1 v_2}{L}$$" }, { "id": 650, "tag": "MECHANICS", "content": "Two parallel light strings, each with length $L$, are horizontally separated by a distance $d$. Their upper ends are connected to the ceiling, and the lower ends are symmetrically attached to a uniformly distributed smooth semicircular arc. The radius of the semicircle is $R$, and its mass is $m$. The gravitational acceleration is $g$. A small ball with the same mass $m$ is placed at the bottom of the semicircle. The entire system undergoes small oscillations in the plane in some manner, with a total energy of $E_{0}$. The potential energy zero point is chosen such that the energy of the system is zero when it is vertically stable. Find the maximum $d$, denoted as $d_{max}$, that allows the system to have zero tension in one of the strings at some moment.\n", "solution": "Let the angle between the rope and the vertical be $\\theta$, and the angle between the line connecting the small ball and the center of the circle and the vertical be $\\varphi$. First, consider the kinetic equations. The kinetic energy $$ T={\\frac{1}{2}}m L^{2}{\\dot{\\theta}}^{2}+{\\frac{1}{2}}m(L{\\dot{\\theta}}+R{\\dot{\\varphi}})^{2} $$ The potential energy $$ V=m g L\\theta^{2}+{\\frac{1}{2}}m g R\\varphi^{2} $$ Thus, the equations of motion are $$ \\begin{array}{c}{{2m L^{2}\\ddot{\\theta}+m L R\\ddot{\\varphi}+2m g L\\theta=0}}{{\\phantom{-}}}{{m R^{2}\\ddot{\\varphi}+m L R\\ddot{\\theta}+m g R\\varphi=0}}\\end{array} $$ Consider the torque at the midpoint of the endpoints of the two ropes on the ceiling $$ \\tau=2m g L\\theta+m g R\\varphi+m L^{2}\\ddot{\\theta}+m L(L-\\left(1-\\frac{2}{\\pi}\\right)R)\\ddot{\\theta}+m L R\\ddot{\\varphi} $$ Substituting the equations of motion, simplifying gives $$ \\tau=2m g\\left(1-\\frac{2}{\\pi}\\right)R\\theta+\\frac{2}{\\pi}m g R\\varphi $$ To require that one rope has no tension, then $\\tau=m g d$. Keeping $E_{0}$ constant and maximizing $d$ is equivalent to minimizing $E_{0}$ under the condition. Obviously, at this time one should have $$ \\dot{\\theta}=\\dot{\\varphi}=0 $$ because the velocity terms do not contribute to the torque and only increase the energy. Therefore, at the critical point $$ E_{0}=m g L\\theta^{2}+{\\frac{1}{2}}m g R\\varphi^{2} $$ According to the Cauchy inequality $$ \\theta^{2}+\\frac{1}{2}m g R\\varphi^{2}\\Bigg)\\left(\\left(\\frac{2m g\\left(1-\\frac{2}{\\pi}\\right)R}{\\sqrt{m g L}}\\right)^{2}+\\left(\\frac{\\frac{2}{\\pi}m g R}{\\sqrt{\\frac{1}{2}m g R}}\\right)^{2}\\right)\\geq\\left(2m g\\left(1-\\frac{2}{\\pi}\\right)R\\theta+\\frac{2}{\\pi}m g R\\varphi\\right)^{2} $$ Substituting the torque, the energy has: $$ d_{\\operatorname*{max}}=\\frac{2}{\\pi}\\sqrt{\\frac{E_{0}R}{m g}}\\sqrt{(\\pi-2)^{2}\\frac{R}{L}+2} $$\n", "answer": "$$\\frac{2}{\\pi}\\sqrt{\\frac{E_0 R}{m g}}\\sqrt{(\\pi-2)^2\\frac{R}{L}+2}$$" }, { "id": 332, "tag": "MECHANICS", "content": "There is a uniform thin spherical shell, with its center fixed on a horizontal axis and able to rotate freely around this axis. The spherical shell has a mass of $M$ and a radius of $R$. There is also a uniform thin rod, with one end smoothly hinged to the axis at a distance $d$ from the center of the sphere, and the other end resting on the spherical shell. The thin rod has a mass of $m$ and a length of $L$ ($d-R 0$ is electrolyte 1, and the region where $y < 0$ is electrolyte 2. The conductivities of the two dielectrics are $\\sigma_{1}, \\sigma_{2}$, and the dielectric constants are $\\varepsilon_{1}, \\varepsilon_{2}$, respectively. On the $xOz$ interface of the two dielectrics, two cylindrical holes with a radius $R$ are drilled in the $z$ direction, spaced $2D (D > R, R, D \\ll L)$ apart, with centers located on the interface as long straight cylindrical holes. Two cylindrical bodies $\\pm$ are inserted into the holes, with the type of the cylinders given by the problem text below.\n\nThe cylindrical bodies $\\pm$ are metal electrodes filling the entire cylinder. Initially, the system is uncharged, and at $t=0$, a power source with an electromotive force $U$ and internal resistance $r_{0}$ is used to connect the electrodes. Find the relationship between the current through the power source and time, denoted as $i(t)$.", "solution": "Given the potential difference $u$, it can be seen:\n\n$$\n\\varphi_{+}=u/2,\\varphi_{-}=-u/2,\\lambda=\\frac{2\\pi(\\varepsilon_{1}+\\varepsilon_{2})\\varphi_{+}}{2\\xi_{+}}=\\frac{\\pi(\\varepsilon_{1}+\\varepsilon_{2})u}{\\operatorname{arccosh}(\\mathrm{D/R})}\n$$ \n\nSelect a surface encapsulating the cylindrical surface and examine Gauss's theorem. For the positive electrode, it is easy to see:\n\n$$\n\\iint\\vec{E}\\cdot d\\vec{S}=L\\oint\\vec{E}\\cdot\\hat{n}d l=\\frac{\\lambda L}{(\\varepsilon_{1}+\\varepsilon_{2})/2}=\\frac{2\\pi u L}{2\\mathrm{arccosh}(D/R)}\n$$ \n\nSince the above potential distribution is deemed directly applicable for the calculation of current, the total current flowing out of the positive electrode is:\n\n$$\nI=\\iint\\sigma\\vec{E}\\cdot d\\vec{S}=\\frac{\\sigma_{1}+\\sigma_{2}}{2}\\times\\frac{2\\pi u L}{2\\mathrm{arccosh}(D/R)}\n$$ \n\nGiven the current $i$ passing through the power source, this current changes the net charge:\n\n$$\n{\\frac{d(\\lambda L)}{d t}}=i-I=i-{\\frac{2\\pi u L}{2\\mathrm{arccosh}(D/R)}}={\\frac{\\pi(\\varepsilon_{1}+\\varepsilon_{2})L}{2\\mathrm{arccosh}(D/R)}}{\\frac{d u}{d t}}\n$$ \n\nAccording to the loop voltage drop equation:\n\n$$\n\\begin{array}{l l}{{U=r_{0}i+u\\rightarrow u=U-r_{0}i}}\\ {{\\rightarrow i-{\\displaystyle\\frac{\\pi(\\sigma_{1}+\\sigma_{2})L}{2\\mathrm{arccosh}(D/R)}}}(U-r_{0}i)=-{\\displaystyle\\frac{\\pi(\\varepsilon_{1}+\\varepsilon_{2})L}{2\\mathrm{arccosh}(D/R)}}r_{0}{\\displaystyle\\frac{d i}{d t}}}\\ {{\\rightarrow{\\displaystyle\\frac{d i}{d t}}={\\displaystyle\\frac{(\\sigma_{1}+\\sigma_{2})U}{r_{0}(\\varepsilon_{1}+\\varepsilon_{2})}}-\\left(\\frac{\\sigma_{1}+\\sigma_{2}}{\\varepsilon_{1}+\\varepsilon_{2}}+{\\displaystyle\\frac{2\\mathrm{arccosh}(D/R)}{\\pi r_{0}L(\\varepsilon_{1}+\\varepsilon_{2})}}\\right)i}}\\end{array}\n$$ \n\nAt time $t=0$, all current should preferentially enter the capacitor. At this time, the initial current is $U/r_{0}$, and this differential equation yields:\n\n$$\ni(t)=\\frac{U}{r_{0}\\left(1+\\frac{2\\mathrm{arccosh}(D/R)}{\\pi r_{0}L(\\sigma_{1}+\\sigma_{2})}\\right)}\\left\\{2+\\frac{2\\mathrm{arccosh}(D/R)}{\\pi r_{0}L(\\sigma_{1}+\\sigma_{2})}-\\exp{\\left[-\\left(\\frac{\\sigma_{1}+\\sigma_{2}}{\\varepsilon_{1}+\\varepsilon_{2}}+\\frac{2\\mathrm{arccosh}(D/R)}{\\pi r_{0}L(\\varepsilon_{1}+\\sigma_{2})}\\right)\\right]}\\right\\},\n$$", "answer": "$$\ni(t)=\\frac{U}{r_0\\left(1+\\frac{2\\arccosh(D/R)}{\\pi r_0 L(\\sigma_1+\\sigma_2)}\\right)}\\left(2+\\frac{2\\arccosh(D/R)}{\\pi r_0 L(\\sigma_1+\\sigma_2)}-\\exp\\left[-\\left(\\frac{\\sigma_1+\\sigma_2}{\\varepsilon_1+\\varepsilon_2}+\\frac{2\\arccosh(D/R)}{\\pi r_0 L(\\varepsilon_1+\\varepsilon_2)}\\right)t\\right]\\right)\n$$" }, { "id": 409, "tag": "ELECTRICITY", "content": "In electromagnetism, we often study the problem of electromagnetic field distribution in a region without charge or current distribution. In such cases, the electromagnetic field will become a tubular field. The so-called tubular field is named because of the nature of the velocity field of an incompressible fluid at every instant. Its characteristic can be described using the language of vector analysis as the field's divergence being equal to zero. Another perspective is that if we take the field lines of a vector field $\\pmb{F}$ (the tangent line at each point being the direction of field strength) and select a flux tube along a set of field lines passing through a certain cross-section, then at any cross-section of the flux tube, the field flux:\n\n$$\n\\int F\\cdot\\mathrm{d}S=\\Phi\n$$ \n\nwill be a conserved quantity. This problem will investigate a special case of a tubular field that is rotationally symmetric around the $z$ axis: using oblate spheroidal coordinates to construct a unique tubular field as an electric field or magnetic field. The so-called oblate spheroidal coordinate system is similar to spherical coordinates, and it is generated by the following coordinate transformations:\n\n$$\nx= a \\mathrm{ch}\\mu \\cos \\nu \\cos \\varphi\n$$ \n\n$$\ny= a \\mathrm{ch}\\mu \\cos \\nu \\sin \\varphi\n$$ \n\n$$\nz=a\\mathrm{sh}\\mu\\sin\\nu\n$$ \n\nWe want the shape of the field lines of an electric field or magnetic field to precisely follow its generatrix direction, i.e., the tangent direction of the curve where field strength changes with $\\mu$ while $\\nu,\\varphi$ remain fixed (upwards when $\\relax z>0$). \n\nFind the charge distribution in the $xy$ plane that can produce this electric field distribution. Assume the field strength near the origin is $E_0$. The dielectric constant is $\\varepsilon_{0}$.", "solution": "First, we determine the equation of the constant-$\\mu$ line. Now, using the trigonometric identity $\\sin^{2}\\nu+\\cos^{2}\\nu=1$, we eliminate the parameter $\\nu$:\n\n$$\n{\\frac{x^{2}}{a^{2}{\\mathrm{ch}}^{2}\\mu}}+{\\frac{z^{2}}{a^{2}{\\mathrm{sh}}^{2}\\mu}}=1\n$$\n\nIts physical significance is that, for an electric field, it forms a surface of equal potential. This is based on the fact that equipotential surfaces should always be perpendicular to the electric field lines. The constant-$\\mu$ lines, being ellipses, satisfy this condition: their tangent directions are aligned with the external bisectors of angles formed by lines connecting the two foci, and they are perpendicular to the tangent directions of the hyperbolic electric field lines, which correspond to the internal bisectors.\n\nNext, we determine an inverse transformation. For a point in the first quadrant with coordinates $(x, z)$, the distances to the two foci are:\n\n$$\nr_{+}=\\sqrt{(x+a)^{2}+z^{2}}\\quad,\\quad r_{-}=\\sqrt{(x-a)^{2}+z^{2}}\n$$\n\nThus, the semi-major axis lengths of the ellipse and hyperbola are respectively:\n\n$$\nr_{+}+r_{-}=2a\\mathrm{ch}\\mu\n$$\n\n$$\nr_{+}-r_{-}=2a\\cos\\nu\n$$\n\nThese two equations help convert Cartesian coordinates into generalized coordinates $\\mu$ and $\\nu$. Next, we utilize the following property: the infinitesimal displacement caused by an increase in $\\mu$ between equipotential surfaces is the derivative of the coordinates with respect to $\\mu$:\n\n$$\n\\mathrm{d}x=a\\mathrm{sh}\\mu\\cos\\nu\\mathrm{d}\\mu\\quad,\\quad\\mathrm{d}z=a\\mathrm{ch}\\mu\\sin\\nu\\mathrm{d}\\mu\n$$\n\nWe proceed to calculate the length of this displacement:\n\n$$\n\\mathrm{d}s={\\sqrt{\\mathrm{d}x^{2}+\\mathrm{d}z^{2}}}=a{\\sqrt{\\mathrm{ch}^{2}\\mu-\\cos^{2}\\nu}}\\mathrm{d}\\mu\n$$\n\nNext, we consider the electric potential difference caused by displacements between one ellipse and another. This difference is equal to the electric field strength multiplied by the displacement length. Because the ellipses represent equipotential surfaces, this product is constant everywhere. In particular, taking $\\nu=90^{\\circ}$ places this point exactly on the $z$-axis, and the electric field strength at this location has already been computed in (2). We obtain the following equation:\n\n$$\n{\\frac{E_{0}a^{2}}{a^{2}+a^{2}\\mathrm{{sh}}^{2}\\mu\\sin^{2}{90^{\\circ}}}}\\cdot a\\mathrm{{ch}}\\mu\\mathrm{{d}}\\mu=E\\cdot a{\\sqrt{\\mathrm{{ch}}^{2}\\mu-\\cos^{2}\\nu}}\\mathrm{{d}}\\mu\n$$\n\nThis gives the distribution of electric field strength. To determine the direction of electric field lines, we only need to locate the bisector of the angle formed by lines connecting the two foci:\n\n$$\n\\theta={\\frac{\\theta_{+}+\\theta_{-}}{2}}\n$$\n\n$$\n\\tan\\theta_{\\pm}={\\frac{z}{x\\pm a}}\n$$\n\nWe organize the result:\n\n$$\nf(x,z)={\\frac{2a^{2}}{\\left[{\\sqrt{(x+a)^{2}+z^{2}}}+{\\sqrt{(x-a)^{2}+z^{2}}}\\right]^{2}}}\\left[\\left[{\\frac{(x+a)^{2}+z^{2}}{(x-a)^{2}+z^{2}}}\\right]^{\\frac{1}{4}}+\\left[{\\frac{(x-a)^{2}+z^{2}}{(x+a)^{2}+z^{2}}}\\right]^{\\frac{1}{4}}\\right]~.\n$$\n\nWith $z=0$, $x=r$, we obtain:\n\n$$\nf(r)=\\frac{2a^{2}}{(r+a+|r-a|)\\sqrt{|r^{2}-a^{2}|}}\n$$\n\nFor the electric field, a charged circular disk with radius $a$ is required to generate it. In this case, the field strength at $r0\n$$ \n\nThat is \n\n$$\nR>{\\frac{a^{2}}{b}}\n$$ \n\nThe moment of inertia of the elliptical cylinder about its central axis is \n\n$$\nI={\\frac{1}{4}}m(a^{2}+b^{2})\n$$ \n\nThus, the kinetic energy of the system is \n\n$$\nE_{k}=\\frac{1}{2}(I+m b^{2})\\dot{\\varphi}^{2}=\\frac{1}{2}\\frac{m(a^{2}+5b^{2})}{4}(\\frac{b R}{a^{2}}-1)^{2}\\dot{\\theta}^{2}\n$$ \n\nTherefore, the frequency of oscillation is \n\n$$\nf={\\frac{1}{2\\pi}}{\\sqrt{\\frac{4g(b+R-{\\frac{b^{2}R}{a^{2}}})}{(a^{2}+5b^{2})({\\frac{b R}{a^{2}}}-1)}}}\n$$", "answer": "$$\nf=\\frac{1}{2\\pi}\\sqrt{\\frac{4g(b+R-\\frac{b^2R}{a^2})}{(a^2+5b^2)\\left(\\frac{bR}{a^2}-1\\right)}}\n$$" }, { "id": 88, "tag": "MODERN", "content": "In a certain atom $A$, there are only two energy levels: the lower energy level $A_{0}$ is called the ground state, and the higher energy level $A^{\\star}$ is called the excited state. The energy difference between the excited state and the ground state is $E_{0}$. When the atom is in the ground state, its rest mass is $m_{0}$; when it is in the excited state, due to its intrinsic instability, it will transition to the ground state while emitting photons externally. It is known that the probability of transition from the excited state to the ground state in unit time in the atom's rest reference frame is $\\lambda$, and the probability of emitting photons in all directions is equal. At the initial moment, the total number of atoms is $N_{0}(N_{0}\\gg1)$, and the experimental reference frame $S$ and the atom's proper reference frame $S^{\\prime}$ are time-synchronized. All $N_{0}$ atoms are in the excited state and have a common velocity $\\pmb{v}$ in the $+\\hat{\\pmb{x}}$ direction relative to the $S$ frame.\n\nConsidering $m_{0}c^{2}\\gg E_{0}$, at this time the recoil effect on the atom due to photon emission can be ignored. Try to solve the angular distribution of the light emission power $w(\\theta)$ from the atom at time $\\pmb{t}$ in the laboratory reference frame $S$. (Calculate the emission power per unit solid angle in the direction of angle $\\pmb{\\theta}$ relative to the $+\\hat{\\pmb{x}}$ axis, rather than the received power at an infinite distance. The speed of light in vacuum is $c$ (known). The answer should be expressed using $\\lambda, N_0, E_0, v, \\theta, \\pi, t, c$. Please check and output the final answer.", "solution": "Consider the reference frame $S^{\\prime}$:\n\n$$\nh\\nu_{0}=E_{0}\n$$ \n\nLet the remaining number of atoms at time $t^{\\prime}$ be $N^{\\prime}$:\n\n$$\n-\\mathrm{d}N^{\\prime}=\\lambda N^{\\prime}\\mathrm{d}t^{\\prime}\n$$ \n\nSolving for $N^{\\prime}$, we get:\n\n$$\nN^{\\prime}=N_{0}e^{-\\lambda t^{\\prime}}\n$$ \n\nThe total power emitted by photons:\n\n$$\nP^{\\prime}=-\\dot{N}^{\\prime}h\\nu_{0}=\\lambda N_{0}E_{0}e^{-\\lambda t^{\\prime}}\n$$ \n\nConsider the relativistic transformation of energy, momentum, and time between $s^{\\prime}$ and $s$:\n\n$$\n\\mathrm{d}t=\\gamma\\mathrm{d}t^{\\prime}\n$$ \n\n$$\n\\mathrm{d}E_{l i g h t}=\\gamma(\\mathrm{d}E_{l i g h t}^{\\prime}+v\\mathrm{d}p_{l i g h t}^{\\prime})=\\gamma\\mathrm{d}E_{l i g h t}^{\\prime}\n$$ \n\nUsing the relations:\n\n$$\n\\begin{array}{r}{\\mathrm{d}E_{l i g h t}=P\\mathrm{d}t}\\ {\\mathrm{d}E_{l i g h t}^{\\prime}=P^{\\prime}\\mathrm{d}t^{\\prime}}\\end{array}\n$$ \n\nWe obtain:\n\n$$\nP=P^{\\prime}=\\lambda N_{0}E_{0}e^{-\\lambda\\frac{\\epsilon}{\\gamma}}\n$$ \n\nWhere:\n\n$$\n\\gamma={\\frac{1}{\\sqrt{1-{\\frac{v^{2}}{c^{2}}}}}}\n$$ \n\nConsidering that number density is Lorentz invariant, let the number of photons emitted in $S$, within $\\theta \\sim \\theta + \\mathrm{d}\\theta$, be $\\dot{n}$. \n\nFrom the transformation of the photon energy and momentum, the relationship between $\\pmb\\theta^{\\prime}$ and $\\pmb\\theta$ can be expressed as:\n\n$$\n\\dot{n}\\mathrm{d}t\\times2\\pi\\sin\\theta\\mathrm{d}\\theta=\\frac{\\dot{N}^{\\prime}}{4\\pi}\\mathrm{d}t^{\\prime}\\times2\\pi\\sin\\theta^{\\prime}\\mathrm{d}\\theta^{\\prime}\n$$ \n\n$$\n\\cos\\theta^{\\prime}=\\frac{\\cos\\theta-\\beta}{1-\\beta\\cos\\theta}\n$$ \n\nAnd the transformation of time:\n\n$$\n\\mathrm{d}t^{\\prime}=\\sqrt{1-\\beta^{2}}\\mathrm{d}t\n$$ \n\nSubstituting into the equations, we get:\n\n$$\n\\dot{n}=\\frac{\\dot{N}^{\\prime}}{4\\pi}\\cdot\\frac{\\mathrm{d}\\cos\\theta^{\\prime}}{\\mathrm{d}\\cos\\theta}\\cdot\\frac{\\mathrm{d}t^{\\prime}}{\\mathrm{d}t}=\\frac{\\dot{N}^{\\prime}}{4\\pi}\\cdot\\frac{(1-\\beta^{2})^{\\frac{3}{2}}}{(1-\\beta\\cos\\theta)^{2}}\n$$ \n\nConsidering the frequency transformation of photons:\n\n$$\n\\nu=\\frac{\\sqrt{1-\\beta^{2}}}{1-\\beta\\cos\\theta}\\nu_{0}\n$$ \n\nWe further obtain the power distribution of the emitted light:\n\n$$\nw={\\frac{\\mathrm{d}P}{\\mathrm{d}\\Omega}}={\\dot{n}h\\nu}={\\frac{P}{4\\pi}}{\\frac{(1-\\beta^{2})^{2}}{(1-\\beta\\cos\\theta)^{3}}}={\\frac{\\lambda N_{0}E_{0}}{4\\pi}}{\\frac{(1-\\beta^{2})^{2}}{(1-\\beta\\cos\\theta)^{3}}}\\cdot e^{-\\lambda{\\frac{4}{7}}}\n$$ \n\nRewriting $\\beta$ in terms of velocity:\nIn the relativistic framework, $\\beta = \\frac{v}{c}$ (where $v$ is the velocity and $c$ is the speed of light in a vacuum). Replacing $\\beta$ and $\\lambda$ with velocity-related quantities, the formula becomes:\n\n$$\nw=\\frac{\\mathrm{d}P}{\\mathrm{d}\\Omega}=\\dot{n}h\\nu=\\frac{P}{4\\pi}\\frac{(1 - (\\frac{v}{c})^{2})^{2}}{(1 - \\frac{v}{c}\\cos\\theta)^{3}}=\\frac{\\lambda N_{0}E_{0}}{4\\pi}\\frac{(1 - (\\frac{v}{c})^{2})^{2}}{(1 - \\frac{v}{c}\\cos\\theta)^{3}}\\cdot e^{-\\lambda\\frac{4}{7}}\n$$", "answer": "$$\nw=\\frac{\\lambda N_0 E_0}{4\\pi}\\frac{\\left(1-\\left(\\frac{v}{c}\\right)^2\\right)^2}{\\left(1-\\frac{v}{c}\\cos\\theta\\right)^3}e^{-\\frac{4\\lambda}{7}}\n$$" }, { "id": 105, "tag": "ELECTRICITY", "content": "This problem aims to guide everyone to discover a very interesting way to understand electromagnetic fields. It is known that the Maxwell equations in vacuum are \n\n$$\n\\nabla\\cdot{\\pmb{E}}=\\frac{\\rho}{\\varepsilon_{0}}\n$$\n\n$$\n\\nabla\\times{\\pmb{{E}}}=-\\frac{\\partial{\\pmb{{B}}}}{\\partial t}\n$$\n\n$$\n\\nabla\\cdot\\pmb{B}=0\n$$\n\n$$\n\\nabla\\times\\boldsymbol{B}=\\mu_{0}\\boldsymbol{j}+\\varepsilon_{0}\\mu_{0}\\frac{\\partial\\boldsymbol{E}}{\\partial t}\n$$\n\nTo make the equations of electromagnetic fields more symmetrical, we introduce magnetic charge. Magnetic charge $\\pmb{g}$ will generate a static magnetic field, and the motion of magnetic charge will generate a static electric field. It takes the following form: \n\n$$\n\\boldsymbol{B}={\\frac{g \\boldsymbol e_{r}}{4\\pi r^{2}}}\n$$\n\n$$\n\\pmb{{E}}=\\frac{\\lambda g\\pmb{{v}}\\times\\pmb{\\boldsymbol{e}}_{r}}{4\\pi c r^{2}}\n$$\n\nFind $\\lambda$ by conservation of magnetic charge.", "solution": "$$\n\\pmb{\\varepsilon}\\rightarrow\\pmb{\\varepsilon}\n$$\n\nIt can be obtained:\n\n$$\n{\\pmb B}\\rightarrow\\mu_{0}{\\pmb B}c\n$$\n\n$$\n\\nabla\\cdot\\pmb{\\cal E}=\\rho\n$$\n\n$$\n\\nabla\\times\\pmb{\\mathcal{E}}=-\\frac{1}{c}\\frac{\\partial\\pmb{\\mathcal{B}}}{\\partial t}\n$$\n\n$$\n\\nabla\\cdot\\pmb{B}=0\n$$\n\n$$\n\\nabla\\times\\pmb{B}=\\frac{1}{c}\\pmb{j}+\\frac{1}{c}\\frac{\\partial\\pmb{E}}{\\partial t}\n$$\n\nFollowing the position where the charge density term appears, add the divergence term of the magnetic field and the curl term of the electric field, we obtain:\n\n$$\n\\begin{array}{c}{\\boldsymbol{\\nabla\\cdot E}=\\rho}\\ {\\boldsymbol{\\nabla\\times E}=\\lambda\\frac{k}{c}-\\frac{1}{c}\\frac{\\partial\\boldsymbol{B}}{\\partial t}}\\ {\\boldsymbol{\\nabla\\cdot B}=\\omega}\\ {\\boldsymbol{\\nabla\\times B}=\\frac{1}{c}\\boldsymbol{j}+\\frac{1}{c}\\frac{\\partial E}{\\partial t}}\\end{array}\n$$\n\nThe conservation of magnetic charge implies: \n\n$$\n\\frac{\\partial\\omega}{\\partial t}+\\nabla\\cdot\\boldsymbol{k}=0\n$$\n\nSince the curl of the electric field does not have divergence, the identity is obtained: \n\n$$\n\\lambda\\frac{\\nabla\\cdot\\pmb{k}}{c}-\\frac{1}{c}\\frac{\\partial\\nabla\\cdot\\pmb{B}}{\\partial t}=0\n$$\n\nSubstituting the divergence expression of the magnetic field and comparing it with the conservation of magnetic charge, we find: \n\n$$\n\\lambda=-1\n$$", "answer": "$$\\lambda = -c$$" }, { "id": 617, "tag": "MECHANICS", "content": "Consider each domino as a uniform, smooth slender rod with height $h$ and mass $m$ (neglecting thickness and width), connected to the ground with a smooth hinge. All rods are initially vertical to the ground and arranged evenly in a straight line, with the distance between adjacent rods being $l$, where $l \\ll h$. The coefficient of restitution between the rods is $e = 0$. For convenience, assume that forces between the rods only occur at the instant of collision, and are otherwise negligible. Assume there are infinitely many rods, all initially standing upright. At a certain moment, a disturbance is applied to the first rod, causing it to fall with a certain initial angular velocity. The angular velocity of each rod at the instant it is hit by the previous rod may vary, but as the dominoes collectively fall forward, this post-collision angular velocity will tend toward a certain value. Determine this value $\\omega_{\\infty}$.\n", "solution": "Suppose the initial angular velocity of the $n$th rod when it falls is $\\omega _{n}$. From the beginning of the fall until it collides with the next rod, the angle through which the rod rotates is denoted as $\\theta$. Since $l\\ll h$, we have $$\\theta\\approx\\tan\\theta\\approx\\sin\\theta=\\frac lh$$ Rigid body dynamics: $$\\frac12mgh\\theta=\\frac13mh^2\\ddot{\\theta}$$ Thus: $\\ddot{\\theta}-\\rho^2\\theta=0$ where $\\rho=\\sqrt{\\frac{3g}{2h}}$ The general solution is: $\\theta(t)=C_1\\cosh\\rho t+C_2\\sinh\\rho t$ With initial conditions $$\\theta(0)=0,\\quad\\dot{\\theta}(0)=\\omega_n$$ Therefore, $$\\theta(t)=\\frac{\\omega_n}{\\rho}\\sinh\\rho t$$ The angle rotated by the $n$th rod before colliding with the $n+1$th rod: $$\\theta_0=\\frac{l}{h}$$ Time required: $$t_n=\\dfrac{1}{\\rho}\\sinh^{-1}\\dfrac{\\rho l}{h\\omega_n}$$ Using properties of hyperbolic functions $$\\cosh^2x-\\sinh^2x=1$$ We obtain $$\\dot{\\theta}(t_n)=\\sqrt{\\omega_n^2+\\frac{\\rho^2l^2}{h^2}}$$ At the moment of collision between the $n$th rod and the $n+1$th rod, the force and impulse must be along the horizontal direction. Let $\\omega_n^{\\prime}$ be the angular velocity of the $n$th rod after the collision. Angular momentum: $$\\dot{\\theta}(t_n)=\\omega_n^{\\prime}+\\omega_{n+1}$$ Inelastic collision: $\\sqrt{h^2-l^2}\\omega_{n+1}-h\\omega^{\\prime}\\cdot\\frac{\\sqrt{h^2-l^2}}h=e\\left[h\\dot{\\theta}(t_n)\\cdot\\frac{\\sqrt{h^2-l^2}}h-0\\right]=0$ Solving yields $$\\omega_{n+1}^{2}=\\frac14\\omega_{n}^{2}+\\frac14\\frac{\\rho^{2}l^{2}}{h^{2}}\\\\\\omega_{n+1}^{2}-\\frac13\\frac{\\rho^{2}l^{2}}{h^{2}}=\\frac14\\left(\\omega_{n}^{2}-\\frac13\\frac{\\rho^{2}l^{2}}{h^{2}}\\right)$$ Solving this sequence gives the angular velocity of the $n$th rod just after collision $$\\omega_n=\\sqrt{\\frac{\\omega_0^2}{4^n}+\\frac{\\rho^2l^2}{3h^2}\\left(1-\\frac1{4^n}\\right)}$$ In this formula, $\\omega_0$ is the initial angular velocity when the first rod starts to fall. Let $n\\to\\infty$ to obtain $$\\omega_\\infty=\\frac{\\rho l}{\\sqrt{3}h}=\\frac lh\\sqrt{\\frac g{2h}}$$", "answer": "$$\\omega_\\infty = \\frac{l}{h} \\sqrt{\\frac{g}{2h}}$$" }, { "id": 702, "tag": "ELECTRICITY", "content": "To establish a rectangular coordinate system in an infinitely large three-dimensional space, there are two uniformly positively charged infinite lines at $(a,0)$ and $(-a,0)$, parallel to the $z$-axis, with a charge line density of $\\lambda$; and two uniformly negatively charged lines at $(0,a)$ and $(0,-a)$, also parallel to the $z$-axis, with a charge line density of $-\\lambda$. For each line, the point at a distance $a$ from the line can be taken as where the electric potential is zero. If a positive ion with a mass of $m$ and a charge of $q$ is placed near the origin, and it is restricted to move only within the $xy$ plane, and somehow all four charged lines are made to rotate counterclockwise around the origin with an angular velocity $\\Omega$, with the angular velocity vector directed along the $z$-axis, given that the vacuum permittivity is $\\varepsilon_{0}$, find the minimum value of $\\Omega$, denoted as $\\Omega_{\\min}$, required for the ion to maintain stable equilibrium near the origin.", "solution": "Near the origin, the electric potential can be expressed as $$ \\begin{array}{r l} U&=-\\displaystyle\\frac\\lambda{2\\pi\\varepsilon_{0}}\\ln\\displaystyle\\frac{\\sqrt{(a-x)^{2}+y^{2}}}{a}-\\displaystyle\\frac\\lambda{2\\pi\\varepsilon_{0}}\\ln\\displaystyle\\frac{\\sqrt{(a+x)^{2}+y^{2}}}{a}\\\\ &+\\displaystyle\\frac\\lambda{2\\pi\\varepsilon_{0}}\\ln\\displaystyle\\frac{\\sqrt{x^{2}+(a-y)^{2}}}{a}+\\displaystyle\\frac\\lambda{2\\pi\\varepsilon_{0}}\\ln\\displaystyle\\frac{\\sqrt{x^{2}+(a+y)^{2}}}{a}\\end{array} $$ Using the formula given in the problem, it can be considered that $x\\cdot y$ are both small quantities, thus we have $$ \\begin{array}{r l} \\frac{\\lambda}{2\\pi\\varepsilon_{0}}\\ln{\\frac{\\sqrt{(a-x)^{2}+y^{2}}}{a}} & =\\frac{\\lambda}{4\\pi\\varepsilon_{0}}\\ln\\frac{a^{2}-2a x+x^{2}+y^{2}}{a^{2}} \\\\ & =\\frac{\\lambda}{4\\pi\\varepsilon_{0}}\\ln\\left(1+{\\frac{x^{2}+y^{2}-2a x}{a^{2}}}\\right) \\\\ & =\\frac{\\lambda}{4\\pi\\varepsilon_{0}}\\left[{\\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\\frac{1}{2}}\\left({\\frac{x^{2}+y^{2}-2a x}{a^{2}}}\\right)^{2}\\right] \\\\ & =\\frac{\\lambda}{4\\pi\\varepsilon_{0}}\\left({\\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\\frac{2x^{2}}{a^{2}}}\\right) \\\\ & =\\frac{\\lambda\\left(-x^{2}+y^{2}-2a x\\right)}{4\\pi\\varepsilon_{0}a^{2}} \\end{array} $$ By proceeding similarly, we obtain the approximate results of the other terms, thus we have $$ {\\begin{array}{r l}\\\\&{U(x,y)=-{\\frac{\\lambda\\left(-x^{2}+y^{2}-2a x\\right)}{4\\pi\\varepsilon_{0}a^{2}}}-{\\frac{\\lambda\\left(-x^{2}+y^{2}+2a x\\right)}{4\\pi\\varepsilon_{0}a^{2}}}}\\\\ &{\\qquad+{\\frac{\\lambda\\left(x^{2}-y^{2}-2a y\\right)}{4\\pi\\varepsilon_{0}a^{2}}}+{\\frac{\\lambda\\left(x^{2}-y^{2}+2a y\\right)}{4\\pi\\varepsilon_{0}a^{2}}}}\\\\ &{\\qquad={\\frac{\\lambda\\left(x^{2}-y^{2}\\right)}{\\pi\\varepsilon_{0}a^{2}}}}\\end{array}} $$ After obtaining the electric potential, we can find the components of the electric field intensity along the $x$ and $y$ directions, obtaining $$ {\\begin{array}{r}{E_{x}=-{\\cfrac{\\partial U(x,y)}{\\partial x}}=-{\\cfrac{2\\lambda x}{\\pi\\varepsilon_{0}a^{2}}}}\\ {E_{y}=-{\\cfrac{\\partial U(x,y)}{\\partial y}}={\\cfrac{2\\lambda y}{\\pi\\varepsilon_{0}a^{2}}}}\\end{array}} $$ Therefore, the equation of motion for the positive ion is $$ \\begin{array}{l}{{m{\\ddot{x}}+\\displaystyle\\frac{2q\\lambda}{\\pi\\varepsilon_{0}a^{2}}x=0}}\\ {{m{\\ddot{y}}-\\displaystyle\\frac{2q\\lambda}{\\pi\\varepsilon_{0}a^{2}}y=0}}\\end{array} $$ It is obvious that in the $x$ direction it is stable, whereas in the $y$ direction it is unstable. Choose the $\\Omega$ system that rotates with the four lines, in this system, the four lines are stationary, so the form of the electric force near the origin remains unchanged. Parameterize the electric force, letting $$ \\omega_{0}^{2}={\\frac{2q\\lambda}{\\pi\\varepsilon_{0}a^{2}m}} $$ Considering the Coriolis force and inertial centrifugal force, the particle's dynamic equation in this system is $$ \\begin{array}{c}{m\\ddot{x}=m\\left(-\\omega_{0}^{2}+\\Omega^{2}\\right)x+2m\\Omega\\dot{y}}\\ {m\\ddot{y}=m\\left(\\omega_{0}^{2}+\\Omega^{2}\\right)x-2m\\Omega\\dot{x}}\\end{array} $$ Which can be simplified to $$ \\begin{array}{r}{\\ddot{x}-2\\Omega\\dot{y}+\\left({\\omega_{0}^{2}-\\Omega^{2}}\\right)x=0}\\ {\\ddot{y}+2\\Omega\\dot{x}-\\left({\\omega_{0}^{2}+\\Omega^{2}}\\right)x=0}\\end{array} $$ To discuss the stability of this two-degree-of-freedom system, consider the normal solution, thus we have $$ \\begin{array}{r}{x=A e^{i\\omega t}}\\ {y=B e^{i\\omega t}}\\end{array} $$ Substitute into the above equations to obtain $$ \\begin{array}{r}{\\left(\\omega_{0}^{2}-\\Omega^{2}-\\omega^{2}\\right)A-2i\\omega\\Omega B=0}\\ {2i\\omega\\Omega A-\\left(\\omega_{0}^{2}+\\Omega^{2}+\\omega^{2}\\right)B=0}\\end{array} $$ The characteristic equation is $$ -\\left(\\omega_{0}^{2}-\\Omega^{2}-\\omega^{2}\\right)\\left(\\omega_{0}^{2}+\\Omega^{2}+\\omega^{2}\\right)-4\\omega^{2}\\Omega^{2}=0 $$ Solving for the angular frequencies of the normal modes gives $$ \\begin{array}{r}{\\omega_{1}=\\sqrt{\\Omega^{2}-\\omega_{0}^{2}}}\\ {\\omega_{2}=\\sqrt{\\Omega^{2}+\\omega_{0}^{2}}}\\end{array} $$ The existing normal modes of the system must all be stable for the system to be stable. Therefore, the condition is $$ \\Omega>\\omega_{0}=\\sqrt{\\frac{2q\\lambda}{\\pi\\varepsilon_{0}a^{2}m}} $$", "answer": "$$\\sqrt{\\frac{2q\\lambda}{\\pi\\varepsilon_0 a^2 m}}$$" }, { "id": 367, "tag": "MECHANICS", "content": "There is a smooth elliptical track, and the equation of the track satisfies $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$). On the track, there is a small charged object that can move freely along the track, with mass $m$ and charge $q$. Now, a point charge with charge $Q$ (same sign as $q$) is placed at the origin. Find the period $T$ of the small oscillations of the charged object around its stable equilibrium position.", "solution": "It is evident that at this point, the small object is in a stable equilibrium position at the two vertices of the major axis of the elliptical orbit. When the small object experiences a slight deviation along the orbit from the stable equilibrium position, as shown in the figure below, the distance from the small object to the origin O is \\( r \\), and the angle relative to the x-axis is \\( \\theta \\) (\\( \\theta << 1 \\)).\n\nFrom the elliptical orbit equation, differentiating both sides, we have:\n\\[ \\frac{2x dx}{a^2} + \\frac{2y dy}{b^2} = 0 \\]\n\nThis leads to the elliptic tangent slope at the point:\n\\[ \\tan \\alpha = \\frac{a^2}{b^2} \\tan \\theta \\]\n\nThe angle between the direction of the electrostatic force on the small object and the normal is:\n\\[ \\beta = \\alpha - \\theta \\]\n\nSince the distance of the small object deviating from the stable equilibrium position is very small, we have \\(\\theta, \\alpha, \\beta << 1\\), which implies:\n\\[ \\beta = \\alpha - \\theta \\approx \\left( \\frac{a^2}{b^2} - 1 \\right) \\theta \\]\n\nMagnitude of the electrostatic force on the small object:\n\\[ F = \\frac{1}{4\\pi \\epsilon_0} \\frac{Qq}{r^2} \\]\n\nWhere \\( r \\) satisfies:\n\\[ \\frac{r^2 \\cos^2 \\theta}{a^2} + \\frac{r^2 \\sin^2 \\theta}{b^2} = 1 \\]\n\nThus, we obtain:\n\\[ r = \\sqrt{\\frac{a^2 b^2}{b^2 + (a^2 - b^2) \\sin^2 \\theta}} = a \\left[ 1 + \\frac{a^2 - b^2}{b^2} \\sin^2 \\theta \\right]^{-\\frac{1}{2}} \\approx a \\left[ 1 + \\frac{a^2 - b^2}{b^2} \\theta^2 \\right]^{\\frac{1}{2}} \\]\n\\[ \\approx a \\left( 1 - \\frac{a^2 - b^2}{2b^2} \\theta^2 \\right) \\]\n\nAnd the tangential component of the electrostatic force:\n\\[ F_t = -F \\sin \\beta \\approx -\\frac{Qq}{4\\pi \\epsilon_0 a^2} \\left( 1 - \\frac{a^2 - b^2}{2b^2} \\theta^2 \\right)^{-2} \\cdot \\left( \\frac{a^2}{b^2} - 1 \\right) \\theta \\]\n\\[ \\approx -\\frac{Qq (a^2 - b^2)}{4\\pi \\epsilon_0 a^2 b^2} \\theta \\]\n\nVelocity of the small object:\n\\[ v = \\dot{r} + r \\dot{\\theta} = -\\frac{a (a^2 - b^2)}{2b^2} 2 \\theta \\dot{\\theta} + a \\left( 1 - \\frac{a^2 - b^2}{2b^2} \\theta^2 \\right) \\dot{\\theta} \\approx a \\dot{\\theta} \\]\n\n\\[ F_t = m a_t = m \\dot{v} = m a \\ddot{\\theta} \\]\n\n\\[ ma \\ddot{\\theta} = -\\frac{Qq (a^2 - b^2)}{4\\pi \\epsilon_0 a^2 b^2} \\theta \\]\n\nThis is the standard equation of simple harmonic motion, with the angular frequency of motion being:\n\\[ \\omega = \\sqrt{\\frac{Qq (a^2 - b^2)}{4\\pi \\epsilon_0 m a^3 b^2}} \\]\n\nPeriod of the small vibrations:\n\\[ T = \\frac{2\\pi}{\\omega} = 2\\pi \\sqrt{\\frac{4\\pi \\epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} \\]", "answer": "$$ T = \\frac{2\\pi}{\\omega} = 2\\pi \\sqrt{\\frac{4\\pi \\epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} $$" }, { "id": 497, "tag": "MECHANICS", "content": "\n\n---\n\nThere is a uniform spring with a spring constant of \\(k\\) and an original length of \\(L\\), with a mass of \\(m\\). One end is connected to a wall, and the other end is connected to a mass \\(M\\) block, placed on a horizontal smooth surface for simple harmonic motion. It is known that \\(m \\ll M\\).\n\nThe spring, after being used for a long time, experiences corrosion, causing the spring constant to decrease and the mass to increase. It is approximately assumed that the spring constant and mass uniformly increase or decrease throughout the spring.\n\nA new spring initially has an amplitude of \\(A_0\\). After a period of time, the spring mass becomes \\(\\gamma_1\\) times its original mass, and the spring constant becomes \\((1 - \\gamma_2\\frac{m}{M})\\) times its original value. Under the following model, find the change in amplitude \\(\\Delta A\\).\n\nAssume that the additional mass added to the spring during corrosion moves at the same speed as the spring, meaning the added mass has the same velocity as the point where it is added.\n\n---", "solution": "First, based on the analysis of the change in the system's mechanical energy, through the integration of energy relationships and separation of variables, the relationship between the energy ratio and the stiffness coefficient as well as mass is obtained: \n\\(\\ln\\frac{E}{E_0} = \\frac{1}{2}\\ln\\frac{k'}{k} + \\frac{1}{2}\\ln\\frac{M'}{M_0'}\\), taking the exponential on both sides gives \\(\\frac{E}{E_0} = \\sqrt{\\frac{k'}{k}\\frac{M'}{M_0'}}\\). \nFrom \\(E = \\frac{1}{2}k'A^2\\) and \\(E_0 = \\frac{1}{2}kA_0^2\\), it follows that \\(\\frac{\\frac{1}{2}k'A^2}{\\frac{1}{2}kA_0^2} = \\sqrt{\\frac{k'}{k}\\frac{M'}{M_0'}}\\). \nSubstituting \\(k' = (1 - \\gamma_2\\frac{m}{M})k\\), \\(M' = M + \\frac{1}{3}\\gamma_1 m\\), \\(M_0' = M + \\frac{1}{3}m\\), and approximating using \\(m \\ll M\\): \n\\(\\frac{A^4}{A_0^4} = \\frac{M + \\frac{1}{3}\\gamma_1 m}{M + \\frac{1}{3}m} \\cdot \\frac{1}{1 - \\gamma_2\\frac{m}{M}} \\approx \\left(1 + \\frac{\\gamma_1 - 1}{3}\\frac{m}{M}\\right)\\left(1 + \\gamma_2\\frac{m}{M}\\right) \\approx 1 + \\left(\\frac{\\gamma_1 - 1}{3} + \\gamma_2\\right)\\frac{m}{M}\\). \nExpanding \\(A^4 \\approx A_0^4\\left[1 + \\left(\\frac{\\gamma_1 - 1}{3} + \\gamma_2\\right)\\frac{m}{M}\\right]\\) to the fourth root gives \\(A \\approx A_0\\left[1 + \\frac{1}{4}\\left(\\frac{\\gamma_1 - 1}{3} + \\gamma_2\\right)\\frac{m}{M}\\right]\\). \nThus, the change in amplitude \\(\\Delta A = A - A_0\\) is: \n\\[\n\\Delta A = \\frac{m A_0}{4M}\\left[\\frac{1}{3}(\\gamma_1 - 1) + \\gamma_2\\right]\n\\]", "answer": "\\[\n\\Delta A = \\frac{m A_0}{4M}\\left[\\frac{1}{3}(\\gamma_1 - 1) + \\gamma_2\\right]\n\\] " }, { "id": 457, "tag": "MECHANICS", "content": "A regular solid uniform $N$-sided polygonal prism, with a mass of $m$ and the distance from the center of its end face to a vertex as $l$, is resting on a horizontal table. The axis of the prism is horizontal and points forward. A constant horizontal force $F$ acts on the center of the prism, perpendicular to its axis and large enough to cause the prism to start rotating. As a result, the prism will roll to the right while undergoing completely inelastic collisions with the table. It is known that the coefficient of static friction with the ground is sufficiently large. After a sufficiently long time, the angular velocity after each collision becomes constant. Find this angular velocity.", "solution": "For a regular $N$-sided polygon with the distance from its center to a vertex being $l$, the moment of inertia about the centroid is: \n\n$$\nI_{o}=\\frac{m l^{2}}{2}(\\frac{1}{3}\\sin^{2}\\frac{\\pi}{N}+\\cos^{2}\\frac{\\pi}{N})\n$$ \n\nThe moment of inertia about a vertex is: \n\n$$\nI=\\frac{m l^{2}}{2}(\\frac{1}{3}\\sin^{2}\\frac{\\pi}{N}+\\cos^{2}\\frac{\\pi}{N})+m l^{2}\n$$ \n\nAfter sufficient time has passed: \n\n$$\n\\frac{1}{2}I(\\Omega^{2}-\\omega^{2})=F\\cdot 2l\\sin\\theta\n$$ \n\n$$\nI_{o}\\Omega+m\\Omega a\\cos{2\\theta} a=\\mathbf{I}\\omega\n$$ \n\nWhere:\n\n$$\n\\theta = \\frac{\\pi}{N}\n$$\n\nAfter collision, \n\n$$\n{\\omega}=\\sqrt{\\frac{8F\\sin\\displaystyle\\frac{\\pi}{N}(9-5\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2}}{m l(2+\\displaystyle\\frac{1}{3}\\sin^{2}\\displaystyle\\frac{\\pi}{N}+\\cos^{2}\\displaystyle\\frac{\\pi}{N})((9+7\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2}-(9-5\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2})}}\n$$", "answer": "$$\n{\\omega}=\\sqrt{\\frac{8F\\sin\\displaystyle\\frac{\\pi}{N}(9-5\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2}}{m l(2+\\displaystyle\\frac{1}{3}\\sin^{2}\\displaystyle\\frac{\\pi}{N}+\\cos^{2}\\displaystyle\\frac{\\pi}{N})((9+7\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2}-(9-5\\tan^{2}\\displaystyle\\frac{\\pi}{N})^{2})}}\n$$ " }, { "id": 432, "tag": "ELECTRICITY", "content": "A homogeneous sphere with a mass of $m$ and a radius of $R$ carries a uniform charge of $Q$ and rotates around the $z$-axis (passing through the center of the sphere) with a constant angular velocity $\\omega$. The formula for calculating the magnetic moment is $\\sum_{i} I_{i} S_{i}$, where $I_{i}$ represents the current in a current loop and $S_{i}$ is the area vector of the current loop. The sphere is placed on an infinitely large superconducting plate, with the $z$-axis oriented vertically and perpendicular to the plane of the plate. The gravitational acceleration is $g$. When the system reaches a stable state (the sphere rotates but its center remains stationary), determine the distance $h$ from the center of the sphere to the surface of the plate. Assume the parameters satisfy the conditions for stable levitation, and neglect dissipative effects such as air resistance.\n\nThe vacuum permittivity is given as $\\epsilon_0$, and the vacuum permeability is given as $\\mu_0$.", "solution": "Using the method of electric imaging and magnetic imaging, the electric force is given by\n\n$$\nF_{e}={\\frac{-Q^{2}}{4\\pi\\varepsilon_{0}(2h)^{2}}}={\\frac{-Q^{2}}{16\\pi\\varepsilon_{0}h^{2}}}\n$$\n\nMagnetic force is given by\n\n$$\nF_{m}=-{\\frac{d}{d x}}\\left({\\frac{2\\mu_{0}\\mu^{2}}{4\\pi x^{3}}}\\right)\\mid_{x=2h}={\\frac{3\\mu_{0}Q^{2}R^{4}\\omega^{2}}{800\\pi h^{4}}}\n$$\n\nForce balance condition is\n\n$$\nF_{e}+F_{m}=m g\n$$\n\nThat is\n\n$$\n\\frac{3\\mu_{0}Q^{2}R^{4}\\omega^{2}}{800\\pi h^{4}}=m g+\\frac{Q^{2}}{16\\pi\\varepsilon_{0}h^{2}}\n$$\n\nSolving for \\( h \\), we get\n\n$$\nh={\\sqrt{\\frac{-Q^{2}+{\\sqrt{Q^{4}+{\\frac{96\\pi\\varepsilon_{0}^{2}m g\\mu_{0}Q^{2}R^{4}\\omega^{2}}{25}}}}}{32\\pi\\varepsilon_{0}m g}}}\n$$", "answer": "$$\nh={\\sqrt{\\frac{-Q^{2}+{\\sqrt{Q^{4}+{\\frac{96\\pi\\varepsilon_{0}^{2}m g\\mu_{0}Q^{2}R^{4}\\omega^{2}}{25}}}}}{32\\pi\\varepsilon_{0}m g}}}\n$$\n" }, { "id": 284, "tag": "OPTICS", "content": "Fiber-optic communication has greatly advanced our information technology development. Below, we will briefly calculate how the light carrying information propagates through a rectangular optical fiber. Consider a two-dimensional waveguide (uniform in the direction perpendicular to the paper), extending along the $x$ direction, with variations in refractive index in the $y$ direction. The middle layer is called the core, with a thickness of $b$ and refractive index $n$; the top and bottom layers are called the cladding, with a refractive index of $n+\\delta n$. Assume the light entering the fiber is a monochromatic wave with a wavelength of $\\lambda$.\n\nWhen reflecting at the waveguide interface, the light transitioning from $n_{1}$ to $n_{2}$ has a reflection coefficient:\n\n$$\nr={\\frac{n_{1}\\cos\\theta_{1}-n_{2}\\cos\\theta_{2}}{n_{1}\\cos\\theta_{1}+n_{2}\\cos\\theta_{2}}}\n$$ \n\nwhere $\\theta_1$ is the angle of incidence and $\\theta_2$ is the angle of refraction. According to theoretical calculations, determine the phase shift $\\varphi$ of the light after its first reflection, given that the angle of incidence is $\\theta$, assuming total internal reflection occurs. Assume the plane wave phase is $\\omega t-kx$.", "solution": "Considering that the optical signal should propagate without loss, the light transmission in the optical fiber should involve total internal reflection at the interface. Assume the incident electric field is represented by ${\\widetilde{E}}_{i}$, and the reflected electric field by $\\widetilde{E}_{r}$; we have $\\widetilde{E_{r}}=r\\cdot\\widetilde{E_{i}}$. By substituting Snell's law \n\n$$\nn_{1}\\sin\\theta_{1}=n_{2}\\sin\\theta_{2}\n$$ \n\ninto the expression for the reflection coefficient, we obtain: \n\n$$\nr={\\frac{n_{1}\\cos\\theta_{1}-{\\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\\sin^{2}\\theta_{1}}}}{n_{1}\\cos\\theta_{1}+{\\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\\sin^{2}\\theta_{1}}}}}\n$$ \n\nTotal internal reflection condition (no loss) requires: \n\n$$\nn_{2}\\leqslant n_{1}\\sin\\theta_{1}\n$$ \n\nThus, the reflection coefficient can be expressed in the form of a complex number: \n\n$$\nr={\\frac{n_{1}\\cos\\theta_{1}-j{\\sqrt{{n_{1}}^{2}\\sin^{2}\\theta_{1}-{n_{2}}^{2}}}}{n_{1}\\cos\\theta_{1}+j{\\sqrt{{n_{1}}^{2}\\sin^{2}\\theta_{1}-{n_{2}}^{2}}}}}={\\frac{{n_{1}}^{2}\\cos2\\theta+{n_{2}}^{2}}{{n_{1}}^{2}-{n_{2}}^{2}}}-j{\\frac{2n_{1}\\cos\\theta_{1}{\\sqrt{{n_{1}}^{2}\\sin^{2}\\theta_{1}-{n_{2}}^{2}}}}{{n_{1}}^{2}-{n_{2}}^{2}}}\n$$ \n\nUsing the double angle formula of tangent, the phase shift of the reflection coefficient $\\widetilde{r}=r_{0}\\cdot e^{j\\varphi}$ can be obtained: \n\n$$\n\\varphi=2\\arctan\\left[{\\frac{\\sqrt{\\sin^{2}\\theta_{1}-\\left(n_{2}/n_{1}\\right)^{2}}}{\\cos\\theta_{1}}}\\right]\n$$ \n\nSubstituting the refractive index distribution of the waveguide, the phase shift resulting from total internal reflection at the upper and lower interfaces can be obtained as: \n\n$$\n\\varphi=2\\arctan\\left[\\frac{\\sqrt{\\sin^{2}\\theta-\\left(1+\\delta n/n\\right)^{2}}}{\\cos\\theta}\\right]\n$$", "answer": "$$\\varphi=2\\arctan\\left(\\frac{\\sqrt{\\sin^2\\theta-\\left(1+\\frac{\\delta n}{n}\\right)^2}}{\\cos\\theta}\\right)$$" }, { "id": 39, "tag": "MECHANICS", "content": "A rectangular wooden block of height $H$ and density $\\rho_{1}$ is gently placed on the water surface. The density of the water is $\\rho_{2}$, where $\\rho_{1} < \\rho_{2}$. The gravitational acceleration is $g$. Consider only the translational motion of the wooden block in the vertical direction, neglecting all resistance in the direction of motion and assuming the height of the water surface remains unchanged. Gently press down on the wooden block and then release it. Afterward, the wooden block is neither fully submerged nor does it detach from the water surface. Determine the period of motion of the wooden block $T$;", "solution": "Taking the top at equilibrium as the origin, establish the coordinate system as shown.\n\nLet the position of the top of the block be $y$. The net force on the block is $-\\rho_{1}H S g + \\rho_{2}(H-h_{0}-y)S g = -\\rho_{2}S g y$. The net force on the block is a linear restoring force, and its motion is simple harmonic motion. The equation of motion for the block is\n$$\n\\rho_{1}H S{\\ddot{y}}+\\rho_{2}S g y=0\n$$\nTherefore, the period of the block is\n$$\nT=2\\pi{\\sqrt{\\frac{\\rho_{1}H}{\\rho_{2}g}}}\n$$", "answer": "$$\nT = 2\\pi \\sqrt{\\frac{\\rho_1 H}{\\rho_2 g}}\n$$" }, { "id": 124, "tag": "MECHANICS", "content": "A wedge-shaped block with an inclination angle of $\\theta$, having a mass of $M$, is placed on a smooth tabletop. Another small block of mass $m$ is attached to the top of the wedge using a spring with a spring constant $k$. The natural length of the spring is $L_{0}$, and the surfaces between the two blocks are frictionless. Now, the small block is released from rest at a position $L_{0}$ away from the top of the wedge, allowing it to slide freely downward. Find the oscillation period of the small block $m$.", "solution": "Let's set the vertical and horizontal accelerations of the small wooden block $m$ as $a_{y}$ and $a\\_x$, respectively. Assume that the horizontal acceleration of the wedge-shaped wooden block $M$ is $A_{x}$. We obtain:\n\n$$\n\\begin{array}\n{r l}{m a_{x}+M A_{x}=0.}&{(1)}\\\\\n\n\\ \\frac{a_{x}-A_{x}}{a_{y}}=\\cot \\theta&{(2)}\\\\\n\n\\ {-N\\sin\\theta+F\\cos\\theta=M A_{x}}&{(3)}\\\\\n\n\\ {m g-N\\cos\\theta-F\\sin\\theta=m a_{y}}&{(4)}\\\\\n\n\\ {F=k(l-l_{0})=\\frac{k x}{\\cos\\theta}\\Bigl(1+\\frac{m}{M}\\Bigr)}&{(5)}\n\\end{array}\n$$\nFrom equations (1) and (2), we get:\n\n$$\na_{y}=(a_{x}-A_{x})\\tan\\theta=a_{x}\\left(1+\\frac{m}{M}\\right)\\tan\\theta\n$$ \n\nSubstitute into equation (4),\n\n$$\n\\begin{array}\n{r l}m g-N\\cos\\theta-F\\sin\\theta=m a_{x}\\left(1+\\frac{m}{M}\\right)\\tan\\theta &{(6)}\n\\end{array}\n$$ \n\nFrom equations (3) and (6), we can eliminate $N$, resulting in:\n\n$$\nm g\\sin\\theta-F=m a_{x}\\left(1+\\frac{m}{M}\\right)\\frac{\\sin^{2}\\theta}{\\cos\\theta}-M A_{x}\\cos\\theta=m a_{x}\\left[\\left(1+\\frac{m}{M}\\right)\\frac{\\sin^{2}\\theta}{\\cos\\theta}+\\cos\\theta\\right]\n$$ \n\nSubstitute $F$ from equation (5),\n\n$$\nm g\\sin\\theta-\\frac{k x}{\\cos\\theta}\\Bigl(1+\\frac{m}{M}\\Bigr)=m a_{x}\\left[\\Bigl(1+\\frac{m}{M}\\Bigr)\\frac{\\sin^{2}\\theta}{\\cos\\theta}+\\cos\\theta\\right]\n$$ \n\nWe find that there is a term $a_{x}\\propto-x$ in the expression. Therefore, the angular frequency is:\n\n$$\n\\begin{array}{c c c}{\\displaystyle{\\omega^{2}=\\frac{\\frac{k}{\\cos\\theta}\\left(1+\\frac{m}{M}\\right)}{m\\left[\\left(1+\\frac{m}{M}\\right)\\frac{\\sin^{2}\\theta}{\\cos\\theta}+\\cos\\theta\\right]}=\\frac{k\\left(1+\\frac{m}{M}\\right)}{m[\\left(1+\\frac{m}{M}\\right)\\sin^{2}\\theta+\\cos^{2}\\theta]}}}\\ {\\displaystyle{\\Rightarrow\\omega^{2}=\\frac{k(M+m)}{m[M+m\\sin^{2}\\theta]}}}\\ \n\\end{array}\n$$\nFinally, we get the period:\n\n$$\nT=2\\pi\\sqrt{\\frac{m(M+m\\sin^{2}\\theta)}{k(M+m)}}\n$$", "answer": "$$\nT=2\\pi\\sqrt{\\frac{m(M+m\\sin^2\\theta)}{k(M+m)}}\n$$" }, { "id": 55, "tag": "ELECTRICITY", "content": "In a vacuum, there is an infinitely long, uniformly charged straight line fixed in place, with a charge line density of $\\lambda$. Additionally, there is a dust particle with mass $m$, which can be considered as an isotropic, uniform dielectric sphere with a volume $V$, and a relative permittivity $\\varepsilon_{r}$. It is given that the volume $V$ of the dielectric sphere is very small, the permittivity of vacuum is $\\varepsilon_{0}$, and the following factors can be neglected: gravity, electromagnetic radiation caused by the motion of charges, and relativistic effects. Study the motion of the dust particle under the influence of the charged straight line: \nFind the force acting on the particle and provide its magnitude.", "solution": "Since the volume of the medium sphere $V$ is very small, we can approximate the external electric field $E$ at the medium sphere as a uniform field. Therefore, the medium sphere is uniformly polarized, and let the polarization intensity inside the sphere be $P$. The polarization surface charge density follows the cosine distribution as below (where $\\alpha=0$ corresponds to the direction of the external field $E$): \n\n$$\n\\sigma_{p}(\\alpha)=P\\cos{\\alpha}\n$$ \n\nThe depolarization field generated by the polarized charges inside the medium sphere is a uniform field (the negative sign indicates the opposite direction to the external field $E$): \n\n$$\nE^{\\prime}=-{\\frac{P}{3\\varepsilon_{0}}}\n$$ \n\nThe total electric field inside the sphere is $E+E^{\\prime}$, and according to the polarization law, we have \n\n$$\nP=(\\varepsilon_{r}-1)\\varepsilon_{0}(E+E^{\\prime})=(\\varepsilon_{r}-1)\\varepsilon_{0}\\left(E-\\frac{P}{3\\varepsilon_{0}}\\right)\n$$ \n\nAccording to Gauss's theorem, the field strength generated by a charged line at the medium sphere is \n\n$$\nE(r)=\\frac{\\lambda}{2\\pi\\varepsilon_{0}r}\n$$ \n\nBy combining the above equations, we can solve for the required polarization intensity \n\n$$\nP(r)=\\frac{\\varepsilon_{r}-1}{\\varepsilon_{r}+2}\\frac{3\\lambda}{2\\pi r}\n$$ \n\nThe electric dipole moment of the medium sphere is \n\n$$\np(r)=P(r)\\cdot V=\\frac{\\varepsilon_{r}-1}{\\varepsilon_{r}+2}\\frac{3\\lambda V}{2\\pi r}\n$$ \n\nThus, the force on the particle is \n\n$$\nF(r)=p(r){\\frac{\\mathrm{d}E}{\\mathrm{d}r}}=-{\\frac{3\\left(\\varepsilon_{r}-1\\right)V\\lambda^{2}}{4\\pi^{2}\\left(\\varepsilon_{r}+2\\right)}}{\\frac{1}{r^{3}}}\n$$", "answer": "$$\n-\\frac{3(\\varepsilon_r-1)V\\lambda^2}{4\\pi^2(\\varepsilon_r+2)r^3}\n$$" }, { "id": 419, "tag": "OPTICS", "content": "Consider a thin layer with refractive index $ n_1 $ and thickness $ d $, sandwiched between a medium with refractive index $ n_2 $ on both sides (for simplicity, let $ n = \\frac{n_1}{n_2} < 1 $). Now, suppose a beam of light with wavelength $ \\lambda $ (wavelength inside $ n_2 $) is incident at an angle $ i_2 $, with a refraction angle $ i_1 $. Through multiple reflections and refractions, and given that the surface of the thin layer exhibits total internal reflection, determine the total reflectance $ R_s $ of the light intensity in this scenario, considering only the s-polarized (perpendicular to the plane of incidence) light intensity. Express the result using $ \\lambda $, $ d $, $ i_2 $, and $ n$.", "solution": "Calculate the amplitude after infinite reflections:\n\n\\[\n\\begin{align*}\nE_{r}&=E_{0}\\left(r + tr't\\mathrm{e}^{\\mathrm{j}2\\delta}+t(r')^{3}t\\mathrm{e}^{\\mathrm{j}4\\delta}+\\cdots +t(r')^{2n - 1}t\\mathrm{e}^{\\mathrm{j}(2n - 2)\\delta}+\\cdots\\right)\\\\\n&=E_{0}\\left(r+tr't\\mathrm{e}^{\\mathrm{j}2\\delta}\\left(1 + r'^{2}\\mathrm{e}^{\\mathrm{j}2\\delta}+\\cdots\\right)\\right)\\\\\n&=E_{0}\\left(r + tr't\\mathrm{e}^{\\mathrm{j}2\\delta}\\frac{1}{1 - r'^{2}\\mathrm{e}^{\\mathrm{j}2\\delta}}\\right)\\\\\n&=E_{0}\\left(r-\\frac{(1 - r'^{2})r\\mathrm{e}^{\\mathrm{j}2\\delta}}{1 - r'^{2}\\mathrm{e}^{\\mathrm{j}2\\delta}}\\right)\\\\\n&=E_{0}\\left(\\frac{r(1 - \\mathrm{e}^{\\mathrm{j}2\\delta})}{1 - r'^{2}\\mathrm{e}^{\\mathrm{j}2\\delta}}\\right)\n\\end{align*}\n\\]\n\nSubstitute the complex refractive index to obtain:\n\n\\[\nR_{s}=\\left(1+\\left(\\frac{2\\mathrm{e}^{-\\frac{2\\pi d}{\\lambda}\\sqrt{\\mathrm{sin}i_2^2-n^2}}}{1 - \\mathrm{e}^{-\\frac{4\\pi d}{\\lambda}\\sqrt{\\mathrm{sin}i_2^2-n^2}}}\\frac{2\\sqrt{(1 - \\sin^{2}i_{2})(\\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\\right)^{2}\\right)^{-1}\n\\]", "answer": "\\[\nR_{s}=\\left(1+\\left(\\frac{2\\mathrm{e}^{-\\frac{2\\pi d}{\\lambda}\\sqrt{\\mathrm{sin}i_2^2-n^2}}}{1 - \\mathrm{e}^{-\\frac{4\\pi d}{\\lambda}\\sqrt{\\mathrm{sin}i_2^2-n^2}}}\\frac{2\\sqrt{(1 - \\sin^{2}i_{2})(\\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\\right)^{2}\\right)^{-1}\n\\] " }, { "id": 458, "tag": "ADVANCED", "content": "The incompressible viscous fluid satisfies the Navier-Stokes equations:\n\n$$\n\\frac{\\partial \\vec{v}}{\\partial t} + (\\vec{v} \\cdot \\nabla) \\vec{v} = -\\frac{1}{\\rho} \\nabla p + \\frac{\\mu}{\\rho} \\Delta \\vec{v}\n$$\n\nwhere $\\eta$ is the viscosity of the viscous fluid, $\\rho$ is the density of the viscous fluid, and:\n\nUsing the Navier-Stokes equations, solve the following problem: An incompressible viscous fluid flows through a regular triangular pipe with a side length of $a$ and a length of $l$, with a pressure difference of $\\Delta p$ between the two ends. Determine the volumetric flow rate $Q$.", "solution": "Hypothesis:\n\n$ v = \\frac{\\Delta p}{l} \\frac{2}{\\sqrt{30\\eta}} h_1 h_2 h_3 $\n\nIt can be proved that:\n\n$ \\Delta(h_1 h_2 h_3) = -(h_1 + h_2 + h_3) = -\\frac{\\sqrt{3}}{2} a $\n\nTherefore, the hypothesis satisfies the Navier-Stokes equations and boundary conditions and constitutes a solution. It is evident that the solution is unique, thus $ v = \\frac{\\Delta p}{l} \\frac{2}{\\sqrt{30\\eta}} h_1 h_2 h_3 $ is the only correct solution.\n\nIntegrating yields the flow rate:\n\n$ Q = \\frac{\\sqrt{3} a^4 \\Delta p}{320 l} $", "answer": "$ Q = \\frac{\\sqrt{3} a^4 \\Delta p}{320\\eta l} $" }, { "id": 206, "tag": "ELECTRICITY", "content": "In a zero-gravity space, two coaxial cone surfaces $A$ and $B$ are placed. Assume their common vertex is located at the origin of the coordinate system. The cylindrical coordinate equations are given as: \n\n$$\nA: r = z \\tan\\alpha_{1} \\quad,\\quad B: r = z \\tan\\alpha_{2}\n$$ \n\nwhere $\\alpha_{2} > \\alpha_{1} (\\alpha$ is the angle between the line connecting a point in space to the origin and the positive direction of the common axis $\\hat{z}$). \nAnalysis of the electric potential in the space: It is known that the electric potentials on surfaces $A$ and $B$ satisfy $V_{A} = 0$, $V_{B} = V_{0}$. Solve for the potential distribution $V(\\alpha)$ between the cone surfaces.", "solution": "**Electric Field Distribution Analysis:**\n\nBased on symmetry, select the cone vertex as the origin, and establish spherical coordinates $(R,\\alpha,\\theta)$ to solve the spatial electric field potential distribution. Here, $R=\\sqrt{r^{2}+z^{2}}$ and $R\\sin\\alpha=r$, which correspond to cylindrical coordinates. The spherical coordinate form of Laplace's equation $\\nabla^{2}V=0$ is written as:\n\n$$\n\\frac{1}{R^{2}}\\frac{\\partial}{\\partial R}(R^{2}\\frac{\\partial V}{\\partial R})+\\frac{1}{R^{2}\\sin\\alpha}\\frac{\\partial}{\\partial\\alpha}(\\sin\\alpha\\frac{\\partial V}{\\partial\\alpha})+\\frac{1}{R^{2}\\sin^{2}\\alpha}\\frac{\\partial^{2}V}{\\partial\\theta^{2}}=0\n$$ \n\nConsidering symmetry, ${\\mathrm{{.}}}V=V(\\alpha)$, the equation simplifies to:\n\n$$\n\\frac{1}{R^{2}\\sin\\alpha}\\frac{\\partial}{\\partial\\alpha}(\\sin\\alpha\\frac{\\partial V}{\\partial\\alpha})=0\n$$ \n\nThat is:\n\n$$\n\\sin\\alpha\\frac{\\partial V}{\\partial\\alpha}=A\n$$ \n\nWhere $A$ is a constant. Integrating both sides:\n\n$$\nV=\\int{\\frac{\\mathrm{d}\\alpha}{\\sin\\alpha}}=A\\ln\\tan{\\frac{\\alpha}{2}}+C\n$$ \n\nConsidering boundary conditions:\n\n$$\nV(\\alpha_{1})=0,V(\\alpha_{2})=V_{0}\n$$ \n\nSubstituting these into the solution yields:\n\n$$\nV=\\frac{V_{0}}{\\ln\\frac{\\tan\\frac{\\alpha_{2}}{2}}{\\tan\\frac{\\alpha_{1}}{2}}}\\mathrm{ln}\\frac{\\tan\\frac{\\alpha}{2}}{\\tan\\frac{\\alpha_{1}}{2}}\n$$", "answer": "$$\n\\frac{V_0}{\\ln\\left(\\frac{\\tan\\left(\\frac{\\alpha_2}{2}\\right)}{\\tan\\left(\\frac{\\alpha_1}{2}\\right)}\\right)}\\ln\\left(\\frac{\\tan\\left(\\frac{\\alpha}{2}\\right)}{\\tan\\left(\\frac{\\alpha_1}{2}\\right)}\\right)\n$$" }, { "id": 204, "tag": "ELECTRICITY", "content": "In modern plasma physics experiments, two methods are commonly used to confine negatively charged particles. In the following discussion, relativistic effects and contributions such as delayed potentials are not considered. \n\nIn space, uniformly charged rings with a radius of $R$ and a charge $Q_{0}$ are placed on planes $z=l$ and $z=-l$, respectively. Additionally, a particle with a charge of $-q$ and a mass of $m$ is located at the origin of the coordinate system. The permittivity of vacuum is given as $\\varepsilon_{0}$. \n\nFind the angular frequency $\\omega_{z}$ corresponding to the perturbation stability of the particle in the $\\hat{z}$ direction.", "solution": "Analyzing the electric field at a small displacement $z$ from the equilibrium position along the $z$ axis\n\nwe obtain\n\n$$\n\\vec{E_{z}}=\\frac{Q}{4\\pi\\varepsilon_{0}}\\left(\\frac{l+z}{[R^{2}+(l+z)^{2}]^{\\frac{3}{2}}}-\\frac{l-z}{[R^{2}+(l-z)^{2}]^{\\frac{3}{2}}}\\right)\n$$ \n\n$$\n\\vec{E_{z}}=\\frac{Q}{2\\pi\\varepsilon_{0}}\\frac{\\left(R^{2}-2l^{2}\\right)}{\\left(R^{2}+l^{2}\\right)^{\\frac{5}{2}}}\\vec{z}\n$$ \n\nThe equation of motion is\n\n$$\nm{\\ddot{z}}=-{\\frac{Q q}{2\\pi\\varepsilon_{0}}}{\\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\\frac{5}{2}}}}z\n$$ \n\nFor stability, the condition of the simple harmonic oscillator equation must be satisfied\n\n$$\nR^{2}>2l^{2}\n$$ \n\nSolving gives\n\n$$\n\\omega_{z}=\\sqrt{\\frac{Q q}{2\\pi\\varepsilon_{0}m}\\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\\frac{5}{2}}}}\n$$", "answer": "$$\n\\sqrt{\\frac{Q q}{2 \\pi \\varepsilon_0 m} \\frac{R^2 - 2l^2}{(R^2+l^2)^{5/2}}}\n$$" }, { "id": 478, "tag": "ELECTRICITY", "content": "In an infinitely large, isotropic, linear dielectric medium, there exists a uniform external electric field $\\vec{E}_{0}$. The vacuum permittivity is given as $\\varepsilon_{0}$. \n\nThe dielectric medium is a liquid dielectric with a relative permittivity of $\\varepsilon_{r}$. A solid, ideal conducting sphere with a radius of $R$ is placed inside the dielectric medium. The net charge carried by the conducting sphere is $Q$. It is assumed that the conducting sphere and the dielectric medium are in close contact, such that when considering forces, the free charges on the surface of the conducting sphere and the polarization charges at the interface of the dielectric medium must be regarded as a whole.\n\nDetermine the magnitude of the electrostatic force $F$ acting on the conducting sphere.", "solution": "Spatial Electric Field Distribution\n\n$$\n\\vec{E}=0,~0R\n$$\n\nTotal Surface Charge Density on the Conductor Sphere (including free charge and polarization charge):\n\n$$\n\\sigma(\\theta)=\\frac{Q}{4\\pi\\varepsilon_{r} R^2}+3\\varepsilon_{0}E_{0}\\cos\\theta\n$$\n\nUtilizing the technique where the electric field experienced by the surface charge is the average of the electric fields on both sides, the total electrostatic force obtained by integrating over the sphere's surface is:\n\n$$\nF=\\frac{1}{\\varepsilon_{r}}QE_{0}\n$$", "answer": "$$\nF=\\frac{1}{\\varepsilon_{r}}QE_{0}\n$$ " }, { "id": 71, "tag": "ELECTRICITY", "content": "Given a particle with charge $q$ and mass $m$ moving in an electric field $\\pmb{E}=E_{x}\\pmb{x}+E_{z}\\pmb{z}$ and a magnetic field $\\pmb{B}=B\\pmb{z}$. The initial conditions are: position $(x_{0},y_{0},z_{0})$ and velocity $(v_{\\perp}\\cos\\delta,v_{\\perp}\\sin\\delta,v_{z})$.\n\nWe know that a particle in a uniform magnetic field undergoes circular Larmor gyration, and the center of this rotation is called the guiding center. The drift velocity of this guiding center due to the electric field can be expressed using $\\pmb{E},\\pmb{B}$ and their magnitudes.\n\nProblem: We discuss the case where the magnetic field is uniform, but the electric field is non-uniform. For simplicity, we assume $\\pmb{E}$ is in the $\\pmb{x}$ direction and varies sinusoidally in the $\\pmb{y}$ direction.\n\n$$\n\\pmb{E} \\equiv E_{0}\\cos(k y)\\hat{\\pmb{x}}\n$$\n\nIn reality, such a charge distribution can occur in a plasma during wave propagation. Find the corresponding guiding center drift velocity. It is known that the electric field is very weak, and the particle's initial position is $(x_{0},y_{0},z_{0})$. Approximate to the lowest order that can distinguish from the uniform electric field case.", "solution": "This problem has been modified; the original problem had four questions.\n\nIf there is an electric field present, we find that the motion of the particle will be a combination of two movements: the normal circular Larmor gyration and a drift towards the center of guidance. We can choose the $\\pmb{x}$ axis along the direction of $\\pmb{E}$, so $\\pmb{E}_{y}=0$, and the velocity components related to transverse components can be treated separately. The equation of motion is \n\n$$\nm{\\frac{d v}{d t}}=q({\\pmb{E}}+v\\times{\\pmb{B}})\n$$ \n\nIts $\\pmb{z}$ component is \n\n$$\n\\frac{d v_{z}}{d t}=\\frac{q}{m}E_{z}\n$$ \n\nIntegrating yields \n\n$$\nv_{\\tau}=\\frac{q E_{z}}{m}t+v_{z0}\n$$ \n\nThis is a simple motion along the direction of $\\pmb{B}$. The transverse components of the previous equation are \n\n$$\n\\frac{d\\upsilon_{x}}{d t}=\\frac{q}{m}E_{x}\\pm\\omega_{c}\\upsilon_{y}\n$$ \n\n$$\n\\frac{d\\upsilon_{y}}{d t}=0\\pm\\omega_{c}\\upsilon_{x}\n$$ \n\nwhere $\\omega_{c}\\equiv{\\frac{|g|B}{m}}$ is the Larmor gyration angular frequency. Differentiating these two equations gives \n\n$$\n\\ddot{v}_{x}=-\\omega_{c}^{2}v_{x}\n$$ \n\n$$\n\\ddot{v}_{y}=-\\omega_{c}^{2}(v_{y}+\\frac{E_{x}}{B})\n$$ \n\nWe can rewrite the above equations as \n\n$$\n\\frac{d^{2}}{d t^{2}}(v_{y}+\\frac{E_{x}}{B})=-\\omega_{c}^{2}(v_{y}+\\frac{E_{x}}{B})\n$$ \n\nTherefore, if we use $\\begin{array}{r}{v_{y}+\\frac{E_{x}}{B}}\\end{array}$ instead of ${\\pmb v_{\\pmb y}}$, the equation simplifies to the situation when the electric field is zero. Thus, the two equations can be replaced by \n\n$$\nv_{x}=v_{\\perp}e^{i(\\omega_{\\mathrm{e}}t+\\delta)}\n$$ \n\n$$\n\\upsilon_{y}=-i\\upsilon_{\\perp}e^{i(\\omega_{c}t+\\delta)}-\\frac{E_{x}}{B}\n$$ \n\nwhere $\\delta$ is the angle between the component of the particle's initial velocity in the $_{xy}$ plane ${\\pmb v}_{\\bot}$ and the $\\pmb{x}$ axis (counter-clockwise direction). Further integration and taking the real part yields the equations of motion \n\n$$\nx=x_{0}+\\frac{v_{\\perp}}{\\omega_{c}}\\sin(\\omega_{c}t+\\delta)\n$$ \n\n$$\ny=y_{0}-\\frac{E_{x}}{B}t+\\frac{v_{\\perp}}{\\omega_{c}}[1-\\cos(\\omega_{c}t+\\delta)]\n$$ \n\n$$\nz=\\frac{q E_{z}}{2m}t^{2}+v_{z0}t+z_{0}\n$$ \n\nFrom the equations of motion, it is observed that the Larmor motion of the particle is identical to the case without an electric field, but there is a drift superimposed in the $-y$ direction towards the center of guidance ${\\pmb v}_{{\\pmb g}{\\pmb c}}$ (for $E_{z}>0)$.\n\nTo obtain the general formula for ${\\pmb v}_{{\\pmb g}{\\pmb c}}$, we can solve the equation (1) using vector form. Since we already know that the term $\\textstyle m{\\frac{d v}{d t}}$ only gives rise to circular motion with frequency $\\omega_{c}$, this term can be ignored in equation (1). Thus, equation (1) becomes \n\n$$\n\\pmb{{ E}}+\\pmb{{\\upsilon}}\\times\\pmb{{ B}}=0\n$$ \n\nBy taking the cross product of $\\pmb{B}$ with the above expression, we get \n\n$$\n\\pmb{E}\\times\\pmb{B}=\\pmb{B}\\times(\\pmb{v}\\times\\pmb{B})=\\pmb{v}\\pmb{B}^{2}-\\pmb{B}(\\pmb{v}\\cdot\\pmb{B})\n$$ \n\nThe transverse component of this equation is \n\n$$\nv_{\\perp q c}=E\\times B/B^{2}\\equiv v_{E}\n$$ \n\nWe define this transverse component as $\\pmb{v}_{E}$, which is the electric field drift towards the center of guidance.\n\nThe equation of motion for the particle is \n\n$$\nm{\\frac{d{\\boldsymbol{v}}}{d t}}=q[{\\boldsymbol{E}}({\\boldsymbol{y}})+{\\boldsymbol{v}}\\times{\\boldsymbol{B}}]\n$$ \n\nDecomposing into scalar form, we have \n\n$$\n\\dot{v}_{x}=\\frac{q B}{m}v_{y}+\\frac{q}{m}E_{x}(y)\n$$ \n\n$$\n\\dot{v}_{y}=-\\frac{q B}{m}v_{x}\n$$ \n\nSimplifying yields \n\n$$\n\\ddot{\\upsilon}_{x}=-\\omega_{c}^{2}\\upsilon_{x}\\pm\\omega_{c}\\frac{\\dot{E}_{x}(y)}{B}\n$$ \n\n$$\n\\ddot{v}_{y}=-\\omega_{c}^{2}v_{y}-\\omega_{c}^{2}\\frac{E_{x}(y)}{B}\n$$ \n\nHere, $\\pmb{E_{x}}(\\pmb{y})$ is the electric field at the particle's location. To compute this value, we need to know the particle's trajectory, which is precisely what we attempt to solve initially. If the electric field is weak, as a first approximation, we can estimate $E_{x}(y)$ using the undisturbed trajectory.\n\nThe trajectory in the absence of an electric field is given by \n\n$$\ny=y_{0}\\pm r_{L}\\cos\\omega_{c}t\n$$ \n\nwhere $\\begin{array}{r}{r_{L}=\\frac{m v_{\\perp}}{|q|B}}\\end{array}$ is the Larmor gyration radius, and we obtain \n\n$$\n\\ddot{v}_{y}=-\\omega_{c}^{2}v_{y}-\\omega_{c}^{2}\\frac{E_{0}}{B}\\cos k(y_{0}\\pm r_{L}\\cos\\omega_{c}t)\n$$ \n\nWe seek a solution that is the sum of the gyration at $\\omega_{c}$ and a stationary drift $_{v_{E}}$. Since we are interested in expressing $v_{E}$, we can average over one cycle to eliminate the gyration motion. Thus, the equation yields \n\n$$\n\\bar{v}_{x}=0\n$$ \n\nIn the above equation, the average of the oscillating term $\\ddot{v}_{y}$ is clearly zero, giving us \n\n$$\n\\overline{{\\ddot{v}}}_{y}=0=-\\omega_{c}^{2}\\overline{{v}}_{y}-\\omega_{c}^{2}\\frac{E_{0}}{B}\\overline{{\\cos k(y_{0}\\pm r_{L}\\cos\\omega_{c}t)}}\n$$ \n\nUpon expanding the small approximation and averaging, we find \n\n$$\n\\overline{{v}}_{y}=-\\frac{E_{x}(y_{0})}{B}(1-\\frac{1}{4}k^{2}r_{L}^{2})\n$$ \n\nThus, due to non-uniformity, the usual $\\pmb{{E}}\\times\\pmb{{B}}$ drift is modified to \n\n$$\n\\boxed{v_{E}={\\frac{E\\times B}{B^{2}}}(1-{\\frac{1}{4}}{(k \\frac{m v_{\\perp}}{\\vert{q}\\vert B})}^2)}\n$$ \n\nThe correction term represents the finite Larmor radius effect under the sinusoidal electric field distribution.", "answer": "$$\nv_E = \\frac{E \\times B}{B^2} \\left(1 - \\frac{1}{4} \\left(k \\frac{m v_\\perp}{|q| B}\\right)^2 \\right)\n$$" }, { "id": 259, "tag": "MODERN", "content": "In the inertial frame \\(S\\), at time \\(t = 0\\), four particles simultaneously start from the origin and move in the directions of \\(+x, -x, +y, -y\\), respectively, with velocity \\(v\\). Consider another inertial frame \\(S'\\), which moves relative to \\(S\\) along the positive x-axis with velocity \\(u\\). At the initial moment, the two reference frames satisfy \\(t = t' = 0\\), where \\(t'\\) represents the time in the \\(S'\\) reference frame, and at the initial moment, the origins of the two reference frames coincide. Derive the relationship between the area of the quadrilateral formed by connecting the four particles in reference frame \\(S'\\) and time \\(t'\\). Consider the effects of special relativity, with the speed of light given as \\(c\\).", "solution": "In the inertial frame \\( S \\), four particles start from the origin at \\( t = 0 \\), moving with velocity \\( v \\) in the \\( +x, -x, +y, -y \\) directions, respectively. Another inertial frame \\( S' \\) moves relative to \\( S \\) in the positive \\( x \\)-direction with velocity \\( u \\). At the initial moment, the origins of the two reference frames coincide, and their clocks are synchronized. We need to find the relationship between the area of the quadrilateral formed by connecting the four particles' positions and time \\( t' \\) in frame \\( S' \\), taking into account relativistic effects as per special relativity.\n\n---\n\n1. **Lorentz Transformation**:\n - The coordinate transformation equations:\n \\[\n x' = \\gamma \\left( x - ut \\right), \\quad t' = \\gamma \\left( t - \\frac{ux}{c^2} \\right)\n \\]\n where \\( \\gamma = \\frac{1}{\\sqrt{1 - u^2/c^2}} \\).\n\n---\n\n2. **Coordinate Transformation of Particles**:\n - **Particle 1 (in the \\( +x \\) direction)**:\n \\[\n x_1' = \\gamma \\frac{(v - u)t}{1 - uv/c^2}, \\quad y_1' = 0\n \\]\n Time \\( t_1' = \\gamma t \\left(1 - \\frac{uv}{c^2}\\right) \\)\n\n - **Particle 2 (in the \\( -x \\) direction)**:\n \\[\n x_2' = \\gamma \\frac{-(v + u)t}{1 + uv/c^2}, \\quad y_2' = 0\n \\]\n Time \\( t_2' = \\gamma t \\left(1 + \\frac{uv}{c^2}\\right) \\)\n\n - **Particle 3 (in the \\( +y \\) direction)**:\n \\[\n x_3' = -\\gamma ut, \\quad y_3' = \\frac{vt}{\\gamma}\n \\]\n Time \\( t_3' = \\gamma t \\)\n\n - **Particle 4 (in the \\( -y \\) direction)**:\n \\[\n x_4' = -\\gamma ut, \\quad y_4' = -\\frac{vt}{\\gamma}\n \\]\n Time \\( t_4' = \\gamma t \\)\n\n---\n\n3. **Coordinates at the Same Time \\( t' \\)**:\n - Coordinates of Particle 1 at time \\( t' \\) in frame \\( S' \\):\n \\[\n x_1' = \\frac{(v - u)t'}{1 - uv/c^2}, \\quad y_1' = 0\n \\]\n\n - Coordinates of Particle 2 at time \\( t' \\) in frame \\( S' \\):\n \\[\n x_2' = \\frac{-(v + u)t'}{1 + uv/c^2}, \\quad y_2' = 0\n \\]\n\n - Coordinates of Particles 3 and 4 at time \\( t' \\) in frame \\( S' \\):\n \\[\n x_3' = -ut', \\quad y_3' = \\frac{vt'}{\\gamma}\n \\]\n \\[\n x_4' = -ut', \\quad y_4' = -\\frac{vt'}{\\gamma}\n \\]\n\n---\n\n4. **Calculation of the Area**:\n - Coordinates of the four vertices of the quadrilateral:\n - \\( A \\left( \\frac{(v - u)t'}{1 - uv/c^2}, 0 \\right) \\)\n - \\( B \\left( -ut', \\frac{vt'}{\\gamma} \\right) \\)\n - \\( C \\left( -ut', -\\frac{vt'}{\\gamma} \\right) \\)\n - \\( D \\left( \\frac{-(v + u)t'}{1 + uv/c^2}, 0 \\right) \\)\n\n - Length of the base (distance from \\( A \\) to \\( D \\)):\n \\[\n \\text{Base} = \\left| \\frac{(v - u)t'}{1 - uv/c^2} - \\frac{-(v + u)t'}{1 + uv/c^2} \\right| = 2vt' \\frac{(1 - u^2/c^2)}{1 - (uv/c)^2}\n \\]\n\n - Height (distance from \\( B \\) to the \\( x \\)-axis):\n \\[\n \\text{Height} = \\frac{vt'}{\\gamma}\n \\]\n\n - Area formula:\n \\[\n \\text{Area} = \\text{Base} \\times \\text{Height} = 2vt' \\frac{(1 - u^2/c^2)}{1 - (uv/c)^2} \\times \\frac{vt'}{\\gamma}\n \\]\n Substituting \\( \\gamma = \\frac{1}{\\sqrt{1 - u^2/c^2}} \\) and simplifying:\n \\[\n \\text{Area} = \\frac{2v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}\n \\]", "answer": "$$\\frac{2 v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}$$" }, { "id": 766, "tag": "MECHANICS", "content": " A small ring $A$ with mass $m$ is placed on a smooth horizontal fixed rod and connected to a small ball $B$ with mass $m$ by a thin string of length $l$. Initially, the string is pulled to a horizontal position, and then the system is released from rest. Find: When the angle between the string and the horizontal rod is $\\theta$, what is the tension in the string?\n", "solution": "【Solution】Let the angle between the rope and the rod be $\\theta$, the velocity of the small ring A be $\\boldsymbol{v}_{A}$, and the velocity of the small ball B relative to the small ring be $v'$. Then the velocity of the small ball B relative to the bottom surface $\\boldsymbol{v}_{B}$ is given by the relative motion formula, that is $$ \\boldsymbol{v}_{B} = \\boldsymbol{v'} + \\boldsymbol{v}_{A} \\tag{1} $$ $$ \\upsilon_{Br} = \\upsilon_{A} - \\upsilon^{\\prime} \\sin\\theta \\tag{2} $$ $$ \\upsilon_{B} = \\sqrt{(\\upsilon_{A} - \\upsilon^{\\prime} \\sin\\theta)^2 + (\\upsilon^{\\prime} \\cos\\theta)^2} \\tag{3} $$ By horizontal momentum conservation and mechanical energy conservation for the system, we have $$ m \\upsilon_{A} + m \\upsilon_{Br} = 0 \\tag{4} $$ $$ \\frac{1}{2} m \\upsilon_{A}^2 + \\frac{1}{2} m \\upsilon_{B}^2 - m g l \\sin\\theta = 0 \\tag{5} $$ Where the zero point of gravitational potential energy is taken at the release point of ball B. Substitute equations (2) and (3) into equations (4) and (5) respectively to obtain $$ m\\upsilon_{A}+ m (\\upsilon_{A} - \\upsilon^{\\prime} \\sin\\theta) = 0 \\tag{6} $$ $$ \\frac{1}{2} m \\upsilon_{A}^2 + \\frac{1}{2} m \\left[(\\upsilon_{A} - \\upsilon^{\\prime} \\sin\\theta)^2 + (\\upsilon^{\\prime} \\cos\\theta)^2\\right] - m g l \\sin\\theta = 0 \\tag{7} $$ Jointly solving equations (6) and (7), we obtain $$ \\upsilon^{\\prime 2} = \\frac{4g l \\sin\\theta}{1 + \\cos^2\\theta} \\tag{8} $$ $$ \\upsilon_{A} = \\frac{1}{2} \\sin\\theta \\sqrt{\\frac{4g l \\sin\\theta}{1 + \\cos^2\\theta}} \\tag{9} $$ Let the angle between $ u_{B}$ and $ u^{\\prime}$ be $a$. According to the geometric relationship shown in the figure, using the sine theorem, we have $$ \\frac{\\upsilon_{A}}{\\sin a} = \\frac{\\upsilon^{\\prime}}{\\sin\\left(\\frac{\\pi}{2} + \\theta - a\\right)} \\tag{10} $$ Substituting equations (8) and (9) into equation (10), we can obtain $$ \\frac{\\upsilon^{\\prime}}{\\upsilon_{A}} = \\frac{\\cos(\\theta - a)}{\\sin a} = \\frac{\\cos\\theta \\cos a + \\sin\\theta \\sin a}{\\sin a} = \\frac{2}{\\sin\\theta} \\tag{11} $$ Therefore, we can solve to get $$ \\cot a = \\frac{1 + \\cos^2\\theta}{\\sin\\theta \\cos\\theta} $$ That is $$ \\left\\{ \\begin{array}{l} \\sin a = \\frac{\\sin\\theta \\cos\\theta}{\\sqrt{1 + 3\\cos^2\\theta}} \\\\ \\cos a = \\frac{1 + \\cos^2\\theta}{\\sqrt{1 + 3\\cos^2\\theta}} \\end{array} \\right. \\tag{12} $$ Let the tension in the rope at this time be $T$ and the acceleration of the small ring A be $a$, then we have $ T \\cos\\theta = m a \\tag{13} $ Taking the small ring A as the reference frame, in this non-inertial frame, the forces acting on the small ball B are shown in the figure. In the figure $f_i = m a$, from the figure, we obtain the normal equation for the circular motion of the small ball B as $$ T + m a \\cos\\theta - m g \\sin\\theta = m \\frac{\\upsilon^{\\prime 2}}{l} \\tag{14} $$ Substituting equation (7) into equation (13), we obtain $$ T = \\frac{5 + \\cos^2\\theta}{1 + \\cos^2\\theta} m g \\sin\\theta - m a \\cos\\theta \\tag{15} $$ Jointly solving equations (12) and (14), we obtain $$ T = \\frac{5 + \\cos^2\\theta}{(1 + \\cos^2\\theta)^2} m g \\sin\\theta \\tag{16} $$\n", "answer": "$$\nT = \\frac{5 + \\cos^2 \\theta}{(1 + \\cos^2 \\theta)^2} m g \\sin \\theta\n$$" }, { "id": 716, "tag": "MECHANICS", "content": "The principle of a rotational speed measurement and control device is as follows. At point O, there is a positive charge with an electric quantity of Q. A lightweight, smooth-walled insulating thin tube can rotate around a vertical axis through point O in the horizontal plane. At a distance L from point O inside the tube, there is a photoelectric trigger control switch A. A lightweight insulating spring with a free length of L/4 is fixed at the O end, and the other end of the spring is connected to a small ball with mass m and positive charge q. Initially, the system is in static equilibrium. The thin tube rotates about a fixed axis under the action of an external torque, allowing the small ball to move within the thin tube. When the rotational speed $\\omega$ of the thin tube gradually increases, the small ball reaches point A in the thin tube and just achieves radial equilibrium relative to the thin tube, triggering the control switch. The external torque instantaneously becomes zero, thus limiting excessive rotational speed; at the same time, the charge at point O becomes an equal amount of negative charge -Q. By measuring the position B of the radial equilibrium point of the small ball relative to the thin tube thereafter, the rotational speed can be determined. If the distance OB is measured to be $L/2$, determine the rotational speed $\\omega$ of the thin tube when the ball is at point B. Express the result using the following physical quantities: Electric charge $Q$, ball's electric charge $q$, mass $m$, length $L$, and Coulomb's constant $k$.\n", "solution": "Let the angular velocity of the thin tube be $\\omega_A$. When the small ball is in equilibrium at point A relative to the thin tube, we have: $$ k_0 \\cdot \\frac{3}{4}L - \\frac{k q Q}{L^2} = m L \\omega_A^2 \\tag{1} $$ When the small ball is in equilibrium at point B ($OB = L/2$), with angular velocity $\\omega_B$, we have: $$ k_0 \\cdot \\frac{1}{4}L + \\frac{k q Q}{L^2 / 4} = \\frac{L}{2} m \\omega_B^2 \\tag{2} $$ Conservation of angular momentum gives: $$ m L^2 \\omega_A = m \\cdot \\frac{L^2}{4} \\omega_B \\tag{3} $$ From this, we obtain: $$ \\omega_B = 4 \\omega_A $$ Substituting the above into equation (2): $$ k_0 \\cdot \\frac{1}{4}L + \\frac{k q Q}{L^2 / 4} = 8 L m \\omega_A^2 \\tag{4} $$ Simultaneously solving equations (1) and (4), we find the spring constant: $$ k_0 = \\frac{48 k q Q}{23 L^3} \\tag{5} $$ Substituting equation (5) into equation (2), we can find the angular velocity of the thin tube at this time: $$ \\omega_B = 4 \\sqrt{ \\frac{13 k q Q}{23 m L^3} } \\tag{6} $$\n", "answer": "$$\\omega_B = 4 \\sqrt{\\frac{13 k q Q}{23 m L^3}}$$" }, { "id": 103, "tag": "THERMODYNAMICS", "content": "Solving physics problems involves many techniques and methods: analogy, equivalence, diagrams, and so on. A smart person like you can definitely use these techniques and methods to solve the following problem: \n\nIn space, there is an infinite series of nodes, numbered in order as $\\cdots -3, -2, -1, 0, 1, 2, 3, \\cdots.$ From each node, there are three thermal resistances connected. The connection method is: for node ${\\pmb n}$, there are thermal resistances $R$ to nodes ${\\pmb n} \\pm 1$ respectively. Additionally, if $m$ is even, ${n}$ is odd, and $n = m + 3$, there is a thermal resistance $R$ between node $m$ and node $n$. Find the equivalent thermal resistance $R_{04}$ between node 0 and node 4 when all other nodes remain adiabatic.", "solution": "Noticing that it can be arranged in a zigzag pattern, the diagram is shown above:\n\nFirst, consider $R_{02}$, which is relatively simple. Due to symmetry, the two nodes divide the network into two parts, each side being equivalent to $\\scriptstyle{R^{\\prime}}$. The self-similarity of $\\scriptstyle{R^{\\prime}}$ provides the equivalence shown on the right of the diagram above. Therefore, the 35 resistors are also $\\scriptstyle{R^{\\prime}}$, leading to:\n\n$$\nR^{\\prime}=R+R//(R+R^{\\prime})\n$$\n\nDiscarding the negative root gives:\n\n$$\n\\smash{R^{\\prime}=\\sqrt{3}R}\n$$\n\nThus:\n\n$$\nR_{02}=\\frac{\\sqrt{3}}{2}R\n$$\n\nBy treating $\\scriptstyle{R^{\\prime}}$ as equivalent to the series connection of $\\scriptstyle{R^{\\prime}-R}$ and $\\scriptstyle{R}$, the equivalent network between 0123 is shown in the lower-left diagram above. Consequently:\n\n$$\nR_{01}=(R^{\\prime}-R)//(R^{\\prime}-R+R+R)\n$$\n\n$$\nR_{03}=R//(R^{\\prime}-R+R^{\\prime}-R+R)\n$$\n\nThe calculations yield:\n\n$$\nR_{01}=\\frac{\\sqrt{3}}{3}R\n$$\n\n$$\nR_{03}=\\left(1-\\frac{\\sqrt{3}}{6}\\right)R\n$$\n\nFinally, the equivalent circuit diagram for $R_{04}$ is shown in the lower-right diagram above, which is a balanced bridge. Thus:\n\n$$\nR_{04}=2R//2R^{\\prime}\n$$\n\nResulting in:\n\n$$\nR_{04}=\\left(3-\\sqrt{3}\\right)R\n$$", "answer": "$$\nR_{04} = (3 - \\sqrt{3})R\n$$" }, { "id": 310, "tag": "MODERN", "content": "Consider an ideal mirror moving at relativistic velocity, with mass $m$ and area $S_{\\circ}$. (The direction of photon incidence is the same as the direction of the mirror's motion.)\n\nNow consider the case where the mirror is moving with an initial velocity $\\beta_{0}c$. In this situation, the mirror is unconstrained by external forces, and photons are incident on it with constant power for a certain period of time, with energy $E$. Assuming the mirror's velocity after irradiation is $\\beta_{1}\\mathfrak{c}$, find the expression for $\\beta_{1}$.", "solution": "List the conservation of energy and momentum:\n\n$$\nE+{\\frac{m c^{2}}{\\sqrt{1-{\\beta_{0}}^{2}}}}=E^{\\prime}+{\\frac{m c^{2}}{\\sqrt{1-{\\beta_{1}}^{2}}}}\n$$ \n\n$$\n\\frac{E}{c}+\\frac{m c\\beta_{0}}{\\sqrt{1-\\beta_{0}{}^{2}}}=\\frac{m c\\beta_{1}}{\\sqrt{1-\\beta_{1}{}^{2}}}-\\frac{E^{\\prime}}{c}\n$$ \n\nSolving, we get:\n\n$$\n\\beta_{1}=\\frac{\\left(\\sqrt{\\displaystyle\\frac{1+\\beta_{0}}{1-\\beta_{0}}}+\\frac{2E}{m c^{2}}\\right)^{2}-1}{\\left(\\sqrt{\\displaystyle\\frac{1+\\beta_{0}}{1-\\beta_{0}}}+\\frac{2E}{m c^{2}}\\right)^{2}+1}\n$$", "answer": "$$\\frac{\\left(\\sqrt{\\frac{1+\\beta_0}{1-\\beta_0}}+\\frac{2E}{mc^2}\\right)^2 - 1}{\\left(\\sqrt{\\frac{1+\\beta_0}{1-\\beta_0}}+\\frac{2E}{mc^2}\\right)^2 + 1}$$" }, { "id": 110, "tag": "MECHANICS", "content": "A homogeneous picture frame with a light string, string length $2a$, frame mass $m$, length $2c$, and width $2d$, is hanging on a nail. Ignoring friction, with gravitational acceleration $g$. The mass of the light string is negligible, and it is inextensible, with its ends connected to the two vertices of one long side of the picture frame. Objects other than the picture frame, nail, and light string are not considered. Find the angle $\\alpha$ between the long side of the picture frame and the horizontal line when in equilibrium, where $\\alpha \\in (0, \\frac{\\pi}{2}], c^4 > a^2d^2 - c^2d^2$.", "solution": "As shown in the figure, the trajectory of the nail is an ellipse $\\begin{array}{r}{\\frac{x^{2}}{\\alpha^{2}}+\\frac{y^{2}}{b^{2}}=1}\\end{array}$ where $b={\\sqrt{a^{2}-c^{2}}}$. It is easy to know from geometric relationships that the angle between the line connecting the nail and the center of mass of the picture frame and the $y$-axis is $\\alpha$.\n\nWrite the parametric equations of the ellipse:\n\n$$\n\\begin{array}{r}{x=a\\sin\\theta}\\ {}\\ {y=b\\cos\\theta}\\end{array}\n$$\n\nObtain the distance between the center of mass of the picture frame and the nail, and $\\pmb{\\alpha}$:\n\n$$\nh={\\sqrt{x^{2}+(y+d)^{2}}}={\\sqrt{(a\\sin\\theta)^{2}+(b\\cos\\theta+d)^{2}}}\n$$\n\n$$\n\\alpha=\\arctan({\\frac{x}{y+d}})=\\arctan({\\frac{a\\sin\\theta}{b\\cos\\theta+d}})\n$$\n\nBy the equilibrium of moments, the line connecting the center of mass of the picture frame and the nail is vertical, so the potential energy is:\n\n$$\nE_{p}=-m g h=-m g\\sqrt{(a\\sin\\theta)^{2}+(b\\cos\\theta+d)^{2}}\n$$\n\nCondition for extremum of potential energy:\n\n$$\n{\\frac{d E_{p}}{d\\theta}}=-m g{\\frac{c^{2}\\sin\\theta\\cos\\theta-b d\\sin\\theta}{\\sqrt{(a\\sin\\theta)^{2}+(b\\cos\\theta+d)^{2}}}}=0\n$$\n\nSolve to obtain:\n\n$$\n\\theta_{2}=\\operatorname{arccos}(\\frac{b d}{c^{2}})\n$$\n\nThe condition for the existence of ${\\theta_{2}}$ is:\n\n$$\nc^{2}>{\\sqrt{a^{2}-c^{2}}}d\n$$\n\nSubstitute into the previous equation to obtain:\n\n$$\n {{\\alpha=\\arctan({\\frac{\\sqrt{c^{4}-d^{2}(a^{2}-c^{2})}}{a d}})}}\n$$", "answer": "$$\n\\alpha = \\arctan\\left(\\frac{\\sqrt{c^4 - d^2(a^2 - c^2)}}{ad}\\right)\n$$" }, { "id": 256, "tag": "MODERN", "content": "Two relativistic particles X, each with rest mass $M$, experience a short-range attractive force $F(r) = \\alpha/r^2$ (where $\\alpha$ is a positive constant) in the zero momentum reference frame C, and are bound by this short-range attractive force to form a pair $\\mathrm{X_{2}}$. The speed of light in a vacuum is $c$, and the reduced Planck constant is $\\hbar = h / (2\\pi)$.\n\nIf a bound pair is stable in the ground state $n=1$, the maximum value of $\\alpha$, $\\alpha_{c}$, can be calculated.\n\nTo bombard $\\mathrm{X_{2}}$ and separate $\\mathrm{X}$, neglecting the rest mass and utilizing a particle with ultra-high speed and energy $E$, the experimental setup involves bombarding $\\mathrm{X_{2}}$ at rest in the laboratory frame $\\mathrm{L}$. Set $\\alpha = \\alpha_{c}/2$. If the ultra-high-speed particle causes the transition of $\\mathrm{X_{2}}$ and emit a photon, find the minimum value $E_{1}$ of $E$.", "solution": "In the center-of-mass system, the momentum of the two particles X is the same, denoted as $p$, and at this time, there exists the angular momentum quantization condition: \n\n$$\np\\times{\\frac{r}{2}}\\times2=p r=n\\hbar,n\\in\\mathbb{N}\n$$ \n\nAccording to the dynamics equation, the angular velocity $\\omega=2v/r$. Based on Newton's second law: \n\n$$\np\\omega={\\frac{\\alpha}{r^{2}}}\\rightarrow\\alpha=p\\times{\\frac{2v}{r}}\\times r^{2}=2p r v=2n\\hbar v\n$$ \n\nNote that the attractive potential energy is $V(r)=-\\alpha/r$. Introducing the kinetic energy $T$, and noting that the dynamic mass is $T/c^{2}$, it is straightforward to write the system's energy as: \n\n$$\nE=-{\\frac{\\alpha}{r}}+2T=-{\\frac{\\alpha p}{n\\hbar}}+2T=-{\\frac{\\alpha T v}{n\\hbar c^{2}}}+2T=2T\\left(1-\\left({\\frac{\\alpha}{2n\\hbar c}}\\right)^{2}\\right)\n$$ \n\nBy calculating the system energy, we obtain: \n\n$$\nT={\\frac{M c^{2}}{\\sqrt{1-\\left({\\frac{\\alpha}{2n\\hbar c}}\\right)^{2}}}}\\to E_{n}=2M c^{2}{\\sqrt{1-\\left({\\frac{\\alpha}{2n\\hbar c}}\\right)^{2}}}\n$$ \n\nSince the square root must be meaningful, we establish: \n\n$$\n\\left({\\frac{\\alpha}{2\\hbar c}}\\right)^{2}\\leq1\\rightarrow\\alpha_{c}=2\\hbar c\n$$ \n\nWriting out the binding energy for the ground states with $n=1,2$: \n\n$$\n\\begin{array}{r}{E_{n=1}=2M c^{2}\\sqrt{1-\\left(\\cfrac{\\hbar c}{2\\hbar c}\\right)^{2}}=\\sqrt{3}M c^{2}}\\ {E_{n=2}=2M c^{2}\\sqrt{1-\\left(\\cfrac{\\hbar c}{4\\hbar c}\\right)^{2}}=\\cfrac{\\sqrt{15}}{2}M c^{2}}\\end{array}\n$$ \n\nThe invariant modulus squared of the initial state is: \n\n$$\n\\left(\\sum E\\right)^{2}-\\left(\\sum P c\\right)^{2}=\\left(E+{\\sqrt{3}}M c^{2}\\right)^{2}-E^{2}=3\\left(M c^{2}\\right)^{2}+2{\\sqrt{3}}M c^{2}E\n$$ \n\nIf the system successfully ensures a transition, then in the critical state, the system's total energy in the zero-momentum frame must be no less than the binding energy of the transition to $n=2$: \n\n$$\n3\\left(M c^{2}\\right)^{2}+2{\\sqrt{3}}M c^{2}E\\geq\\left({\\sqrt{15}}M c^{2}\\right)^{2}/4\n$$ \n\nSolving this gives: \n\n$$\nE_{1}={\\frac{15/4-3}{2{\\sqrt{3}}}}M c^{2}={\\frac{\\sqrt{3}}{8}}M c^{2}\n$$", "answer": "$$\nE_1 = \\frac{\\sqrt{3}}{8}Mc^2\n$$" }, { "id": 112, "tag": "ELECTRICITY", "content": "Initially, a conductive dielectric sphere with a free charge of 0 is placed in a vacuum. It is known that the radius of the conducting sphere is $R$, its relative permittivity is $\\varepsilon_{r}$, and its conductivity is $\\sigma$. At the moment $t=0$, a uniform external field ${{\\vec{E}}_{0}}$ is applied around the conducting sphere, and free charge begins to accumulate on the surface of the conductor. The permittivity of the vacuum is $\\varepsilon_0$. Try to determine the total Joule heat generated by the system from the initial state to the steady state.", "solution": "Considering the moment when $\\ell\\to0^{+}$, the polarization properties of the dielectric are prioritized over the conductive properties. At this time, the potential distribution is equivalent to the polarization of a dielectric sphere with a relative dielectric constant of $\\varepsilon_{\\mathsf{r}}$ in a uniform external field.\n$$\n\\sigma_{i}=P\\cos\\theta\n$$\n\nFurther,\n\n$$\n\\bar{P}=(\\varepsilon_{\\mathrm{r}}-1)\\varepsilon_{0}\\left(E_{0}-\\frac{P}{3\\varepsilon_{0}}\\right)\n$$\n\nSolving,\n\n$$\nP=3\\varepsilon_{0}\\frac{\\varepsilon_{r}-1}{\\varepsilon_{r}+2}E_{0}\n$$\n\nEffective electric dipole outside the sphere,\n\n$$\n\\mathscr{P}=\\frac{4}{3}\\pi\\mathscr{R}^{3}P=\\frac{4\\pi\\varepsilon_{0}(\\varepsilon_{r}-1)}{\\varepsilon_{r}+2}E_{0}R^{3}\n$$\n\nSolving,\n\n$$\nU_{i n.}(t\\rightarrow0^{+})=-{\\frac{3}{\\varepsilon_{r}+2}}E_{0}r\\cos\\theta\n$$\n\n$$\n\\zeta_{c x}(t\\rightarrow0^{+})=-E_{0}r\\cos\\theta+\\frac{\\varepsilon_{r}-1}{\\varepsilon_{r}+2}E_{0}\\frac{R^{3}}{r^{2}}\\cos\\theta\n$$\n\nConsidering the moment when it reaches a steady state, the conductive dielectric sphere fully exhibits the properties of an ideal conductor $(\\varepsilon_{r}\\to+\\infty)$, solving,\n\n$$\n\\bar{U}_{i n}(t;\\to\\infty)=0\n$$\n\n$$\nU_{\\mathrm{cx}}(t\\rightarrow\\infty)=-E_{0}\\tau\\cos\\theta+E_{0}\\frac{R^{3}}{r^{2}}\\cos\\theta\n$$\n\nConsidering the moment $t$, when the interior of the conductive dielectric sphere is free from charge, its free charges should be distributed sinusoidally on the surface, assuming the spatial potential distribution satisfies,\n\n$$\nU=\\left\\{\\begin{array}{l l}{-E_{0}r\\cos\\theta+A(t)\\frac{\\cos\\theta}{r^{2}},}&{\\mathrm{if~}RR$ above the ground. Establish a Cartesian coordinate system where the $z$-axis is vertical, the ground is at $z=0$, and the sphere's center is located at $(0,0,H)$. The sphere rotates uniformly with an angular velocity vector along the $x$ direction, and the magnitude is $\\omega$. Consider a droplet of water being flung off the surface of the sphere, and then follows a trajectory under the influence of gravity.\n\nIf the angular velocity is too small, water droplets cannot be flung off the top, so there's a minimum angular velocity $\\omega_{1}$.\n\nWhen the angular velocity $\\omega>\\omega_{1}$, there is still a region on the sphere where droplets cannot be normally flung off. Find the equation of the curve projecting the boundary of this region onto the $xy$-plane (express this as an equation involving $y^2$ in terms of $x$).", "solution": "Take the $x$-axis as the polar axis, and the $z$-axis as the reference axis for the azimuthal angle, i.e.: \n\n$$\n\\begin{aligned}{x}&={R\\cos{\\theta}}\\\\ {y}&={R\\sin{\\theta}\\sin{\\varphi}}\\\\{z}&={H+R\\sin{\\theta}\\cos{\\varphi}}\\end{aligned}\n$$ \n\nThen the velocity of the circular motion at the point: \n\n$$\nv=\\omega r=\\omega R\\sin\\theta\n$$ \n\nThe radius of curvature of the parabolic path of the water droplet follows: \n\n$$\ng\\cos\\varphi={\\frac{v^{2}}{\\rho}}\n$$ \n\nIn order to eject the water drop, it needs to satisfy: \n\n$$\n\\rho>r=R\\sin\\theta\n$$ \n\nSetting the equality gives the boundary condition: \n\n$$\n\\omega^{2}R\\sin\\theta=g\\cos\\varphi\n$$ \n\nExpressing $\\theta, \\varphi$ in terms of $x, y$ and simplifying: \n\n$$\n\\boxed{y^{2}=\\frac{\\omega^{4}}{g^{2}}(R^{2}-x^{2})\\left(x^{2}-R^{2}+\\frac{g^{2}}{\\omega^{4}}\\right)}\n$$", "answer": "$$\n\\boxed{y^{2}=\\frac{\\omega^{4}}{g^{2}}(R^{2}-x^{2})\\left(x^{2}-R^{2}+\\frac{g^{2}}{\\omega^{4}}\\right)}\n$$ " }, { "id": 331, "tag": "OPTICS", "content": "Building the experimental setup for diffraction with a steel ruler: Establish a spatial Cartesian coordinate system, with the positive $x$ direction pointing vertically downward and the positive $z$ direction pointing perpendicularly towards the wall. The angle between the steel ruler and the $z$-axis in the horizontal plane is $\\phi$, and the angle between the steel ruler and the horizontal plane is $\\xi$. A laser is directed along the positive $z$ direction onto the surface of the steel ruler, with the horizontal distance from the point of contact to the wall being $L$. The graduations on the steel ruler are evenly distributed with a spacing of $d$, oriented perpendicular to the direction of the ruler.\n\nBy considering the wave vector continuous along the direction of the scale markings, one can derive the equation of the curve where the diffraction pattern appears (general property); by considering the strong interference between the graduations, one can determine the locations of the primary maxima in the diffraction pattern (specific property).\n\nFor simplicity, assume the steel ruler is completely horizontal, setting $\\xi=0$. Find: the spacing of the primary maxima near the zeroth-order maximum $\\Delta y$.", "solution": "Let the coordinates of the diffraction point be $(x,y)$, we can derive the incident wave vector and the reflected wave vector \n\n$$\n\\left\\{\\begin{array}{c}{\\displaystyle\\boldsymbol{k}=\\frac{2\\pi}{\\lambda}\\hat{\\boldsymbol{z}}}\\ {\\displaystyle\\boldsymbol{k}^{\\prime}=\\frac{2\\pi}{\\lambda}\\frac{x\\hat{\\boldsymbol{x}}+y\\hat{\\boldsymbol{y}}+L\\hat{\\boldsymbol{z}}}{\\sqrt{x^{2}+y^{2}+L^{2}}}}\\end{array}\\right.\n$$ \n\nThe wave vector is continuous along the direction of the scale mark \n\n$$\n({\\pmb k}^{\\prime}-{\\pmb k})\\cdot{\\widehat{\\pmb x}}=0\n$$ \n\nThis yields \n\n$$\nx=0\n$$ \n\nThe interference between scales is extremely strong (assume the order is $n$) \n\n$$\n(k^{\\prime}-k)\\cdot d(\\sin\\phi\\widehat{\\pmb{y}}+\\cos\\phi\\widehat{\\pmb{z}})=2n\\pi\n$$ \n\nSolving, we obtain \n\n$$\n(\\frac{n\\lambda}{d}+\\cos\\phi)^{2}x^{2}+\\left[(\\frac{n\\lambda}{d}+\\cos\\phi)^{2}-\\sin\\phi^{2}\\right]y^{2}-2\\sin\\phi\\cos\\phi L y+\\left[(\\frac{n\\lambda}{d})^{2}+2\\frac{n\\lambda}{d}\\cos\\phi\\right]L^{2}=0\n$$ \n\nThe solution for $y$ is \n\n$$\ny_{n}=\\frac{\\sin\\phi\\cos\\phi+\\sqrt{\\sin\\phi^{2}\\cos\\phi^{2}-\\left[(\\frac{n\\lambda}{d}+\\cos\\phi)^{2}-\\sin\\phi^{2}\\right]\\left[(\\frac{n\\lambda}{d})^{2}+2\\frac{n\\lambda}{d}\\cos\\phi\\right]}}{(\\frac{n\\lambda}{d}+\\cos\\phi)^{2}-\\sin\\phi^{2}}L\n$$ \n\nAccording to the problem statement \n\n$$\n\\frac{n\\lambda}{d}\\ll1\n$$ \n\nThen \n\n$$\ny_{n}\\approx\\tan2\\phi L-\\frac{1+\\tan2\\phi^{2}}{\\sin\\phi}\\frac{n\\lambda}{d}L\n$$ \n\nThe difference in $y$ is \n\n$$\n\\Delta y=\\frac{1+\\tan2\\phi^{2}}{\\sin\\phi}\\frac{\\lambda L}{d}\n$$", "answer": "$$\n\\Delta y = \\frac{1 + \\tan^2(2\\phi)}{\\sin\\phi} \\frac{\\lambda L}{d}\n$$" }, { "id": 498, "tag": "OPTICS", "content": "The interference phenomenon in variable refractive index systems is a new issue of concern in the field of optics in recent years. There is a thin film of non-dispersive variable refractive index medium with a constant thickness \\(d\\), where the refractive index within the film changes linearly with the distance from the film's surface. The refractive index at the lower surface is \\(n_a\\), and at the upper surface is \\(n_b > n_a\\). A beam of parallel light enters the film from the lower surface at an incident angle \\(i\\). Taking the incident point as the origin, the positive direction of the \\(x\\)-axis is along the lower surface of the film to the right, and the positive direction of the \\(y\\)-axis is perpendicular to the film surface upward.\n\nIf the incident light is monochromatic light with wavelength \\(\\lambda\\), and the refractive index of air for this wavelength of light is \\(n_0 = 1\\), the distance between the incident points of two adjacent beams of incident light is \\(2S\\). The incident point of the first beam on the lower surface is denoted as Q, and the point from which it exits from the lower surface again after its first reflection from the upper surface is denoted as P, with OP = \\(2S\\). It is important to note that \\(S\\) is unknown and can be determined through other physical quantities; the final answer should not contain \\(S\\). Ignoring the phase change caused by reflection, find the phase difference \\(\\varphi\\) between the two beams of light.", "solution": "The phase difference is \\(\\varphi = \\frac{2\\pi}{\\lambda} \\Delta L\\), where \\(\\Delta L = L - L_0\\).\n\\(L\\) is the total optical path from the incident point Q through reflection to the exit point P in the medium.\n\\(L_0 = 2S \\sin i\\) is the optical path difference in air for two adjacent incident beams, which can also be understood as: if the light travels a horizontal distance \\(2S\\) in air (with a refractive index of 1), it is the projection of the optical path along the direction parallel to the incident light.\nThis definition of \\(\\Delta L\\) describes the additional optical path difference caused by light traveling in the medium relative to traveling the same equivalent distance in air, which is often used to describe the formation of interference patterns.\n\n1. **Calculate the optical path \\(L\\) in the medium**:\n We have already calculated:\n \\(L = 2 \\int_0^d \\frac{n(y)^2}{\\sqrt{n(y)^2 - \\sin^2 i}} dy\\)\n Substituting variables \\(u = n(y) = n_a + ky\\), \\(dy = \\frac{d}{n_b - n_a} du\\).\n \\(L = 2 \\int_{n_a}^{n_b} \\frac{u^2}{\\sqrt{u^2 - \\sin^2 i}} \\frac{d}{n_b - n_a} du = \\frac{2d}{n_b - n_a} \\int_{n_a}^{n_b} \\frac{u^2}{\\sqrt{u^2 - \\sin^2 i}} du\\)\n Using the integration formula \\(\\int \\frac{x^2 dx}{\\sqrt{x^2 - A^2}} = \\frac{x}{2} \\sqrt{x^2 - A^2} + \\frac{A^2}{2} \\ln|x + \\sqrt{x^2 - A^2}| + C\\), where \\(A = \\sin i\\).\n \\(L = \\frac{2d}{n_b - n_a} \\left[ \\frac{u}{2} \\sqrt{u^2 - \\sin^2 i} + \\frac{\\sin^2 i}{2} \\ln(u + \\sqrt{u^2 - \\sin^2 i}) \\right]_{n_a}^{n_b}\\)\n \\(L = \\frac{d}{n_b - n_a} \\left[ u \\sqrt{u^2 - \\sin^2 i} + \\sin^2 i \\ln(u + \\sqrt{u^2 - \\sin^2 i}) \\right]_{n_a}^{n_b}\\)\n \\(L = \\frac{d}{n_b - n_a} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} + \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right]\\)\n\n2. **Calculate the horizontal displacement \\(2S\\)**:\n We have already calculated:\n \\(2S = 2 \\int_0^d \\frac{\\sin i}{\\sqrt{n(y)^2 - \\sin^2 i}} dy\\)\n \\(2S = 2 \\sin i \\int_0^d \\frac{dy}{\\sqrt{n(y)^2 - \\sin^2 i}}\\)\n \\(2S = 2 \\sin i \\left[ \\frac{d}{n_b - n_a} \\int_{n_a}^{n_b} \\frac{du}{\\sqrt{u^2 - \\sin^2 i}} \\right]\\)\n \\(2S = \\frac{2d \\sin i}{n_b - n_a} \\left[ \\ln(u + \\sqrt{u^2 - \\sin^2 i}) \\right]_{n_a}^{n_b}\\)\n \\(2S = \\frac{2d \\sin i}{n_b - n_a} \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right)\\)\n\n3. **Calculate the optical path difference \\(L_0\\) in air**:\n \\(L_0 = 2S \\sin i\\)\n \\(L_0 = \\left[ \\frac{2d \\sin i}{n_b - n_a} \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right] \\sin i\\)\n \\(L_0 = \\frac{2d \\sin^2 i}{n_b - n_a} \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right)\\)\n\n4. **Calculate the total optical path difference \\(\\Delta L\\)**:\n \\(\\Delta L = L - L_0\\)\n \\(\\Delta L = \\frac{d}{n_b - n_a} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} + \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right] - \\frac{2d \\sin^2 i}{n_b - n_a} \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right)\\)\n \\(\\Delta L = \\frac{d}{n_b - n_a} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} + (\\sin^2 i - 2\\sin^2 i) \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right]\\)\n \\(\\Delta L = \\frac{d}{n_b - n_a} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} - \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right]\\)\n\n5. **Calculate the phase difference \\(\\varphi\\)**:\n \\(\\varphi = \\frac{2\\pi}{\\lambda} \\Delta L\\)\n \\[ \\varphi = \\frac{2\\pi d}{\\lambda (n_b - n_a)} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} - \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right] \\]\n Or written in the form of arccosh:\n \\[ \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) = \\text{arccosh}\\left(\\frac{n_b}{\\sin i}\\right) - \\text{arccosh}\\left(\\frac{n_a}{\\sin i}\\right) \\]\n Therefore:\n \\[ \\varphi = \\frac{2\\pi d}{\\lambda (n_b - n_a)} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} - \\sin^2 i \\left( \\text{arccosh}\\left(\\frac{n_b}{\\sin i}\\right) - \\text{arccosh}\\left(\\frac{n_a}{\\sin i}\\right) \\right) \\right] \\]\n\nFinal answer:\nThe phase difference \\(\\varphi\\) between the two beams is:\n\\[ \\varphi = \\frac{2\\pi d}{\\lambda (n_b - n_a)} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} - \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right] \\]", "answer": "\\[ \\varphi = \\frac{2\\pi d}{\\lambda (n_b - n_a)} \\left[ n_b\\sqrt{n_b^2 - \\sin^2 i} - n_a\\sqrt{n_a^2 - \\sin^2 i} - \\sin^2 i \\ln \\left( \\frac{n_b + \\sqrt{n_b^2 - \\sin^2 i}}{n_a + \\sqrt{n_a^2 - \\sin^2 i}} \\right) \\right] \\]" }, { "id": 737, "tag": "ELECTRICITY", "content": "There is a centrally symmetric magnetic field in space that is directed inward perpendicular to the paper. The magnitude of the magnetic field varies with distance $r$ from the center O, and is given by the formula ${\\mathbf{B}(\\mathbf{r})=\\mathbf{B}_{0}\\left(\\frac{r}{R}\\right)^{n}}$. A charged particle with charge $q$ and mass $m$ moves in uniform circular motion of radius $R$ around O in the plane perpendicular to the magnetic field. If the particle is given a small radial disturbance, it will oscillate slightly along the radial direction. Find the period of these small oscillations.", "solution": "(1) According to Newton's second law: $$ {\\mathfrak{q B}}_{0}v_{0}={\\frac{m v_{0}^{2}}{R}} $$ We obtain: $$ \\mathrm{v}_{0}={\\frac{q B_{0}R}{m}} $$ (2) The principle of angular momentum: $$ \\frac{d\\mathrm{L}}{d t}=\\mathrm{B}_{0}\\Big(\\frac{r}{R}\\Big)^{n}r q\\dot{r} $$ Rearrange terms: $$ dL-\\frac{B_0q}{R^n}r^{n+1}dr=0 $$ Integrate and substitute the result from question (1): $$ \\mathrm{L}-\\frac{\\mathrm{B}_{0}q}{(n+2)R^{n}}r^{n+2}=q B_{0}R^{2}(1-\\frac{1}{n+2}) $$ (3) Conservation of energy: $$ {\\frac{{\\mathbf{L}}^{2}}{2m r^{2}}}+{\\frac{1}{2}}m{\\dot{r}}^{2}=c $$ That is: $$ \\frac{(q B_{0}R^{2}\\Big(1-\\frac{1}{n+2}\\Big)+\\frac{{\\bf B}_{0}q}{(n+2)R^{n}}r^{n+2})^{2}}{2m r^{2}}+\\frac{1}{2}m{\\dot{r}}^{2}=c $$ Differentiate the above equation with respect to time: $$ \\mathrm{m}\\ddot{r}+\\frac{(n+1)B_{0}^{2}q^{2}r^{2n+1}}{m(n+2)^{2}R^{2n}}+\\frac{q^{2}B_{0}^{2}n(n+1)r^{n-1}}{(n+2)^{2}R^{n-2}m}-\\frac{q^{2}B_{0}^{2}R^{4}(n+1)^{2}}{m(n+2)^{2}r^{3}}=0 $$ Let: $$ {\\bf{r}}={\\bf{R}}+\\delta{\\bf{r}} $$ Expand the above equation for small quantities and get: $$ \\mathrm{m}\\ddot{\\delta{r}}+\\frac{(n+1)B_{0}^{2}q^{2}(2n+1)\\delta r}{m(n+2)^{2}}+\\frac{q^{2}B_{0}^{2}n(n+1)(n-1)\\delta r}{(n+2)^{2}m}+\\frac{3 q^{2}B_{0}^{2}R^{4}(n+1)^{2}\\delta r}{m(n+2)^{2}}=0 $$ That is: $$ \\mathrm{m}\\ddot{\\delta{r}}+(n+1)\\frac{q^2B_{0}^2\\delta r}{m}=0 $$ The above equation is the standard formula for simple harmonic motion, yielding: $$ \\displaystyle\\mathbb{T}=\\frac{2\\pi}{\\sqrt{n+1}}\\frac{m}{q B_{0}} $$", "answer": "$$\n\\frac{2\\pi}{\\sqrt{n+1}} \\frac{m}{q B_0}\n$$" }, { "id": 636, "tag": "ELECTRICITY", "content": "Establish a Cartesian coordinate system Oxyz, with a hypothetical sphere of radius $R$ at the origin. Place $n$ rings of radius $R$ along the meridional circles, all passing through the points (0, 0, R) and (0, 0, -R). The angle between any two adjacent rings is $\\frac{\\pi}{n}$, and each ring is uniformly charged with positive charge $Q$. The setup is stationary, and a negative charge $−q$ with mass $m$ is performing circular motion on the equatorial plane at a distance $r_{0}$ from the center of the sphere ($r_0\\gg R$). Now, give the charge a radial disturbance, and attempt to find the difference between its radial oscillation period and angular revolution period. The vacuum permittivity is known as $\\epsilon_0$.", "solution": "Consider the case of a single circular loop, establishing a spherical coordinate system $(r, \\alpha)$ with the loop axis as the polar axis. Then: $$ V = {\\frac{1}{4\\pi\\varepsilon_{0}}}\\int_{0}^{2\\pi}{\\frac{Q d\\theta}{2\\pi}}{\\frac{1}{\\sqrt{r^{2}+R^{2}-2R r \\sin\\alpha \\cos\\Theta}}}={\\frac{Q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{1}{r}}+{\\frac{R^{2}}{r^{3}}}{\\frac{1-3\\cos^{2}\\alpha}{2}}\\right)\\#\\left({\\frac{1}{r}}+{\\frac{R^{2}}{r^{2}}}\\right). $$ Next, consider the superposition of multiple circular loops. Based on geometric relations: $$ V_{i}={\\frac{Q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{1}{r}}+{\\frac{R^{2}}{r^{3}}}{\\frac{1-3\\sin^{2}\\theta\\cos^{2}\\left(\\varphi-{\\frac{\\mathrm{i}\\pi}{\\mathrm{n}}}\\right)}{2}}\\right) $$ Summing results in: $$ V={\\frac{n Q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{1}{r}}-{\\frac{R^{2}}{r^{3}}}{\\frac{1-3\\cos^{2}\\theta}{4}}\\right) $$ On the equatorial plane, $\\textstyle{\\theta={\\frac{\\pi}{2}}}$ $$ V={\\frac{n Q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{1}{r}}-{\\frac{R^{2}}{4r^{3}}}\\right) $$ Assuming the angular momentum is $L=m\\omega_{\\theta}r_{0}^{2}$, the effective potential energy is $$ \\mathrm{V}_{\\mathrm{eff}}=-{\\frac{n Q q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{1}{r}}-{\\frac{R^{2}}{4r^{3}}}\\right)+{\\frac{L^{2}}{2m r^{2}}} $$ The first derivative set to zero: $$ -{\\frac{n Q q}{4\\pi\\varepsilon_{0}}}{\\left({\\frac{-1}{r^{2}}}+{\\frac{3R^{2}}{4r^{4}}}\\right)}-{\\frac{L^{2}}{m r^{3}}}=0 $$ Solving gives: $$ \\omega_{\\theta}^{2}=\\frac{n Q q}{4\\pi\\varepsilon_{0}m r_{0}}\\left(\\frac{1}{r_{0}^{2}}-\\frac{3R^{2}}{4r_{0}^{4}}\\right) $$ The second derivative: $$ -{\\frac{n Q q}{4\\pi\\varepsilon_{0}}}\\left({\\frac{2}{r^{3}}}-{\\frac{3R^{2}}{r^{5}}}\\right)+{\\frac{3L^{2}}{m r^{4}}}=m\\omega_{r}^{2} $$ Solving gives: $$ \\omega_{r}^{2}=\\frac{n Q q}{4\\pi\\varepsilon_{0}m r_{0}}\\left(\\frac{1}{r_{0}^{2}}+\\frac{3R^{2}}{4r_{0}^{4}}\\right) $$ $$ T_{r}-T_{\\oplus}=\\frac{2\\pi}{\\omega^{2}}(\\omega_{\\theta}-\\omega_{r})=-\\frac{3\\pi R^{2}}{2r_{0}^{2}}\\sqrt{\\frac{4\\pi\\varepsilon_{0}m r_{0}^{3}}{n Q q}} $$", "answer": "$$-\\frac{3\\pi R^2}{2r_0^2}\\sqrt{\\frac{4\\pi\\varepsilon_0 m r_0^3}{n Q q}}$$" }, { "id": 420, "tag": "MECHANICS", "content": "B and C are two smooth fixed pulleys with negligible size, positioned on the same horizontal line. A and D are two objects both with mass $m$, connected by a light and thin rope that passes over the fixed pulleys. Initially, the system is stationary, and the distances between AB and CD are both in the direction of gravity, where the distance between AB is $x_{0}$ and the distance between CD is $L$, which is sufficiently large such that D will not touch C within the time frame discussed in the problem. At this moment, ball A is pulled so that the line AB deviates from the vertical line by a small angle $\\theta_{0}$ (without changing the length of AB), and the system begins to move. Taking gravitational acceleration as $\\pmb{g}$, and assuming that A descends very slowly so that we can approximately consider the length of AB remains unchanged, A moves around B with a pendulum-like motion. Try to solve for the amplitude of the oscillation angle $\\theta$ of A when $AB=x$.\n", "solution": "Let the tension in the rope be $\\intercal$. According to Newton's second law for block D, $T-mg=m{\\ddot{x}}$ (1). Using the given assumptions, write the relation of the pendulum's oscillation angle with time: $$ \\theta_{t} = \\theta \\cos(\\sqrt{\\frac{g}{x}}t + \\phi) $$ For $\\mathsf{A}$, write down the expression of Newton's second law: $$ mg-T+m\\dot{\\theta}_{t}^{2}x = m\\ddot{x} $$ Combining (2) and (3), we get: $$ mg-T+{\\textstyle{\\frac{1}{2}}}mg\\theta^{2}=m\\ddot{x} $$ Since the work done on A by the tension in the rope and the gravity of A is equal to the increment in A's kinetic energy, and A's kinetic energy can be separately written as the kinetic energy of vertical motion and the kinetic energy of oscillation (as the oscillation angle is small, the two can be considered independent): $$ (mg-T)\\dot{x}=\\frac{d}{dt}(\\frac{1}{2}m\\dot{x}^{2}+\\frac{1}{2}mg x\\theta^{2}) $$ Combining (1), (4), and (5), we get: $$ g\\theta^{2}=4\\ddot{x} \\quad -\\ddot{x}\\dot{x}=\\frac{d}{dt}(\\frac{1}{2}\\dot{x}^{2}+2x\\ddot{x}) $$ $\\textstyle{\\bar{\\mathcal{F}}}\\psi/{\\bar{\\mathcal{H}}}{\\ddot{x}}dx={\\frac{1}{2}}d{\\dot{x}}^{2}$, simplifying the above equation, we get: $$ \\dot{x}^{2}+2x\\ddot{x}=const $$ Using the initial conditions at $\\mathtt{t}{=}0$: $$ x=x_{0} \\quad \\ddot{x}=g\\theta_{0}^{2}/4 $$ We can rewrite equation (8) as: $$ \\dot{x}^{2}+2x\\ddot{x}=g\\theta_{0}^{2}x_{0}/2 $$ Multiplying both sides of the above equation by ${\\mathsf{dx}}$, and using $\\ddot{\\bf x}dx=\\frac{1}{2}d\\dot{x}^{2}$, we get: $$ \\dot{x}^{2}dx+xd\\dot{x}^{2}=d(x\\dot{x}^{2})=\\frac{1}{2}g\\theta_{0}^{2}x_{0}dx $$ Integrating, we obtain: $$ \\dot{x}^{2}=\\frac{1}{2}g\\theta_{0}^{2}x_{0}(x-x_{0})/x $$ Differentiating both sides with respect to time, eliminating the first-order derivative of $\\mathsf{x}$ with respect to $\\mathtt{t}$, we get: $$ \\ddot{x}=\\frac{1}{4}g\\theta_{0}^{2}\\frac{x_{0}^{2}}{x^{2}} $$ Combining with equation (6), we can solve for the amplitude of the oscillation angle $\\theta$: $$ \\theta=\\theta_{0}\\frac{x_{0}}{x} $$\n", "answer": "$$\n\\theta=\\theta_{0}\\frac{x_{0}}{x}\n$$ " }, { "id": 228, "tag": "MECHANICS", "content": "AB is a uniform thin rod with mass $m$ and length $l_{2}$. The upper end B of the rod is suspended from a fixed point O by an inextensible soft and light string, which has a length of $l_{1}$. Initially, both the string and the rod are hanging vertically and at rest. Subsequently, all motion occurs in the same vertical plane, with all angles rotating counterclockwise being positive.\n\nAssume at a certain moment, the angles between the string, the rod, and the vertical direction are $\\theta_{1}$ and $\\theta_{2}$ respectively. The angular velocities of the string around the fixed point O and the rod around its center of mass are $\\omega_{1}$ and $\\omega_{2}$ respectively. Determine the angular acceleration $\\alpha_{2}$ of the rod around its center of mass, expressing the answer only in terms of the physical quantities given in the problem and the gravitational acceleration $g$.", "solution": "In the vertical plane where points O, B, and A are located, establish a plane coordinate system with point O as the origin, the horizontal ray to the right as the $\\mathbf{X}$ axis, and the vertical ray upward as the $\\mathbf{y}$ axis. The acceleration of the pole's center of mass C, denoted as $(\\boldsymbol{a}_{\\mathrm{G}x},\\boldsymbol{a}_{\\mathrm{G}y})$, satisfies the center of mass motion theorem\n\n$$\nm a_{\\mathrm{c}x}=-T\\sin\\theta_{1},m a_{\\mathrm{c}y}=-m g+T\\cos\\theta_{1}\\textcircled{7}\n$$ \n\nIn this equation, $T$ is the magnitude of the tension in the rope. At the same time, the pole rotates around its center of mass under the moment caused by the rope tension $T$ relative to the center of mass. By the rotational theorem, we have\n\n$$\n\\frac{1}{12}m l_{2}^{2}\\alpha_{2}=-T\\frac{1}{2}l_{2}\\sin(\\theta_{2}-\\theta_{1})\\textcircled{8}\n$$ \n\nFrom the geometric relationship, we have\n\n$$\nx_{\\mathrm{{B}}}(t)=x_{\\mathrm{{C}}}(t)-{\\frac{1}{2}}l_{2}\\sin\\theta_{2}(t),y_{\\mathrm{{B}}}(t)=y_{\\mathrm{{C}}}(t)+{\\frac{1}{2}}l_{2}\\cos\\theta_{2}(t)\n$$ \n\nDifferentiating both sides of the above equations with respect to time $t$, the velocity of point B satisfies the conditions\n\n$$\n{\\boldsymbol{v}}_{\\mathrm{{B}}x}(t)={\\boldsymbol{v}}_{\\mathrm{{C}}x}(t)-{\\frac{1}{2}}{\\boldsymbol{\\omega}}_{2}(t){\\boldsymbol{l}}_{2}\\cos\\theta_{2}(t),\\quad{\\boldsymbol{v}}_{\\mathrm{{B}}y}(t)={\\boldsymbol{v}}_{\\mathrm{{C}}y}(t)-{\\frac{1}{2}}{\\boldsymbol{\\omega}}_{2}(t){\\boldsymbol{l}}_{2}\\sin\\theta_{2}(t),\n$$ \n\nDifferentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B satisfies the conditions\n\n$$\na_{\\mathrm{B}x}=a_{\\mathrm{C}x}-\\frac{1}{2}\\alpha_{2}l_{2}\\cos\\theta_{2}+\\frac{1}{2}\\omega_{2}^{2}l_{2}\\sin\\theta_{2},a_{\\mathrm{B}y}=a_{\\mathrm{C}y}-\\frac{1}{2}\\alpha_{2}l_{2}\\sin\\theta_{2}-\\frac{1}{2}\\omega_{2}^{2}l_{2}\\cos\\theta_{2}\n$$ \n\nMeanwhile, point B rotates around point O with a fixed axis under the condition of an inextensible rope, thus\n\n$$\nx_{\\mathrm{{B}}}(t)=l_{\\mathrm{{1}}}\\sin\\theta_{\\mathrm{{1}}}(t),y_{\\mathrm{{B}}}(t)=-l_{\\mathrm{{1}}}\\cos\\theta_{\\mathrm{{1}}}(t)\n$$ \n\nDifferentiating both sides of the above equations with respect to time $t$, the velocity of point B also satisfies the conditions\n\n$$\nV_{_{\\mathrm{B}x}}(t)=\\omega_{1}(t)l_{1}\\cos\\theta_{1}(t),~V_{_{\\mathrm{B}y}}(t)=\\omega_{1}(t)l_{1}\\sin\\theta_{1}(t)\n$$ \n\nDifferentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B also satisfies the conditions\n\n$$\na_{\\scriptscriptstyle{\\mathrm{B}x}}=\\alpha_{1}l_{1}\\cos\\theta_{1}-\\alpha_{1}^{2}l_{1}\\sin\\theta_{1},a_{{\\scriptscriptstyle{\\mathrm{B}y}}}=\\alpha_{1}l_{1}\\sin\\theta_{1}+\\omega_{1}^{2}l_{1}\\cos\\theta_{1}\n$$ \n\nThe angular accelerations of the rope around the suspension point and the pole around the center of mass can be solved as\n\n$$\n\\begin{array}{r l}&{\\alpha_{2}=\\cfrac{3\\sin(\\theta_{1}-\\theta_{2})\\left[2g\\cos\\theta_{1}+2l_{1}\\omega_{1}^{2}+l_{2}\\omega_{2}^{2}\\cos(\\theta_{1}-\\theta_{2})\\right]}{l_{2}\\left[1+3\\sin^{2}(\\theta_{1}-\\theta_{2})\\right]}}\\end{array}\n$$", "answer": "$$\n\\frac{3\\sin(\\theta_1-\\theta_2)\\left[2g\\cos\\theta_1+2l_1\\omega_1^2+l_2\\omega_2^2\\cos(\\theta_1-\\theta_2)\\right]}{l_2\\left[1+3\\sin^2(\\theta_1-\\theta_2)\\right]}\n$$" }, { "id": 450, "tag": "MECHANICS", "content": "Xiao Ming discovered an elliptical plate at home with semi-major and semi-minor axes of \\( A \\) and \\( B \\), respectively. Using one focus \\( F \\) as the origin, a polar coordinate system was established such that the line connecting the focus and the vertex closest to the focus defines the polar axis direction. Through extremely precise measurements, it was found that the mass surface density satisfies the equation \\(\\sigma = \\sigma_{0}(1 + e\\cos\\varphi)^3\\), where \\( e \\) is the eccentricity of the ellipse. When the plate's major axis is positioned vertically and released from rest on a sufficiently rough tabletop, the plate moves only within the plane it lies in. Find the angular acceleration \\(\\beta\\) of the plate when the major axis becomes horizontal.", "solution": "At this time, the angular velocity is $\\omega$, the angular acceleration is $\\beta$, and the acceleration of the center of mass is $a_{x}, a_{y}$.\n\nThe moment of inertia about the instantaneous center $\\mathrm{P}$ is $I_{P}=I+m A^{2}$.\n\nFrom the conservation of energy, the contact condition at point P gives the acceleration \n$a = \\omega^2 \\rho = \\omega^2 \\frac{A^2}{B}$.\n\nConsidering the rotational theorem about the instantaneous center: \n$I_{P} \\vec{\\beta} = \\vec{M} - m \\vec{r}_{c} \\times \\vec{a}$\n\nWe obtain: \n\n$$\nI_{P} \\beta = m \\sqrt{A^2 - B^2} \\bigg(g + \\omega^{2} \\frac{A^{2}}{B} \\bigg)\n$$\n\n$$\n\\beta = 4A \\sqrt{A^2 - B^2} g \\frac{2A^{4} + 2A^{3} \\sqrt{A^2 - B^2} - A^{3}B + B^{4}}{(2A^{3} + B^{3})^{2}B}\n$$", "answer": "$$\n\\beta=4A \\sqrt{A^2 - B^2} g\\frac{2A^{4}+2A^{3}\\sqrt{A^2 - B^2}-A^{3}B+B^{4}}{(2A^{3}+B^{3})^{2}B}\n$$" }, { "id": 708, "tag": "OPTICS", "content": "\\\"Choose a sodium lamp for Young's double-slit interference experiment. The wavelengths of the sodium lamp's double yellow lines are $\\lambda_{1}$ and $\\lambda_{2}$ respectively. Due to a very small wavelength difference, the higher-order interference fringes on the screen will become blurred. Neglecting the width of the slits themselves, introduce the contrast function $\\gamma \\equiv\\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$ to measure the clarity of the interference. When $\\gamma = 0$, the fringes will become blurred. Additionally, if the background light intensity becomes $1/e$ of the system's maximum possible light intensity, it will also cause the observer to subjectively perceive the interference fringes as too dim, leading to blurriness. If the width of the double slits is $d$ and the horizontal distance from the screen to the double slits is $L$. Use a filter to filter out the light with a wavelength of $\\lambda_{1}$. Due to the broadening of the spectral line caused by the thermal motion of molecules in the sodium lamp, if the temperature of the sodium lamp is $T$ and the mass of the sodium atom is $m$, try to calculate the position where the fringes become blurred for the first time. Assume that the sodium atoms follow the Maxwell velocity distribution law (and that the problem can be solved using the far-field condition).", "solution": "Examine the effect of monochromatic light on the screen: $$ U = U_0 \\left( e^{i k \\cdot \\frac{d x}{L}} + 1 \\right) $$ The resulting light intensity is: $$ I = U \\cdot U^* = U_0^2 \\left(2 + 2\\cos\\left(k \\cdot \\frac{d x}{L}\\right)\\right) = I_0 \\left(1 + \\cos\\left(k \\cdot \\frac{d x}{L}\\right)\\right) $$ For Doppler frequency shift due to thermal motion, examine the Doppler effect of electromagnetic waves: $$ \\frac{k}{k_0} = \\frac{\\sqrt{1 - \\beta^2}}{1 - \\beta \\cos\\theta} \\approx 1 + \\beta \\cos\\theta = 1 + \\frac{v_x}{c} \\quad $$ For particles with horizontal velocity in the interval $v - v + \\mathrm{d}v$ (in terms of emitted light intensity), we have: $$ \\frac{\\mathrm{d}I}{I_0} = f_x(v) \\, \\mathrm{d}v \\quad $$ For light intensity on the screen: $$ I \\propto \\int f_x(v)\\,\\mathrm{d}v \\cos\\frac{k_0 \\left(1 + \\frac{v_x}{c}\\right) d x}{L}\\ = \\int f_x(v)\\,\\mathrm{d}v \\left( \\cos\\frac{k_0 d x}{L}\\cos\\frac{v k_0 d x}{L c}\\ - \\sin\\frac{k_0 d x}{L}\\sin\\frac{v k_0 d x}{L c} \\right) \\quad $$ Notice that the second term is an odd function, and its integral equals 0. The integral of the first term is calculated as: $$ s_1 = \\operatorname{Re} \\int_{-\\infty}^{\\infty} \\sqrt{\\frac{m}{2\\pi k T}} e^{- \\frac{m v^2}{2k T} + i \\frac{v k_0 d x}{L c}} \\, \\mathrm{d}v = e^{\\frac{\\frac{k_0 d x^2}{L c}}{\\frac{2m}{k T}}} \\quad $$ When it decreases to $\\frac{1}{e}$, the background light will become blurred, thus: $$ \\frac{\\frac{k_0 d x^2}{L c}}{\\frac{2m}{k T}} = 1 \\quad $$ Solve for: $$ x = \\sqrt{\\frac{2m}{k T}} \\cdot \\frac{L c \\lambda_2}{2\\pi d} \\quad $$", "answer": "$$\nx = \\sqrt{\\frac{2m}{k T}} \\cdot \\frac{L c \\lambda_2}{2\\pi d}\n$$" }, { "id": 649, "tag": "ELECTRICITY", "content": "In modern plasma physics experiments, negative particles are often constrained in two ways. In the following discussion, we do not consider relativistic effects or retarded potentials. Uniformly charged rings with radius $R$ are placed on planes $z=l$ and $z=-l$ in space, respectively. The rings are perpendicular to the $z$ axis, and each carries a charge of $Q_{0}$. A particle with charge $-q$ and mass $m$ is placed at the origin of the coordinate system. Given that $Q_0, q > 0$, the vacuum permittivity is $\\epsilon_{0}$ and the vacuum permeability is $\\mu_{0}$. To keep the point charge stable in both the $\\hat{z}$ and $\\hat{r}$ directions, we consider rotating the two charged rings in the $\\hat{z}$ direction with a constant angular velocity $\\Omega$. Given a small disturbance to the point charge at the origin, provide the minimum $\\Omega$ required for the particle to remain stable in all directions.", "solution": "Analyzing the magnetic field at a small displacement $z$ along the $z$-axis away from the coordinate origin $$ \\vec{B_{z}}={\\frac{\\mu_{0}Q\\Omega R}{4\\pi}}\\left({\\frac{R}{[R^{2}+(l+z)^{2}]^{\\frac{3}{2}}}}+{\\frac{R}{[R^{2}+(l-z)^{2}]^{\\frac{3}{2}}}}\\right) $$ yields $$ \\vec{B_{z}}=\\frac{\\mu_{0}Q\\Omega}{2\\pi}\\frac{R^{2}}{(R^{2}+l^{2})^{\\frac{3}{2}}}\\hat{z} $$ Observing that near the origin, $\\vec{B}$ can be considered as a uniform field along the $\\hat{z}$ direction, denoted as $B_{0}$, consider a particle starting from the origin. By the angular momentum theorem $$ \\frac{\\mathrm{d}\\vec{L}}{\\mathrm{d}t}=-q B_{0}r\\dot{r}\\vec{\\varphi} $$ $$ L+\\frac{1}{2}q B_{0}r^{2}=0 $$ $$ v_{\\varphi}=-\\frac{q B_{0}r}{2m} $$ Substitute into the conservation of energy equation to obtain the system's effective potential energy $$ V_{e f f}=[{\\frac{\\mu_{0}^{2}q^{2}Q^{2}\\Omega^{2}}{32m\\pi^{2}}}{\\frac{R^{4}}{(R^{2}+l^{2})^{3}}}-{\\frac{q Q}{8\\pi\\varepsilon_{0}}}{\\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\\frac{5}{2}}}}]r^{2}+{\\frac{q Q}{4\\pi\\varepsilon_{0}}}{\\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\\frac{5}{2}}}}z^{2} $$ For the system to be stable along $\\hat{r},\\hat{z}$, the coefficients should both be greater than 0, solving gives $$ R^{2}>2l^{2} $$ $$ \\Omega>\\frac{2\\pi}{\\mu_{0}R^{2}}\\sqrt{\\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\\frac{1}{2}}}{\\pi\\varepsilon_{0}Q q}} $$ That is, the minimum value is $$ \\frac{2\\pi}{\\mu_{0}R^{2}}\\sqrt{\\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\\frac{1}{2}}}{\\pi\\varepsilon_{0}Q q}} $$", "answer": "$$\n\\frac{2\\pi}{\\mu_0 R^2} \\sqrt{\\frac{m(R^2 - 2l^2)(R^2 + l^2)^{1/2}}{\\pi \\varepsilon_0 Q q}}\n$$" }, { "id": 688, "tag": "ELECTRICITY", "content": "In space, there is an axisymmetric magnetic field, with the direction of the magnetic field pointing outward perpendicular to the plane, and its magnitude depends only on the distance from the center of symmetry, $B(r) = B_{0} \\left(\\frac{r}{r_{0}}\\right)^{n}$. A particle with mass $m$ and charge $q$ moves in a circular motion with radius $r_{0}$ around the center of symmetry under the influence of the magnetic field. Find: if the charged particle is given a small radial disturbance, determine the period $T$ of the small radial oscillations.", "solution": "According to symmetry, the system in this problem has conserved canonical angular momentum with respect to the magnetic field's center of symmetry. To find the canonical angular momentum, we list the corresponding rate of change equation: $$ \\frac{d L}{d t} = q v_{r} B_{0} \\Big(\\frac{r}{r_{0}}\\Big)^{n} r $$ Rearranging terms and integrating yields: $$ d L - \\frac{q B_{0}}{r_{0}^{n}} r^{n+1} d r = 0 \\to L - \\frac{q B_{0}}{(n+2) r_{0}^{n}} r^{n+2} = C = \\frac{n+1}{n+2} q B_{0} r_{0}^{2} $$ (Conservation quantity 6 points) Thus, we obtain the function of angular momentum and $r$: $$ L = L(r) = \\frac{q B_{0}}{n+2} \\biggl(\\frac{r^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\\biggr) $$ The energy conservation of the charged particle gives: $$ \\frac{1}{2} m \\dot{r}^{2} + \\frac{1}{2} m v_{\\theta}^{2} = \\frac{1}{2} m \\dot{r}^{2} + \\frac{L(r)^{2}}{2m r^{2}} = E $$ (Energy conservation 6 points) Give the particle a radial perturbation, let $r = r_{0} + \\delta r$, $\\dot{r} = (\\dot{\\delta r})$, resulting in: $$ \\frac{1}{2} m (\\dot{\\delta r})^{2} + \\frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} \\frac{\\left(\\frac{(r_{0}+\\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\\right)^{2}}{(r_{0}+\\delta r)^{2}} = E $$ To analyze the oscillation, we need to expand the potential energy and retain terms up to the second order, so we first look at the denominator: $$ \\begin{array}{r l} &\\left(\\frac{(r_{0}+\\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\\right)^{2} = \\left(r_{0}^{2}\\Bigl(1+\\frac{\\delta r}{r_{0}}\\Bigr)^{n+2} + (n+1) r_{0}^{2}\\right)^{2} \\\\ &\\qquad = r_{0}^{4} \\Biggl(1 + (n+2) \\frac{\\delta r}{r_{0}} + \\frac{(n+2)(n+1)}{2} \\Bigl(\\frac{\\delta r}{r_{0}}\\Bigr)^{2} + (n+1) \\Biggr)^{2} \\\\ &\\qquad = r_{0}^{4} (n+2)^{2} \\Biggl(1 + \\frac{\\delta r}{r_{0}} + \\frac{n+1}{2} \\Bigl(\\frac{\\delta r}{r_{0}}\\Bigr)^{2} \\Biggr)^{2} = r_{0}^{4} (n+2)^{2} \\left(1 + 2 \\frac{\\delta r}{r_{0}} + (n+2) \\frac{\\delta r}{r_{0}}\\right)^{2} \\end{array} $$ In the above equation, we repeatedly used the small quantity approximation formula $\\begin{array}{r} (1+x)^{n} = 1 + n x + \\frac{n(n-1)}{2}x^{2} + \\ldots, \\end{array}$ so the potential energy term can be written as: $$ \\frac{\\left(\\displaystyle\\frac{(r_{0}+\\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\\right)^{2}}{(r_{0}+\\delta r)^{2}} = r_{0}^{2} (n+2)^{2} \\left(1 + 2 \\frac{\\delta r}{r_{0}} + (n+2) \\Big(\\frac{\\delta r}{r_{0}}\\Big)^{2}\\right) \\left(1 + \\frac{\\delta r}{r_{0}}\\right)^{-2} $$ $$ = r_{0}^{2} (n+2)^{2} \\left(1 + 2 \\frac{\\delta r}{r_{0}} + (n+2) \\Big(\\frac{\\delta r}{r_{0}}\\Big)^{2}\\right) \\left(1 - 2 \\frac{\\delta r}{r_{0}} + 3 \\Big(\\frac{\\delta r}{r_{0}}\\Big)^{2}\\right) $$ Thus the second-order terms of potential energy are: $$ \\frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} r_{0}^{2} (n+2)^{2} (n+1) \\bigg(\\frac{\\delta r}{r_{0}}\\bigg)^{2} = \\frac{q^{2} B_{0}^{2}}{2m} (n+1) \\delta r^{2} $$ (Correctly expanding small terms 12 points) Thus, the energy conservation equation is: $$ \\frac{1}{2} m (\\dot{\\delta r})^{2} + \\frac{q^{2} B_{0}^{2}}{2m} (n+1) \\delta r^{2} = C $$ According to the formula for simple harmonic oscillation, we know this represents an oscillation period of: $$ T = \\frac{2 \\pi m}{q B_{0} \\sqrt{n+1}} $$", "answer": "$$T = \\frac{2\\pi m}{q B_0 \\sqrt{n+1}}$$" }, { "id": 118, "tag": "ELECTRICITY", "content": "A regular dodecahedron resistor network is given. Except for $R_{BC} = 2r$, the resistance between all other adjacent vertices is $r$. Points $B$ and $C$ are the two endpoints of one edge of the dodecahedron. Find the resistance between points $B$ and $C$.", "solution": "Boldly introducing negative resistance, $BC$ is equivalent to a parallel combination of $\\pmb{r}$ and $-2r$: \n\n$$\n\\frac{1}{r}+\\frac{1}{-2r}=\\frac{1}{2r}\n$$\n\nAfter extracting $-2r$, the remaining resistance value can be constructed using the forced current method. Inject $19I_{1}$ into node $B$, and each of the other nodes emits a current of 1. Then consider injecting $I$ into each of the other nodes while node $C$ emits $191$. Thus, the current flowing through the branch: \n\n$$\nI_{BC}=20I\n$$\n\nVoltage: \n\n$$\nU_{BC}=\\frac{19Ir}{3}\\cdot2\n$$\n\nThus: \n\n$$\nU_{BC}=I_{BC}R_{BC}^{\\prime}\n$$\n\nResulting in: \n\n$$\nR_{BC}^{\\prime}=\\frac{19}{30}r\n$$\n\nThen, parallel it with $-2r$: \n\n$$\n\\frac{1}{R_{BC}}=\\frac{1}{R_{BC}^{\\prime}}+\\frac{1}{-2r}\n$$\n\nFinally obtained: \n\n$$\nR_{BC}=\\frac{38}{41}r\n$$", "answer": "$$\nR_{BC} = \\frac{38}{41}r\n$$" }, { "id": 318, "tag": "MECHANICS", "content": "Consider a small cylindrical object with radius $R$ and height $h$. Determine the expression for the force $F$ acting on the cylinder when a sound wave passes through it. The axial direction of the cylinder is the direction of wave propagation. In the sound wave, the displacement of a particle from its equilibrium position is $\\psi = A\\cos kx\\cos 2\\pi ft$, the ambient pressure is $P_0$, and the adiabatic index of air is $\\gamma$.", "solution": "The force on the cylinder will be \n\n$$\nF = -\\pi R^{2}(p(y+h)-p(y)) = -\\pi R^{2}h\\frac{d p}{d y}\n$$ \n\nFor a traveling wave, substitute \n\n$$\n\\Delta p(y,t) = -\\gamma p_{0}A k\\cos(k y-2\\pi f t)\n$$ \n\n$$\n\\frac{d p}{d y} = \\gamma p_{0}A k^{2}\\sin(k y-2\\pi f t)\n$$ \n\nTherefore, the force is \n\n$$\nF = -\\pi R^2 h\\gamma p_0 A k^2\\sin(ky-2\\pi ft)\n$$\n\nHowever, for a standing wave, we have \n\n$$\n\\psi = A\\cos{k x}\\cos{2\\pi}f t\n$$ \n\nThus, \n\n$$\n\\Delta p = \\gamma P_{0}k A\\sin k x\\cos2\\pi f t\n$$ \n\n\nTherefore, \n\n$$\n\\frac{d p}{d x} = \\gamma P_{0}k^{2}A\\cos k x\\cos2\\pi f t\n$$ \n\nThus, the force is \n\n$$\nF = -\\pi R^2 h\\gamma P_0 k^2 A\\cos kx \\cos 2\\pi ft\n$$", "answer": "$$\nF = -\\pi R^2 h \\gamma P_0 k^2 A \\cos(kx) \\cos(2\\pi ft)\n$$" }, { "id": 62, "tag": "ADVANCED", "content": "In certain solids, ions have spin angular momentum and can be regarded as a three-dimensional real vector $\\vec{S}$ with a fixed length under the semiclassical approximation, where the length $\\vert\\vec{S}\\vert = S$ is a constant. The magnetic moment $\\overrightarrow{M}$ of the ions is usually proportional to the spin vector, ${\\overrightarrow{M}} = \\gamma{\\overrightarrow{S}}$, where $\\gamma(>0)$ is a constant. The spin-spin interaction between ions $i$ and $j$ is usually the Heisenberg interaction, $-J_{i,j}\\vec{S}_{i} \\cdot \\vec{S}_{j}$. Under a static uniform external magnetic field $\\overrightarrow{B}$, the total energy (Hamiltonian) of the system is typically given by $\\begin{array}{r}{H = -\\sum_{(i,j)}J_{i,j}\\vec{S}_{i}\\cdot\\vec{S}_{j} - }\\end{array}$ $\\begin{array}{r}{\\gamma\\overrightarrow{B}\\cdot\\sum_{i}\\overrightarrow{S}_{i}}\\end{array}$. Here, $(i,j)$ indicates a sum over all distinct pairs without order, meaning $(i,j)$ and $(j,i)$ represent the same pair and are not counted twice. The semiclassical time evolution of the spin vector satisfies the Heisenberg equation of motion (here $\\times{}$ represents the vector cross product):\n$$\n\\frac{\\mathrm{d}}{\\mathrm{d}t}\\vec{S}_{i}=\\frac{\\partial H}{\\partial \\vec{S}_{i}}\\times\\vec{S}_{i}\n$$\nConsider two spins, $H = -J\\vec{S}_{1}\\cdot\\vec{S}_{2} - \\gamma\\vec{B}\\cdot\\left(\\vec{S}_{1}+\\vec{S}_{2}\\right)$, where $J>0$. Without loss of generality, let $\\overrightarrow{B}$ be along the $\\mathbf{Z}$ direction, $\\vec{B}=B\\hat{\\bf z}$, $B \\geq 0$. The ground state (lowest energy state) of the system is $\\vec{S}_{1}=\\vec{S}_{2}=S\\hat{\\mathbf{z}}$, and this state does not evolve over time. Consider states close to the ground state, $\\vec{S}_{1}=S\\big(x_{1},y_{1},\\sqrt{1-{x_{1}}^{2}-{y_{1}}^{2}}\\big)$, $\\vec{S}_{2}=S\\big(x_{2},y_{2},\\sqrt{1-{x_{2}}^{2}-{y_{2}}^{2}}\\big)$, where $\\left|x_{1,2}\\right|\\ll1, \\left|y_{1,2}\\right|\\ll1$. Find the product of the final normal mode frequencies.", "solution": "Solution: The Heisenberg equation of motion is\n$$\n\\begin{array}{r}{\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}\\vec{S}_{1}=-\\left(J\\vec{S}_{2}+\\gamma\\vec{B}\\right)\\times\\vec{S}_{1}}\\ {\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}\\vec{S}_{2}=-\\left(J\\vec{S}_{1}+\\gamma\\vec{B}\\right)\\times\\vec{S}_{2}}\\end{array}\n$$\nNote that $\\vec{S}_{\\pm}=\\vec{S}_{1}\\pm\\vec{S}_{2}$ can be defined, then $\\begin{array}{r}{H=\\frac{J}{2}\\big|\\overrightarrow{S}_{+}\\big|^{2}-J S^{2}-\\gamma\\overrightarrow{B}\\cdot\\overrightarrow{S}_{+}}\\end{array}$,\n$$\n\\frac{\\mathrm{d}}{\\mathrm{d}t}\\vec{S}_{+}=-\\gamma\\vec{B}\\times\\vec{S}_{+}\n$$\n$$\n\\frac{\\mathrm{d}}{\\mathrm{d}t}\\vec{S}_{-}=-\\left(\\vec{J S}_{+}+\\gamma\\vec{B}\\right)\\times\\vec{S}_{-}\n$$\nThe equations of motion for $x_{1,2},y_{1,2}$ in the lowest order approximation are\n$$\n\\begin{array}{c}{{\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}(x_{1}+x_{2})=\\gamma B\\cdot(y_{1}+y_{2})}}\\ {{\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}(y_{1}+y_{2})=-\\gamma B\\cdot(x_{1}+x_{2})}}\\ {{\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}(x_{1}-x_{2})\\approx(2J S+\\gamma B)\\cdot(y_{1}-y_{2})}}\\ {{\\displaystyle\\frac{\\mathrm{d}}{\\mathrm{d}t}(y_{1}-y_{2})\\approx-(2J S+\\gamma B)\\cdot(x_{1}-x_{2})}}\\end{array}\n$$\nThe product of the characteristic frequencies is\n$\\gamma B(2J S+\\gamma B)$", "answer": "$$\\gamma B(2JS + \\gamma B)$$" }, { "id": 143, "tag": "MECHANICS", "content": "Consider an elastic soft rope with original length $a$, and elastic coefficient $k$. With one end fixed, the other end is attached to a particle with a mass of $m$. And the rope remains horizontal. The particle moves on a smooth horizontal surface. Initially, the rope is stretched to a length of $a+b$, and then the particle is released. Determine how much time it takes for the particle to return to the original release position.", "solution": "First consider \\( x > a \\). At this time, the particle only experiences the elastic force \\( -k(x-a) \\) in the horizontal direction. According to Newton's second law, the equation of motion for the particle is\n\n$$\nm{\\ddot{x}} = -k(x-a),\n$$\n\nwhich is the equation of simple harmonic motion. To solve equation (1), make the substitution \\(\\xi = x - a\\), then equation (1) becomes\n\n$$\nm\\ddot{\\xi} = -k\\xi.\n$$\n\nFurther, with the transformation \\(\\ddot{\\xi} = \\frac{\\mathrm{d}^{2}\\xi}{\\mathrm{d}t^{2}} = \\frac{\\mathrm{d}\\xi}{\\mathrm{d}t} \\frac{\\mathrm{d}}{\\mathrm{d}\\xi}\\left(\\frac{\\mathrm{d}\\xi}{\\mathrm{d}t}\\right) = \\frac{1}{2} \\frac{\\mathrm{d}\\dot{\\xi}^{2}}{\\mathrm{d}\\xi}\\), equation (2) becomes\n\n$$\n\\mathrm{d}\\dot{\\xi}^{2} = -2\\omega^{2}\\xi\\mathrm{d}\\xi,\n$$\n\nwhere \\(\\omega = \\sqrt{k/m}\\). Integrating the above, and setting the initial conditions \\(\\xi = \\xi_{0}, \\dot{\\xi} = v_{0} \\) at \\(t=0\\), yields\n\n$$\n\\dot{\\xi}^{2} = v_{0}^{2} - \\omega^{2}(\\xi^{2} - \\xi_{0}^{2}) = (v_{0}^{2} + \\omega^{2}\\xi_{0}^{2}) - \\omega^{2}\\xi^{2},\n$$\n\nwhich implies\n\n$$\n\\frac{\\mathrm{d}\\xi}{\\mathrm{d}t} = \\pm\\sqrt{(v_{0}^{2} + \\omega^{2}\\xi_{0}^{2}) - \\omega^{2}\\xi^{2}} = \\pm\\omega\\sqrt{A^{2} - \\xi^{2}},\n$$\n\nwhere \\({\\cal A} = \\sqrt{\\frac{v_{0}^{2}}{\\omega^{2}} + \\xi_{0}^{2}}\\). Further integration and considering the initial conditions gives\n\n$$\n\\xi = A\\sin(\\pm\\omega t + \\arcsin \\frac{\\xi_{0}}{A}) = A\\cos(\\omega t + \\alpha),\n$$\n\nwhere \\(\\alpha = \\pm\\arcsin{\\frac{\\xi_{0}}{A}} \\mp {\\frac{\\pi}{2}}\\). Thus, the general solution of equation (1) is\n\n$$\nx = a + A\\cos(\\omega t + \\alpha).\n$$\n\nAccording to the problem, at \\(t=0\\), \\(x=a+b, \\dot{x}=0\\). Substituting into equation (3) determines the constants of integration to be \\(A=b, \\alpha=0\\). Substituting back into equation (3), we obtain\n\n$$\nx = a + b\\cos\\omega t.\n$$\n\nThis shows that the motion of the particle in the domain \\(x > a\\) is periodic. The time for the particle to move from \\(x = a + b\\) to \\(x = a\\) and back is the period, which is\n\n$$\nt_{0} = {\\frac{1}{2}}\\frac{2\\pi}{\\omega} = \\pi\\sqrt{\\frac{m}{k}}.\n$$\n\nWhen \\(x < -a\\), the particle experiences an elastic force of \\(-k(x+a)\\) in the horizontal direction. Note that because \\(x < -a\\), \\( -(x+a) > 0\\), so the elastic force points in the positive \\(x\\) direction, which is consistent with the actual situation. The equation of motion is\n\n$$\nm{\\ddot{x}} = -k(x+a).\n$$\n\nUsing a similar method to the above, the general solution of the equation is\n\n$$\nx = -a + B\\cos(\\omega t + \\beta),\n$$\n\nwhere \\(B, \\beta\\) are constants of integration. According to analysis in \\(\\textcircled{2}\\), \\(\\dot{x}|_{x=a} = \\dot{x}|_{x=-a}\\), we determine \\(B=b\\). Equation (5) indicates that the motion of the particle in the region \\(x < -a\\) is also simple harmonic, with the same frequency as in the region \\(x > a\\). The time for the particle to move from \\(x=-a\\) to \\(x=-(a+b)\\) and back is also \\(t_{0}\\).\n\nIn summary, the total time taken for the particle to move from \\(x = a + b \\) to \\(x = a\\) and from \\(x = -a\\) to \\(x = -(a + b)\\) and back is\n\n$$\nt_{1} = 2t_{0} = 2\\pi\\sqrt{\\frac{m}{k}}.\n$$\n\nThe previous discussion also shows that considering just \\(x > a\\) is sufficient to determine the characteristics of the particle's motion in the interval \\(|x| > a\\), especially concerning the motion's time.\n\n\\(\\textcircled{2}\\) When \\(|x|1$ for the dielectric, enters the metal (which at this frequency has an equivalent relative permittivity $\\varepsilon_{2}<0$) with a wavevector $k_{x}=\\alpha$, $k_{y}=-\\mathrm{i}\\beta$ ($\\alpha$ and $\\beta$ are real numbers), the electric field lies in the $xy$ plane, and the interface is defined as the $y=0$ plane, with the positive $y$-axis pointing from the dielectric towards the metal, there exists a specific horizontal wavevector $k_{x}$ such that no reflected light is generated. Instead, a unique optical mode satisfying Maxwell’s equations is formed solely by the incident and refracted light—this corresponds to the surface plasmon polariton mode. \n\nHowever, in this problem, we need to consider the case where the metal actually has a relative complex permittivity $\\widetilde{\\varepsilon}_{2}=\\varepsilon_{21}+\\mathrm{i}\\varepsilon_{22}$, where $\\varepsilon_{21}<0$ and the imaginary part $\\varepsilon_{22}$ is a small quantity. Under these conditions, the surface plasmon polariton mode not only propagates but also attenuates. Please express the wavelength $\\lambda$ of the surface plasmon polariton mode in terms of the dielectric relative permittivity $\\varepsilon_{1}$, the metal’s relative complex permittivity, and the vacuum wavelength $\\lambda_{0}$. Approximate to the lowest order.", "solution": "The magnetic field strength in the $z$ direction is tangentially continuous:\n\n$$\nH_{1z}=H_{2z}=H\n$$\n\nAccording to Maxwell's equations:\n\n$$\n\\nabla\\times\\pmb{H}=\\varepsilon\\frac{\\partial\\pmb{E}}{\\partial t}\n$$\n\nFurthermore, since the electric field is tangentially continuous:\n\n$$\nE_{1x}=E_{2x}\n$$\n\nThe relationship obtained is:\n\n$$\n\\frac{k_{1y}}{\\varepsilon_{1}}=\\frac{-\\mathrm{i}\\beta_{1}}{\\varepsilon_{1}}=\\frac{\\mathrm{i}\\beta_{2}}{\\varepsilon_{2}}=\\frac{k_{2y}}{\\varepsilon_{2}}\n$$\n\n$$\n\\begin{array}{r}{k_{1}^{2}=\\alpha^{2}-\\beta_{1}^{2}=\\varepsilon_{1}k_{0}^{2}>0}\\\\ {k_{2}^{2}=\\alpha^{2}-\\beta_{2}^{2}=\\varepsilon_{2}k_{0}^{2}<0}\\end{array}\n$$\n\nSolving for $k_{1x}=k_{2x}=k_{x}=\\alpha$, we find:\n\n$$\nk_{x}=\\frac{\\omega}{c}\\sqrt{\\frac{\\varepsilon_{1}\\varepsilon_{2}}{\\varepsilon_{1}+\\varepsilon_{2}}}\n$$\n\nSquaring the above expression and rewriting, we get:\n\n$$\n(k_{x1}+\\mathrm{i}k_{x2})^{2}=k_{0}^{2}\\frac{\\varepsilon_{1}(\\varepsilon_{21}+\\mathrm{i}\\varepsilon_{22})}{\\varepsilon_{1}+\\varepsilon_{21}+\\mathrm{i}\\varepsilon_{22}}\\approx k_{0}^{2}\\frac{\\varepsilon_{1}\\varepsilon_{21}}{\\varepsilon_{1}+\\varepsilon_{21}}\\left(1+\\mathrm{i}\\frac{\\varepsilon_{22}}{\\varepsilon_{21}}\\right)\\left(1-\\mathrm{i}\\frac{\\varepsilon_{22}}{\\varepsilon_{1}+\\varepsilon_{21}}\\right)\n$$\n\nBy comparing the real and imaginary parts and making approximations:\n\n$$\n\\begin{array}{c}{{k_{x1}^{2}=k_{0}^{2}\\displaystyle\\frac{\\varepsilon_{1}\\varepsilon_{21}}{\\varepsilon_{1}+\\varepsilon_{21}}}}\\\\ {{2k_{x1}k_{x2}=k_{0}^{2}\\displaystyle\\frac{\\varepsilon_{1}\\varepsilon_{21}}{\\varepsilon_{1}+\\varepsilon_{21}}\\cdot\\displaystyle\\frac{\\varepsilon_{1}\\varepsilon_{22}}{\\varepsilon_{21}\\left(\\varepsilon_{1}+\\varepsilon_{21}\\right)}}}\\end{array}\n$$\n\nSubstituting into:\n\n$$\nk_{0}=\\frac{2\\pi}{\\lambda_{0}}\\quad,\\quad k_{x1}=\\frac{2\\pi}{\\lambda}\\quad,\\quad k_{x2}=\\frac{2\\pi}{d}\n$$\n\nFinally, we obtain:\n\n$$\n\\lambda=\\sqrt{\\frac{\\varepsilon_{1}+\\varepsilon_{21}}{\\varepsilon_{1}\\varepsilon_{21}}}\\lambda_{0}\n$$", "answer": "$$\n\\lambda=\\sqrt{\\frac{\\varepsilon_{1}+\\varepsilon_{21}}{\\varepsilon_{1}\\varepsilon_{21}}}\\lambda_{0}\n\n$$" }, { "id": 115, "tag": "ADVANCED", "content": "Consider a charged rod undergoing relativistic rotation. Within an infinitely long cylinder with radius $R$, positive charges with number density $n$ and charge $q$ revolve around the axis with a uniform angular velocity of $\\omega$ in a relativistic motion. The original reference frame is $S$, and now we switch to a reference frame $S^{\\prime}$ that moves parallel to the axis with a velocity $\\beta c$ relative to frame $S$. We consider the radiation from the accelerating charges. In the relativistic case, the instantaneous outgoing radiated electromagnetic power of a charge $q$ with arbitrary velocity $\\pmb{v}$ and arbitrary acceleration $\\pmb{a}$ follows the Larmor formula: \n$$\nP={\\frac{\\gamma^{6}q^{2}}{6\\pi\\varepsilon_{0}c^{3}}}\\left[a^{2}-{\\frac{(v\\times a)^{2}}{c^{2}}}\\right]\n$$\n\nwhere $\\gamma=1/\\sqrt{1-v^{2}/c^{2}}$. Assume all charge radiation is incoherent. Without using energy or power transformation formulas, calculate directly the radiation power per unit length of the rod in frame $S^{\\prime}$ through the charge radiation method. The speed of light in a vacuum is $c$.", "solution": "Charge and current distribution:\n\n$$\n\\rho=n q\n$$\n\n$$\nj_{\\theta}=\\rho v=\\Im\\Im q\\omega\\tau\n$$\n\nNote that the above equation is valid only for $rR)}\\end{array}\\right.}\n$$\n\n$$\n\\mathcal{B}_{z}(r)=\\left\\{\\begin{array}{l l}{\\frac{\\mu_{0}n q\\omega(R^{2}-r^{2})}{2}}&{(rR)}\\end{array}\\right.\n$$\n\nIn a new frame of reference, the number density scales directly:\n\n$$\nn^{\\prime}=\\frac{n}{\\sqrt{1-\\beta^{2}}}\n$$\n\nThe radius remains unchanged:\n\n$$\n\\boldsymbol{r}^{\\prime}=\\boldsymbol{r}\n$$\n\nBut velocity acquires a new component:\n\n$$\nv_{z}^{\\prime}=-\\beta c\n$$\n\nThe original components also change. Since one loop in the original reference frame corresponds to proper time, the angular velocity of the new helical motion becomes:\n\nThus, the new charge and current distribution is:\n\n$$\nv_{\\theta}^{\\prime}\\cdot T^{\\prime}=v_{\\theta}T\n$$\n\n$$\nT^{\\prime}=\\frac{T}{\\sqrt{1-\\beta^{2}}}\n$$\n\n$$\n\\rho^{\\prime}=\\pi^{\\prime}q=\\frac{n q}{\\sqrt{1-\\beta^{2}}}\n$$\n\n$$\nj_{\\theta}^{\\prime}=\\rho^{\\prime}v_{\\theta}^{\\prime}=n q\\omega r\n$$\n\n$$\nj_{z}^{\\prime}=\\rho^{\\prime}v_{z}^{\\prime}=-\\frac{n q\\beta c}{\\sqrt{1-\\beta^{2}}}\n$$\n\nThus, the original two electromagnetic fields are scaled proportionally as described above:\n\n$$\nE_{r}^{\\prime}(r)=\\left\\{\\begin{array}{l l}{\\displaystyle\\frac{n q r}{2\\varepsilon_{0}\\sqrt{1-\\beta^{2}}}}&{(rR)}\\end{array}\\right.\n$$\n\n$$\nB_{\\natural}^{\\prime}(r)=\\left\\{{\\begin{array}{l l}{{\\frac{\\mu_{0}n q\\omega r^{2}}{2}}}&{(rR)}\\end{array}}\\right.\n$$\n\nHowever, an axial current is also produced, leading to a surrounding magnetic field, expressed as:\n\n$$\nB_{\\theta}(r)=\\int_{0}^{r}\\frac{\\mu_{0}\\mathrm{d}I}{2\\pi r}\\quad,\\quad\\mathrm{d}I=j_{z}^{\\prime}(r^{\\prime})\\cdot2\\pi r^{\\prime}\\mathrm{d}r^{\\prime}\n$$\n\nThe integration yields the result:\n\n$$\nB_{\\theta}^{\\prime}(r)=\\left\\{\\begin{array}{l l}{-\\frac{\\mu_{0}n q\\beta c r}{2\\sqrt{1-\\beta^{2}}}}&{(rR)}\\end{array}\\right.\n$$\n\n(3)\n\nRadiation power of a single particle in the original frame:\n\n$$\nP_{1}=\\frac{q^{2}}{6\\pi\\varepsilon_{0}c^{3}}\\cdot\\frac{\\omega^{4}r^{2}\\left(1-\\omega^{2}r^{2}/c^{2}\\right)}{\\left(1-\\omega^{2}r^{2}/c^{2}\\right)^{3}}\n$$\n\nParticle number per unit length:\n\n$$\n\\mathrm{d}N=n\\cdot2\\pi\\tau\\mathrm{d}r=\\pi n\\mathrm{d}\\left(r^{2}\\right)\n$$\n\nIntegration:\n\n$$\nP=\\int_{0}^{R}P_{1}\\mathrm{d}N\n$$\n\nGives:\n\n$$\nP=\\frac{n q^{2}c}{6\\varepsilon_{0}}\\left[\\frac{\\omega^{2}R^{2}/c^{2}}{1-\\omega^{2}R^{2}/c^{2}}+\\ln\\left(1-\\frac{\\omega^{2}R^{2}}{c^{2}}\\right)\\right]\n$$\n\nRadiation power of a single particle in the new frame:\n\n$$\nP_{1}^{\\prime}=\\frac{q^{2}}{6\\pi\\varepsilon_{0}c^{3}}\\cdot\\frac{\\left(v_{\\theta}^{\\prime2}/r\\right)^{2}\\cdot\\left[1-(v_{\\bar{s}}^{\\prime2}+v_{\\theta}^{\\prime2})/c^{2}\\right]}{[1-(v_{\\bar{s}}^{\\prime2}+v_{\\bar{\\varepsilon}}\\cdot\\bar{\\mathbf{\\xi}}}\n$$\n\nReorganized as:\n\n$$\n\\mathcal{P}_{1}^{\\prime}=\\mathcal{P}_{1}\n$$\n\nHowever, the number of particles per unit length increases by a factor of $1/\\sqrt{1-\\beta^{2}}$. Hence, the total power:\n\n$$\nP^{\\prime}=\\frac{P}{\\sqrt{1-\\beta^{2}}}\n$$\n\nGives:\n\n$$\nP^{\\prime}=\\frac{n q^{2}c}{6\\varepsilon_{0}\\sqrt{1-\\beta^{2}}}\\left[\\frac{\\omega^{2}R^{2}/c^{2}}{1-\\omega^{2}R^{2}/c^{2}}+\\ln\\left(1-\\frac{\\omega^{2}R^{2}}{c^{2}}\\right)\\right]\n$$", "answer": "$$\nP'=\\frac{n q^2 c}{6 \\varepsilon_0 \\sqrt{1-\\beta^2}} \\left[\\frac{\\omega^2 R^2 / c^2}{1-\\omega^2 R^2 / c^2}+\\ln\\left(1-\\frac{\\omega^2 R^2}{c^2}\\right)\\right]\n$$" }, { "id": 493, "tag": "ELECTRICITY", "content": "There is a uniform rigid current-carrying circular ring with mass $M$ and radius $R$, carrying a current of $I$. At a distance $h$ from the center of the ring, there is a magnetic dipole with a magnetic dipole moment of $m$, which is fixed and cannot rotate. The line connecting the magnetic dipole and the center of the ring is perpendicular to the plane of the ring. The ring can only rotate about a certain diameter axis passing through the center of the ring. Initially, the plane of the ring is perpendicular to the direction of the magnetic dipole moment. Now, the ring is slightly rotated by a small angle. Find the period $T_1$ of the simple harmonic vibration of the ring after release (assuming that all parameters satisfy the stable equilibrium condition, and in the initial state, the direction of the magnetic field produced by the current on the ring at the position of the magnetic dipole is the same as that of the magnetic dipole). Given the vacuum permeability is $\\mu_0$.", "solution": "When the ring is rotated by an angle \\( \\theta \\), a cylindrical coordinate system is established, and the coordinates corresponding to the position of \\( \\vec{m} \\) are\n\n\\[\n\\begin{cases} \n\\rho = h \\sin \\theta \\\\\nz = h \\cos \\theta \n\\end{cases}\n\\]\n\nThe magnetic field component \\( B_z(0, z) \\) is given by\n\n\\[\nB_z(0, z) = \\frac{\\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}\n\\]\n\nThe expression for the energy of simple harmonic motion involves the second-order term of \\( \\theta \\), so both \\( B_z \\) and \\( B_\\rho \\) are retained up to the second-order term in \\( \\rho \\) (where \\( B_{\\rho,0} \\), \\( B_{z,1} \\), and \\( B_{\\rho,2} \\) are all zero):\n\n\\[\n\\begin{cases} \nB_{z,0} = B_z(0, z) = \\frac{\\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \\\\\nB_{\\rho,1} = -\\frac{\\rho}{2} \\frac{\\mathrm{d}B_z(0, z)}{\\mathrm{d}z} = \\frac{3 \\mu_0 I R^2 z \\rho}{4(R^2 + z^2)^{5/2}} \\\\\nB_{z,2} = -\\frac{\\rho^2}{4} \\frac{\\mathrm{d}^2 B_z(0, z)}{\\mathrm{d}z^2} = \\frac{3 \\mu_0 I R^2 (R^2 - 4z^2) \\rho^2}{8(R^2 + z^2)^{7/2}}\n\\end{cases}\n\\]\n\nUsing the small-angle approximation \\( \\sin \\theta \\approx \\theta \\) and \\( \\cos \\theta \\approx 1 - \\frac{1}{2} \\theta^2 \\):\n\n\\[\n\\begin{cases} \nB_{z,0} \\approx \\frac{\\mu_0 I R^2}{2(R^2 + h^2)^{3/2}} \\left(1 + \\frac{3}{2} \\frac{h^2 \\theta^2}{h^2 + R^2}\\right) \\\\\nB_{\\rho,1} \\approx \\frac{3 \\mu_0 I R^2 h^2 \\theta}{4(R^2 + h^2)^{5/2}} \\\\\nB_{z,2} \\approx \\frac{3 \\mu_0 I R^2 h^2 \\theta^2 (R^2 - 4h^2)}{8(R^2 + h^2)^{7/2}}\n\\end{cases}\n\\]\n\nThe potential energy at the angular displacement \\( \\theta \\) is\n\n\\[\nW = -\\vec{m} \\cdot \\vec{B} = -m(B_z \\cos \\theta + B_\\rho \\sin \\theta) = -m(B_{z,0} \\cos \\theta + B_{z,2} \\cos \\theta + B_{\\rho,1} \\sin \\theta)\n\\]\n\n\\[\n= -m \\left[ \\frac{\\mu_0 I R^2}{2(R^2 + h^2)^{3/2}} \\left(1 + \\frac{3}{2} \\frac{h^2 \\theta^2}{h^2 + R^2}\\right) \\left(1 - \\frac{\\theta^2}{2}\\right) + \\frac{3 \\mu_0 I R^2 h^2 \\theta^2 (R^2 - 4h^2)}{8(R^2 + h^2)^{7/2}} \\left(1 - \\frac{\\theta^2}{2}\\right) + \\frac{3 \\mu_0 I R^2 h^2 \\theta^2}{4(R^2 + h^2)^{5/2}} \\right]\n\\]\n\n\\[\n\\approx -\\frac{m \\mu_0 I R^2}{2(R^2 + h^2)^{3/2}} + \\frac{1}{2} \\frac{\\mu_0 m I R^2 (2h^4 + 2R^4 - 11R^2 h^2)}{4(h^2 + R^2)^{7/2}} \\theta^2\n\\]\n\nIn the above derivation, all terms higher than the second order in \\( \\theta \\) are neglected.\n\nThe rotational kinetic energy of the ring is\n\n\\[\nK = \\frac{1}{2}\\left(\\frac{1}{2}MR^{2}\\right)\\dot{\\theta}^{2}\n\\]\n\nAccording to the principle of determining simple harmonic motion by energy methods, the resonant period is\n\n\\[\nT_{1}=2\\pi\\sqrt{\\frac{\\frac{1}{2}MR^{2}}{\\frac{\\mu_{0}mIR^{2}(2h^{4}+2R^{4}-11R^{2}h^{2})}{4(h^{2}+R^{2})^{7/2}}}} = 2\\pi\\sqrt{\\frac{2M(h^{2}+R^{2})^{7/2}}{\\mu_{0}mI(2h^{4}+2R^{4}-11R^{2}h^{2})}}\n\\]", "answer": "\\[\nT_{1}= 2\\pi\\sqrt{\\frac{2M(h^{2}+R^{2})^{7/2}}{\\mu_{0}mI(2h^{4}+2R^{4}-11R^{2}h^{2})}}\n\\] " }, { "id": 69, "tag": "MECHANICS", "content": "In 2023, the team of astronomers led by Jerome Oros at San Diego State University in the United States released their latest research findings, discovering a new binary star system called \"Kepler-49\". Meanwhile, astronomical observations indicate that nearby planets are influenced by this binary star system, resulting in orbital precession. This problem aims to explain this phenomenon concisely.\n\n\"Kepler-49\" is a binary star system mainly composed of two stars (Saka and Juipte) with masses $M_{S}$ and $M_{J}$ (both known). Under the mutual gravitational interaction, they perform stable circular motion around the center of mass. With an angular frequency $\\Omega$, the respective motion radii $R_{S}$, $R_{J}$ (unknown quantities) and the distance between them $R_{0}$ (unknown quantity) can be determined. The gravitational constant $G$ is known. \n\nThe rotation center of the two stars is set as the origin of coordinates, and a polar coordinate system is established with the closest point as the radial distance to describe the movement of the planet. Experimental observation shows that the scale of the Saka and Juipte system is much smaller than the distance to the planet (i.e., $R_{0} \\ll r$). Meanwhile, during one period of $\\Omega$, the planet can be approximately considered as not moving. The modified potential energy $V_{\\mathrm{eff}}$ (unknown quantity) of the planet can be determined, retaining the lowest order correction term $\\delta V$.\n\nIt is known that the energy of the planet is $E < 0$, the orbital angular momentum is $\\bar{L}$, and the mass is $m$. Only considering the zeroth-order term $V_{0}$ of the gravitational potential energy, the motion equation of the planet $r(\\theta)$ (unknown quantity) can be determined.\n\nQuestion: Consider now the form of potential energy $V = V_{0} + \\delta V$ for the planet. In this case, $\\vec{B}$ is no longer conserved. Solve for the orbital precession angle $\\Delta{\\alpha}$ (express the answer in terms of $M_{J}, M_{S}, G, \\Omega, m, L$).\n\nHint 1: The Laplace-Runge-Lenz vector is introduced in the Kepler problem, and in this problem, it is defined as\n\n$$\n\\vec{B} = \\vec{v} \\times \\vec{L} - G(M_{J} + M_{S})m\\hat{r}.\n$$\n\nConsider the planet moving only under the potential energy $V = V_{0}$, and in this case, let $\\vec{B}_{0} = \\vec{B}$. It can be proved that $\\vec{B}_{0}$ is conserved (the proof method is unknown), and the expression for $|\\vec{B}_{0}|$ can be determined (unknown quantity).\n\nHint 2: The precession angle is defined as the rotation angle of the closest point during a single period of the planet. Consider the LRL vector pointing to the closest point, retaining the lowest order term:\n\n$$\n\\Delta\\alpha = \\left|\\frac{\\vec{B}_{0}}{|\\vec{B}_{0}|} \\times \\left(\\frac{\\vec{B}}{|\\vec{B}|} - \\frac{\\vec{B}_{0}}{|\\vec{B}_{0}|}\\right)\\right| \\approx \\frac{|\\vec{B}_{0} \\times \\Delta\\vec{B}|}{|\\vec{B}_{0}|^{2}}.\n$$", "solution": "From \n\n$$\n\\frac{G M_{J}M_{S}}{(R_{J}+R_{S})^{2}}=M_{J}R_{J}\\Omega^{2}=M_{S}R_{S}\\Omega^{2}\n$$ \n\nwe solve \n\n$$\n\\begin{array}{c}{{R_{J}=\\displaystyle\\frac{{M}_{S}}{{ M}_{S}+{ M}_{J}}\\left(\\frac{{ G}({ M}_{S}+{ M}_{J})}{\\Omega^{2}}\\right)^{\\frac{1}{3}}}}\\ {{{}}}\\ {{R_{S}=\\displaystyle\\frac{{ M}_{J}}{{ M}_{S}+{ M}_{J}}\\left(\\frac{{ G}({ M}_{S}+{ M}_{J})}{\\Omega^{2}}\\right)^{\\frac{1}{3}}}}\\end{array}\n$$ \n\nFurther \n\n$$\nR_{0}=R_{J}+R_{S}=\\left({\\frac{G(M_{S}+M_{J})}{\\Omega^{2}}}\\right)^{\\frac{1}{3}}\n$$ \n\nAs stated in the problem, during a single period of rotation of the binary system, it is approximately assumed that the planet's position does not change; at time $t$, the potential energy is \n\n$$\nV=-{\\frac{G M_{J}m}{\\sqrt{r^{2}+R_{J}^{2}-2r R_{J}\\cos\\left(\\Omega t-\\varphi_{0}\\right)}}}-{\\frac{G M_{S}m}{\\sqrt{r^{2}+R_{S}^{2}+2r R_{S}\\cos\\left(\\Omega t-\\varphi_{0}\\right)}}}\n$$ \n\nExpand the gravitational potential in a power series in $\\scriptstyle{\\frac{R_{0}}{r}}$: \n\n$$\nV=-{\\frac{G M_{J}m}{r}}\\left(1+\\cos\\left(\\Omega t-\\varphi_{0}\\right){\\frac{R_{J}}{r}}+{\\frac{3\\cos^{2}\\left(\\Omega t-\\varphi_{0}\\right)-1}{2}}{\\frac{R_{J}^{2}}{r^{2}}}\\right)\\ldots\n$$ \n\n$$\n\\ldots-{\\frac{G M_{s}m}{r}}\\left(1-\\cos\\left(\\Omega t-\\varphi_{0}\\right){\\frac{R_{s}}{r}}+{\\frac{3\\cos^{2}\\left(\\Omega t-\\varphi_{0}\\right)-1}{2}}{\\frac{R_{s}^{2}}{r^{2}}}\\right)\n$$ \n\nConsider the time-averaged result and retain the lowest-order correction term, yielding $V_{\\mathbf{eff}}$: \n\n$$\nV_{\\mathrm{eff}}={\\bar{V}}={\\frac{\\Omega}{2\\pi}}\\int_{0}^{\\frac{2\\pi}{n}}V\\mathrm{d}t=V_{0}+\\delta V=-{\\frac{G(M_{J}+M_{S})m}{r}}-{\\frac{G M_{J}M_{S}m}{4(M_{J}+M_{S})}}\\cdot{\\frac{R_{0}^{2}}{r^{3}}}\n$$ \n\nWhen considering only the zero-order term of the potential energy \n\n$$\nV_{0}(r)=-\\frac{G(M_{S}+M_{J})m}{r}\n$$ \n\nFor $\\pmb{E}<\\mathbf{0}$, the planet is on an elliptical orbit; provided without proof (to be memorized as fundamental knowledge): \n\n$$\ne=\\sqrt{1+\\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}\n$$ \n\n$$\np={\\frac{L^{2}}{G(M_{J}+M_{S})m^{2}}}\n$$ \n\nAs stated in the problem, taking the position near the center as the polar radius, we solve \n\n$$\nr(\\theta)=\\frac{\\frac{L^{2}}{G(M_{J}+M_{S})m^{2}}}{1+\\sqrt{1+\\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}\\cos\\theta}\n$$ \n\n(4) \n\nConsidering the form of the potential energy at this time \n\n$$\nV=V_{0}=-\\frac{G(M_{J}+M_{S})m}{r}\n$$ \n\n$\\vec{B}$ conservation equivalently means it does not change over time. Consider the conservation of angular momentum \n\n$$\n{\\frac{\\mathrm{d}{\\vec{B}}}{\\mathrm{d}t}}={\\frac{1}{m}}({\\vec{F}}\\times L)-G(M_{J}+M_{S})m{\\dot{\\theta}}{\\hat{\\theta}}={\\frac{1}{m}}\\cdot{\\frac{G(M_{J}+M_{S})m}{r^{2}}}\\cdot m r^{2}{\\dot{\\theta}}{\\hat{\\theta}}-G(M_{J}+M_{S})m{\\dot{\\theta}}{\\hat{\\theta}}=0\n$$ \n\nFor computational convenience, select the special position where the planet is near the center at $\\theta=0$: \n\n$$\n\\vec{B}_{0}=\\frac{L^{2}(1+e)}{p}-G(M_{J}+M_{S})m\\hat{r}=G(M_{J}+M_{S})m\\sqrt{1+\\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}\\hat{r}\n$$ \n\nwhere $\\hat{r}$ is the direction of the planet near the center. \n\n(5) \n\nFrom the above derivation, we know that in the case where only the zero-order term of the potential energy is considered, $\\vec{B}$ is conserved and its direction points to the near-center point. Since $\\vec{B}$ reflects information about the orbit's shape and orientation, consider the calculation of $\\Delta\\vec{B}$ at this time \n\n$$\n\\frac{\\mathrm{d}\\vec{B}}{\\mathrm{d}t}=\\frac{1}{m}(\\vec{F}\\times L)-G(M_{J}+M_{S})m\\dot{\\theta}\\hat{\\theta}=-\\frac{1}{m}\\frac{\\partial(\\delta V)}{\\partial r}\\hat{r}\\times\\vec{L}=\\frac{3G M_{J}M_{S}m}{4(M_{J}+M_{S})}\\cdot\\frac{R_{0}^{2}}{r^{2}}\\dot{\\theta}\\hat{\\theta}\n$$ \n\nIntegrate over a single period \n\n$$\n\\Delta\\vec{B}=\\int_{0}^{T}\\frac{\\mathrm{d}\\vec{B}}{\\mathrm{d}t}\\cdot\\mathrm{d}t=\\int_{0}^{2\\pi}\\frac{3G M_{J}M_{S}m}{4(M_{J}+M_{S})}\\cdot\\frac{R_{0}^{2}}{r^{2}}\\mathrm{d}\\theta\\hat{\\theta}\n$$ \n\nFor the lowest-order correction term, use the elliptical orbit equation obtained in (3): \n\n$$\nr={\\frac{p}{1+e\\cos\\theta}}\n$$ \n\n$$\n\\Delta\\vec{B}=\\int_{0}^{2\\pi}\\frac{3G M_{J}M_{S}m R_{0}^{2}}{4(M_{J}+M_{S})}\\cdot\\frac{(1+e\\cos\\theta)^{2}}{p^{2}}(-\\sin\\theta\\hat{x}+\\cos\\theta\\hat{y})\\mathrm{d}\\theta=\\frac{3G M_{J}M_{S}m R_{0}^{2}}{2(M_{J}+M_{S})}\\cdot\\frac{e\\pi}{p^{2}}\\hat{y}\n$$ \n\nUsing the hint in the problem \n\n$$\n\\Delta\\alpha={\\frac{|\\vec{B_{0}}\\times\\Delta\\vec{B}|}{|\\vec{B_{0}}|^{2}}}={\\frac{3\\pi}{2}}\\cdot{\\frac{G^{\\frac{8}{3}}M_{J}M_{S}(M_{S}+M_{J})^{\\frac{2}{3}}m^{4}}{\\Omega^{\\frac{4}{3}}L^{4}}}\n$$", "answer": "$$\\Delta\\alpha = \\frac{3\\pi}{2} \\cdot \\frac{G^{8/3} M_J M_S (M_S + M_J)^{2/3} m^4}{\\Omega^{4/3} L^4}$$" }, { "id": 65, "tag": "ELECTRICITY", "content": "In the upper half-space, there is a uniform magnetic field with magnitude \\( B \\) directed vertically upwards, and the ground is sufficiently rough.\n\nNow, there is a solid insulating sphere with uniformly distributed mass \\( m \\) and radius \\( R \\) . On its surface at the equator, there are 3 mutually orthogonal wires with negligible mass and thickness and a unit length resistance of \\( \\lambda \\). These wires are connected at the intersection point. The sphere is placed on the ground so that the circular surface formed by one of the wires is perpendicular to the direction of the magnetic field. It is known that the sphere's center has acquired a velocity \\( v_0 \\) in the horizontal plane, and its initial angular velocity is such that it is in a pure rolling state. What is the subsequent change in velocity?", "solution": "By noticing that the cutting of the magnetic field by the wire induces an electromotive force, which in turn generates Ampere force and torque, the motion state of the sphere changes. Now, due to the completeness of the circuit, the magnitude of the Ampere force is zero, and we only need to consider torque. Since the magnetic field here is uniform, the problem is to determine the magnetic moment $\\mu$. \n\nWe use a fundamental yet concise method to solve this problem: we move to the rest frame of the sphere to analyze the problem, and establish the coordinate system as shown in the figure. In this frame, the magnetic field appears to rotate. It should be noted that the focus here is on solving the current distribution. This change of frame is purely mathematical and does not involve relativistic effects. \n\nBelow, we only consider the variation of the magnetic field in the $\\pmb{x}$ direction, denoted as $\\dot{B_{x}}$, while the variations in the $y/z$ directions are analyzed analogously. Since the relationships between electromotive force, current, and magnetic moment are linear, superposition can be performed in the end! Here, due to the symmetry introduced by the circuit, the electromotive forces and resistances for the eight mesh loops in the front and back are exactly the same, meaning all the loop equations for each mesh are identical. This results in no current through the cross-shaped wires in the center. \n\nOnly the circular arcs perpendicular to the ${\\pmb x}$ direction have current, with the electromotive force given by:\n\n$$\n\\varepsilon=-\\pi R^{2}\\dot{B_{x}}\n$$ \n\nThus, the current is:\n\n$$\nI={\\frac{\\varepsilon}{2\\pi R\\lambda}}\n$$ \n\nThe $\\pmb{\\mathscr{x}}$ component of the total magnetic moment is:\n\n$$\n\\mu_{x}=I S=-\\frac{\\pi R^{3}}{2\\lambda}\\dot{B_{x}}\n$$ \n\nTherefore, the total magnetic moment is:\n\n$$\n\\vec{\\mu}=-\\frac{\\pi R^{3}}{2\\lambda}\\dot{\\vec{B}}\n$$ \n\nIn the rest frame, the magnetic field rotates at an angular velocity of $-\\alpha$ (referring to the angular velocity of the sphere's rotation). Hence, $\\dot{\\vec{B}}=-\\vec{\\omega}\\times\\vec{B}$. In the ground frame, we arrive at a remarkably elegant result:\n\n$$\n\\vec{\\mu}=\\frac{\\pi\\mathcal{R}^{3}}{2\\lambda}\\vec{\\omega}\\times\\vec{B}\n$$ \n\nThis result is remarkably coincidental and symmetric, allowing us to utilize a simple vector method for mechanical analysis. Following the coordinate system established in the first question:\n\nThe theorem of the motion of the center of mass is:\n\n$$\nm{\\frac{d{\\vec{v}}}{d t}}={\\vec{f}}\n$$ \n\nRotational motion law:\n\n$$\n\\frac{2}{5}m R^{2}\\frac{d\\vec{\\omega}}{d t}=R\\vec{n}\\times\\vec{f}+\\vec{\\mu}\\times\\vec{B}\n$$ \n\nPure rolling constraint:\n\n$$\n\\vec{v}+R\\vec{\\omega}\\times\\vec{n}=0\n$$ \n\nBy taking the cross product with $\\vec{n}$ on both sides of the rotational motion law, we get:\n\n$$\n\\frac{2}{5}m R^{2}\\frac{d\\vec{n}\\times\\vec{\\omega}}{d t}=-R\\vec{f}+\\frac{\\pi R^{3}}{2\\lambda}\\vec{n}\\times\\left[\\left(\\vec{B}\\cdot\\vec{\\omega}\\right)\\vec{B}-B^{2}\\vec{\\omega}\\right]\n$$ \n\nNoting that $\\vec{n}\\times\\vec{B}=0$, the second term on the right-hand side of the above equation is omitted. Substituting the pure rolling constraint yields:\n\n$$\n\\frac{2}{5}m\\frac{d\\overrightarrow{v}}{d t}=-\\overrightarrow{f}-\\frac{\\pi R B^{2}}{2\\lambda}\\overrightarrow{v}\n$$ \n\nCombining the theorem of the motion of the center of mass:\n\nDirect integration gives:\n\n$$\n\\frac{7m}{5}\\frac{d\\overrightarrow{v}}{d t}=-\\frac{\\pi R B^{2}}{2\\lambda}\\overrightarrow{v}\n$$ \n\n$$\n\\vec{v}=\\vec{v_{0}}e^{-\\frac{5\\pi R B^{2}}{14m\\lambda}t}\n$$ \n\nThis result indicates that while the direction of velocity of the sphere does not change, its magnitude continually decreases.", "answer": "$$\n\\vec{v} = \\vec{v_0} e^{-\\frac{5\\pi R B^2}{14m\\lambda} t}\n$$" }, { "id": 131, "tag": "ELECTRICITY", "content": "Place a small magnetic needle with a magnetic moment \\( m \\) directly above a superconducting sphere with a radius \\( R \\) at a distance \\( a \\). It is known that the magnetic field outside the superconducting sphere can be regarded as the superposition of the magnetic field of \\( m \\) and a magnetic field produced in space by an internal magnetic moment \\( m' \\) inside the superconducting sphere. Find the magnitude of \\( m' \\). (It is known that the magnetic moment \\( m \\) has a magnetic induction field distribution in space as: \\( \\mathbf{B}={ \\frac{ \\mu_{0}m}{4 \\pi r^{3}}}{ \\left(2 \\cos \\theta \\mathbf{e}_{r}+ \\sin \\theta \\mathbf{e}_{ \\theta} \\right)} \\))", "solution": "Based on symmetry, the form is definitely on the line connecting with the center of the sphere. As shown in the figure, assume \\(m' \\) is at a height \\(r' \\) above point O, in the downward direction \\( \\boldsymbol{r^{ \\prime}} \\) (if the direction is upward, the calculated value of \\(m \\) would be negative). The normal component of the magnetic induction intensity on the surface of a superconductor is zero, allowing us to calculate the normal component of the magnetic field at points P and Q. At point P: $$ B_{p}= \\frac{ \\mu_{0}}{4 \\pi} \\frac{2m}{ \\left(a-R \\right)^{3}}- \\frac{ \\mu_{0}2m}{4 \\pi \\left(R-b \\right)^{3}}=0 $$ At point Q: $$ B_{ \\varrho}= \\frac{ \\mu_{0}}{4 \\pi} \\frac{2m}{ \\left(a+R \\right)^{3}}- \\frac{ \\mu_{0}2m^{ \\prime}}{4 \\pi(R+b)^{3}}=0 $$ Solving equations (1) and (2) simultaneously, we get $$ \\begin{array}{c} \\ {{m^{ \\prime}= \\displaystyle \\frac{R^{3}}{a^{3}}m}} \\end{array} $$", "answer": "$$m' = \\frac{R^3}{a^3} m$$" }, { "id": 709, "tag": "MECHANICS", "content": "In a zero-gravity space, there is a ring made of a certain material, with a radius of $R$, density $\\rho$, Young's modulus $E$, and a circular cross-section with a radius of $r$ ($r \\ll R$). The center of the ring is stationary in a rotating reference frame with an angular velocity $\\Omega$, and the normal of the ring is in the same direction as $\\Omega$. $\\Omega$ is much smaller than the natural frequency of the ring's vibration. A perturbation is applied, causing vibrations on the ring. Considering a standing wave solution of order $m$, it is known that the antinodes of the standing wave will precess. Find the precession angular velocity $\\delta\\omega$. The positive direction is defined to be the same as $\\Omega$. Ignore motion perpendicular to the plane of the ring; the order $m$ indicates that the displacement has the form of $\\cos(m\\theta)$ or $\\sin(m\\theta)$.\n", "solution": "Assume the radial and angular displacements of a particle on the ring are $w, v$, and the kinetic energy, potential energy line density are given as $$ T = \\frac{1}{2}\\rho S\\left((\\dot v + \\Omega (R + w))^2 + (\\dot w - \\Omega v)^2\\right) $$ $$ V = \\frac{E I}{2 R^4}\\left(w + \\partial_\\theta^2w\\right)^2 $$ The incompressibility constraint $$ \\partial_\\theta v + w = 0 $$ Thus, the Lagrangian density is $$ \\mathcal{L} = \\frac{1}{2}\\rho S\\left((\\dot v + \\Omega (R + w))^2 + (\\dot w - \\Omega v)^2\\right) - \\frac{E I}{2 R^4}\\left(w + \\partial_\\theta^2w\\right)^2 - \\lambda (\\partial_\\theta v + w) $$ Substituting into the Lagrange's equations, we get $$ \\begin{cases} \\ddot v + 2 \\Omega \\dot w + \\partial_\\theta \\lambda = 0, \\\\ \\ddot w - 2 \\Omega \\dot v + \\frac{EI}{\\rho S R}\\left(\\partial_\\theta^4 + 2\\partial_\\theta^2 + 1\\right)w - \\lambda = 0, \\\\ \\partial_\\theta v + w = 0 \\end{cases} $$ Eliminate variables to obtain $$ \\partial_t^2\\partial_\\theta^2 w - \\partial_t^2 w + 4 \\Omega \\partial_t \\partial_\\theta w + \\frac{EI}{\\rho S R}(\\partial_\\theta^6 + 2\\partial_\\theta^4 + \\partial_\\theta^2)w = 0 $$ Substituting the $m$-th order traveling wave solution $w = e^{i(\\omega t + m\\theta)}$ gives $$ (m^2+1)\\omega^2-4m\\Omega\\omega-\\frac{EI}{\\rho S R}(m^6-2m^4+m^2)=0 $$ Solving gives $$ \\omega = \\frac{4m\\Omega\\pm\\sqrt{16m^2\\Omega^2+\\frac{4EI}{\\rho S R}(m^6-2m^4+m^2)}}{2(m^2+1)}=\\omega_1 \\pm \\omega_{0} $$ The precession standing wave solution is $$ w = \\cos((\\omega_1 + \\omega_{0})t+m\\theta)+\\cos((\\omega_1 - \\omega_{0})t+m\\theta) = 2 \\cos(\\omega_1 t + m\\theta)\\cos(\\omega_0 t) $$ It can be observed that the wave crest will retreat with a precession angular velocity, according to our sign convention, $$ \\delta\\omega = - \\frac{2}{m^2+1}\\Omega $$ Of course, it can also be solved using Newtonian mechanics methods, but it is a bit more tedious.\n", "answer": "$$\n\\delta \\omega = -\\frac{2}{m^2+1} \\Omega\n$$" }, { "id": 423, "tag": "THERMODYNAMICS", "content": "One mole of a substance is a simple $p,v,T$ system. The coefficient of body expansion in any case is\n\n$$\n\\alpha=3/T\n$$\n\nThe adiabatic equation in any case is\n\n$$\np v^{2}=C\n$$\n\nThe isobaric heat capacity of the system in the arbitrary case is exactly: $$ \n\n$$\nc_{p}\\propto \\frac{p v}{T}\n$$ \n\nThe scale factor is a state-independent dimensionless constant, and\nTo determine the entropy change $$S_D-S_A$$ of the D state compared to the A state, two states $$A,D$$ on the $$p-v$$ graph are chosen, knowing that $$p_{A}$$,$$v_{A}$$,$$T_{A}$$,$$p_{D}$$,$$v_{D}$$ and that the entropy change of the D state compared to the A state is $$S_D-S_A$$.\n", "solution": "First of all\n\n$$\n\\alpha={\\frac{1}{v}}\\left({\\frac{\\partial v}{\\partial T}}\\right)_{p}={\\frac{3}{T}}\n$$\n\nThe equation of state can be obtained\n\n$$\nv=T^{3}h(p)\n$$\n\nWrite the full differential of entropy and utilize Maxwell's relation:\n\n$$\nd S={\\frac{\\partial S}{\\partial T}}d T+{\\frac{\\partial S}{\\partial p}}d p\n$$\n\n$$\n\\frac{\\partial S}{\\partial p}=-\\frac{\\partial v}{\\partial T}=-3T^{2}h\n$$\n\nDuring adiabatic process $$ d S=0$$\n\n$$\n{\\frac{d p}{d T}}={\\frac{\\frac{\\partial S}{\\partial T}}{3T^{2}h}}}\n$$\n\nAnd the adiabatic process is\n\n$$\np v^{2}=T^{6}p h^{2}=C\n$$\n\nThe logarithmic differentiation yields\n\n$$\n\\frac{d p}{d T}=-\\frac{6p h}{T(h+2p h^{\\prime})}\n$$\n\nComparing the two derivatives gives\n\n$$\n\\frac{\\partial S}{\\partial T}=-\\frac{18T p h^{2}}{h+2p h^{\\prime}}}\n$$\n\nDue to the properties of full differentiation, the two partial derivatives continue to conform to\n\n$$\n\\frac{\\partial}{\\partial p}\\left(-\\frac{18T p h^{2}}{h+2p h^{\\prime}}\\right)=\\frac{\\partial}{\\partial T}\\left(-3T^{2}h\\right)\n$$\n\nGet the differential equation\n\n$$\n\\left({\\frac{3p h^{2}}{h+2p h^{\\prime}}\\right)^{\\prime}=h\n$$\n\nSimplify it like this\n\n$$\nh=\\left[\\frac{3p h^{3}}{(p h^{2})^{\\prime}}\\right]^{\\prime}=-3p h^{3}\\cdot\\frac{(p h^{2})^{\\prime\\prime}}{[(p h^{2})^{\\prime}]^{2}}}+3\\ frac{(p h^{2})^{\\prime}h+p h^{2}h^{\\prime}}{(p h^{2})^{\\prime}}\n$$\n\n$$\n{\\frac{(p h^{2})^{\\prime}}{p h^{3}}}\\cdot h={\\frac{(p h^{2})^{\\prime}}{p h^{2}}}=-3{\\frac{(p h^{2})^{\\prime\\prime}}{(p h^{2})^{\\prime}}}}}+ 3{\\frac{(p h^{2})^{\\prime}}{p h^{2}}}}+3{\\frac{h^{\\prime}}{h}}}\n$$\n\nThe solution is\n\n$$\n(p h^{2})^{-2/3}(p h^{2})^{\\prime}=C h=C p^{-1/2}\\cdot(p h^{2})^{1/2}\n$$\n\nFurther moving the right-hand side $(p h^{2})^{1/2}$ to the left-hand side and integrating over $P$ finally yields\n\n$$\nh={\\frac{1}{C_{1}{\\sqrt{p}}({\\sqrt{p}}+C_{2})^{3}}}}\n$$\n\nAnd thus the final equation of state\n\n$$\nT=C_{1}^{\\frac{1}{3}}p^{\\frac{1}{6}}(\\sqrt{p}+C_{2})v^{\\frac{1}{3}}\n$$\n\nSince the isobaric heat capacity is \n\n$$\nc_{P}=T{\\frac{\\partial S}{\\partial T}}\n$$ \n\nSubstituting gives the result \n\n$$\nc_{P}=-{\\frac{18T^{2}p h^{2}}{h+2p h^{\\prime}}}}={\\frac{6T^{2}}{C_{1}({\\sqrt{p}}+C_{2})^{2}}}}\n$$ \n\nSince \n\n$$\nv=T^{3}\\cdot{\\frac{1}{C_{1}{\\sqrt{p}}({\\sqrt{p}}+C_{2})^{3}}}}\n$$ \n\nSubstituting the relationship yields \n\n$$\n\\lambda=6\\left(1+{\\frac{C_{2}}{\\sqrt{p}}}\\right)\n$$ \n\nIn order to make $\\lambda$ constant we deduce that \n\n$$\nC_{2}=0\n$$ \n\nThus, only \n\nSince $$ C_{2}=0 $$ the previous conclusion reduces to \n\n$$\n\\lambda=6\n$$ \n\nSince $C_{2}=0$ simplifies to\n\n$$\nh={\\frac{1}{C_{1}p^{2}}}\n$$\n\n$$\nT^{3}=C_{1}p^{2}v\n$$\n\n$$\nd S=\\frac{6T}{C_{1}p}d T-\\frac{3T^{2}}{C_{1}p^{2}}d p\n$$\n\nThus, according to the equation of state\n\n$$\nC_{1}=\\frac{3T_A^{3}}{p_{A}v_{A}^2}\n$$\n\n$$\n\\left({\\frac{T_{D}}{T_{A}}}\\right)^{3}=\\left({\\frac{p_{D}}{p_{A}}}\\right)^{2}\\left({\\frac{v_{D}}}{v_{A}}}\\right)\n$$\n\nGet\n\n$$\nT_{D}=T_{A}\\left(\\frac{p_D^2v_{D}}{p_A^2v_{A}}\\right)^{\\frac{1}{3}}\n$$\n\nThe entropy integral gives\n\n$$\nS={\\frac{3T^{2}}{C_{1}p}}+C^{\\prime}=3{\\sqrt{\\frac{T v}{C_{1}}}}+C^{\\prime}\n$$\n\nSince the third law of thermodynamics\n\n$$\nC^{\\prime}=0\\quad,\\quad S=\\frac{3T^{2}}{C_{1}p}\n$$\n\nThus entropy changes\n\n$$\nS_{D}-S_{A}={\\frac{3T_{D}^{2}}{C_{1}p_{D}}}-{\\frac{3T_{A}^{2}}{C_{1}p_{A}}}=\\frac{3p_Av_A}{T_A}\\left(\\left(\\frac{p_Dv_D^2}{p_Av_A^ 2}\\right)^{\\frac{1}{3}}-1\\right)\n$$\n", "answer": "$$\nS_{D}-S_{A}=\\frac{3p_Av_A}{T_A}\\left(\\left(\\frac{p_Dv_D^2}{p_Av_A^2}\\right)^{\\frac{1}{3}}-1\\right)\n$$\n" }, { "id": 36, "tag": "MECHANICS", "content": "Assume that the Earth is initially a solid sphere with a uniform density of $\\rho$ and a volume of $V={\\frac{4}{3}}\\pi R^{3}$, and is incompressible. The gravitational constant $G$ is known.\n\nFirst, consider the effect of Earth's rotation. Earth rotates around its polar axis with a constant angular velocity $\\omega$, forming an oblate spheroid under the action of inertial centrifugal forces.\n\nAlso consider the tidal forces exerted by Earth's satellite—the Moon. For simplicity, we establish the following simplified model: consider the Moon as a homogeneous ring with mass line density $\\textstyle{\\lambda={\\frac{m}{2\\pi r}}}$, located in the Earth's equatorial plane and centered at the Earth's center with a radius of $r$ ($r\\gg R$). Find the eccentricity of the Earth when stable $\\begin{array}{r}{e=\\frac{\\sqrt{a^{2}-b^{2}}}{a}}\\end{array}$, where $a$ is the semi-major axis and $b$ is the semi-minor axis.", "solution": "The incompressibility condition is given as \n\n$$\n{\\frac{4\\pi a^{2}b}{3}}={\\frac{4\\pi R^{3}}{3}}\n$$ \n\nThus, \n\n$$\na={\\frac{R}{(1-e^{2})^{\\frac{1}{6}}}}\n$$ \n\nA uniformly dense ellipsoid can be divided into many thin ellipsoid shells with the same eccentricity. It can be proven that the surface mass density distribution of each ellipsoid shell is identical to the surface charge density distribution of ellipsoid conductors in electrostatics. The gravitational potential inside an ellipsoid shell is zero, so the self-energy of each ellipsoid shell can be calculated and summed. The total gravitational self-energy is \n\n$$\nV_{g}=\\int_{0}^{a}-{\\frac{4\\pi}{3}}{\\sqrt{1-e^{2}}}a^{3}{\\frac{4\\pi G\\rho a{\\sqrt{1-e^{2}}}}{e}}\\mathrm{sin}^{-1}e d a\n$$ \n\nAfter integration, \n\n$$\nV_{g}=-{\\frac{16\\pi^{2}}{15}}G\\rho^{2}R^{5}{\\frac{(1-e^{2})^{\\frac{1}{6}}}{e}}\\sin^{-1}e\n$$ \n\nThe total centrifugal potential energy is \n\n$$\nV_{i}=\\iiint-{\\frac{1}{2}}\\rho\\omega^{2}(x^{2}+y^{2})d V\n$$ \n\nAfter integration, \n\n$$\nV_{g}=-\\frac{4\\pi}{15}\\rho\\omega^{2}R^{5}\\frac{1}{(1-e^{2})^{\\frac{1}{3}}}\n$$ \n\nThe tidal potential energy per unit mass is \n\n$$\nv_{t}=\\int_{0}^{2\\pi}-\\frac{G\\lambda r d\\varphi}{[(x-r\\cos\\varphi)^{2}+(y-r\\sin\\varphi)^{2}+z^{2}]^{\\frac{1}{2}}}\n$$ \n\nAfter approximation, \n\n$$\nv_{t}=-{\\frac{G m}{r}}+{\\frac{G m}{4r^{3}}}(2z^{2}-x^{2}-y^{2})\n$$ \n\nThe total tidal potential energy is \n\n$$\nV_{t}=\\iiint[-{\\frac{G m\\rho}{r}}+{\\frac{G m\\rho}{4r^{3}}}(2z^{2}-x^{2}-y^{2})]d V=-{\\frac{4\\pi G m\\rho R^{3}}{3}}-{\\frac{2\\pi G m\\rho R^{5}}{15}}{\\frac{e^{2}}{r^{3}(1-e^2)^{1/6}}}\n$$ \n\nThe total potential function is the sum of gravitational self-energy, centrifugal potential energy, and tidal potential energy. After approximation, \n\n$$\nV=-\\frac{16\\pi^{2}}{15}G\\rho^{2}R^{5}-\\frac{4\\pi}{15}\\rho\\omega^{2}R^{5}-\\frac{4\\pi}{3}\\frac{G m\\rho R^{3}}{r}-\\frac{4\\pi}{45}\\rho\\omega^{2}R^{5}e^{2}-\\frac{2\\pi}{15}\\frac{G m\\rho R^{5}}{r^{3}}e^{2}+\\frac{16\\pi^{2}}{675}G\\rho^{2}R^{5}e^{4}\n$$ \n\nWhen stable, the potential function reaches a minimum, \n\n$$\ne={\\sqrt{\\frac{30\\omega^{2}+45{\\frac{G m}{r^{3}}}}{16\\pi\\rho G}}}\n$$ \n\n**[Solution Method 2] (Alternative Method: Surface Equipotential Approach)** \n\nThe surface equation for a rotational ellipsoid with a short axis in spherical coordinates is \n\n$$\nr^{\\prime}(\\theta)^{2}=\\frac{a^{2}b^{2}}{a^{2}\\cos\\theta^{2}+b^{2}\\sin\\theta^{2}}\n$$ \n\nFrom equation $\\textcircled{2}$, \n\n$$\nr^{\\prime}(\\theta)\\approx R+\\frac{R}{6}(1-3\\cos\\theta^{2})\n$$ \n\nUsing the multipole expansion method in electrostatics and expanding to the quadrupole moment of a rotational ellipsoid, the gravitational potential per unit mass is \n\n$$\nv_{g}=-{\\frac{4\\pi G\\rho R^{3}}{3r^{\\prime}}}-{\\frac{2\\pi G\\rho R^{5}}{15{r^{\\prime}}^{5}}}{\\frac{e^{2}}{(1-e^{2})^{\\frac{1}{3}}}}{({x^{\\prime}}^{2}+{y^{\\prime}}^{2}-2{z^{\\prime}}^{2})}\n$$ \n\nThe centrifugal potential per unit mass is \n\n$$\nv_{i}=-\\frac{1}{2}\\omega^{2}({x^{\\prime}}^{2}+{y^{\\prime}}^{2})\n$$ \n\nThe tidal potential per unit mass is \n\n$$\nv_{t}=-{\\frac{G m}{r}}+{\\frac{G m}{4r^{3}}}(2{z^{\\prime}}^{2}-{x^{\\prime}}^{2}-{y^{\\prime}}^{2})\n$$ \n\nThe total surface potential per unit mass, which is the sum of the three potentials, is approximated as \n\n$$\nv(\\theta)=-\\frac{4\\pi G\\rho R^{2}e^{2}}{45}(3\\cos\\theta^{2}-1)-\\frac{1}{2}\\omega^{2}R^{2}\\sin\\theta^{2}-\\frac{G m}{r}+\\frac{G m R^{2}}{4r^{3}}(3\\cos\\theta^{2}-1)\n$$ \n\nThe surface equipotential condition requires $v$ to be independent of $\\theta$, yielding \n\n$$\ne={\\sqrt{\\frac{30\\omega^{2}+45{\\frac{G m}{r^{3}}}}{16\\pi\\rho G}}}\n$$", "answer": "$$\n\\sqrt{\\frac{30\\omega^2 + \\frac{45Gm}{r^3}}{16\\pi\\rho G}}\n$$" }, { "id": 333, "tag": "ELECTRICITY", "content": "A dielectric cylinder with a radius $R$, mass $m$, and uniform length $L \\gg R$ is permanently polarized, where the magnitude of the polarization intensity vector at any point a distance $r$ from the center is $P_{0}e^{k r}$, directed radially outward. Initially, the cylinder is at rest, and a constant external torque $M$ is applied to the cylinder. Find the angular acceleration $\\beta$ of the cylinder. The final result is expressed in terms of $M, m, k, R, L, P_0, \\mu_0$.", "solution": "When the angular velocity of the cylinder is $\\omega$, the surface current density is \n\n$$\ni=\\sigma\\omega R=P_{0}\\omega R e^{k R}\n$$\n\nThe volume current density in region C is \n\n$$\nj_{2}=\\rho r\\omega=-(1+k r)P_{0}\\omega e^{k r}\n$$\n\nFrom Ampère's circuital theorem: \n\n$$\n\\begin{array}{l}{{\\displaystyle{\\cal B}=\\int_{r}^{R}\\mu_{\\scriptscriptstyle0}j d r+\\mu_{\\scriptscriptstyle0}i}}\\ {{\\mathrm{}}}\\ {{\\displaystyle{\\cal B}=\\mu_{\\scriptscriptstyle0}P_{0}\\omega r e^{k r}}}\\end{array}\n$$\n\nMagnetic flux \n\n$$\n\\Phi=\\int_{0}^{r}\\boldsymbol{B}\\cdot2\\pi r d r\n$$\n\n$$\n\\Phi=2\\pi\\mu_{\\scriptscriptstyle0}\\omega P_{\\scriptscriptstyle0}(\\frac{r^{\\prime}{}^{2}}k e^{k r}-\\frac r{k^{2}}e^{k r}+\\frac2{k^{3}}e^{k r})\n$$\n\nElectromagnetic induction \n\n$$\n\\begin{array}{l}{E=-\\displaystyle\\frac{1}{2\\pi r}\\frac{\\partial}{\\partial t}\\Phi}\\ {E=-\\displaystyle\\mu_{0}\\beta P_{0}e^{\\displaystyle k r}(\\frac{r}{k}{-}\\frac{2}{k^{2}}{+}\\frac{2}{k^{3}r})}\\end{array}\n$$\n\nElectromagnetic torque \n\n$$\nM_{c}=\\int_{0}^{R}E\\cdot\\rho L\\cdot2\\pi r^{2}d r+E R\\sigma\\cdot2\\pi R L_{\\mathrm{~/~}}\n$$\n\nAfter integrating \n\nFrom the momentum theorem: \n\n$$\nM+M_{e}=\\frac{1}{2}m R^{2}\\beta\\stackrel{\\_}{\\_}.\n$$\n\nSolving for $\\beta$ \n\n$$\n\\beta = \\frac{2M}{mR^2+4\\mu_0P_0^2\\pi L[\\frac 3{8k^4}+(\\frac{R^3}{2k}-\\frac{3R^2}{4k^2}+\\frac{3R}{4k^3}-\\frac 3 {8k^4})e ^{2kR}}]\n$$", "answer": "$$\n\\beta = \\frac{2M}{mR^2 + 4\\mu_0 P_0^2 \\pi L \\left(\\frac{3}{8k^4} + \\left(\\frac{R^3}{2k} - \\frac{3R^2}{4k^2} + \\frac{3R}{4k^3} - \\frac{3}{8k^4}\\right)e^{2kR}\\right)}\n$$" }, { "id": 50, "tag": "THERMODYNAMICS", "content": "A cylindrical container with a cross-sectional area $S$ is fixed on the ground. The container walls are thermally insulated, and the top is sealed with a thermally insulated piston of mass $p_{0}S/g$ (where $g$ is the gravitational acceleration). The lower surface of the piston is parallel to the bottom surface of the container. The exterior of the container is a vacuum, and the interior contains a diatomic ideal gas with the mass of a single molecule being $m$ and the total number of molecules $N$. The temperature of the gas throughout the container is uniform at $T_{0}$. \n\nSlowly sprinkle a total mass of powder $(p_{1}-p_{0})S/g$ onto the piston. Neglect all friction, and consider the effect of gravity on the distribution of gas molecules. Find the final temperature of the gas inside the container, $T_{1}$.", "solution": "In equilibrium in a gravitational field, the molecular number density at temperature $T$ follows the Boltzmann distribution:\n\n$$\nn(z) = n_{0} \\exp\\left(-\\frac{m g z}{k T}\\right)\n$$\n\nIn the formula, $n_{0}$ is the molecular number density at $z=0$ (the bottom of the container), which is an unknown to be determined; $k$ is the Boltzmann constant. The total number of molecules in the container remains constant, always equal to $N$:\n\n$$\nN = \\int_{0}^{H} S n(z) \\operatorname{d}z = S n_{0} \\int_{0}^{H} \\exp\\left(-\\frac{m g z}{k T}\\right) \\operatorname{d}z = \\frac{S n_{0} k T}{m g} \\left[1 - \\exp\\left(-\\frac{m g H}{k T}\\right)\\right]\n$$\n\nSince the system is adiabatic and the powder is spread slowly, it can be assumed that the gas inside the container undergoes a quasi-static adiabatic process. At any moment during the process, let the temperature of the gas inside the container be $T$, the pressure at the lower surface of the piston inside the container be $p$, and the height from the lower surface of the piston to the bottom of the container be $H$. According to the first law of thermodynamics, for the gas inside the container, we have\n\n$$\n-p S \\mathrm{d}H = \\frac{5}{2}N k \\mathrm{d}T + \\mathrm{d}E_{p}\n$$\n\nIn this formula, $\\frac{5}{2}N k \\mathrm{d}T$ represents the increase in internal energy of a diatomic ideal gas, and $\\mathrm{d}E_{p}$ is the increase in gravitational potential energy of the gas in the container. Taking $z=0$ as the zero point of gravitational potential energy, we have\n\n$$\n\\begin{array}{l}{E_{p} = \\int_{0}^{H} m g z S n(z) \\mathrm{d}z = m g S n_{0} \\int_{0}^{H} z \\exp\\left(-\\frac{m g z}{k T}\\right) \\mathrm{d}z } \\\\ { = -k T S H n_{0} \\exp\\left(-\\frac{m g H}{k T}\\right) + k T S n_{0} \\cdot \\frac{k T}{m g} - k T S n_{0} \\exp\\left(-\\frac{m g H}{k T}\\right) \\cdot \\frac{k T}{m g}} \\end{array}\n$$\n\nAt the lower surface of the piston, according to the ideal gas pressure formula, we have\n\n$$\np = n(H) \\cdot k T = n_{0} \\exp\\left(-\\frac{m g H}{k T}\\right) k T\n$$\n\nSubstituting into equation $\\textcircled{4}$ gives\n\n$$\nE_{p} = -p S H + (p S + N m g) \\frac{k T}{m g} - p S \\frac{k T}{m g} = -p S H + N k T\n$$\n\nHere, $p + \\frac{N m g}{S} = n_{0} k T$ is the pressure at the bottom of the container. Taking the differential of the above equation gives $\\mathrm{d}E_{p} = -p S \\mathrm{d}H - S H \\mathrm{d}p + N k \\mathrm{d}T$, substituting into equation $\\textcircled{3}$ gives\n\n$$\nS H \\mathrm{d}p = \\frac{7}{2} N k \\mathrm{d}T\n$$\n\nCombining equations $\\textcircled{2}$ and $\\textcircled{5}$ gives\n\n$$\nH = \\frac{k T}{m g} \\ln\\frac{p S + N m g}{p S}\n$$\n\nSubstituting into equation $\\textcircled{7}$ gives\n\n$$\n\\begin{array}{c}{{\\displaystyle{\\frac{S}{m g}}\\ln\\frac{p S + N m g}{p S} \\mathrm{d}p = \\frac{7}{2} N \\frac{\\mathrm{d}T}{T}}} \\\\ {\\mathrm{}} \\\\ {\\displaystyle{\\ln(p S + N m g) \\mathrm{d}(p S) - \\ln(p S) \\mathrm{d}(p S) = \\frac{7}{2} N m g \\frac{\\mathrm{d}T}{T}}} \\end{array}\n$$\n\nAccording to piston equilibrium, initial state $T = T_{0}$, $p = p_{0}$, final state $T = T_{1}$, $p = p_{1}$, integrating the above equation gives\n\n$$\n\\frac{p_{1} S}{N m g} \\ln\\left(1 + \\frac{N m g}{p_{1} S}\\right) - \\frac{p_{0} S}{N m g} \\ln\\left(1 + \\frac{N m g}{p_{0} S}\\right) + \\ln\\frac{p_{1} S + N m g}{p_{0} S + N m g} = \\frac{7}{2} \\ln\\frac{T_{1}}{T_{0}}\n$$\n\nTherefore, the temperature we seek is\n\n$$\nT_{1} = T_{0} \\frac{\\left(1 + \\frac{N m g}{p_{1} S}\\right)^{\\frac{2 p_{1} S}{7 N m g}}}{\\left(1 + \\frac{N m g}{p_{0} S}\\right)^{\\frac{2 p_{0} S}{7 N m g}}} \\left(\\frac{p_{1} S + N m g}{p_{0} S + N m g}\\right)^{\\frac{2}{7}}\n$$", "answer": "$$\nT_0 \\frac{\\left(1+\\frac{Nmg}{p_1 S}\\right)^{\\frac{2p_1 S}{7Nmg}}}{\\left(1+\\frac{Nmg}{p_0 S}\\right)^{\\frac{2p_0 S}{7Nmg}}} \\left(\\frac{p_1 S+Nmg}{p_0 S+Nmg}\\right)^{\\frac{2}{7}}\n$$" }, { "id": 221, "tag": "MODERN", "content": "A three-dimensional relativistic oscillator moves in a space filled with uniform \"dust.\" During motion, \"dust\" continuously adheres to the oscillator, which is assumed to increase the rest mass of the sphere without altering its size. The collision is adiabatic, and the \"dust\" quickly replenishes the region the sphere just passed. The original length of the spring is vanishing, and the potential energy $V$ can be expressed as $V=\\frac{1}{2}k x^2$, where $k$ is a known constant. The cross-sectional area of the sphere is $A$, the density of the \"dust\" is $\\rho$, the initial rest mass of the sphere is $m_{0}$, and the speed of light in vacuum is $c$. Initially, the sphere performs uniform circular motion with a radius $R_{0}$. Find the time $t$ required when the circular motion radius changes to $R$. Assume $R^2 \\gg A$ and $m_{0} \\gg \\rho A R$.", "solution": "From the problem statement, it can be assumed that the sphere is constantly maintaining uniform circular motion \n\n$$\nv=\\omega r\n$$ \n\n$$\n\\omega={\\sqrt{\\frac{k}{\\gamma m}}}\n$$ \n\nAlso, angular momentum conservation \n\n$$\n\\gamma m v r=c o n s t\n$$ \n\nWhere $\\gamma = \\begin{array}{r}{\\gamma=\\frac{1}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}}\\end{array}$ \n\n$$\n\\gamma m r^{4}=c o n s t=B\n$$ \n\nNext, solve for the initial $B$. From \n\n$$\n{\\frac{1}{\\gamma^{2}}}=1-{\\frac{v^{2}}{c^{2}}}=1-{\\frac{k r^{2}}{\\gamma m c^{2}}}\n$$ \n\nSimplify to obtain \n\n$$\n\\gamma^{2}-{\\frac{k r^{2}}{m c^{2}}}\\gamma-1=0\n$$ \n\nTherefore \n\n$$\n\\gamma=\\frac{\\frac{k r^{2}}{m c^{2}}+\\sqrt{\\left(\\frac{k r^{2}}{m c^{2}}\\right)^{2}+4}}{2}\n$$ \n\nSubstitute initial conditions, we have \n\n$$\n\\gamma_{0}=\\frac{\\frac{k R_{0}^{2}}{m_{0}c^{2}}+\\sqrt{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}\\right)^{2}+4}}{2}\n$$ \n\n$$\nB=\\frac{\\frac{k R_{0}^{2}}{m_{0}c^{2}}+\\sqrt{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}\\right)^{2}+4}}{2}m R_{0}^{4}\n$$ \n\nAdiabatic process implies total energy conservation, and mass changes with time \n\n$$\n{\\frac{d}{d t}}\\gamma m=\\rho A v-{\\frac{k r}{c^{2}}}{\\frac{d r}{d t}}\n$$ \n\nSubstitute angular momentum conservation \n\n$$\nd(\\frac{B}{r^{4}})=\\rho A\\sqrt{\\frac{k}{B}}r^{3}d t-d(\\frac{k r^{2}}{2c^{2}})\n$$ \n\nResulting in \n\n$$\n\\Delta(-\\frac{k}{c^{2}r}+\\frac{4B}{7r^{7}})=\\rho A\\sqrt{\\frac{k}{B}}t\n$$ \n\nSimplify to get \n\n$$\nt=\\frac{1}{\\rho A}\\sqrt{\\frac{B}{k}}(\\frac{k}{c^{2}}(-\\frac{1}{R}+\\frac{1}{R_{0}})+\\frac{4}{7}B(\\frac{1}{R^{7}}-\\frac{1}{R_{0}^{7}}))\n$$", "answer": "$$\nt=\\frac{1}{\\rho A}\\sqrt{\\frac{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}+\\sqrt{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}\\right)^{2}+4}\\right)m R_{0}^{4}}{2k}}\\left(\\frac{k}{c^{2}}\\left(-\\frac{1}{R}+\\frac{1}{R_{0}}\\right)+\\frac{4}{7}\\frac{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}+\\sqrt{\\left(\\frac{k R_{0}^{2}}{m_{0}c^{2}}\\right)^{2}+4}\\right)m R_{0}^{4}}{2}\\left(\\frac{1}{R^{7}}-\\frac{1}{R_{0}^{7}}\\right)\\right)\n$$" }, { "id": 120, "tag": "THERMODYNAMICS", "content": "Consider an ideal gas with a known fixed constant $\\gamma$ undergoing a rectangular cycle on a $p-V$ diagram, which consists of two isochoric processes and two isobaric processes forming a positive cycle. It is known that the area of this cycle is fixed as $W$ (where $W$ is sufficiently small, negating the possibility of negative pressure or volume). The rectangle must contain the point $(p_0,V_0)$. Find the maximum efficiency of this cycle.", "solution": "If the center shifts to $\\mathcal{p}^{\\prime}, V^{\\prime}$, then the heat absorbed is: \n\n$$\nQ={\\frac{W}{2}}+{\\frac{\\Delta p V^{\\prime}+\\gamma p^{\\prime}\\Delta V}{\\gamma-1}}\n$$\n\nLet us first consider the translation of a fixed rectangular shape with ${\\Delta p,\\Delta V=W/\\Delta p}$ in the $\\mathcal{P}V$ diagram. Clearly, minimizing the heat absorbed requires minimizing the values of the center's $p,V$.Thus, it becomes optimal only when the rectangle is translated so that the $(p_0, V_0)$ point is at the upper-right corner of the rectangle: \n\n$$\np^{\\prime}=p_0-{\\frac{\\Delta p}{2}}\\quad,\\quad V^{\\prime}=V_0-{\\frac{\\Delta V}{2}}\n$$\n\nSubstituting back, the heat absorbed becomes: \n\n$$\nQ={\\frac{\\Delta p V+\\gamma p\\Delta V}{\\gamma-1}}-{\\frac{W}{\\gamma-1}}\n$$\n\nFrom the same inequality, the maximum efficiency is: \n\n$$\n\\eta=\\frac{\\gamma-1}{2\\sqrt{\\gamma p_0 V_0/W}-1}\n$$", "answer": "$$\n\\eta=\\frac{\\gamma-1}{2\\sqrt{\\gamma p_0 V_0/W}-1}\n$$" }, { "id": 121, "tag": "MODERN", "content": "Two spacecraft are traveling in a space medium that is flowing uniformly at a constant velocity $u$ with respect to an inertial frame $S$. The spacecraft are moving through the medium with equal relative velocities $v$ with respect to the medium. Neither of the velocities is known. Ultimately, the velocities of the two spacecraft with respect to the inertial frame $S$ are $v_{1}$ and $v_{2}$, and the angle between the directions of these velocities is an acute angle $\\alpha$. These three quantities are given. The speed of light is $c$. Considering relativistic effects, determine the minimum possible value of $u$.", "solution": "First, let's derive a byproduct of velocity transformation. The velocity transformation is:\n\n$$\n v_{x}^{\\prime}={\\frac{v_{x}-u}{1-u v_{x}/c^{2}}}\n$$\n\n$$\nv_{y,z}^{\\prime}=v_{y,z}\\cdot\\frac{\\sqrt{1-u^{2}/c^{2}}}{1-u v_{x}/c^{2}}\n$$\n\nHence, we can calculate:\n\n$$\n \\begin{array}{l}{\\gamma=\\cfrac{1}{\\sqrt{1-(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})/c^{2}}}}\\ {=\\cfrac{1}{\\sqrt{1-(v_{x}^{\\prime2}+v_{y}^{\\prime2}+v_{z}^{\\prime2})/c^{2}}}}\\end{array}\n$$\n\nObtain the transformation:\n\n$$\n \\gamma^{\\prime}=\\gamma\\cdot\\frac{1-{u v_{x}}/{c^{2}}}{\\sqrt{1-{u^{2}}/{c^{2}}}}\n$$\n\nThe minimum value of $v$ can be calculated by first determining the relative velocity of two planes:\n\n$$\n \\gamma_{12}=\\frac{1-\\beta_{1}\\beta_{2}\\cos\\alpha}{\\sqrt{(1-\\beta_{1}^{2})(1-\\beta_{2}^{2})}}\n$$\n\nConsider a system where two spaceships have equal but opposite velocities, this velocity $v=\\beta c$ is what we seek. At this moment, the relative velocity of the two planes can be obtained:\n\nThus, jointly obtain:\n\n$$\n \\gamma_{12}=\\frac{1+\\beta^{2}}{\\sqrt{(1-\\beta^{2})(1-\\beta^{2})}}=\\frac{1+\\beta^{2}}{1-\\beta^{2}}\n$$\n\n$$\nv=c{\\sqrt{\\frac{c^{2}-v_{1}v_{2}\\cos\\alpha-{\\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}{c^{2}-v_{1}v_{2}\\cos\\alpha+{\\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}}}\n$$\n\nThe minimum value of $u=\\beta_{\\mathsf{u}}c$ can be set such that its magnitude and direction in the original reference system are between angles $\\varphi_{1},\\varphi_{2}$ of both velocities. Therefore, after transformation, shared velocity:\n\n$$\n =\\gamma_{i}\\frac{1-\\beta_{u}\\beta_{\\mathrm{{i}c o s\\varphi_{i}}}}{\\sqrt{1-\\beta_{u}^{2}}}\n$$\n\n$$\n\\gamma_{1}^{\\prime}=\\gamma_{2}^{\\prime}\n$$\n\nObtain velocity:\n\n$$\n \\beta_{u}=\\frac{\\gamma_{1}-\\gamma_{2}}{\\gamma_{1}\\beta_{1}\\cos\\varphi_{1}-\\gamma_{2}\\beta_{2}\\cos\\varphi_{2}}\n$$\n\nThe denominator can be expanded using the auxiliary angle formula to find extremes:\n\n$$\n\\cdots=(\\gamma_{1}\\beta_{1}-\\gamma_{2}\\beta_{2}\\cos\\alpha)\\cos\\varphi_{1}-\\gamma_{2}\\beta_{2}\\sin\\alpha\\sin\\varphi_{1}\\leq{\\sqrt{(\\gamma_{1}\\beta_{1}-\\gamma_{2}\\beta_{2}\\cos\\alpha)^{2}+(\\gamma_{2}\\beta_{2}\\sin\\alpha)^{2}}}\n$$\nSubstitute to find the extreme value:\n\n$$\nu=c^{2}{\\frac{\\left|{\\sqrt{c^{2}-v_{1}^{2}}}-{\\sqrt{c^{2}-v_{1}^{2}}}\\right|}{{\\sqrt{c^{2}\\left(v_{1}^{2}+v_{2}^{2}\\right)-2v_{1}^{2}v_{2}^{2}-2v_{1}v_{2}\\cos\\alpha{\\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}}}}\n$$", "answer": "$$\nu = c^2 \\frac{\\left| \\sqrt{c^2 - v_1^2} - \\sqrt{c^2 - v_1^2} \\right|}{\\sqrt{c^2 (v_1^2 + v_2^2) - 2 v_1^2 v_2^2 - 2 v_1 v_2 \\cos \\alpha \\sqrt{(c^2 - v_1^2)(c^2 - v_2^2)}}}\n$$" }, { "id": 648, "tag": "MECHANICS", "content": "In a vacuum, there are concentric spherical shells made of insulating material with radii $R$ and $2R$. The inner shell is uniformly charged with a total charge of $Q(Q>0)$, while the outer shell is uncharged. There is also a small insulating sphere with a charge of $+q(q>0)$ and a mass $m$, which can be launched from a point on the inner shell with a certain velocity. The direction can be adjusted, and its motion is confined between the inner and outer shells. Collisions with the inner and outer shells are perfectly elastic, and the charges do not change during collisions. It is known that the sphere can return to its starting point after making one full orbit around the center. If the sphere is launched from point $A$ on the inner shell and it collides with the inner shell a total of $n$ times (counting the final return to the starting point as one collision), determine the minimum launch speed $v$, ignoring gravity.\n", "solution": "Let the distance from the center of the sphere to the position of the ball at the extreme point be $r_{1}$, and the speed be $ u_{\\mathrm{_I}}$. According to the conservation of angular momentum and energy: $$ m u_{0}R\\sin\\theta = m u_{1}r_{1} = L $$ $$ {\\frac{1}{2}}{m{ u_{0}}^{2}}+{\\frac{k Q q}{R}} = {\\frac{1}{2}}{m{ u_{1}}^{2}}+{\\frac{k Q q}{r_{1}}} = E = {\\frac{k Q q}{2a}} $$ In an appropriate polar coordinate system, we have: $$ p = \\frac{L^{2}}{kQq m}, \\quad e = \\sqrt{1+\\frac{2E}{m}\\frac{L^{2}}{k^{2}{Q}^{2}q^{2}}} $$ When $Q>0$, the charged ball undergoes hyperbolic motion. Each time it collides with the inner sphere shell, it also collides with the outer sphere shell, and returns to point $A$. The angular displacement around the center during inner and outer sphere collisions is: $$ \\alpha = {\\frac{2\\pi}{2n}} = {\\frac{\\pi}{n}} $$ The curve $AB$ is its trajectory, with point $O$ as one focus of the hyperbola, and $C$ as the other focus. According to conic section knowledge, the trajectory of the hyperbola to focus $O$ is: $$ r = \\frac{p}{e\\cos\\theta^{\\prime}-1} $$ where $\\theta^{\\prime}$ is the angle between the trajectory of the ball and its polar axis. Considering that: $$ 2a = {\\overline{{A O}}} - {\\overline{{A C}}} = R - {\\overline{{A C}}} = {\\overline{{B O}}} - {\\overline{{B C}}} = 2R - {\\overline{{B C}}} $$ $$ {\\overline{B C}} - {\\overline{{A C}}} = R $$ Thus the possible trajectory of point $C$ is a hyperbola, and since $r$ continuously increases from $A$ to $\\boldsymbol{B}$, the polar axis of the hyperbola must not pass through the trajectory $AB$. To minimize the launch speed $ u_{0}$, and thus minimize energy $E$, $2a$ should be maximized, making $AC$ minimal, satisfying the condition when focus $C$ is at the extended line $OA$ and intersects the asymptote of the hyperbola at point $D$. At this point: $$ \\theta = \\frac{\\pi}{2} $$ The speed is minimal, and according to geometric relations: $$ a+c = R $$ Additionally, according to previous equations: $$ \\theta^{\\prime} = 0,\\quad r = {\\frac{p}{e\\cos0-1}} = R $$ $$ \\begin{array}{c c}{{\\theta^{\\prime}=\\alpha,}}&{{r=\\displaystyle\\frac{p}{e\\cos{\\alpha}-1}}=2R} {{}}&{{}} \\end{array} $$ $$ {{\\displaystyle a=\\frac{2\\cos{\\alpha}-1}{2\\cos{\\alpha}}R}} $$ Substituting into previous equations: $$ u_{0} = \\sqrt{\\frac{2k Q q}{m R}\\frac{1-\\cos\\frac{\\pi}{n}}{2\\cos\\frac{\\pi}{n}-1}} $$\n", "answer": "$$\n\\sqrt{\\frac{2kQq}{mR} \\cdot \\frac{1-\\cos\\left(\\frac{\\pi}{n}\\right)}{2\\cos\\left(\\frac{\\pi}{n}\\right)-1}}\n$$" }, { "id": 72, "tag": "THERMODYNAMICS", "content": "At time $t=0$, an adiabatic cylinder with a cross-sectional area $S$ is divided into two equal parts with a volume of $V_{0}$ by an adiabatic, thin and lightweight movable piston. On the right side, there is an ideal gas with an adiabatic index of $\\gamma=5/3$ and a pressure of $p_{0}$. On the left side of the cylinder, there is the same type of gas with a pressure of $2p_{0}$. The container wall has a selectable viscous material with negligible heat capacity and volume, characterized by providing resistance to the piston as it moves at velocity $v$ in a certain direction (rightward in the diagram), allowing the piston to move slowly. Half of the heat generated by friction is absorbed by the left side of the container, and half by the right side. The piston is released from rest; calculate the displacement $l$ after the piston stops moving (express the result as a coefficient with two significant figures), shown in the form:\n$$\nl=x \\frac{V_{0}}{S} (x \\text{ unknown, to two decimal places})\n$$ \nThis problem assumes that all processes are quasistatic processes.", "solution": "From the proportional relationship between work done and heat absorbed:\n\n$$\n\\frac{1}{2}\\left(p_{1}\\mathrm{d}V_{1}+p_{2}\\mathrm{d}V_{2}\\right)=T_{1}\\mathrm{d}S_{1}=T_{2}\\mathrm{d}S_{2}\n$$ \n\nFrom the first law of thermodynamics:\n\n$$\n\\mathrm{d}U_{i}=T_{i}\\mathrm{d}S_{i}-p_{i}\\mathrm{d}V_{i}\n$$ \n\nAnd the equation of state:\n\n$$\nU_{i}=\\frac{3}{2}\\nu_{i}R T_{i}=\\frac{3}{2}p_{i}V_{i}\n$$ \n\nWe obtain:\n\n$$\n\\mathrm{d}U_{1}={\\frac{3}{2}}\\mathrm{d}\\left(p_{1}V_{1}\\right)=-{\\frac{1}{2}}p_{1}\\mathrm{d}V_{1}+{\\frac{1}{2}}p_{2}\\mathrm{d}V_{2}\n$$ \n\n$$\n\\mathrm{d}U_{2}={\\frac{3}{2}}\\mathrm{d}\\left(p_{2}V_{2}\\right)=-{\\frac{1}{2}}p_{2}\\mathrm{d}V_{2}+{\\frac{1}{2}}p_{1}\\mathrm{d}V_{1}\n$$ \n\nTo solve these two differential equations, we note that due to energy conservation, the total internal energy remains unchanged:\n\n$$\nU_{1}+U_{2}={\\frac{3}{2}}\\left(p_{1}V_{1}+p_{2}V_{2}\\right)={\\frac{9}{2}}p_{0}V_{0}\n$$ \n\nThus, for the above two equations, we define:\n\n$$\nV_{1}=(1+x)V_{0}\\quad,\\quad V_{2}=(1-x)V_{0}\n$$ \n\nSimplifying the first equation:\n\n$$\n\\frac{4}{3}p_{1}V_{2}\\mathrm{d}V_{1}+\\frac{1}{3}p_{1}V_{1}\\mathrm{d}V_{2}+V_{1}V_{2}\\mathrm{d}p_{1}=p_{0}V_{0}\\mathrm{d}V_{2}\n$$ \n\nMultiplying both sides by $V_{1}^{1/3}V_{2}^{-2/3}$, the left side conveniently forms a total differential:\n\n$$\n\\mathrm{d}\\left[p_{1}(1+x)^{\\frac{1}{3}}(1-x)^{\\frac{1}{3}}\\right]=-p_{0}{\\frac{(1+x)^{1/3}}{(1-x)^{2/3}}}\\mathrm{d}x\n$$ \n\nDue to conservation of internal energy, the pressure at final equilibrium is equal on both sides:\n\n$$\np\\cdot2V_{0}=3p_{0}V_{0}\\quad,\\quad p={\\frac{3}{2}}p_{0}\n$$ \n\nThus, we finally obtain the integral equation:\n\n$$\n2-\\frac{3}{2}(1+y)^{\\frac{1}{3}}(1-y)^{\\frac{1}{3}}=\\int_{0}^{y}\\frac{(1+x)^{1/3}}{(1-x)^{2/3}}\\mathrm{d}x\n$$ \n\nSolution by numerical methods:\n\n$$\ny=0.20\n$$ \n\nThat is:\n\n$$\nl=0.20\\frac{V_{0}}{S}\n$$", "answer": "$$\nl=0.20\\frac{V_0}{S}\n$$" }, { "id": 603, "tag": "ELECTRICITY", "content": "The characteristics of a medium in an electric field are described as follows: $D=\\varepsilon E=\\varepsilon_0 E+P$ where $E$ and $D$ are the electric field intensity and electric displacement, respectively, $\\varepsilon$ is the dielectric constant of the medium, $P$ is the polarization intensity (electric dipole moment per unit volume), and $\\varepsilon_0$ is the permittivity of free space. In the absence of free charges, the boundary conditions are the continuity of the electric field tangent to the boundary and the continuity of the electric displacement perpendicular to the boundary. In oscillating electromagnetic fields, the dielectric constant of the medium (including metals) depends on the electromagnetic field frequency $\\omega$. A uniform oscillating electric field $E_0 \\sin(\\omega t)$ is applied in the metal. Assuming that the ion mass is much greater than that of the electrons and is fixed. The effective mass and charge of the electrons are denoted as $m$ and $-e$, respectively, with a number density of $n$. Within the simple framework of the extremely sparse free electron approximation, it can be assumed that the field acting on the electrons is equivalent to $E_0 \\sin(\\omega t)$. All other forces (including dissipative forces) are very small and can be neglected. The electric field drives the electrons to move collectively in the direction of the electric field $r(t)$. Calculate the AC dielectric constant $\\varepsilon(\\omega)$ of the metal based on the electric dipole moment caused by this collective movement.", "solution": "Write the Newton's second law for the motion of the electron: \\[ m\\frac{d^2x}{dt^2}=eE_0 \\sin(\\omega t) \\] We obtain \\[ r=\\frac{eE_0}{m\\omega^2}\\sin(\\omega t) \\] Since the motion of ions is not considered, they do not contribute to the total polarizability. The electric dipole moment of the electron and the total polarizability are \\[ p=ex \\] \\[ P=np=-\\frac{ne^2E_0}{m\\omega^2}\\sin(\\omega t) \\] The formula for calculating the dielectric constant is \\[ \\varepsilon \\varepsilon_0 E=\\varepsilon_0 E+P \\] We obtain the dielectric constant \\[ \\varepsilon =1-\\frac{ne^2}{\\varepsilon_0 m\\omega^2} \\]", "answer": "$$\\varepsilon = 1 - \\frac{ne^2}{\\varepsilon_0 m \\omega^2}$$" }, { "id": 670, "tag": "ELECTRICITY", "content": "Consider a plasma system composed of protons and electrons, where the equilibrium number density of positive and negative charges is \\( n_0 \\). A point charge \\( q \\) is placed in this plasma. Due to the Coulomb interaction, the point charge will attract opposite charges and repel like charges, causing a change in the charge distribution within the plasma: within a certain spatial range around the point charge, the densities of positive and negative charges are no longer equal, thereby weakening the electric field of the point charge. This phenomenon is known as Debye shielding. It is known that in spherical coordinates, the Poisson equation for the electrostatic field is: \\[ \\nabla^2 \\varphi = \\frac{1}{r^2} \\frac{d}{dr} \\left( r^2 \\frac{\\mathrm{d}\\varphi}{\\mathrm{d}r} \\right) = -\\frac{\\rho(r)}{\\varepsilon_0}, \\] where \\( \\varphi(r) \\) is the electric potential distribution. Hint: try assuming a trial solution of the form \\( \\varphi(r) = \\frac{u(r)}{r} \\). Assume the plasma satisfies the following conditions: At equilibrium, the number density \\( n(\\varphi) \\) of protons and electrons follows the Boltzmann distribution; The plasma is overall electrically neutral; \\( e\\varphi \\ll kT \\) (\\( e \\) is the elementary charge, \\( k \\) is the Boltzmann constant, \\( T \\) is the plasma temperature). The Debye radius \\( \\lambda_D \\) is defined as the distance at which the potential decays to \\( \\frac{1}{e} \\) of the vacuum point charge potential. Derive the expression for \\( \\lambda_D \\).", "solution": "The number density of positive and negative ions are respectively: $$ n_{i}(\\varphi)=n_{i0}e^{-\\frac{e\\varphi}{k T}} $$ $$ n_{e}(\\varphi)=n_{e0}e^{\\frac{e\\varphi}{k T}} $$ Charge density: $$ \\rho=n_{i}q+n_{e}(-q)=n_{i0}e\\cdot e^{-{\\frac{e\\varphi}{k T}}}-n_{e0}e\\cdot e^{\\frac{e\\varphi}{k T}} $$ The plasma is electrically neutral, i.e., $$ n_{i0}=n_{e0}=n_{0} $$ Thus, we can get: $$ \\rho=-2\\frac{e^{2}\\varphi}{k T}n_{0} $$ Then the Poisson's equation becomes: $$ \\frac{1}{r^{2}}\\frac{\\mathrm{d}}{\\mathrm{d}r}(r^{2}\\frac{\\mathrm{d}\\varphi}{\\mathrm{d}r})=\\frac{2e^{2}\\varphi n_{0}}{k T\\varepsilon_{0}} $$ Let $\\varphi={\\frac{u}{r}}$, substitute and simplify to obtain: $$ \\frac{\\mathrm{d}^{2}u}{\\mathrm{d}r^{2}}=\\frac{2e^{2}u n_{0}}{k T\\varepsilon_{0}} $$ $\\lambda_{D}=\\sqrt{\\frac{k T\\varepsilon_{0}}{2e^{2}n_{0}}}$ From the original equation, it is easy to get: $$ u=C_{1}e^{\\frac{r}{\\lambda_{D}}}+C_{2}e^{-\\frac{r}{\\lambda_{D}}} $$ Considering the actual situation, the plasma is electrically neutral at infinity, so the potential must converge, thus $C_{1}=0$, and the potential tends towards that of a point charge at infinitesimal distances, hence $$ \\varphi(r)={\\frac{u}{r}}={\\frac{q}{4\\pi\\varepsilon_{0}r}}e^{-{\\frac{r}{\\lambda_{D}}}} $$ According to the problem statement, the characteristic scale of the Debye shielding is given directly as: $$ \\lambda_{_D}=\\sqrt{\\frac{k T\\varepsilon_{0}}{2e^{2}n_{0}}} $$", "answer": "$$\n\\lambda_D = \\sqrt{\\frac{k T \\varepsilon_0}{2 e^2 n_0}}\n$$" }, { "id": 140, "tag": "MECHANICS", "content": "Three identical homogeneous balls are placed on a smooth horizontal surface, touching each other and are close enough to each other. A rope is wrapped around the spheres at the height of their centers, tying them together. A fourth identical sphere is placed on top of the three spheres. Find the tension $T$ in the rope. It is given that the weight of each sphere is $P$.", "solution": "In the text, $P$ represents the gravity on sphere $O$.\n\nSince the spheres are completely identical, the string forms part of an equilateral triangle, so the tension $T$ along the direction $CG$ for sphere $C$ should equal the component of $N$ in that direction, i.e.,\n\n$$\n2T\\cos{\\frac{\\pi}{6}} = N\\cos\\theta.\n$$\n\nFrom equation (1) and equation (2), we have\n\n$$\nT = {\\frac{P}{3{\\sqrt{3}}}}\\cot\\theta.\n$$\n\nSince $G$ is the center of the equilateral triangle $ABC$ formed by the centers of spheres $A$, $B$, and $C$ (Figure 1.11(c)), from geometric relationships, we know\n\n$$\nCG = \\frac{2}{3}(2R)\\cos{\\frac{\\pi}{6}} = \\frac{2R}{\\sqrt{3}},\n$$\n\nthus,\n\n$$\n\\cos\\theta = {\\frac{CG}{CO}} = {\\frac{1}{\\sqrt{3}}}, \\quad \\cot\\theta = {\\frac{1}{\\sqrt{2}}}.\n$$\n\nSubstituting equation (4) into equation (3), we get\n\n$$\nT = {\\frac{\\sqrt{6}}{18}}P.\n$$", "answer": "$$\\frac{\\sqrt{6}}{18}P$$" }, { "id": 653, "tag": "ELECTRICITY", "content": "A massless insulating string of length $\\alpha$ is fixed at one end, with a conductor ring of mass $m$ and radius $r$ ($r \\ll \\alpha$) suspended on the other end, forming a simple pendulum. Initially, the pendulum string is vertical, and the ring is just inside a sector of a uniform magnetic field. The center of the sector is at the suspension point, with its edges just tangent to the ring. The magnetic field is perpendicular to the plane of the ring and directed inward, with a magnetic flux density of $B$. If the ring is given an initial horizontal velocity (perpendicular to the magnetic field) with a magnitude of $v$, what is the minimum initial velocity $v_{0}$ required for the pendulum string to reach the horizontal direction? The gravitational acceleration is $g$, the magnetic field is sufficiently strong, and the torque due to the Ampere force is much greater than the torque due to gravity when the ring moves within the magnetic field. The ring is a superconducting ring with a self-inductance coefficient $L$.", "solution": "If the ring is a superconducting ring, when it moves in a magnetic field, due to the properties of the superconductor, the magnetic flux in the ring remains unchanged, i.e.: $$ L I + B S = B \\cdot \\pi r^2 $$ Thus, the current in the ring is: $$ \\begin{array}{l}{I = \\displaystyle\\frac{B}{L}(\\pi r^{2} - S)}\\ = \\displaystyle\\frac{B}{L}\\left\\{r^{2}\\arcsin\\left[\\frac{a(\\theta_{0}-\\theta)}{r}\\right] - \\sqrt{2r a\\theta-a^{2}\\theta^{2}} \\cdot a(\\theta_{0}-\\theta)\\right\\}\\end{array} $$ $$ \\phi = \\operatorname{arccos}\\left[\\frac{a(\\theta_{0}-\\theta)}{r}\\right] $$ We have: $$ \\cos\\phi = \\frac{a(\\theta_{0}-\\theta)}{r}, \\sin\\phi = \\sqrt{1-\\cos^{2}\\phi} = \\sqrt{1-\\frac{a^{2}(\\theta_{0}-\\theta)^{2}}{r^{2}}} $$ Thus: $$ I = \\frac{B}{L}(r^{2}\\phi - r^{2}\\sin\\phi\\cos\\phi), l = 2r\\sin\\phi $$ The torque about the suspension point due to the Ampère force acting on the part of the ring within the magnetic field is approximately: $$ \\tau = I B l a = \\frac{2B^{2}r a}{L}(r^{2}\\phi - r^{2}\\sin\\phi\\cos\\phi)\\sin\\phi $$ The dynamic equation of the ring is: $$ m a^{2}\\frac{d\\omega}{dt} = -\\frac{2B^{2}r a}{L}(r^{2}\\phi - r^{2}\\sin\\phi\\cos\\phi)\\sin\\phi $$ Since $\\frac{d\\omega}{dt} = \\omega\\frac{d\\omega}{d\\theta}$, we have: $$ d\\theta = \\frac{r}{a} \\sin (\\phi) d\\phi $$ Thus: Integrating both sides, given that initially when $\\theta = 0$, $\\phi = 0$, and $\\omega \\approx \\frac{v_{0}}{a}$, we have: $$ m a^{2}\\int_{\\frac{v_{0}}{a}}^{\\omega}\\omega d\\omega = -\\frac{2B^{2}r^{4}}{L}\\int_{0}^{\\phi}\\left(\\phi - \\sin\\phi\\cos\\phi\\right)\\sin^{2}\\phi d\\phi $$ We obtain: $$ \\frac{1}{2}m a^{2}\\left(\\omega^{2}-\\frac{v_{0}^{2}}{a^{2}}\\right) = -\\frac{B^{2}r^{4}}{2L}\\left[\\phi^{2} - \\phi\\sin2\\phi - \\frac{1}{2}(\\cos2\\phi - 1) - \\sin^{4}\\phi\\right] $$ Thus: $$ \\omega^{2} = \\frac{v_{0}^{2}}{a^{2}} - \\frac{B^{2}r^{4}}{m a^{2}L}\\left[\\phi^{2} - \\phi\\sin2\\phi - \\frac{1}{2}(\\cos2\\phi - 1) - \\sin^{4}\\phi\\right] $$ When the ring has just completely left the magnetic field area, $\\theta = 2\\theta_{0}$, corresponding to $\\phi = \\pi$, substituting into the above equation, we get: $$ \\omega_{2}^{2} = \\frac{v_{0}^{2}}{a^{2}} - \\frac{\\pi^{2}B^{2}r^{4}}{m a^{2}L} $$ To make the pendulum reach the horizontal direction, it should satisfy: $$ v_{0} = \\sqrt{2g a + \\frac{\\pi^{2}B^{2}r^{4}}{m L}} $$", "answer": "$$\\sqrt{2g a+\\frac{\\pi^2 B^2 r^4}{m L}}$$" }, { "id": 719, "tag": "THERMODYNAMICS", "content": "The martian atmosphere can be considered as composed only of very thin $CO_{2}$. The molar mass of this atmosphere is denoted by $\\mu$, and the atmosphere at the same height can be considered as an ideal gas in equilibrium. The mass of Mars is $M_{m}$ (far greater than the total mass of the martian atmosphere), and its radius is $R_{m}$. Assume the relationship between the atmospheric mass density and the height $h$ above the surface of Mars is given by $$ \\alpha(h)=\\rho_{\\mathrm{o}}(1+\\frac{h}{R_{m}})^{1-n} $$ where $\\rho_{\\mathrm{0}}$ is a constant, and $\\textit{n}(n>4)$ is also a constant. Derive the expression for the temperature $T(h)$ of the martian atmosphere as a function of the height $h$. Express the result using the following physical quantities: the mass of Mars $M_{m}$, radius $R_{m}$, molar mass of the atmosphere $\\mu$, constant $n$, gravitational constant $G$, and ideal gas constant $R$.", "solution": "(1) Basic Assumptions and Ideal Gas Law In the atmosphere at a certain height \\( h \\) on Mars, a thin horizontally placed box-shaped region is defined. The volume of this region is \\( V \\), the mass of the atmosphere inside is \\( M \\), the pressure is \\( P(h) \\), and the temperature is \\( T(h) \\). Because the box is very thin, the density, pressure, and temperature of the gas can be considered constant. According to the ideal gas law: $$ P(h)V = n R T(h) $$ The volume of the atmosphere inside the box is: $$ V = \\frac{M}{\\rho(h)} $$ The number of moles of the Mars atmosphere inside the box is: $$ n = \\frac{M}{\\mu} $$ By combining equations (1), (2), and (3), we get: $$ T(h) = \\frac{P(h) \\mu}{\\rho(h) R} $$ (2) Determine the Pressure \\( P(h) \\) Consider a thin spherical cap-shaped atmospheric layer from height \\( x \\) to \\( x + dx \\), subtending a solid angle \\( \\Delta \\Omega \\). Let \\( \\Delta \\Omega \\) be very small, so that the radial lines on the side of the spherical cap can be considered parallel to each other. From the force analysis, we know: $$ [P(x + dx) - P(x)] \\cdot \\Delta \\Omega (R_m + x)^2 + \\rho(x) g(x) \\cdot \\Delta \\Omega (R_m + x)^2 dx = 0 $$ That is: $$ \\frac{dP}{dx} = - \\rho(x) g(x) $$ Where the gravitational acceleration from the surface of Mars to height \\( x \\) is: $$ g(x) = \\frac{G M_m}{(R_m + x)^2} $$ Substituting into equation (6) and integrating, we obtain: $$ P(h) = G M_m \\int_h^{\\infty} \\frac{\\rho(x)}{(R_m + x)^2} dx $$ (3) Substitute and Determine the Temperature Expression Substitute equation (7′) into equation (4), and use the density distribution expression provided in the problem: $$ \\rho(x) = \\rho_0 \\left(1 + \\frac{x}{R_m}\\right)^{1 - n} $$ Finally, we get: $$ T(h) = \\frac{\\mu G M_m}{n R (R_m + h)} $$ Final Result (Temperature Distribution of Martian Atmosphere) $$ \\boxed{T(h) = \\frac{\\mu G M_m}{n R (R_m + h)}} $$ This result indicates that the temperature of Mars' atmosphere decreases inversely with height \\( h \\).", "answer": "$$T(h) = \\frac{\\mu G M_m}{n R (R_m + h)}$$" }, { "id": 211, "tag": "MODERN", "content": "In the ground frame S, two trains with an intrinsic length of $L_{0}$ are stationary along the $x$ axis, with the left side of train A located at $x=0$. The gap between the left side of train B and the right side of train A is $D$. At time $t=0$ in the ground frame S, both trains A and B synchronize their carriage times to $t_{A}=t_{B}=0$, and the clocks are located on the left side of each train. At this moment, the two trains start moving with constant accelerations relative to themselves, $a_{A}$ and $a_{B}$.\n\nLet the distance between train carriages AB in frame S be $D(t)$. Express the value of $a_{B}$ when $D(t)\\equiv D$ in terms of $L_{0}$, $a_{A}$, and the speed of light $c$.", "solution": "Since the left end of the train is aligned with the ground clock, the left end of the train can be regarded as a point mass, and thus we can study uniform accelerated motion.\n\nLet the velocity of the left end at time $t$ in the ground frame be $v(t)$, at time $t+d t$, first perform a Lorentz transformation to the rest frame with velocity $v(t)$. In the rest frame, obtain an infinitesimal velocity $adτ$:\n\nUsing the proper time formula:\n\n$$\nv+dv={\\frac{v+adτ}{1+{\\frac{v\\times adτ}{c^{2}}}}}\\rightarrow dv=adτ\\left(1-{\\frac{v^{2}}{c^{2}}}\\right)\n$$\n\n$$\ndτ={\\sqrt{1-{\\frac{v^{2}}{c^{2}}}}}dt \\to dv=a{\\sqrt{1-{\\frac{v^{2}}{c^{2}}}^{3}}}dt\n$$ \n\nSeparate variables on both sides, yielding:\n\n$$\nat=\\int_{0}^{v}{\\frac{dV}{1-{V}^{2}/{c}^{2}}}={\\frac{v}{1-{v}^{2}/{c}^{2}}}\n$$\n\nSolving for the velocity function:\n\n$$\nv(t)={\\frac{at}{\\sqrt{1+\\left({\\frac{at}{c}}\\right)^{2}}}}\\to x(t)=x_{0}+\\int_{0}^{t}v(t)dt=x_{0}+{\\frac{c^{2}}{a}}\\left({\\sqrt{1+\\left({\\frac{at}{c}}\\right)^{2}}}-1\\right)\n$$\n\nSelect special cases A, B to write the displacement function:\n\n$$\n\\left\\{\\begin{array}{l}{\\displaystyle x_{l A}=\\frac{c^{2}}{a_{A}}\\left(\\sqrt{1+\\left(\\frac{a_{A}t}{c}\\right)^{2}}-1\\right)}\\ {\\displaystyle x_{l B}=L_{0}+D+\\frac{c^{2}}{a_{B}}\\left(\\sqrt{1+\\left(\\frac{a_{B}t}{c}\\right)^{2}}-1\\right)}\\end{array}\\right.\n$$\n\nUsing the motion on the left side, we can calculate the proper time:\n\n$$\nτ=\\int_{0}^{t}{\\sqrt{1-\\left({\\frac{v}{c}}\\right)^{2}}}dt=\\int_{0}^{t}{\\frac{dt}{\\sqrt{1+\\left({\\frac{at}{c}}\\right)^{2}}}}={\\frac{c}{a}}\\mathrm{arcsinh}{\\frac{at}{c}}\n$$\n\nApply to special cases A, B:\n\n$$\n\\left\\{\\begin{array}{l l}{{\\displaystyleτ_{A}=\\frac{c}{a_{A}}\\mathrm{arcsinh}\\frac{a_{A}t}{c}}}\\ {{\\displaystyleτ_{B}=\\frac{c}{a_{B}}\\mathrm{arcsinh}\\frac{a_{B}t}{c}}}\\end{array}\\right.\n$$\n\nTaking train A as an example, given the proper time $τ_{A}$, the speed of A at this time is:\n\n$$\nv_{A}(τ)={\\frac{a_{A}t}{\\sqrt{1+\\left({\\frac{a_{A}t}{c}}\\right)^{2}}}}=c\\operatorname{tanh}{\\frac{a_{A}τ}{c}}\n$$\n\nThe left side of the train system $(0,τ)$ and the ground frame $(x_{l A},t)$ are aligned, allowing for easy writing of the Lorentz transformation for the right end time and coordinates in the ground frame:\n\n$$\nt_{r,A}=t+{\\frac{v_{A}L_{0}/c^{2}}{1-v_{A}^{2}/c^{2}}}=t+\\cosh{\\frac{a_{A}τ}{c}}\\left({\\frac{L_{0}}{c}}\\operatorname{tanh}{\\frac{a_{A}τ}{c}}\\right)=t+{\\frac{L_{0}}{c}}\\sinh{\\frac{a_{A}τ}{c}}=\\left(1+{\\frac{a_{A}L_{0}}{c^{2}}}\\right)t\n$$\n\n$$\nx_{r,A}=x_{l,A}+\\frac{L_{0}}{1-v_{A}^{2}/c^{2}}=\\left(\\frac{c^{2}}{a_{A}}+L_{0}\\right)\\sqrt{1+\\left(\\frac{a_{A}t}{c}\\right)^{2}}-\\frac{c^{2}}{a_{A}}\n$$\n\nSince $x_{r,A}$ occurs at event S frame $t=t_{r,A}$, the correct $x_{r,A}(t)$ function should replace $t$:\n\n$$\nx_{r,A}={\\frac{c^{2}}{{\\frac{a_{A}}{1+{\\frac{a_{A}L_{0}}{c^{2}}}}}}}{\\sqrt{1+\\left({\\frac{a_{A}t/c}{1+{\\frac{a_{A}L_{0}}{c^{2}}}}}\\right)^{2}}}-{\\frac{c^{2}}{a_{A}}\n$$\n\nAt this moment, the distance between the right side of A and the left side of B is:\n\n$$\n=x_{l,B}-x_{r,A}=L_{0}+D+{\\frac{c^{2}}{a_{B}}}\\left(\\sqrt{1+\\left({\\frac{a_{B}t}{c}}\\right)^{2}}-1\\right)+{\\frac{c^{2}}{a_{A}}}-{\\frac{c^{2}}{{\\frac{a_{A}}{1+{\\frac{a_{A}L_{0}}{c^{2}}}}}}}\\sqrt{1+\\left({\\frac{a_{A}t/c}{1+{\\frac{a_{A}L_{0}}{c^{2}}}}}\\right)^{2}}\n$$\n\nIf the distance is constant, then naturally:\n\n$$\na_{B}={\\frac{a_{A}}{1+{\\frac{a_{A}L_{0}}{c^{2}}}}}\n$$", "answer": "$$\n\\frac{a_A}{1+\\frac{a_A L_0}{c^2}}\n$$" }, { "id": 139, "tag": "MECHANICS", "content": "Two smooth cylinders, each with a radius of $r_{1}$ but with weights $P_{1}$ and $P_{2}$ respectively, are placed in a smooth cylindrical groove with a radius of $R$. Find the tangent value of the angle $\\varphi$ when the two cylinders are in equilibrium. The angle $\\varphi$ is defined as the angle between the line connecting the center of the higher cylinder and the center of the cylindrical groove, and the horizontal line. Assume $\\cos\\beta = \\frac{r}{R - r}$. Express the result in terms of $\\beta$.", "solution": "Under the action of these forces, cylinders 1 and 2 reach equilibrium. Triangle \\( O O_{1}O_{2} \\) is an isosceles triangle. Let \\(\\beta=\\angle O O_{1}O_{2},\\) then \n\n\\[ \n\\cos \\beta = \\frac{r}{R-r}.\n\\]\n\nFor cylinder 1, the condition for force equilibrium is \n\n\\[ \n\\begin{array}{r l}\n&{P_{1}+f\\sin(\\beta-\\varphi)-N_{1}\\sin(2\\beta-\\varphi)=0,}\\\\\n&{f\\cos(\\beta-\\varphi)-N_{1}\\cos(2\\beta-\\varphi)=0.}\n\\end{array}\n\\]\n\nBy eliminating \\( N_{1} \\) from the two equations in expression (2), we obtain \n\n\\[ \nP_{1}\\cos(2\\beta-\\varphi)-f\\sin\\beta=0.\n\\]\n\nEliminating \\( f \\), we obtain \n\n\\[ \nP_{1}\\cos(\\beta-\\varphi)-N_{1}\\sin\\beta=0.\n\\]\n\nFor cylinder 2, at equilibrium we have \n\n\\[ \n\\begin{array}{r l}\n&{P_{2}-f\\sin(\\beta-\\varphi)-N_{2}\\sin\\varphi=0,}\\\\\n&{f\\cos(\\beta-\\varphi)-N_{2}\\cos\\varphi=0.}\n\\end{array}\n\\]\n\nBy eliminating \\( N_{2} \\) from the two equations in expression (5), we obtain \n\n\\[ \nP_{2}\\cos\\varphi-f\\sin\\beta=0.\n\\]\n\nEliminating \\( f \\), we obtain \n\n\\[ \nP_{2}\\cos(\\beta-\\varphi)-N_{2}\\sin\\beta=0.\n\\]\n\nBy eliminating \\( f \\) from equations (3) and (6), we obtain \n\n\\[ \n\\tan\\varphi = \\frac{P_{2}-P_{1}\\cos(2\\beta)}{P_{1}\\sin(2\\beta)}.\n\\]\n\nThis is the relationship that \\(\\varphi\\) satisfies when the system is at equilibrium.", "answer": "$$\\varphi = \\arctan \\frac{P_2 - P_1 \\cos(2\\beta)}{P_1 \\sin(2\\beta)}$$" }, { "id": 106, "tag": "MECHANICS", "content": "Two thin rods, each with mass $m$ and length $l$, are centrally connected by a thin string (with negligible mass). The upper ends are also connected by a very short flexible string (with negligible length). The other end of each rod is free to slide across a horizontal table without friction. The plane formed by the rods and strings is perpendicular to the tabletop, with gravitational acceleration $g$. Initially, the rods are stationary, forming an angle $\\theta_{0}$ with the tabletop. When the thin string at the center is cut, the rods fall vertically until they hit the table. What is the velocity $v$ of the rods just before the ends connected at the table touch the tabletop?", "solution": "Solution: Using the conservation of energy:\nInitial mechanical energy: $\\frac{1}{2}mgl\\sin\\theta_{0}$ \nMechanical energy while in motion (angular velocity is $\\dot{\\theta}$): translational kinetic energy $^{+}$ rotational kinetic energy $^+$ gravitational potential energy:\n$$\n{\\frac{1}{2}}m{\\dot{\\theta}}^{2}{\\frac{l^{2}}{4}}+{\\frac{1}{2}}{\\frac{1}{12}}m l^{2}{\\dot{\\theta}}^{2}+{\\frac{1}{2}}m g l \\sin\\theta={\\frac{1}{6}}m l^{2}{\\dot{\\theta}}^{2}+{\\frac{1}{2}}m g l \\sin\\theta\n$$\nWe obtain the angular velocity $\\dot{\\theta}: \\dot{\\theta} = \\sqrt{3g/l(\\sin\\theta_{0} - \\sin\\theta)}$. The terminal velocity is: $v=-\\sqrt{3gl\\sin\\theta_{0}}$", "answer": "$$v = -\\sqrt{3 g l \\sin \\theta_0}$$" }, { "id": 58, "tag": "ELECTRICITY", "content": "An arbitrary multi-terminal resistor network structure, for example with $N+1$ terminals, sets one specific terminal (0 terminal) as the reference for electric potential, setting its potential at 0. The potentials for other terminals (numbered 1, 2, ..., N) are $v_{1}, v_{2}, \\ldots, v_{N}$, and the current flowing into the network through these corresponding terminals (negative when flowing out) are $I_{1}, I_2, \\ldots, I_n$. Therefore, we have:\n\n$$\n\\left\\{\n\\begin{array}{l}\nV_{1} = a_{11} I_{1} + a_{12} I_{2} + \\cdots + a_{1N} I_{N} \\\\\nV_{2} = a_{21} I_{1} + a_{22} I_{2} + \\cdots + a_{2N} I_{N} \\\\\n\\vdots \\\\\nV_{i} = a_{i1} I_{1} + a_{i2} I_{2} + \\cdots + a_{iN} I_{N} \\\\\n\\vdots \\\\\nV_{N} = a_{N1} I_{1} + a_{N2} I_{2} + \\cdots + a_{NN} I_{N} \n\\end{array}\n\\right.\n$$\n\nDue to symmetry, we know that $a_{ij} = a_{ji}$.\n\nThis means that the properties of the $N+1$ terminal network can be determined by $\\frac{N(N+1)}{2}$ independent parameters: $a_{11}, a_{12}, \\ldots , a_{ij}, \\ldots , a_{NN} (i0$ be known. If the equilibrium is stable, find the vibration period $T$ near the equilibrium position (expressed in terms of $n$, $I_{0}$, $r$, $R$, $L$, $m$), and the answer may include $z_{0}$. It is known that the radius of the metal ring is $r \\ll R$, and the self-inductance is $L$.", "solution": "The equilibrium condition is\n\n$$\nF(z)=m g\n$$\n\n$$\n\\frac{\\pi}{2}\\mu_{0}n I R^{2}r^{2}\\frac{1}{(z^{2}+R^{2})^{3/2}}(\\frac{\\mu_{0}n I\\pi r^{2}}{2L}(1-\\frac{z}{\\sqrt{z^{2}+R^{2}}})-I_{0})=m g\n$$\n\nConsider the stability of the equilibrium. If $\\frac{d F}{d z}$ is negative, then the equilibrium is stable; otherwise, it is unstable.\n\n$$\n\\left.{\\frac{d F}{d z}}\\right|_{z_{0}}={\\frac{\\pi}{2}}\\mu_{0}n I R^{2}r^{2}\\cdot[({\\frac{\\mu_{0}n I\\pi r^{2}}{2L}}-I_{0}){\\frac{-3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}}+{\\frac{\\mu_{0}n I\\pi r^{2}}{2L}}{\\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}}]\n$$\n\nTherefore, the requirement for stability is\n\n$$\n\\frac{3z_{0}^{2}-R^{2}}{3z_{0}\\sqrt{z_{0}^{2}+R^{2}}}<1-\\frac{2I_{0}L}{\\mu_{0}n I\\pi r^{2}}\n$$\n\nIf the equilibrium is stable, then\n\n$$\nT=2\\pi\\sqrt{\\frac{m}{\\left.\\displaystyle-\\frac{d F}{d z}\\right|_{z_{0}}}}\n$$\n\n$$\nT=2\\pi\\sqrt{\\frac{m}{\\frac{\\pi}{2}\\mu_{0}n I R^{2}r^{2}\\cdot[(\\frac{\\mu_{0}n I\\pi r^{2}}{2L}-I_{0})\\frac{3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}-\\frac{\\mu_{0}n I\\pi r^{2}}{2L}\\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}]}}\n$$", "answer": "$$\nT=2\\pi\\sqrt{\\frac{\\frac{1}{(z_0^{2}+R^{2})^{3/2}}(\\frac{\\mu_{0}n I\\pi r^{2}}{2L}(1-\\frac{z_0}{\\sqrt{z_0^{2}+R^{2}}})-I_{0})}{g\\left[(\\frac{\\mu_{0}n I\\pi r^{2}}{2L}-I_{0})\\frac{3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}-\\frac{\\mu_{0}n I\\pi r^{2}}{2L}\\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}\\right]}}\n$$\n\n" }, { "id": 43, "tag": "ELECTRICITY", "content": "Two dielectric materials with base areas both equal to $\\pmb{A}$, thicknesses of $h_{1}$ and $h_{2}$ respectively, dielectric constants of $\\varepsilon_{1}$ and $\\varepsilon_{2}$ respectively, and conductivities of $\\sigma_{1}$ and $\\sigma_{2}$ respectively, are placed tightly between two highly conductive flat plates with the same area $\\pmb{A}$ and a separation distance of $(h_{1}+h_{2})$, forming a capacitor. The system is connected into a circuit , with a power source of electromotive force $U_{0}$. Initially, the capacitor is uncharged, and at $\\scriptstyle t=0$, the switch is turned on. Determine the time-dependent relationship of the current $I(t)$ flowing through the switch for $t > 0$.", "solution": "Let the magnitudes of the electric field intensity in the two media be $E_{1}, E_{2}$ respectively, and according to the relationship between the electric potentials, we have \n\n$$\nE_{1}h_{1} + E_{2}h_{2} = U_{0}\n$$ \n\nThe magnitudes of the current density in the two media are \n\n$$\n\\begin{aligned}\n j_1 &= \\sigma_1 E_1, \\\\\n j_2 &= \\sigma_2 E_2.\n\\end{aligned}\n$$ \n\nLet the free charge surface density at the contact of the two media be $\\sigma_{\\bullet}$, then \n\nFrom this, we obtain \n\n$$\n\\left\\{\\begin{array}{l l}{\\displaystyle{\\frac{\\mathrm{d}\\sigma_{\\circ}}{\\mathrm{d}t}}=j_{1}-j_{2}}\\ {\\sigma_{\\circ}=\\varepsilon_{2}E_{2}-\\varepsilon_{1}E_{1}}\\end{array}\\right.\n$$ \n\n$$\nU_{0} = \\frac{\\mathrm{d}\\sigma_{0}}{\\mathrm{d}t}\\cdot\\left(\\frac{h_{1}\\varepsilon_{2} + h_{2}\\varepsilon_{1}}{\\sigma_{1}\\varepsilon_{2} - \\sigma_{2}\\varepsilon_{1}}\\right) + \\sigma_{0}\\cdot\\left(\\frac{\\sigma_{1}h_{2} + \\sigma_{2}h_{1}}{\\sigma_{1}\\varepsilon_{2} - \\sigma_{2}\\varepsilon_{1}}\\right)\n$$ \n\nThat is \n\n$$\n\\sigma_{0} = \\frac{U_{0}\\left(\\sigma_{1}\\varepsilon_{2} - \\sigma_{2}\\varepsilon_{1}\\right)}{\\sigma_{1}h_{2} + \\sigma_{2}h_{1}}\\cdot\\left(1-\\mathrm{e}^{\\frac{t}{\\tau}}\\right)\n$$ \n\nWhere the time constant is \n\n$$\n\\tau = \\frac{h_{1}\\varepsilon_{2} + h_{2}\\varepsilon_{1}}{h_{1}\\sigma_{2} + h_{2}\\sigma_{1}}\n$$ \n\nThe current in the circuit can be obtained as \n\n$$\nI = A{\\left(\\sigma_{1}E_{1} + \\varepsilon_{1}{\\frac{\\mathrm{d}E_{1}}{\\mathrm{d}t}}\\right)} = A{\\left(\\sigma_{2}E_{2} + \\varepsilon_{2}{\\frac{\\mathrm{d}E_{2}}{\\mathrm{d}t}}\\right)}\n$$ \n\nSubstitute in to get \n\n$$\nI = A U_{0}\\cdot\\left[\\frac{\\sigma_{1}\\sigma_{2}}{\\sigma_{1}h_{2}+\\sigma_{2}h_{1}} + \\mathrm{e}^{\\frac{\\sigma_{1}h_{2}+\\sigma_{2}h_{1}}{\\varepsilon_{1}h_{2}+\\varepsilon_{2}h_{1}}t}\\cdot\\left(\\frac{\\varepsilon_{2}\\sigma_{1}+\\varepsilon_{1}\\sigma_{2}}{\\varepsilon_{1}h_{2}+\\varepsilon_{2}h_{1}} - \\frac{\\sigma_{1}\\sigma_{2}}{\\sigma_{1}h_{2}+\\sigma_{2}h_{1}} - \\frac{\\varepsilon_{1}\\varepsilon_{2}\\left(\\sigma_{1}h_{2}+\\sigma_{2}h_{1}\\right)}{\\left(\\varepsilon_{1}h_{2}+\\varepsilon_{2}h_{1}\\right)^{2}}\\right)\\right]\n$$", "answer": "$$\nI = A U_0 \\left[\\frac{\\sigma_1 \\sigma_2}{\\sigma_1 h_2 + \\sigma_2 h_1} + e^{-\\frac{\\sigma_1 h_2 + \\sigma_2 h_1}{\\varepsilon_1 h_2 + \\varepsilon_2 h_1} t} \\left(\\frac{\\varepsilon_2 \\sigma_1 + \\varepsilon_1 \\sigma_2}{\\varepsilon_1 h_2 + \\varepsilon_2 h_1} - \\frac{\\sigma_1 \\sigma_2}{\\sigma_1 h_2 + \\sigma_2 h_1} - \\frac{\\varepsilon_1 \\varepsilon_2 (\\sigma_1 h_2 + \\sigma_2 h_1)}{(\\varepsilon_1 h_2 + \\varepsilon_2 h_1)^2} \\right)\\right]\n$$" }, { "id": 117, "tag": "MECHANICS", "content": "There is an available cycloidal valley, which can be described using a cycloid (downward along the $y$-axis): \n$$\n\\left\\{{\\begin{array}{l}{x=R(\\theta-\\sin\\theta)}\\\\ {y=R(1-\\cos\\theta)}\\end{array}}\\right.\n$$\n Someone wants to release a cube with a mass of $M$ and with equal length, width, and height of $r$ at rest from one end of the valley at $\\theta = 0$. It is released facing the slope. A layer of snow with a surface density of $\\sigma$ is attached to the valley. The snow is in contact with the ground but does not interact, remaining stationary until it touches the object, at which point it adheres to its surface without altering the object's shape. Assuming gravitational acceleration $g$, there are the approximate conditions $\\sigma r R \\ll M, r \\ll R$ (results are retained up to the first order of $r$). Determine the highest position the cube can reach at the other end (considering no rolling occurs).\n", "solution": "In this sub-question, energy is not conserved, but analyzing its motion still presents certain difficulties. Considering that the problem only requires accuracy to the first-order term, the loss of ascent quotient can be divided into two parts for correction, with zero-order motion considered during the correction process. For the collision process between the square block and the snow, consider the conservation of momentum: $$ M v = (M + \\sigma\\gamma u{\\mathrm{d}}t)(v - {\\mathrm{d}}v) $$ This results in: $$ \\mathrm{d}\\upsilon = -\\frac{\\sigma r v^{2}}{M}\\mathrm{d}t $$ The energy loss during the process is: $$ d E = \\frac{1}{2}(M + \\mathrm{d}M)(\\upsilon + \\mathrm{d}\\upsilon)^{2} - \\frac{1}{2}(M)\\upsilon^{2} = -\\frac{1}{2}\\sigma r\\upsilon^{2}\\upsilon\\mathrm{d}t = -\\frac{1}{2}\\sigma r\\upsilon^{2}\\mathrm{d}s $$ In this case, it can be treated as linear motion, and the equation is not influenced by acceleration. Therefore, it can be corrected independently, with the zero-order velocity expression as: $$ \\upsilon_{0} = \\sqrt{2g R(1-\\cos\\theta)} $$ Substituting into the integral: $$ \\delta W = \\int f\\mathrm{d}s = \\int_{0}^{2\\pi} -g R(1-\\cos\\theta)\\sigma\\boldsymbol{r}\\cdot2R\\sin\\frac{\\theta}{2}\\mathrm{d}\\theta = -\\frac{32}{3}g R^{2}r\\sigma $$ This gives the energy loss caused by collisions, which we convert into a height correction, then combine it with the correction for the gravitational component of the snow from the first question, resulting in the total reached height: $$ \\begin{array}{l} {{\\delta y_{f} = \\displaystyle\\frac{32}{3}\\displaystyle\\frac{\\sigma R^{2}r}{M}}}{{}}\\\\ {{\\delta y_{g} = \\displaystyle\\frac{32}{3}\\displaystyle\\frac{\\sigma R^{2}r}{M}}}{{}}\\\\ {{\\delta y = \\displaystyle\\frac{64}{3}\\displaystyle\\frac{\\sigma R^{2}r}{M}}} \\end{array} $$ The two combined terms are both first-order and do not interfere with each other.\n", "answer": "$$\n\\delta y=\\frac{64}{3}\\frac{\\sigma R^2 r}{M}\n$$" }, { "id": 42, "tag": "OPTICS", "content": "We consider a special rotationally symmetric refractive index distribution $n=n(r)$, such that the light trajectory is $r=a\\cos q\\theta$ (although it is not very realistic for the refractive index to diverge at $r=0$). In the space where $r0)$ and above, set up the moment equilibrium at the $\\pmb{x}$ position: \n\n$$\nM = \\frac{m g}{2}(\\frac{l}{2}-x)\n$$\n\nSubstitute into the previous equation: \n\n$$\ny^{\\prime\\prime} = \\frac{6m g}{E a h^{3}}(\\frac{l}{2}-x)\n$$\n\nInitial conditions: \n\n$$\n\\begin{array}{r}{y(0)=0}\\ {y^{\\prime}(0)=0}\\end{array}\n$$\n\nIntegrating yields: \n\n$$\ny = {\\frac{3m g l}{2E a h^{3}}}x^{2} - {\\frac{m g}{E a h^{3}}}x^{3}\n$$\n\nSubstitute $\\begin{array}{r}{x=\\frac{1}{2}}\\end{array}$ to obtain: \n\n$$\n\\lambda = \\frac{m g l^{3}}{4E a h^{3}}\n$$\n\nBy the results of question (1), the metal beam can be equated to a spring with a stiffness coefficient of: \n\n$$\nk = \\frac{4E a h^{3}}{l^{3}}\n$$\n\nThis gives the angular frequency of vertical vibration: \n\n$$\n\\omega_{1} = {\\sqrt{\\frac{k}{m}}} = {\\sqrt{\\frac{4E a h^{3}}{m l^{3}}}}\n$$\n\nAs shown in the diagram, let the heavy object deviate from the center by $\\delta l$, denote $\\begin{array}{r}{\\varepsilon=\\frac{\\delta\\ell}{\\ l}}\\end{array}$\n\nFrom mechanical equilibrium: \n\n$$\nN_{1} = \\frac{m g}{2}(1-2\\varepsilon)\n$$\n\n$$\nN_{2} = \\frac{m g}{2}(1+2\\varepsilon)\n$$\n\nFor the part above the horizontal axis, establish the moment equilibrium equation at (2 >0): \n\n$$\nM_{1} = {{N}_{2}}\\left[\\frac{l}{2}\\left(1-2\\varepsilon\\right)-x\\right]\n$$\n\nSimilarly, for ${\\mathfrak{x}}<{\\mathfrak{0}}$: \n\n$$\nM_{2} = N_{1}\\left[\\frac{l}{2}\\left(1+2\\varepsilon\\right)+x\\right]\n$$\n\nInitial conditions: \n\n$$\n\\begin{array}{c}{y\\left(0\\right)=0}\\ {y^{\\prime}\\left(0\\right)=k_{0}}\\end{array}\n$$\n\nSubstitute into equation (2) and solve using initial conditions: \n\nSubstitute boundary conditions: \n\n$$\n\\begin{array}{c}{{y_{1}=k_{0}x+\\displaystyle\\frac{3m g l}{2E a h^{3}}\\left(1-4\\varepsilon^{2}\\right)x^{2}-\\displaystyle\\frac{m g}{E a h^{3}}\\left(1+2\\varepsilon\\right)x^{3}(x>0)}} \\\\ {{y_{2}=k_{0}x+\\displaystyle\\frac{3m g l}{2E a h^{3}}\\left(1-4\\varepsilon^{2}\\right)x^{2}+\\displaystyle\\frac{m g}{E a h^{3}}\\left(1-2\\varepsilon\\right)x^{3}(x<0)}}\\end{array}\n$$\n\n$$\ny_{1}\\left(\\frac{l}{2}\\left(1-2\\varepsilon\\right)\\right) = y_{2}\\left(-\\frac{l}{2}\\left(1+2\\varepsilon\\right)\\right) = \\lambda^{\\prime}\n$$\n\nSolve to obtain: \n\n$$\n\\begin{array}{l}{{k_{0}=\\displaystyle\\frac{2m g l^{2}}{E a h^{3}}\\left(1-4\\varepsilon^{2}\\right)\\varepsilon}} \\\\ {{\\lambda^{\\prime}=\\displaystyle\\frac{m g l^{3}}{4E a h^{3}}\\left(1-4\\varepsilon^{2}\\right)^{2}}}\\end{array}\n$$\n\nWhen $\\varepsilon\\ll1$, we have: \n\n$$\n\\lambda^{\\prime} \\approx \\frac{m g l^{3}}{4E a h^{3}}\\left(1-8\\varepsilon^{2}\\right)\n$$\n\nTherefore, the gravitational potential energy of the heavy object is: \n\n$$\nE_{p1} = \\frac{2m^{2}g^{2}l^{3}}{E a h^{3}}\\varepsilon^{2}\n$$\n\nConsider a small segment of the metal beam and its elastic potential energy: \n\n$$\nd E_{p2} = \\int{\\frac{1}{2}}E\\epsilon^{2}d V = \\int_{-{\\frac{h}{2}}}^{{\\frac{h}{2}}}{\\frac{{\\cal E}a y^{\\prime\\prime2}d x}{2}}z^{2}d z = {\\frac{{\\cal E}a h^{3}}{24}}y^{\\prime\\prime2}d x\n$$\n\nIntegrating yields: \n\n$$\nE_{p2} = \\int_{-\\frac{1}{2}(1+2\\varepsilon)}^{0}\\frac{E a h^{3}}{24}y_{2}^{\\prime\\prime}d x + \\int_{0}^{\\frac{1}{2}(1-2\\varepsilon)}\\frac{E a h^{3}}{24}{y_{1}^{\\prime\\prime}}^{2}d x = \\frac{m^{2}g^{2}l^{3}(1-4\\varepsilon^{2})^{2}}{8E a h^{3}} \\approx \\frac{m^{2}g^{2}l^{3}(1-8\\varepsilon^{2})}{8E a h^{3}}\n$$\n\nThe kinetic energy of the ball is: \n\n$$\nE_{k} = {\\frac{1}{2}}m l^{2}{\\dot{\\varepsilon}}^{2}\n$$\n\nTotal energy of the system: \n\n$$\nE = E_{k} + E_{p1} + E_{p2} = {\\frac{1}{2}}m l^{2}{\\dot{\\varepsilon}}^{2} + {\\frac{m^{2}g^{2}l^{3}}{E a h^{3}}}\\varepsilon^{2} + E_{0} = {\\frac{1}{2}}M{\\dot{\\varepsilon}}^{2} + {\\frac{1}{2}}K\\varepsilon^{2} + E_{0}\n$$\n\nGiving the angular frequency of horizontal vibration:\n\n$$\n\\omega_{2} = {\\sqrt{\\frac{K}{M}}} = {\\sqrt{\\frac{2m g^{2}l}{E a h^{3}}}}\n$$", "answer": "$$\n\\sqrt{\\frac{2m g^2 l}{E a h^3}}\n$$" }, { "id": 89, "tag": "ELECTRICITY", "content": "A typhoon is a vortex-like weather system, and as it passes, it is often accompanied by lightning. When the electric field strength in the air reaches a certain value, the air breaks down, generating lightning. The breakdown field strength of air is denoted as $E_{m}$. The typhoon system is modeled as an isothermal atmospheric ideal gas with a temperature denoted as ${T}_{0}$. Under strong ionization conditions, it can be considered as consisting of neutral monoatomic gas (with mass $m_{tot}=m_{N}+m_{e}$) and free electrons (with mass $m_{e}$) released through ionization. It is known that each atom can ionize to release at most one electron, with the total number of monoatomic gas particles being $N$.\n\nThe rotation speed of the typhoon is ${\\overrightarrow{\\omega}}=\\omega{\\hat{z}}$. In this problem, we consider a cylindrical typhoon system with radius ${R}$ and height $H(H \\gg R)$. Both the electrons and the positive ions are confined within this cylindrical volume and do not escape. Interactions between free electrons and free electrons, free electrons and positive ions, and positive ions and positive ions are ignored. Additionally, the effects of internal potential distributions within the typhoon on the distribution of free electrons and positive ions are not considered (calculated results can be simplified using $k=\\frac{\\omega^2}{2k_BT_0}$). The permittivity of free space is $\\varepsilon_0$. \n\nAfter the typhoon has stabilized, researchers need to deploy unmanned aerial vehicles (UAVs) to gather data within the typhoon system. The UAV has a volume of $V(V^{\\frac{1}{3}} \\ll R)$ and mass ${m}$. To ensure stable operation during data collection, the UAV must remain at a radius ${r_{0}}$ while orbiting the typhoon center with the same angular velocity $\\omega$. It is known that the UAV can only exert thrust in the vertical direction. At time $t=0$, the UAV in stable operation experiences a small radial perturbation and acquires an initial radial velocity $\\overrightarrow{v_{0}}$ with $v_{0} \\ll \\omega r_{0}$. Observations show that the UAV exhibits periodic precession. Solve for its precession angular frequency $\\Omega$ (the answer can be expressed in terms of $q_{0}$ and $r_{0}$).", "solution": "According to the problem statement, a single ideal gas under the isothermal model can satisfy the Boltzmann distribution in any continuous potential energy field distribution. By entering the rotating reference frame of the typhoon, centrifugal potential energy is introduced:\n\n$$\nV_{\\text{eff}} = -\\frac{1}{2}m\\omega^2r^2\n$$\n\nThus, the mass density of free electrons and positive ions satisfies:\n\n$$\n\\rho_{-}(r) = \\rho_{-0}e^{m_{\\text{e}}kr^2}, \\rho_{+}(r) = \\rho_{+0}e^{m_{N}kr^2}\n$$\n\nConsider that the total number of electrons and positive ions satisfies the normalization condition:\n\nSolving yields:\n\n$$\nN = \\frac{\\int_{0}^{R}2\\pi r H\\times\\rho_{-}(r)dr}{m_e}, N = \\frac{\\int_{0}^{R}2\\pi r H\\times\\rho_{+}(r)dr}{m_N}\n$$\n\n$$\n\\rho_{-}(r) = \\rho_{-0}e^{-m_{e}kr^2} = \\frac{k m_{e}^2N}{\\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr^2}\n$$\n\n$$\n\\rho_{+}(r) = \\rho_{+0}e^{-m_{N}kr^2} = \\frac{k m_{N}^2N}{\\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr^2}\n$$\n\nFrom\n\n$$\nPV = nRT\n$$\n\nIt follows that\n\n$$\nP(r) = \\left(\\frac{\\rho_{+}}{m_{N}} + \\frac{\\rho_{-}}{m_{e}}\\right)k_{B}T_{0}\n$$\n\n$$\nP(r) = \\left(\\frac{k m_{e}N}{\\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr^2} + \\frac{k m_{N}N}{\\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr^2}\\right)k_{B}T_{0}\n$$\n\nWe prioritize considering the electric field intensity at the maximum electric field, analyzing symmetry $\\vec{E} = E(r)\\hat{r}$\n\n$$\nE(r) = \\frac{\\int_{0}^{r}(e\\rho_{+} - e\\rho_{-})(2\\pi r \\, dr)}{2\\pi r\\varepsilon_{0}H} = \\frac{N e}{2\\pi r H\\varepsilon_{0}}\\left[\\frac{e^{m_{N}kr^2}-1}{e^{m_{N}k R^2}-1} - \\frac{e^{m_{e}kr^2}-1}{e^{m_{e}k R^2}-1}\\right]\n$$\n\nConsidering when $\\pmb{E}$ reaches maximum value, it satisfies:\n\n$$\n\\frac{dE(r)}{dr} = 0\n$$\n\nSimplifying, the $r_{\\text{max}}$ at maximum electric field satisfies the equation when $r = r_{\\text{max}}$:\n\n$$\n\\left[\\frac{e^{m_{N}kr^2}-1}{e^{m_{N}k R^2}-1} - \\frac{e^{m_{e}kr^2}-1}{e^{m_{e}k R^2}-1}\\right] - 2kr^2\\left[m_{N}\\frac{e^{m_{N}kr^2}}{e^{m_{N}k R^2}-1} - m_{e}\\frac{e^{m_{e}kr^2}}{e^{m_{e}k R^2}-1}\\right] = 0\n$$\n\nSubstituting the data, solving yields:\n\n$$\nr_{\\text{max}} = 719.96 \\, \\mathrm{m}\n$$\n\nAt this time:\n\n$$\nE_{\\text{max}} = \\frac{N e}{2\\pi r_{\\text{max}}H\\varepsilon_{0}}\\left[\\frac{e^{m_{N}kr_{\\text{max}}^2}-1}{e^{m_{N}k R^2}-1} - \\frac{e^{m_{e}kr_{\\text{max}}^2}-1}{e^{m_{e}k R^2}-1}\\right] = -9.83 \\times 10^{18} \\, \\mathrm{V/m}\n$$\n\nSince:\n\n$$\n\\left|E_{\\text{max}}\\right| > E_m\n$$\n\nIt can break down air.\n\nFor analyzing the forces on the drone in the ground reference frame, the buoyancy it experiences satisfies:\n\n$$\n\\vec{F}' = (\\rho_{-}(r_{0}) + \\rho_{+}(r_{0}))V g\\hat{z} + (\\rho_{-}(r_{0}) + \\rho_{+}(r_{0}))V\\omega^2r_{0}(-\\hat{r})\n$$\n\nIn the $\\hat{r}$ direction, list the force balance equation:\n\n$$\nE(r_{0})q_{0} - (\\rho_{-}(r_{0}) + \\rho_{+}(r_{0}))V\\omega^2r_{0} = -m\\omega^2r_{0}\n$$\n\nSubstituting specific values, solving gives:\n\n$$\nq_{0} = \\frac{-m\\omega^2r_{0} + \\left(\\frac{k m_{\\circ}^2N}{\\pi H(\\circ^{m_{\\circ}kr_{0}^2}-1)}e^{m_{\\circ}kr_{0}^2} + \\frac{k m_{N}^2N}{\\pi H(\\circ^{m_{N}kr_{0}^2}-1)}e^{m_{N}kr_{0}^2}\\right)V\\omega^2r_{\\circ}}{\\frac{N e}{2\\pi r_{0}H\\varepsilon_{0}}\\left[\\frac{e^{m_{N}kr_{0}^2}-1}{e^{m_{N}kr_{0}^2}-1} - \\frac{e^{m_{e}kr_{0}^2}-1}{e^{m_{\\circ}kr_{0}^2}-1}\\right]}\n$$\n\nFurther consider the force balance equation in the $\\hat{z}$ direction:\n\n$$\nF_{0} + (\\rho_{-}(r_{0}) + \\rho_{+}(r_{0}))V g - m g = 0\n$$\n\nSubstitute specific values, solving yields:\n\n$$\nF_{0} = -(\\rho_{-}(r_{0}) + \\rho_{+}(r_{0}))V g + m g\n$$\n\n$$\nF_{0} = m g - \\left(\\frac{k m_{e}^2N}{\\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr_{0}^2} + \\frac{k m_{N}^2N}{\\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr_{0}^2}\\right)V g\n$$\n\nWith conservation of angular momentum:\n\n$$\nL_{0} = m r_{0}^2\\omega\n$$\n\nAssume a relative stable position $\\boldsymbol{r}_{0}$ with a perturbation distance $\\delta r$, and write out the dynamic equation:\n\n$$\nm\\ddot{\\delta r} = \\frac{L_{0}^2}{m(r_{0}+\\delta r)^3} + E(r_{0}+\\delta r)q_{0} - [\\rho_{-}(r_{0}+\\delta r) + \\rho_{+}(r_{0}+\\delta r)]V\\omega^2(r_{0}+\\delta r)\n$$\n\nAnalyze and calculate the results in each term:\n\nEffective centrifugal force term:\n\n$$\n\\frac{L_{0}^2}{m(r+\\delta r)^3} = \\frac{L_{0}^2}{m r_{0}^3} + \\frac{\\partial\\frac{L_{0}^2}{m r^3}}{\\partial r}\\delta r = \\frac{L_{0}^2}{m r_{0}^3} - 3m\\omega^2\\delta r\n$$\n\nRadial electric field term:\n\n$$\nE(r_{0}+\\delta r)q_{0} = E(r_{0})q + \\frac{N e q_{0}}{2\\pi H\\varepsilon_{0}r_{0}^2}\\left[\\frac{(2k m_{N}r_{0}^2 - 1)e^{m_{N}kr_{0}^2} + 1}{e^{m_{N}k R^2} - 1} - \\frac{(2k m_{e}r_{0}^2 - 1)e^{m_{e}kr_{0}^2} + 1}{e^{m_{e}k R^2} - 1}\\right]\\delta r\n$$\n\nRadial component of buoyancy term:\n\n$$\n\\begin{array}{rl}\n-[\\rho_{-}(r_{0}+\\delta r) + \\rho_{+}(r_{0}+\\delta r)]V\\omega^2(r_{0}+\\delta r) =& -[\\rho_{-}(r_{0}) + \\rho_{+}(r_{0})]V\\omega^2r_{0} - \\ldots\n\\end{array}\n$$\n\n$$\n\\ldots \\frac{k N V\\omega^2}{\\pi H}\\left[\\frac{(2k m_{N}r_{0}^2 + 1)e^{m_{N}kr_{0}^2}}{e^{m_{N}k R^2} - 1}m_{N}^2 + \\frac{(2k m_{e}r_{0}^2 + 1)e^{m_{e}kr_{0}^2}}{e^{m_{e}k R^2} - 1}m_{e}^2\\right]\\delta r\n$$\n\nSince\n\n$$\n\\frac{L_{0}^2}{m r_{0}^3} + E(r_{0})q_{0} - [\\rho_{-}(r_{0}) + \\rho_{+}(r_{0})]V\\omega^2r_{0} = 0\n$$\n\nZero-order term sum is zero, thus only considering the first-order terms, we obtain the precession angular frequency:\n\n$$\n\\boxed{\\Omega^2 = 3\\omega^2 - \\frac{N e q_{0}}{2\\pi m H\\varepsilon_{0}r_{0}^2}\\left[\\frac{(2k m_{N}r_{0}^2 - 1)e^{m_{N}kr_{0}^2} + 1}{e^{m_{N}k R^2} - 1} - \\frac{(2k m_{\\text{e}}r_{0}^2 - 1)e^{m_{\\text{e}}kr_{0}^2} + 1}{e^{m_{\\text{e}}k R^2} - 1}\\right] + \\frac{k N V\\omega^2}{\\pi m H}\\left[\\frac{(2k m_{N}r_{0}^2 + 1)e^{m_{N}kr_{0}^2}}{e^{m_{N}k R^2} - 1}m_{N}^2 + \\frac{(2k m_{e}r_{0}^2 + 1)e^{m_{e}kr_{0}^2}}{e^{m_{e}k R^2} - 1}m_{e}^2\\right]}\n$$", "answer": "$$3\\omega^2 - \\frac{Ne q_0}{2\\pi m H \\varepsilon_0 r_0^2} \\left( \\frac{(2k m_N r_0^2 - 1)e^{m_N k r_0^2} + 1}{e^{m_N k R^2} - 1} - \\frac{(2k m_e r_0^2 - 1)e^{m_e k r_0^2} + 1}{e^{m_e k R^2} - 1} \\right) + \\frac{k N V \\omega^2}{\\pi m H} \\left( \\frac{(2k m_N r_0^2 + 1)e^{m_N k r_0^2}}{e^{m_N k R^2} - 1} m_N^2 + \\frac{(2k m_c r_0^2 + 1)e^{m_c k r_0^2}}{e^{m_c k R^2} - 1} m_c^2 \\right)$$" }, { "id": 194, "tag": "ELECTRICITY", "content": "In the upper half-plane above the $x$-axis, there is a magnetic field with intensity $B$ directed inward, perpendicular to the page. A charged particle with rest mass $m_{0}$ and charge $-q (q>0)$ is shot at high speed into the positive $y$ direction from the origin. It experiences a drag force equal to $q c B$ in magnitude, opposite to the direction of velocity. It is known that the charged particle comes to a stop after turning through an angle $\\theta_{0} (\\theta_{0} > \\frac{\\pi}{2})$ in the direction of velocity, and at point $A$ on its trajectory, the velocity is parallel to the $x$-axis ($\\theta_{A} = \\frac{\\pi}{2}$). Now, a smooth track is laid along the path of the charged particle (taking $\\theta_{0} = \\pi$). Another particle with the same rest mass $m_{0}$ and charge $-q$ is launched from the origin along the inner side of the track with the same initial momentum, experiencing a drag force proportional to velocity, ${\\vec{f}}_{0} = -q B {\\vec{v}}$. What is the normal force $N_{A}$ exerted on the track by the charged ion when it reaches point $A$? Consider relativistic effects, with the speed of light in vacuum as $c$.", "solution": "", "answer": "" }, { "id": 547, "tag": "ELECTRICITY", "content": "For the electromagnetic cannon model, its structure consists of two parallel rails spaced $\\mathit{l}$ apart, with one end connected to a power supply for energy, and the other end connected to a metal rod that can slide freely on the rails to form a circuit. In the situation where the circuit length $x$ is much larger than the spacing $\\mathit{l}$ (but ignoring the delay in circuit signal propagation caused by the length), it can be assumed that the self-inductance coefficient $L$ of the circuit is linearly related to $x$, i.e., $L = A x + B$. $A$ and $B$ are two constants. The current flowing through the metal rod is $I$, and the permeability of vacuum is $\\mu_{0}$. In fact, for different electromagnetic cannon configurations, the value of the Ampere force on the metal rod is actually different. Assume the rail is a thin-walled cylinder with a radius $r \\ll \\mathit{l}$. Under direct current conditions, it can be assumed that the current is uniformly distributed over the surface of the cylinder. Make an appropriate approximation and calculate the specific expression of the Ampere force on the metal rod.", "solution": "", "answer": "" }, { "id": 647, "tag": "MECHANICS", "content": "This problem discusses the motion and forces on a particle in a rapidly oscillating field. First, we discuss the one-dimensional theory in a rapidly oscillating field. Let $q$ denote the coordinate of the particle. For a particle in a potential field $V_{0}(q)$ with mass $m$, if there is a rapidly oscillating force $f( \\boldsymbol{q},t)$ acting on it (the angular frequency $ \\omega$ of $f$ is very high), its equation of motion is: $$ m{ \\ddot{q}}=F_{0}=-{ \\frac{ \\mathrm{d}V_{0}}{ \\mathrm{d}q}}(q)+f(q,t) $$ The core of the theory is to divide the actual motion into fast motion (high-frequency oscillation) and slow motion (average motion): fast motion is denoted as $x(t)$, and slow motion is denoted as $X$. Due to the high frequency, the response of $x$ is very small, that is $x \\ll X$. Expanding the above equation: $$ m \\ddot{X}+m \\ddot{x}=- \\frac{ \\mathrm{d}V_{0}}{ \\mathrm{d}q} \\Big \\vert_{X}- \\frac{ \\mathrm{d}^{2}V_{0}}{ \\mathrm{d}q^{2}} \\Big \\vert_{X} \\cdot x+f(X,t)+ \\frac{ \\partial f}{ \\partial x} \\Big \\vert_{X,t} \\cdot x $$ First, solve for fast motion: $ \\begin{array}{r}{m \\ddot{x}=- \\frac{ \\mathrm{d}^{2}V_{0}}{ \\mathrm{d}q^{2}} \\Big \\vert_{X} \\cdot x+f(X,t)} \\end{array}$, which is evidently a forced oscillation. Considering the high-frequency approximation, that is $ \\begin{array}{r} { m \\omega^{2} \\gg \\frac{ \\mathrm{d}^{2}V_{0}}{ \\mathrm{d}q^{2}} \\bigg|_{X}} \\end{array}$, we obtain $$ x=-{ \\frac{1}{m \\omega^{2}}}f(X,t) $$ Thus, in fact, when studying rapidly oscillating forces, the second term on the right side of the equation does not have any practical effect. Next, solve for the slow motion. Substitute the result of the fast motion into the actual dynamical equation to obtain the slow motion. Extending this situation to three dimensions, if $ abla \\times f=0$, the effective potential $F_{ \\mathrm{eff}}=- abla V_{ \\mathrm{eff}}$ can be written down. Please provide the expression of $V_{ \\mathrm{eff}}$ in terms of some average value of $f$.", "solution": "", "answer": "" }, { "id": 549, "tag": "MECHANICS", "content": "Please solve the following physics problem, and enclose the final answer in \\boxed{}: Consider a regular dodecahedron with side length $a$ in an environment without a gravitational field. At the center of the dodecahedron, there is a mass $m$. Each vertex of the dodecahedron is connected to the mass $m$ by a spring. Each spring has a natural length $l$ ($l$ is equal to the radius of the circumscribed sphere of the dodecahedron), and all springs have the same spring constant $k$. The mass of the springs is negligible. Find the eigenfrequency of small oscillations of the mass $m$.", "solution": "", "answer": "" }, { "id": 267, "tag": "THERMODYNAMICS", "content": "We divide medium molecules into two categories: one is non-polar molecules, such as carbon tetrachloride; the other is polar molecules. For polar molecules, the centers of positive and negative charges are separated, forming an intrinsic dipole moment $\\vec{p}_{i}$, such as in a water molecule. Under the influence of an external field, the positive and negative charge centers of non-polar molecules shift, forming a dipole. This mechanism is called induction polarization. For polar molecules, the intrinsic dipole moment rotates to a stable equilibrium position under the influence of the external field. This process is called orientation polarization. In this problem, we discuss the orientation polarization of the medium.\n\nAssuming the dipole moment of a water molecule is $p_{w}$ and considering a system with a number density of water vapor molecules $n$, the interaction energy between the water dipole moment $p_{w}$ and the applied electric field is assumed to follow the Boltzmann distribution. That is, the isotropic distribution probability is multiplied by the factor $e^{-\\frac{E_{p}}{k T}}$, where $E_{p}$ is the interaction energy between the dipole and the external field, $k$ is the Boltzmann constant, and the system temperature is assumed to be $T$. Derive the polarization vector $P$ of the system under the condition of a weak external field $E$.", "solution": "", "answer": "" }, { "id": 350, "tag": "ADVANCED", "content": "A homogeneous semicircular ring with a radius $R$, bending stiffness $EI$, and mass $m$ is placed with its opening facing upwards on a rigid horizontal surface in a state of rest. Assume all deformations are small, and the bending stiffness $EI$ is defined as: $M=EI(k-k_0)$, where $k$ is the curvature at a point, $k_0$ is the original curvature, and $M$ is the applied external moment. What is the height of its center of mass above the ground? The answer should be retained to the first-order term of $\\frac{R^2 m g}{EI}$.", "solution": "", "answer": "" }, { "id": 756, "tag": "ELECTRICITY", "content": "Unlike the motion of charged particles in an electromagnetic field, the motion of neutral atoms in an electromagnetic field has unique characteristics. By modeling the neutral atom as a homogeneous sphere with a relative dielectric constant of $ \\varepsilon_{r}$ and a radius of $R$, we consider the motion laws of the atom in a specific situation. In a spatial rectangular coordinate system, there is a uniform magnetic field $B$ in the $+z$ direction, where if $B < 0$, it means the magnetic field direction is actually along $-z$. A uniformly charged insulating thin wire with a linear charge density of $+ \\lambda$ is fixed on the $z$ axis. Now a neutral atom is placed at $x = r_{0} \\gg R, y = z = 0$ and released with an initial velocity $v_{0}$ in the $+y$ direction. Ignoring the retardation effect and relativistic effect, and assuming the atom has a uniform mass distribution with a density of $ \\rho$. If the atom moves in a circular motion, try to find the value of $v_{0}$. Given that the vacuum permittivity is $ \\varepsilon_{0}$.", "solution": "", "answer": "" }, { "id": 196, "tag": "MECHANICS", "content": "In the graduate entrance exam of the University of Wisconsin's Department of Physics in 1975, there appeared a seemingly simple rigid ball collision problem: Two rigid spheres of the same material—where the radius of the lower sphere is $2a$, and the radius of the upper sphere is $a$—fall from a height of $h$ (measured from the center of the larger sphere) above the ground. Assume that the centers of both spheres always remain on the same vertical line and that all collisions are elastic. Under these conditions, what is the maximum height the center of the upper sphere can reach? (Hint: Assume the larger sphere first collides with the ground and rebounds before colliding with the smaller sphere.)\n\nIt is now known that the rigid sphere collision model described in this problem satisfies chaotic conditions. Moreover, for this chaotic problem, as long as basic physical laws (such as energy conservation and rigidity assumptions) are satisfied, all states can occur—it is merely a matter of how much time it takes. Determine the highest position $h_{max}$ the smaller upper sphere could possibly reach after a very long period of time following its release.", "solution": "", "answer": "" }, { "id": 340, "tag": "ELECTRICITY", "content": "Consider two infinitely long parallel cylindrical metal conductors located in an isotropic linear dielectric medium characterized by a dielectric constant \\( \\epsilon \\), magnetic permeability \\( \\mu \\), and electrical conductivity \\( \\sigma \\). The radius of each conductor is \\( r \\), and their separation is \\( d \\), where \\( d \\gg r \\). The capacitance per unit length between the two conductors is denoted as \\( C_{l} \\), the inductance per unit length as \\( L_{l} \\), and the electrical conductance per unit length as \\( G_{l} \\) (none of which are known quantities).\n\nNow, consider constructing an infinitely long one-dimensional circuit network: the lower conductor serves as the ground line, while the upper conductor contains an infinite number of inductors \\( L \\). The right end of the \\( n \\)-th inductor and the left end of the \\( (n+1) \\)-th inductor define the \\( n \\)-th node.Each node is connected to the ground line through a capacitor \\( C \\) and conductance \\( G \\). Resistance and conductance are connected in parallel.Let the electric potential at the \\( n \\)-th node be \\( V_{n} \\).\n\nUsing Kirchhoff's equations, a recurrence relation for \\( V_{n} \\) and its time derivative can be derived. Let \\( C = C_{l}dx \\), \\( L = L_{l}dx \\), and \\( G = G_{l}dx \\); by continuous approximation of the recurrence relation, the governing partial differential equation for \\( V(x,t) \\) can be obtained.\n\nWe hypothesize a solution for the decaying electric potential wave:\n\n$$\nV(x,t) = V_{0}\\cos(kx - \\omega t + \\varphi)e^{-px}\n$$\n\nSubstituting this solution into the partial differential equation, \\( k(\\omega) \\) and \\( p(\\omega) \\) can be determined.\n\nThe group velocity is defined as:\n\n$$\nv_{g} \\equiv \\frac{d\\omega}{dk}\n$$\n\nSolve for the group velocity \\( v_{g}(\\omega) \\) of the decaying electric potential wave.", "solution": "", "answer": "" }, { "id": 717, "tag": "OPTICS", "content": "Let the refractive index of the core material of an optical fiber Bragg grating be $n_1 = 1.51$. In the fiber, a one-dimensional optical structure is constructed by periodically changing the refractive index of the core material, with the changed material's refractive index being $n_2 = 1.55$. This structure is made up of alternating layers of two refractive indices: the layers have refractive indices $n_2$ and $n_1$, and thicknesses $d_2$ and $d_1$ respectively, forming a periodic refractive index distribution with a total of N layers. Light is perpendicularly incident on this structure and, after passing through the alternating layers, undergoes reflection and transmission at each layer interface. In each period, the odd layers are medium layers with refractive index $n_1$ and thickness $d_1$, the even layers are medium layers with refractive index $n_2$ and thickness $d_2$, with the arrangement: $$n_1,\\, n_2,\\, n_1,\\, n_2,\\, \\dots,\\, n_1,\\, n_2$$ To simplify the analysis, assume that absorption loss in the medium is neglected during the design process, and only consider single reflections of light at each layer interface. When light is perpendicularly incident from a medium with refractive index $n_1$ to a medium with refractive index $n_2$, it gets partially reflected and partially transmitted at the interface. Specifically, if the incident light's electric field strength is $E_0$, then: The electric field strength of the reflected light is: $E_r = \\frac{n_1 - n_2}{n_1 + n_2} E_0$ The electric field strength of the transmitted light is: $E_t = \\frac{2n_1}{n_1 + n_2} E_0$ Assuming that the wavelength of the incident light in vacuum is $\\lambda = 1.06 \\mu m$, if the device's amplitude reflectivity is required to reach $8\\%$, what is the minimum total number of layers N? Hint: Use the coherent superposition of light without needing to use the Bragg reflector reflectivity approximation.", "solution": "", "answer": "" }, { "id": 726, "tag": "MECHANICS", "content": "A cannon located on the ground can fire shells in any direction at a fixed rate $u$. The horizontal firing direction is the x-axis, and the vertical direction to the ground is the y-axis. The boundary of the safe zone is defined as: if the target is within this boundary, the shell may hit the target; if it is outside this boundary, it cannot be hit regardless of how it is fired. Find the equation of the boundary of the safe zone in the air, written in the form $y = y(x)$. The gravitational acceleration $g$ is known.\n", "solution": "", "answer": "" }, { "id": 149, "tag": "THERMODYNAMICS", "content": "A vertical adiabatic container with a height of $2H$ and a volume of $2V$ is sealed at the bottom by an adiabatic piston. Initially, the container is divided into two equal-volume sections by an adiabatic partition of mass $m$. The partition rests on a support, and the contact point is sealed to prevent any leakage. Each section initially contains helium gas at pressure $p$ and temperature $T$. Using a force, the piston is slowly pushed upward. Consider the following question:\n\nFind the temperature $T_{1}$ of the gas in the upper gas chamber when the piston reaches the support. \nHint: During the process, the partition will leak for a certain period.", "solution": "", "answer": "" }, { "id": 492, "tag": "ELECTRICITY", "content": "Solve the following problem using the principle of virtual work: \nTwo identical superconducting rings have a radius of \\( a \\), each carrying a current \\( I \\) and with self-inductance \\( L \\). These rings are coaxially and parallelly positioned with their centers separated by a distance \\( D \\). The magnetic field generated by the current in each ring within the other ring can be considered as a magnetic dipole field. Assume a virtual displacement \\( \\delta \\) occurs in the distance between the rings, i.e., \\( D \\to D + \\delta \\). Based on the conservation of magnetic flux in superconducting coils, please derive the interaction force between the two rings from the perspective of energy. The vacuum permeability is given as \\( \\mu_0 \\).", "solution": "", "answer": "" }, { "id": 235, "tag": "THERMODYNAMICS", "content": "The molar mass of a certain ideal gas is $\\mu$, and it flows from left to right through a long straight adiabatic horizontal duct with smooth inner walls. The cross-sectional area of the duct is $S$. The internal energy of 1 mole of this gas at an absolute temperature of $T$ is ${\\frac{5}{2}}R T$, where $R$ is the universal gas constant.\n\nA heating device is fixed and placed in the middle of a duct to heat the gas with a constant power. Assume that the resistance of the heating device to the gas flow can be neglected. After the gas flow stabilizes, although the gas in the vicinity of the heating device has nonuniform states, it gradually becomes uniform as the distance from the heating device increases. In the uniform steady-flow region to the left of the heating device, the pressure of the gas is $p_{0}$, the temperature is $T_{0}$, and the flow velocity to the right is $V_{\\mathrm{0}}$. Given that the pressure of the gas in the uniform steady-flow region to the right of the heating device is $p_{1}$, determine the temperature $T_{1}$ of the gas in that region.", "solution": "", "answer": "" }, { "id": 359, "tag": "MECHANICS", "content": "The rotational inertia of a planar object relative to its center of mass is given by $I=\\rho^2 M$. On a smooth horizontal surface, there is a thin plate of arbitrary shape with mass $M$, and point $O$ is the center of mass of the plate, with a radius of gyration $\\rho_0$. There is a beetle $A$ on the plate with mass $m$. A polar coordinate system is established with the center of mass of the plate as the origin, and this coordinate system is fixed relative to the plate. In this reference frame, the motion laws of the beetle $A$ relative to the plate are known, given by $\\rho(t)$ and $\\varphi(t)$. Using the conservation of linear momentum and angular momentum, determine the absolute angular velocity $\\Omega$ of the plate at any given time.\n", "solution": "", "answer": "" }, { "id": 532, "tag": "ELECTRICITY", "content": "Equilibrium Problem of Charged Plates in a Capacitor \n\nThe bottom area of a cylindrical insulating adiabatic container is \\( A \\), and its height is \\( 2L \\). The container stores gas with an initial pressure of \\( p_{0} \\), temperature \\( T_{0} \\), and constant-volume molar heat capacity \\( C_{V} \\). There are 3 identical metal plates, each with the same area as the bottom area of the cylindrical container. Two of them are grounded and fixed at the two bottom surfaces inside the container. The third metal plate is parallel to the bottom surface and positioned in the middle of the container, dividing the container and the gas inside it equally left and right. Now, an electric charge of \\( Q \\) is rapidly introduced to the middle metal plate. It is assumed that the middle plate is neither leaking electricity nor gas, and can freely move along the axis of the container. The relative permittivity of the gas is approximately 1. \\( L \\) is sufficiently small, and the conductor plates conduct heat well enough that the gas on both sides always reaches thermal equilibrium. The ideal gas constant is denoted as \\( R \\).The dielectric constant in a vacuum is \\( \\varepsilon_{0} \\). Let the mass of the middle plate be \\( m \\). If \\( Q > 2\\sqrt{\\varepsilon_{0}p_{0}}A = Q_{c} \\), find the period of small oscillations when the plate is given a small disturbance at the stable equilibrium position.", "solution": "", "answer": "" }, { "id": 414, "tag": "MECHANICS", "content": "When the cup is tilted at a certain angle, if a lateral impulse is applied, it will start to spin in place. Now, let's establish a model to study the situation where the cup spins stably in place. Let the gravitational acceleration be $g$, and model the cup as a cylinder with mass $m$, radius $r$, and height $h$, with its generatrix deviating by an angle $\\theta$ from the vertical line. Place the cylinder on a horizontal surface, with the contact point labeled as A. Assume point A moves in a circular motion on the horizontal surface with an angular velocity $\\Omega$, and the cylinder undergoes pure rolling. Try to study the radius $R$ of the circular motion of point A. For ease of calculation, take $h=2r$, and stipulate that $h$ does not appear in the final result.\n", "solution": "", "answer": "" }, { "id": 207, "tag": "ELECTRICITY", "content": "Two eccentric spherical conductors with radii $R_{1}$ and $R_{2}$ form a capacitor. The centers of the spheres are separated by a distance $d \\ll R_{1}, R_{2}$. The dielectric constant in vacuum is $\\varepsilon_{0}$, and $R_{2} > R_{1}$. Find the reciprocal of the capacitance of the system, retaining only the lowest-order non-zero term that includes $d$, and list relevant boundary conditions for an approximate solution.", "solution": "", "answer": "" }, { "id": 257, "tag": "THERMODYNAMICS", "content": "There is a vertically placed, sealed, adiabatic cylindrical container with a cross-sectional area of \\( A \\). Inside, there is a thermally insulated piston of mass \\( m \\). The thickness of the piston is negligible compared to the length of the cylinder. Initially, the piston is fixed at the center of the cylinder, dividing it into two chambers of equal length \\( l \\). Each chamber contains \\( n \\) moles of monatomic ideal gas at an initial temperature of \\( T_{0} \\). \n\nIn the following problem, you may assume that the friction between the piston and the cylinder wall is negligible, and \\( l \\) is much larger than the displacement of the piston (\\( l \\gg z \\)).\n\nAlthough the friction between the piston and the cylinder wall is small, after a long time it will eventually bring the piston to rest. Furthermore, due to the motion of the piston, the temperatures of the two chambers eventually become equal, and all the heat dissipated due to friction is converted into the internal energy of the gas. Determine the final temperature \\( T_{f} \\) of the gas inside the cylinder when the piston comes to rest. It is assumed that the heat capacities of the cylinder walls and the piston are negligible.", "solution": "", "answer": "" }, { "id": 27, "tag": "MECHANICS", "content": "On an incline, a snow rod made of snow rolls down the slope purely, absorbing the thin layer of snow on the slope into itself, causing the snow rod to become increasingly larger. We assume the snow rod always maintains a cylindrical shape, and its initial volume is very small and can be neglected. The snow is evenly distributed on the slope, with a surface density of $\\sigma$, and will be absorbed as long as the snow rod passes over it. The density of the snow is $\\rho$. It is known that\n$$\n\\sin \\theta = \\frac{3}{7}\n$$\nwhere $\\theta$ is the angle of the incline. Find the relationship between the velocity of the center of the snow rod and time.", "solution": "", "answer": "" }, { "id": 330, "tag": "ELECTRICITY", "content": "In electromagnetism, we often draw analogies between magnetic dipoles and electric dipoles. However, the force patterns in a non-uniform field differ between the two. Consider the force experienced by a charged rigid body moving in a non-uniform magnetic field. In space, there exists a homogeneous solid sphere with a radius of \\( R \\), mass \\( m \\), and uniformly distributed charge \\( q \\). Gravitational effects are neglected. A Cartesian coordinate system is established, and a magnetic field is present in space given by \\( \\vec{B} = (B_0 + \\alpha x) \\hat{z} \\). Using the equation of motion for the center of mass of the sphere, consider the initial state where the sphere's center of mass is located at the origin, its velocity is \\( (v_{x0}, 0, 0) \\), and it has no angular velocity. It is known that the position of the first turning point in the \\( x \\)-direction is \\( x = l \\). Provide the expression for \\( v_{x0} \\).", "solution": "", "answer": "" }, { "id": 643, "tag": "OPTICS", "content": "In this problem, we study a simple \\\"gas-fueled rocket.\\\" The main body of the rocket is a plastic bottle, which can take off after adding a certain amount of fuel gas and igniting it. It is known that the external atmospheric pressure is $P_{0}$, and the initial pressure of the gas in the bottle is $P_{0}$, with a temperature of $T_{0}$. The proportion of the molar amount of fuel gas is $\\alpha$, and the rest is air. Alcohol is chosen as the fuel, with the reaction equation in air as: $C_{2}H_{5}OH(g) + 3O_{2}(g) \\rightarrow 3H_{2}O(g) + 2CO_{2}(g)$. Assume that the air contains only $N_{2}$ and $O_{2}$, all gases are considered as ideal gases, and none of the vibrational degrees of freedom are excited. It is known that under conditions of pressure $P_{0}$ and temperature $T_{0}$, the reaction heat is $-\\lambda$ (enthalpy change for $1\\mathrm{mol}$ of ethanol reacting under isothermal and isobaric conditions). Ignore heat loss, assume the reaction is complete, and that there is air remaining, with the gas volume remaining constant throughout. Find the pressure $P_{1}$ of the gas inside the bottle after the gas has completely reacted upon ignition.", "solution": "", "answer": "" }, { "id": 421, "tag": "OPTICS", "content": "Consider a Young's double-slit interference device and establish a rectangular coordinate system, with the direction of the slit connection as the $x$-axis, the direction of light propagation as the $z$-axis, and the $y$-axis determined by the right-hand rule. The object screen is located at $z=0$, the double slits are located at $(\\pm a,0,0)$, and the light screen is located at $z=D$. The region $0\\le z\\le D$ is filled with a medium with a refractive index distribution of $n\\left(z\\right)=1+\\sigma z$. The wavelength of the incident light is $\\lambda$. The angles at which a light ray is emitted are represented by $\\alpha$, $\\beta$, and $\\gamma$, which are the angles with the $x$, $y$, and $z$ coordinate axes, respectively. \n\nTaking the approximation $\\gamma\\ll1$ (hence $\\alpha,\\beta\\lesssim\\pi/2)$, but with $\\sigma D\\sim1$, $\\sigma a\\ll\\pi/2-\\alpha$, and ignoring square terms of small quantities, write the function expression $f(x,y,p)=0$ on the light screen for the $p$th-order interference bright fringe (defined as the optical path difference being $\\pm p\\lambda$). Provide $f(x, y, p)$. The term with $x$ must be positive with a coefficient of 1; if there is another term where the sign can be either positive or negative, please mark the sign of that term as plus-minus $(\\pm)$.", "solution": "", "answer": "" }, { "id": 427, "tag": "ELECTRICITY", "content": "Establish a 3D right-handed rectangular coordinate system $(x,y,z)$. There is a $4a\\times6a$ resistor network on the xy-plane, with the endpoint $\\mathsf{A}$ having coordinates (0,0,0) and endpoint B having coordinates (6a, 4a, 0). Each small square has a side length of $a$, and the resistivity is uniform. A current $I$ flows in from A, and a current $I$ flows out from B. Now, introduce a magnetic field $\\mathsf{B} = (B_{0},0,B_{0})$ to the entire space. Try to solve for the magnitude of the torque $M$ acting on point A in the resistor network.", "solution": "", "answer": "" }, { "id": 185, "tag": "ELECTRICITY", "content": "The upper and lower plates of the capacitor are semicircular metallic plates $a$,$b$ with a radius of $ \\pmb{R}$. The centers are separated by a distance $d \\ll R$, and the centers of metallic plates $a$ and $b$ are located on the same vertical axis. The lower surface is horizontal, but due to manufacturing processes, the upper metallic plate is not strictly parallel to the lower surface. The straight edge of the upper surface is parallel to the lower surface, and the angle between the two planes is $ \\alpha R \\ll d$. Because the angle is very small, we can approximately consider that in the top view, both metallic plates are still strictly semicircular. There is an axis $c$ at the center of the upper metallic plate, perpendicular to the upper surface, around which the upper surface can freely rotate. In the top view, the angular displacement of the diameters of the circular discs on the upper and lower surfaces of the capacitor is $ \\pmb{ \\theta}$. Determine the capacitance $c( \\theta)$ of this capacitor at this angular displacement (As an approximation, we assume that charges only exist on the directly opposing parts of the upper and lower surfaces, and no charges are present on the remaining parts).", "solution": "", "answer": "" }, { "id": 233, "tag": "MODERN", "content": "In high-energy physics, the detection of photons is a very important and meaningful task. To this end, consider a simplified model: A photon with energy $E_{0}$, which is much smaller than the rest energy $m_{e}c^{2}$ of an electron, will undergo $n$ random Compton scatterings within a material consisting of initially static free electrons, resulting in its energy decreasing to $E$. Try to determine the specific form of energy $E$ as it decreases with the number of scatterings $n$, given that the speed of light in vacuum is $c$.", "solution": "", "answer": "" }, { "id": 334, "tag": "MECHANICS", "content": "Consider the process of a stepladder collapsing and bouncing off the wall. The stepladder can be viewed as two uniform rods, each with length $L$, hinged together at the top $A$ with a smooth, lightweight hinge. The left rod has a mass of $m$, while the mass of the right rod can be neglected. The static and kinetic friction coefficients between the bottom end $B$ of the left rod and the ground are both $\\mu$. The bottom end $C$ of the right rod makes smooth contact with the ground. Initially, the two rods are almost overlapping, with an angle of zero. The ground is horizontal, and the distance from endpoint $C$ to a vertical wall corner $O$ is $\\sqrt{2}L$. The rods are released from rest, and the acceleration due to gravity is $g$.\n\nTo ensure that during the collapse process of the stepladder, the endpoint $C$ continuously slides to the right towards $O$ while endpoint $B$ remains stationary, find the minimum value of the friction coefficient $\\mu$.", "solution": "", "answer": "" }, { "id": 224, "tag": "MECHANICS", "content": "Consider the wheel-slider transmission mechanism. There is a ring with radius $R$ on the horizontal plane, and there is an infinitely long smooth rod passing through the center of the ring. Looking from above, the intersection point of the rod and the right side of the ring is $P$. On the ring and the rod, there are two masses, $A$ and $B$, respectively, with a transmission rod between them. At time $t=0$, mass $A$ starts to rotate counterclockwise with a constant angular speed $\\omega$ from the hinge $P$, while mass $B$ is on the right side of $A$. This movement drives the slider at the $B$ end of the transmission rod $AB$ to reciprocate along the track. The length of the transmission rod $AB$ is $l > R$. Calculate the acceleration of the slider at the $B$ end moving to the left at any time $t$.", "solution": "", "answer": "" }, { "id": 587, "tag": "MECHANICS", "content": "A homogeneous thin rod is placed on the track of a real-axis rotated hyperboloid. Given the rod’s length $L$ and mass $M$. The equation of the hyperbola in the plane is $$ {\\frac{y^{2}}{a^{2}}}-{\\frac{x^{2}}{b^{2}}}=1 $$ where $g$ is the known gravitational acceleration. If the hyperbolic surface is smooth and the rod is confined within the vertical xy-plane, find the slope $k$ of the rod at equilibrium ($k>0$) expressed in terms of $a$, $b$, and $L$ (under the condition that $aL>2b^2$).\n", "solution": "", "answer": "" }, { "id": 512, "tag": "MECHANICS", "content": "Collision is one of the simplest mechanical interactions between objects, representing a process of energy and momentum exchange. The chain reaction caused by one-dimensional collision cascades provides a simplified model for a series of natural phenomena dominated by continuous collisions. When studying such systems, the physical quantities we're interested in are the proportion of energy or momentum transmitted through the collision cascade and its dependence on the restitution coefficient and the mass ratio of the colliding objects. Between the initial ball of mass $M$ (fixed) and the terminal ball of mass $m$ (fixed), $n$ balls ($n \\gg 1$) are placed (with masses $m_{i}$, $i = 1, 2, ..., n$, tunable). Only the first collision between neighboring balls is considered. All collisions are inelastic, characterized by a restitution coefficient $e$ (very close to 1).\n\nRecently, Ricardo and Lee demonstrated that when the mass distribution of the cascading ball chain ensures that the mass of each ball $m_{i}$ is the geometric mean of the masses of its two neighboring balls, $\\sqrt{m_{i-1}m_{i+1}}$, the transmission efficiency of kinetic energy and momentum between the initial and terminal balls is maximized. For an $n$-ball system, please write down the maximum transmission efficiency of kinetic energy between the initial and terminal balls $r_{K n}$;", "solution": "", "answer": "" }, { "id": 776, "tag": "ELECTRICITY", "content": "For conductors, reaching electrostatic equilibrium from a non-equilibrium state takes a certain amount of time, but generally, this time is very short, so this process is often neglected in conventional electrostatic analyses. However, many effects generated by this process are still not negligible. Below, we will make a simple analysis, considering that the charge movement speed is much less than the speed of light. There is a dielectric sphere with a radius of $R$, dielectric constant $ \\epsilon$, and conductivity $ \\alpha$ in a vacuum. At time $t=0$, a uniform electric field $E_0$ is suddenly applied. We assume that the dielectric sphere first reaches electrostatic equilibrium in a non-conductive sense and then begins to leak charge. Determine how much time passes during the leakage process before the current density at all locations is reduced to half of its initial value.", "solution": "", "answer": "" }, { "id": 479, "tag": "ELECTRICITY", "content": "In an infinitely large isotropic linear dielectric medium, there exists a uniform external electric field $\\vec{E}_{0}$. The vacuum permittivity is known to be $\\varepsilon_{0}$.\n\nThe dielectric is a solid with a relative permittivity $\\varepsilon_{r}$. Within the dielectric, a spherical cavity with radius $R$ is carved out, and the cavity can be considered a vacuum. Then, an ideal conductor sphere with a radius slightly smaller than $R$ is placed within the cavity, such that the conductor sphere just does not come into contact with the cavity walls. The conductor sphere carries a net charge of $Q$.\n\nFind the magnitude of the electrostatic force $F$ exerted on the conductor sphere.", "solution": "", "answer": "" }, { "id": 703, "tag": "ELECTRICITY", "content": "A point charge with mass $m$ and negative charge $-q(q>0)$ moves without resistance in a spatial magnetic field $B=B_0a/r (B_0>0, a>0)$ along the +z direction. Here, $r$ is the distance of the particle from the z-axis, $r= \\sqrt{x^2+y^2}$. Assume the coordinate system is right-handed. At $t=0$, the charge is located on the x-axis with coordinates $x=a, y=z=0$, and its velocity is exactly along the +y direction with a magnitude of $v_0 (v_0>0)$. If the particle appears at $t>0$ with its velocity direction perpendicular to $ \\vec r$ (where $ \\vec r$ represents the radial vector from the origin to the particle), find the distance $r$（different from $r_0$） from the particle to the z-axis, assuming $qB_0a-mv_0>0$. Hint: Consider the differential equation of the particle's angular momentum $L$ and radial distance $r$.", "solution": "", "answer": "" }, { "id": 288, "tag": "THERMODYNAMICS", "content": "For electromagnetic radiation within a cavity, the method of studying gases can be used for investigation, and we refer to this as a photon gas. It is known that the internal energy of a photon gas with a volume $V$ and temperature $T$ can be expressed as $U = a V T^{4}$, where $a$ is a known constant. A thermal radiation field consists of radiation with various wavelengths. When a blackbody radiation cavity undergoes adiabatic expansion, not only does the radiation temperature change, but the radiation wavelength also changes correspondingly due to the Doppler effect. Therefore, there must be a certain relationship between the change in radiation wavelength and the change in temperature. Derive the relationship between the radiation wavelength and the temperature within a spherical blackbody radiation cavity undergoing adiabatic expansion. It can be assumed that the expansion rate of the cavity is much less than the speed of light.", "solution": "", "answer": "" }, { "id": 378, "tag": "ELECTRICITY", "content": "A very large metal conductor is shaped into an L-shaped conductor by removing a 90-degree dihedral angle. The edge of the dihedral angle is the z-axis, and its two half planes are the xz (x > 0) half plane and the yz (y > 0) half plane, respectively. The region where x > 0, y > 0 (i.e., the removed 90-degree dihedral angle) is vacuum. The L-shaped conductor is grounded, with its center point at the origin. An infinitely long, thin wire with a radius of r is parallel to the z-axis, located at x = a, y = b, -∞ < z < ∞. The vacuum permittivity is \\(\\varepsilon_0\\). Calculate the capacitance per unit length in the z direction between the thin wire and the L-shaped conductor using \\(\\varepsilon_0, a, b, r\\). Assume that the wire radius r is much smaller than a and b.", "solution": "", "answer": "" }, { "id": 462, "tag": "ELECTRICITY", "content": "### Given\nAn isolated conducting ellipsoid with charge \\(Q\\) and the surface (Cartesian coordinate) equation \\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}+\\frac{z^{2}}{c^{2}} = 1\\). The surface charge distribution at electrostatic equilibrium is \\(\\sigma=\\frac{Q}{4\\pi abc}(\\frac{x^{2}}{a^{4}}+\\frac{y^{2}}{b^{4}}+\\frac{z^{2}}{c^{4}})^{-1/2}\\). The vacuum permittivity is \\(\\epsilon_0\\).\n\n### Problem\nFor a conducting thin circular disk with radius \\(R\\) and charge \\(Q\\), find the system electrostatic energy at electrostatic equilibrium.", "solution": "", "answer": "" }, { "id": 431, "tag": "ELECTRICITY", "content": "Given a superconducting coil with self-inductance $L$ and radius $b$, initially a current $I_0$ is applied at infinity. There is also a fixed superconducting sphere with radius $a$. The coil is moved from infinity towards the sphere such that the normal direction of the coil is consistent with the direction of the line connecting the center of the coil to the center of the sphere and points outward from the sphere's center. Till the distance between the center of the coil and the center of the sphere is $r$. From this point on, the movement process to be considered is as follows: The center of the coil is slowly translated along a circle with the center of the sphere as the center and radius $r$ (without changing the orientation of the coil) until the normal direction of the coil is perpendicular to the line connecting the coil and the sphere, at which point the line connecting the two has rotated by 90 degrees. Find the work done by the external force $A$ during this process. The vacuum permeability is known as $\\mu_0$.", "solution": "", "answer": "" }, { "id": 287, "tag": "THERMODYNAMICS", "content": "When uncharged cells are in close proximity, fluctuations on the surface of the cell membrane can create repulsive forces between the cells. As a simplification, one can ignore the bending of the cell membrane's surface and treat it as a two-dimensional plane. The cell membrane can be approximately considered as an elastic membrane with a surface tension coefficient of $\\alpha$ and a mass per unit area of $\\sigma$. It is known that the cell membrane is in a thermodynamic equilibrium environment at temperature $T$, and the Boltzmann constant is $k_{B}$.\n\nAssume there are two interacting cell membrane samples, each a square membrane of size $L \\times L$, with the membranes parallel and at a distance $D (D \\ll L)$. To describe the surface fluctuations of the cell membrane, use a coordinate system $(x, y)$ with the origin $O$ at the vertex of one cell membrane. Let $h(x, y)$ represent the surface fluctuation of the sample.\n\nIf $D \\to +\\infty$, the cell membrane under thermal equilibrium should undergo free oscillations. The surface vibration mode of the cell membrane should be a two-dimensional standing wave concerning the edges, with the following form:\n\n$$\nh(x,y)=h_{0}(t)\\sin\\left(k_{x}x\\right)\\sin\\left(k_{y}y\\right)\n$$\n\nInvestigate when $k_{x}, k_{y}>0$, the minimum value mode of $|k_{x}|, |k_{y}|$.\n\nThe fluctuations between the two cell membranes must not intrude into each other, and the fluctuation of the lower cell membrane can be considered as $h \\leq D/2$. The following identity is known:\n\n$$\n\\frac{D}{2}=\\sum_{m,n=1;m,n\\text{是奇数}}^{\\infty}\\frac{4D}{mn\\pi^2}\\sin\\left(\\frac{m\\pi x}{L}\\right)\\sin\\left(\\frac{n\\pi y}{L}\\right),~(x,y\\in[0,L])\n$$\n\nAs an estimate, geometric constraints can be independently considered for each $(k_{x}, k_{y})$ component.\n\nFind the repulsive force $F$ between the cell membranes when $k_{B}T\\ll\\alpha D^{2}$.\n\n[Hint] If $X, Y, Z, T, P$ are all state functions and satisfy the differential relationship:\n\n$$\nd X=Y d Z+T d P\n$$\n\nthen:\n\n$$\n\\left({\\frac{\\partial Y}{\\partial P}}\\right)_{Z}=\\left({\\frac{\\partial T}{\\partial Z}}\\right)_{P}\n$$", "solution": "", "answer": "" }, { "id": 774, "tag": "MECHANICS", "content": "In a vacuum far from other celestial bodies, there exists a very thin gas cloud. Under the influence of gravity, the gas accumulates and eventually forms a spherically symmetric distribution upon reaching equilibrium. The density distribution is given by:\n$$\n\\rho = \\rho_0 e^{-r/r_0}\n$$\n\nAfter the gas cloud forms, a small meteorite core (with mass much smaller than the total mass of the gas cloud and density much greater than $\\rho_0$) orbits the center of the gas cloud in a uniform circular motion with radius $R$. Now, the small meteorite core experiences a slight radial perturbation. Determine the precession angular velocity of the orbit after stabilization. Express the final result using the angular velocity $\\omega$ during stable uniform circular motion and the dimensionless parameter $u=\\frac{R}{r_0}$. ", "solution": "", "answer": "" }, { "id": 172, "tag": "ELECTRICITY", "content": "Above an infinitely large grounded conductor plate at a distance $L$, there is a light rod with a length of $2R$, with charges $Q$ and $-Q$ at each end, respectively; both have a mass $m$. The center of the light rod is fixed and it can rotate freely. Consider only effects related to electrostatics. \nPlease analyze the frequency of small oscillations when the rod is vertical.", "solution": "", "answer": "" }, { "id": 525, "tag": "MODERN", "content": "Using the Lorentz transformation, find the velocity of a particle moving with velocity $\\vec{v}$ in the S frame, as observed in the S' frame moving with velocity $\\vec{v_{0}}$ relative to the S frame. Do not represent the result using a cross product.", "solution": "", "answer": "" }, { "id": 304, "tag": "MECHANICS", "content": "A smooth hemispherical bowl with a radius of $R$ is fixed in place.The rim of the bowl is horizontal.A uniform rod with a length of $L$ leans against the edge of the bowl. The lower end of the rod is supported by the bowl, while the upper end extends outside of the bowl. Considering the equilibrium condition, find the angle $\\theta$ that the rod makes with the horizontal when it is in equilibrium.", "solution": "", "answer": "" }, { "id": 66, "tag": "MECHANICS", "content": "Connect a horizontal plane with an inclined plane using an arc with radius $R$. The angle between the inclined plane and the horizontal plane is $\\theta$. A rope with linear density $\\lambda$ and length $L$ is placed on the horizontal plane and the arc, with its right end just at the junction of the arc and the inclined plane. The gravitational acceleration is $g$, and $R \\to 0$ with $R \\ll L$. All surfaces are smooth, and the rope should detach at some position. Let the distance the rope slides down the inclined plane be $x$. First, consider that $R$ exists, calculate the result, and then let $R=0$. Try to determine which part of the rope detaches first (expressed the result with the angle $\\varphi$ between the line connecting the point on the arc and the center of the circle and the vertical line).", "solution": "", "answer": "" }, { "id": 777, "tag": "ELECTRICITY", "content": "Consider two concentric thin metallic spherical shells, with the inner shell having a radius of $r$ and the outer shell having a radius of $R$. The space between the two shells is completely filled with a uniform medium characterized by a dielectric constant $ \\epsilon$, conductivity $ \\alpha$, specific heat capacity $c$, and density $ \\rho$. Initially, the system is uncharged. At time $t=0$, a charge $Q$ is suddenly placed on the inner shell, while the outer shell remains initially uncharged. Assumptions: 1. The metallic shells are ideal conductors. 2. The heat capacity of the metallic shells can be neglected. 3. The system is adiabatic, meaning all Joule heat generated by the current flowing through the medium is completely absorbed by the medium, with no heat lost to the surroundings. 4. The dielectric constant $ \\epsilon$ and conductivity $ \\alpha$ of the medium remain constant throughout the process. Find: After the system reaches the final electrostatic equilibrium state (i.e., after a sufficiently long time, when the current stops flowing), what is the total temperature increase $ \\Delta T$ of the medium?", "solution": "", "answer": "" }, { "id": 375, "tag": "OPTICS", "content": "A double convex thin lens, with both front and back interfaces being spherical surfaces, has curvature radii $R_1$ and $R_2$, respectively. The distance between the spherical surfaces is negligible, and the system is axially symmetric relative to the optical axis. The medium has a refractive index of $n$, and the system is placed in air (taking the refractive index of air as $n=1$). Consider a non-paraxial object point, with the distance between it and the optical center line (\"object distance\") being $u$, and the angle between its line to the optical center and the optical axis being $\\theta$. Consider only the light rays refracted in the vicinity of the optical center that form an image, where the angle between the line to the image point and the optical axis remains $\\theta$. Determine the distance between the image point and the optical center (\"image distance\") $v$, expressed in terms of $R_1, R_2, n,$ and $\\theta$.", "solution": "", "answer": "" }, { "id": 542, "tag": "MECHANICS", "content": "We consider an inhomogeneous cylinder composed of two materials with different densities. The cylinder has radius \\( r \\) and length \\( L \\). The lower half is made of a material with density \\( \\rho \\), while the upper half is made of a material with density \\( c\\rho \\), where \\( c \\) is a parameter satisfying \\( 0 < c < 1 \\). \n\nIn solving the problem, we can utilize the fact that for a semicylinder of radius \\( r \\), its center of mass (CM) is located at a distance \\( \\frac{4r}{3\\pi} \\) from the central axis of the semicylinder. The gravitational acceleration is denoted by \\( g \\). \n\nNow, we assume the cylinder is free to move on a horizontal surface under gravity. The static friction coefficient between the cylinder and the surface is sufficiently large to ensure pure rolling without slipping. At time \\( t = 0 \\), the cylinder is in its equilibrium position and has an initial angular velocity \\( \\omega_0 \\). \n\nIf \\( \\omega_0 \\) is sufficiently small, the cylinder will undergo periodic motion around its stable equilibrium position. For small oscillations, what is the period of oscillation? Express your answer in terms of the given parameters. \n", "solution": "", "answer": "" }, { "id": 323, "tag": "MECHANICS", "content": "A thin steel wire is bent into an ellipse with the polar equation $r={\\frac{p}{1+\\varepsilon\\cos\\theta}}$, where $p\\cdot\\varepsilon$ is a positive constant and $0<\\varepsilon<1$. The major axis of the ellipse intersects the wire at points $A$ and $B$, with $A$ close to the right focus and $B$ close to the left focus. Point charges with charges ${Q}_{1}$ and ${Q}_{2}\\left({Q}_{1}>0,{Q}_{2}>0\\right)$ are fixed at the two foci of the ellipse. Another small ball with mass $m$ and charge $q$ $(q>0)$ is threaded on the wire, insulated from it, and can slide along the wire without friction. Gravity and other resistances are not considered.\n\nFind the angular frequency of small oscillations of the ball around its stable equilibrium position on the wire between $A$ and $B$ (excluding points $A$ and $B$).", "solution": "", "answer": "" }, { "id": 681, "tag": "MODERN", "content": "Consider the following acceleration method: An object with total mass (including the mass of the mirror) \\(m_0\\) and a mirror of area \\(S\\) is placed facing the mirror direction with a nuclear bomb, which is detonated to produce photons that are perfectly reflected on the mirror surface, thereby propelling the mirror forward. Considering that a single nuclear bomb is insufficient to achieve the goal, a series of nuclear bombs are arranged in a straight line within the solar system, and the object is launched along the line. As it passes near each nuclear bomb, the bomb is detonated, further accelerating the object. Assume all nuclear bombs are identical and stationary before explosion, generating a flux density \\(J\\) of energy in the vicinity of the mirror at the moment of explosion (using the stationary bomb as a reference frame). Assume the entire flux consists of photons, which are incident perpendicular to the mirror and last for a duration of \\(t_0\\), with subsequent radiation from the explosion being negligible. Hint: The Doppler effect indicates that when moving away from a light source at speed \\(v\\), the observed frequency ratio compared to the original frequency is \\(\\sqrt{\\frac{c-v}{c+v}}\\). When the mirror velocity is \\(v\\), what is the momentum obtained by the mirror after one nuclear explosion? Express it in terms of \\(J, v, c, t_0\\) (assuming the change in \\(v/c\\) after one explosion is very small).", "solution": "", "answer": "" }, { "id": 396, "tag": "MECHANICS", "content": "A smooth bowl with a radius $R$ is fixed, and the opening plane is horizontal. A smooth, uniform thin rod $AB$ with length $L=4\\sqrt{3}R/3$. End B is located outside the bowl, while end A presses against a point inside the bowl, achieving static equilibrium in the plane passing through the sphere center $O$. The points $D$ and $D^{\\prime}$ on the rod nearly coincide with the support points on the bowl's edge, but $D$ is slightly to the lower left, and $D^{\\prime}$ is slightly to the upper right. Let the angle between the rod and the horizontal plane be $\\theta$.\n\nSuddenly cut the rod at $D^{\\prime}$. Note that after this point, $D^{\\prime}$ will lightly rest on the edge of the bowl. Find the instantaneous angular acceleration of the rod $\\beta^{\\prime}=\\ddot{\\theta}$.", "solution": "", "answer": "" }, { "id": 301, "tag": "MECHANICS", "content": "Assume the Earth is a sphere with radius $R$, mass $M$, and uniformly distributed mass. A small object is launched from a space station at a height $h$ above the Earth's surface. The object moves with an initial velocity relative to the Earth, where the direction of the initial velocity is perpendicular to the line connecting the object to the Earth's center. The gravitational constant is denoted as $G$. If the magnitude of the initial velocity of the object is $v_{0}$, and the object is able to land on the Earth's surface, find the total time it takes from launch to landing.", "solution": "", "answer": "" }, { "id": 745, "tag": "ELECTRICITY", "content": "In the polar coordinate system, there is a logarithmic spiral $r = r_0 \\mathrm{e}^\\theta$. On this trajectory, a point charge is placed starting at $\\theta = 0$, and for each angle $\\theta_0$ turned, another point charge is placed. Their magnitude of the charge is sequentially $-q (q > 0), -2q, 3q, -4q, \\cdots$ and so on to infinity. (The electrostatic constant is known to be $k$.) Find the magnitude of the electric potential $\\varphi$ at the origin.", "solution": "", "answer": "" }, { "id": 788, "tag": "THERMODYNAMICS", "content": "The mechanical properties of the rubber membrane are similar to the surface layer in thermodynamics. In the first approximation, the tension inside the rubber membrane can be described by the surface tension coefficient. A layer of rubber membrane is equivalent to a surface layer (the refractive index of the membrane is the same as that of water, and the membrane is extremely thin). Tension formula: \\[ F = \\sigma l \\] A hollow cylinder, with height \\(L \\), radius \\(R \\), has its two ends covered with a transparent rubber membrane with the same surface tension coefficient, and contains an ideal gas inside at pressure \\(p_0 \\). The cylinder is submerged in a pool to a depth \\(H \\) (assume \\(L \\ll H \\), and the liquid-induced pressure on the upper and lower membranes is the same). When the cylinder is at a distance of \\(H/3 \\) from the water surface, the bottom of the pool is imaged exactly at the water surface. Assume the density of water is \\( \\rho \\), refractive index is \\(n \\), gravitational acceleration is \\(g \\), and atmospheric pressure is \\(p_0 \\). The temperature of the water is independent of depth, and the cylinder has good thermal conductivity. If the cylinder can be treated as a thin lens when submerged in water, but the change in internal gas pressure cannot be ignored, and since the deformation of the membrane is still relatively small (the spherical radius of the membrane \\(r \\gg \\dfrac{R^2}{L} \\)), only first-order small quantities need to be retained. Find the surface tension coefficient \\( \\sigma \\) of the rubber membrane.", "solution": "", "answer": "" }, { "id": 399, "tag": "ELECTRICITY", "content": "Consider a cylindrical dielectric with a radius of $a$ and a length of $l$, where it is known that $l \\gg a$. First, permanently polarize the dielectric so that the direction of the polarization vector at every point is radially outward, and its magnitude is proportional to the distance from the axis. We establish a cylindrical coordinate system $(r, \\theta, z)$, with the cylinder's axis coinciding with the $z$-axis, and appropriately choose the origin so that the cylinder is symmetric with respect to the $xOy$ plane. Thus, the polarization vector is given by $\\vec{P} = \\frac{\\rho_0}{2} r \\hat{r}$. Next, use an external force to rotate the cylinder about the $z$-axis with a constant angular velocity $\\overrightarrow{\\omega} = \\omega \\hat{z}$, while maintaining the polarization strength unchanged. Find the electric field $\\overrightarrow{E}_{f}(r, \\theta, z)$ at a point far from the cylinder $(\\sqrt{r^{2} + z^{2}} \\gg l)$, retaining terms up to the $-4$ power of $r$ and $z$. Possible mathematical knowledge you may use: the expression for divergence in cylindrical coordinates is $\\nabla \\cdot \\vec{A} = \\frac{1}{r} \\frac{\\partial}{\\partial r} (r A_{r}) + \\frac{1}{r} \\frac{\\partial A_{\\theta}}{\\partial \\theta} + \\frac{\\partial A_{z}}{\\partial z}$.", "solution": "", "answer": "" }, { "id": 232, "tag": "MECHANICS", "content": "Two balls, 1 and 2, each with mass $m$, are connected by a massless, rigid, light rod of length $\\ell$, and are resting vertically in the corner of a wall, with ball 1 at the upper end of the rod. Assume both the wall and the ground are frictionless. Initially, give ball 2 a small initial velocity away from the wall. During the motion of the system, at what angle between the rod and the vertical wall does ball 1 begin to detach from the vertical wall?", "solution": "", "answer": "" }, { "id": 254, "tag": "THERMODYNAMICS", "content": "In a vacuum, there exists an electrically neutral plasma cloud composed of electrons and protons. The plasma cloud is in thermal equilibrium within a non-uniform external magnetic field, with an equilibrium temperature of $T$. It is known that the mass of an electron is $m_{e}$, the charge of an electron is $-e$, the mass of a proton is $m_{p}$, the charge of a proton is $e$, the Boltzmann constant is $k_{B}$, and the vacuum permeability is $\\mu_{0}$. The electrostatic effects of the plasma are neglected.\n\nThe distribution function of electrons and protons (distinguished using the index $\\alpha$, which can be used to simplify calculations) in six-dimensional phase space $(x, y, z; v_{x}, v_{y}, v_{z})$ is defined as:\n\n$$\nf(x, y, z; v_{x}, v_{y}, v_{z}):={\\frac{d{\\mathrm{P}}}{d x d y d z d v_{x} d v_{y} d v_{z}}}\n$$\n\nwhere $\\mathrm{P}$ is the probability density of finding a particle at position $(x, y, z)$ with velocity $(v_{x}, v_{y}, v_{z})$. Statistical physics indicates that if there exist additive conserved quantities in the single-particle motion of electrons or protons, such as energy $E$ and momentum $p$, then the thermal equilibrium distribution function $f_{\\alpha}$ is:\n\n$$\nf_{\\alpha} \\propto \\exp\\left(-{\\frac{E_{\\alpha} + V_{\\alpha} p_{\\alpha} + \\dots}{k_{B} T}}\\right)\n$$\n\nHere, $V$ is a velocity-related constant determined by the specific situation.\n\nIf the magnetic field in space can be neglected, and the plasma cloud as a whole begins to rotate about an axis with angular velocity $\\Omega$,\n\nIf the electron density at the center of the rotation axis is $n$, and in the rotating frame relative to the plasma itself, the axial angular velocity of the plasma is $\\Omega_{c}$. The boundary radius of the plasma cloud is $R$ ($R \\ll \\sqrt{k_{B}T / m_{e}(\\Omega + \\Omega_{c})^{2}}$), and the boundary magnetic field is zero. Find the magnetic field distribution $\\boldsymbol{B}(\\boldsymbol{r})$ generated by the plasma in the rotating frame. The answer should be expressed as the product of various factors.", "solution": "", "answer": "" }, { "id": 289, "tag": "OPTICS", "content": "By utilizing crystal materials with non-uniform thickness or non-uniform refractive indices, humans invented convex lenses. Now, let's reverse the thought process: use an ideal convex lens model combination to reconstruct a crystal material with a varying refractive index!\n\nConsider infinitely many ideal semi-convex lenses, each placed equidistantly with an angular spacing of $\\alpha$ between them. Each semi-convex lens is infinitesimally thin, allowing them to be arranged without mutual interference. These lenses are halved along their optical centers, and all optical centers are overlapped at the origin $O$ without crossing one another. Although there are infinitely many semi-convex lenses, $\\alpha$ is an infinitesimal quantity, so the total angular range remains finite and does not extend infinitely in rotation. Finally, the focal length of each convex lens is $f$, which is an infinitely large quantity. However, the quantity $f\\alpha = r_{0}$ is a finite, non-zero characteristic length in the limit. Below, all answers to the subsequent questions will be expressed in terms of $r_{0}$.\n\nTo solve the problem of determining the position and direction of a light ray perpendicular to the first lens upon exiting the last lens, we first perform an equivalence transformation: in fact, by tracing the trajectory of the light ray as it propagates between the lenses, it can be found that this problem is completely equivalent to the refraction of a light ray inside a spherical medium with a refractive index that varies with radius. Determine the refractive index distribution $n(r)$ of the sphere. It is required that $n(r_{0}) = n_{0}$, and it is known that at $r = r_{0}$, the angle between the propagation direction of the light ray and the lens orientation is $\\psi_0$.", "solution": "", "answer": "" }, { "id": 56, "tag": "MECHANICS", "content": "First, establish a polar coordinate system on a plane. Suppose the rabbit starts at the origin, and there is a safe hole at the coordinates \\((\\pi, S)\\). When the rabbit runs into the hole without being caught by the fox, it is considered \"safe.\" Now, the fox starts chasing the rabbit from any starting point \\((\\varphi, r)\\) and always keeps its speed direction along the line connecting it to the rabbit. At the same time, the rabbit starts moving in a straight line towards the safe hole. Assume the speed of the fox is \\(v_2\\), and the speed of the rabbit is \\(v_1 (v_1 < v_2)\\). Suppose the fox is just able to catch the rabbit before it reaches safety. Find the envelope of all starting points of the fox that satisfies the condition of \"catching the rabbit\" as described above; that is, provide the polar equation \\(r = r(\\varphi)\\) for the critical state (just able to catch the rabbit before it is \"safe\").", "solution": "", "answer": "" }, { "id": 665, "tag": "MECHANICS", "content": "A homogeneous cylinder is skewered at an angle on a rigid thin rod (considered as fixed), which passes through the center of the cylinder and intersects perfectly with the circumferences of its two bases. It is constrained by two very small parts, A and B, which do not provide power but only serve to secure the cylinder, and whose inner diameters are very close to the outer diameter of the metal rod. Parts A and B themselves are fixed. Given: the cylinder's mass is $m$, the radius of its base is $R$, and the cylinder's height is $4R$. The mass of the thin rod is negligible. If there is friction at parts A and B, the coefficient of friction is $\\mu$. Initially, the cylinder has an angular velocity $\\omega_{0}$ along its axis. Gravity is not considered. The radius of the thin rod is $r$. Without approximation, solve for the relationship between the magnitude of the cylinder's angular velocity $\\omega$ and time $\\omega(t)$.\n", "solution": "", "answer": "" }, { "id": 429, "tag": "MECHANICS", "content": "There is a non-stretchable, lightweight rope with a length of $2l$, and its two ends are fixed at two points on the ceiling that are $2c$ apart. A smooth small ring with mass $m$ is placed on the rope. The gravitational acceleration is known to be $g$.\n\nDetermine the period of oscillation of the small ring near its equilibrium position as it moves left and right.", "solution": "", "answer": "" }, { "id": 552, "tag": "MECHANICS", "content": "On a slope with an incline angle \\(\\alpha\\), a projectile is located above the slope. The launch point of the projectile is at a height \\(H\\) from the slope directly below it. A projectile is launched with an initial velocity \\(v_0\\) that is much less than the first cosmic velocity, ignoring air resistance, with gravitational acceleration \\(g\\). Find the area of the region on the slope that can be controlled by the projectile.", "solution": "", "answer": "" }, { "id": 383, "tag": "THERMODYNAMICS", "content": "Most of us are familiar with sound waves propagating in the Earth's atmosphere, which are pressure waves. However, within the Sun, we need to consider that the gas density decreases with height due to gravity. We simulate the Sun as an isothermal atmospheric model, where the density $\\rho$ decreases exponentially with height $z$ due to gravity. Let the mass of the atmospheric particles be $m$, the gravitational acceleration at the Sun's surface be $g$ (neglecting its variation with height $z$), Boltzmann constant be $k_B$, and the temperature of the solar atmosphere be $T$. Its adiabatic index (ratio of specific heat at constant pressure to specific heat at constant volume) is $\\gamma$.\n\nWhen sound waves propagate vertically within the atmosphere, particles will undergo small vertical displacements. Let $u(z,t)$ be the vertical displacement of gas particles at time $t$, where $z$ is their equilibrium position. Next, we study a sinusoidal wave with angular frequency $\\omega$. When a wave propagates vertically upwards at a constant speed in the direction of decreasing density $\\rho(z)$, we expect the wave's energy density $w=\\rho \\omega^2 u^2$ to remain unchanged. Thus, we can set $u(z,t)=\\frac{f(z)}{\\sqrt{\\rho(z)}} \\exp(-i\\omega t)$, where $f(z)$ is an unknown function to be determined. When the frequency of sound waves at the Sun's surface is lower than the critical frequency $\\omega_c$, they will be trapped inside the Sun and unable to propagate vertically upwards. Find this frequency $\\omega_c$.", "solution": "", "answer": "" }, { "id": 695, "tag": "MODERN", "content": "As is well known, an ideal black body can be well approximated by a small opening on a cavity. Let the temperature inside the cavity be $T$, the volume be $V$, the Stefan-Boltzmann constant be $\\sigma$, and the black body move along the x-axis with a velocity $v=\\beta c$ relative to a certain inertial frame S, while the small opening of the cavity faces exactly along the x-axis. Try to find in frame S: the black body radiation power density $J$ Express the answer in terms of $\\beta$, $c$, $\\sigma$, $T$, and other basic physical constants.", "solution": "", "answer": "" }, { "id": 590, "tag": "ELECTRICITY", "content": "In a uniformly positively charged large sphere, a small spherical cavity is excavated (the cavity radius is \\(R/2\\), and the distance from the large sphere's center is \\(R/2\\)), such that the total positive charge of the system is \\(Q\\). Now, electrons are emitted from the cavity center \\(o\\) in every direction with the same initial speed \\(v\\), but their kinetic energy is insufficient to reach the opposite end of the cavity diameter. Ignoring gravitational forces, find the envelope equation of all electron trajectories. The final answer should be an explicit formula in terms of \\(x\\) and \\(y\\), where the effective acceleration acting on the electrons is \\[ g=\\frac{eQ}{7\\pi\\varepsilon_0 m R^2} \\] acting to the left.", "solution": "", "answer": "" }, { "id": 548, "tag": "THERMODYNAMICS", "content": "Consider a capillary tube with an inner radius $R$ fixed, vertically placed in a certain liquid, with a surface tension coefficient $\\sigma$, and a contact angle that is an acute angle $\\theta$. The liquid density is $\\rho$. The height the liquid rises in the capillary tube is $H$. In the 17th century, the mathematician Torricelli proposed a model known historically as Torricelli's trumpet, which refers to the following capillary tube: the part above the water surface within a height $H^{\\prime}$ has a radius maintained as $R^{\\prime}$, but at a height $h > H^{\\prime}$, its radius $r$ is inversely proportional to its height, that is, $h r = H^{\\prime} R^{\\prime}$. Find the expression for $H^{\\prime}$ such that the height $h$ to which the liquid rises in the capillary tube exactly equals $H$, assuming the capillary tube is thin enough, approximating the tube wall to be nearly vertical at $h = H$. Express the answer in terms of $R^{\\prime}$, $\\sigma$, $\\theta$, $\\rho$, and the gravitational acceleration $g$.", "solution": "", "answer": "" }, { "id": 329, "tag": "ELECTRICITY", "content": "In a certain type of semiconductor, a class of carriers with charge $q$ and mass $m$ collide with the lattice. Prior to the collision, the carrier moves freely. After the carrier collides with the lattice, the probability $P$ of the next collision occuring and the uncollided motion time $\\Delta t$ satisfy the relation:\n\n$$\nP(\\Delta t)=1-e^{-\\Delta t/\\tau}\n$$\n\n$\\tau$ is a known constant determined by the thermodynamic average. After each collision, the carrier's momentum is reset by the lattice. The gas composed of these carriers can be approximately regarded as a monoatomic ideal gas. Unless otherwise stated, the number density of carriers is uniform and of magnitude $n$, and the thermodynamic temperature of the system is $T$. The Boltzmann constant is known as $k_{B}$.\n\nAssume the average collision time of carriers is $t_{1}$ (unknown), and the characteristic vibration time of the lattice is $t_{2}$. If $t_{1}\\sim t_{2}$, there is a correlation in the impulse components $i, j$ provided by two consecutive collisions, which depends on the time difference $\\Delta t$ of the two impulse actions:\n\n$$\n\\langle p_{0,i}(0)\\cdot p_{0,j}(\\Delta t)\\rangle=\n\\begin{cases}\n\\displaystyle\\frac{\\alpha}{2}\\left[1+\\cos\\left(\\frac{\\Delta t}{t_{2}}\\right)\\right]&{(i=j)}\\\\\n0&{(i\\neq j)}\n\\end{cases}\n$$\n\nwhere $t_{2}$ is a given constant, and the same collision can be regarded as the process $\\Delta t=0$.\n\nWhen there is no electromagnetic field in the semiconductor, the spatial number density has a gradient distribution:\n\n$$\nn(x,y,z)=n_{0}-\\Lambda x\n$$\n\nwhere $\\Lambda$ is a given constant, and $|\\Lambda x|\\ll n_{0}$.\n\nOn this basis, if a uniform magnetic field $\\vec{B}=B\\hat{z}$ is applied in space, the diffusion Hall effect can be observed in the semiconductor. Find the diffusion Hall conductivity $R_{H~D}=j_{D,y}/\\Lambda$, where $j_{D,y}$ is the diffusion current in the unit $y$ direction.", "solution": "", "answer": "" }, { "id": 179, "tag": "ELECTRICITY", "content": "The coaxial cylindrical capacitor with length $l$ consists of a metal rod with radius $a$ and a metal cylindrical shell with an inner diameter of $b$. One end of the capacitor is vertically inserted into a liquid medium with relative permittivity $\\varepsilon_{r}$ and density $\\rho$. Assume the liquid surface is very large, and the cylindrical capacitor is inserted into the liquid to a depth of $h_{0}$ (with $h_{0}\\gg b$, and all edge effects are neglected). A voltage ${V}_{0}$ is applied to the capacitor. Assume the liquid is an insulator, there is no conductive current in the liquid, and the surface tension of the liquid, the adhesive force, and the viscous force at the liquid-metal interface are not considered. The acceleration due to gravity is $g$. Neglect higher-order effects from the liquid flow in the tank and find the angular frequency $\\Omega$ of the micro-vibrations of the liquid inside the cylinder (assuming the liquid surface in the cylinder remains horizontal during the vibrations).", "solution": "", "answer": "" }, { "id": 278, "tag": "MODERN", "content": "The mass of particle X is $m$, and it can decay into two photons with their emission directions uniformly distributed in the center-of-mass system. A large number of X particles are emitted from the origin at high speed $v$ along the positive $z$ axis, and the average lifetime of X particles observed in the laboratory is $\\tau$. Establish a cylindrical coordinate system $\\rho, z, \\varphi$. Construct a photon detector on a cylindrical surface with radius $\\rho=R$. Find the probability distribution function $f(z)$ of the $z$ coordinates of photons detected on the detector after a sufficiently long time. In this problem, it is allowed to express the final result using an integral over angles, without needing to evaluate the integral. Consider the theory of special relativity, and use $c$ to represent the speed of light.", "solution": "", "answer": "" }, { "id": 583, "tag": "ELECTRICITY", "content": "In a sufficiently long cylindrical region with a radius of $a$, the outer side is a conductor with zero resistance, and the interior is vacuum.\n\nInside the cylindrical region, there is an infinitely long straight wire (compared to $a$), with the wire's center at a distance $b$ from the cylinder's axis. The current gradually increases from $0$ to $I$. The radius of the straight wire satisfies $r \\ll a$.\n\nFind the work done by the power source per unit length, $\\begin{array}{r}{\\omega^{\\prime}=\\frac{W^{\\prime}}{L}}\\end{array}$.", "solution": "", "answer": "" }, { "id": 655, "tag": "MECHANICS", "content": "A container filled with water rotates around a vertical axis $OB$ at a constant angular velocity $\\omega$. Inside the container, there is a thin rod $OA$ fixed at an angle $\\theta$ to the horizontal plane. A small ball is threaded onto the rod $OA$ and can slide without friction along the rod. Assume the volume of the ball is $V$, its density is $\\rho$, and the density of the water is $\\rho_{0}(\\rho_{0}>\\rho)$, with gravitational acceleration $g$. Initially, the ball is located at the bottom end $O$ of the rod, and then it starts sliding along the rod from rest. Find: the magnitude of the force exerted by the ball on the rod when the ball's speed in the rotating frame of the container is at its maximum. Assumption: Compared to the displacement caused by the ball, its radius can be neglected; the ball will not touch the walls of the container; the force exerted by the water on the moving ball relative to the water is approximately equal to the force exerted when the ball is at rest relative to the water.\n", "solution": "", "answer": "" }, { "id": 57, "tag": "MECHANICS", "content": "There is a planet with a constant volume \\( V \\) and a constant mass density \\( \\rho \\). What shape should the planet take so that the gravitational force at a specific point on its surface is maximized? Provide the equation for the shape of its surface.", "solution": "", "answer": "" }, { "id": 308, "tag": "MECHANICS", "content": "When a lightweight ball is placed on a horizontally rotating disk, an interesting phenomenon occurs: the object can move on the disk without leaving it. If the ball performs pure rolling without slipping on the disk, then when the ball is placed at a certain position on the disk, the ball exhibits a special motion state. To facilitate the analysis of the ball's motion, a rectangular coordinate system is established that is stationary relative to the experimental frame.\n\nGiven that the angular velocity of the disk is $\\Omega$, the radius of the ball is $a$, the moment of inertia is $kma^2$, point $C$ is the center of mass of the ball, and point $P$ is the position of contact between the ball and the disk in space, and further assuming that the ball does not leave the disk during motion, determine the motion period $T$ of the ball for this special motion.", "solution": "", "answer": "" }, { "id": 230, "tag": "THERMODYNAMICS", "content": "A uniform, infinitely long isothermal pipe is placed vertically in a gravitational field. The gas molecules inside are monoatomic, with a mass of $m$. The number density at the bottom is $n_{0}$, and the area is $S$. At the bottom, there is a piston that does not leak. The gravitational acceleration is known to be $g$, and the temperature is $T$. All the questions below consider local equilibrium, assuming that the relaxation time is extremely short and the local gas is adiabatic when vibrating. Find the change in air vibration over time at height $h$ when the piston vibrates as $x = A \\cos{\\omega t}$. That is, determine the propagation of the sound field, where $A$ is a small quantity. It is known that the (additional) force applied at the piston is $F = f \\cos{\\omega t}$, the adiabatic index of the gas is $\\gamma$, and $4k_{B}T m\\omega^{2}/\\gamma - m^{2}g^{2} < 0$. The Boltzmann constant is $k_{B}$. Find the change in air vibration over time at height $h$ when vibrating.", "solution": "", "answer": "" }, { "id": 189, "tag": "MODERN", "content": "Considering the effects of special relativity, establish a rectangular coordinate system \\((xOy)\\) in space. In a given inertial frame, a particle with mass moves along the \\(x\\)-axis with velocity \\(v_0\\) at time \\(t = 0\\). From \\(t = 0\\), the particle is subjected to a force only in the direction of the \\(y\\)-axis, causing the particle to undergo uniformly accelerated linear motion along the \\(y\\)-axis relative to this reference frame, where the acceleration of the particle in the \\(y\\)-direction is \\(a\\). Before the particle's velocity reaches the velocity of light \\(c\\), provide the trajectory equation \\(x(y)\\) of the particle. There is no need to specify the range of values for \\(x\\) and \\(y\\), with the stipulation that at time \\(t = 0\\), \\(x = y = 0\\).", "solution": "", "answer": "" }, { "id": 370, "tag": "ELECTRICITY", "content": "An electric quadrupole has two forms, one of which is a linear electric quadrupole: it consists of two point charges with charge $+q$ and one point charge with charge $-2q$. The three point charges are arranged in a straight line, with the charge $-2q$ located at the midpoint of the two charges and equidistant from the two $+q$ charges by a distance $l$. All scenarios below are placed in vacuum, with the known electrostatic constant in vacuum being $k$. Place a linear electric quadrupole on the left side outside a grounded conductive sphere with a radius $R$. The center of the electric quadrupole is at a distance $r$ (r > R) from the center of the conductive sphere. The orientation of the electric quadrupole is parallel to the radial vector $r$.Find the magnitude of the force exerted by the conductive sphere on the electric quadrupole.\n", "solution": "", "answer": "" }, { "id": 29, "tag": "MECHANICS", "content": "On a horizontal plane, there is a large homogeneous wooden plank with a surface mass density of $\\sigma$. Above the plank, there exists a static magnetic field $\\begin{array}{r}{\\vec{B}(r)=\\frac{3m v_{0}r}{q r_{0}^{2}}\\vec{e_{z}}}\\end{array}$. A particle with a charge-to-mass ratio of $\\frac{q}{m}$ is launched horizontally from the origin with an initial velocity of $v_{0}$. The particle leaves a dumbbell-shaped trajectory on the wooden plank. Symmetrically cutting out the \"water-droplet-shaped\" half from this trajectory as its boundary, we take this portion as a rigid planar body. Placing this rigid body on a rough horizontal plane and considering the gravitational acceleration $g$, certain configurations of the rigid body can maintain stable equilibrium. Derive the small oscillation period at the point of stable equilibrium.", "solution": "", "answer": "" }, { "id": 666, "tag": "OPTICS", "content": "A student accidentally smeared out one slit on the grating while conducting a grating diffraction experiment and discovered a different \\\"landscape.\\\" The following research is based on this phenomenon. It is known that the wavelength of light used in the experiment is $\\lambda$, the grating constant is ${d}$, and the slit width is $a$. The missing orders phenomenon is not considered. Assume that there are a total of $2N+1$ slits on the grating and due to a manufacturing oversight, one slit is missing in the center. Discuss the distribution of light intensity. Assume the central light intensity of diffraction from one slit of this grating is $I_{0}$", "solution": "", "answer": "" }, { "id": 200, "tag": "OPTICS", "content": "This question explores time coherence from the perspective of randomly emitted wave packets: a light source always emits waves in the form of one wave packet after another. Within the same wave packet, there is a regular variation in the phase between two points, whereas the phase relationship between different wave packets is completely random. At this point (at a given location), time coherence can be understood as \n\n$$\n\\gamma(t_{0})=\\left|\\left\\langle\\frac{E^{*}(t+t_{0})E(t)}{\\left|E(t+t_{0})E(t)\\right|}\\right\\rangle\\right|\n$$ \n\nwhere $<>$ denotes averaging or expectation over $t$. Coherence can be seen as a measure of the correlation between the phase of the electric field at time $t+t_{0}$ and at time $t$. Thus, the electric field magnitude is not discussed here. In this expression, the quantity being averaged is the electric field $E$ divided by its own magnitude. As a result, only the phase information at different times is retained, yielding a dimensionless constant. Additionally, note that averaging is performed first, followed by taking the modulus. \n\nIf there is a light source that emits random wave packets of different lengths, and the number of wave packets of different lengths is distributed according to a probability density, the number of wave packets with a length distributed between $\\tau$ and $\\tau+\\mathrm{d}\\tau$ is \n\n$$\\mathrm{d}n=\\frac{1}{\\tau_{0}}e^{-\\frac{\\tau}{\\tau_{0}}}d\\tau,$$ \n\nwhere $\\tau_{0}$ is a known constant. Determine the time coherence of these randomly emitted wave packets.", "solution": "", "answer": "" }, { "id": 434, "tag": "MODERN", "content": "Action integrals are not only of great significance in quantum theory; they also appear as adiabatic invariants in classical physics. Consider the following problem: A negative charge $-q$ initially moves in a circular orbit around a stationary positive charge $Q$, with angular momentum, radius, radial momentum, and angle denoted as $L$, $r$, $p_{r}$, and $\\theta$ respectively. Now, for some reason, the charge of the positive charge $Q$ changes very slowly to $Q^{\\prime} = \\lambda Q$. This change only affects the electrostatic force on the negative charge without causing other effects. Then, theoretical mechanics can prove that $J_{1} = \\oint p_{r}\\mathrm{d}r$ and $J_{2} = \\oint L\\mathrm{d}\\theta = \\int_{0}^{2\\pi}L\\mathrm{d}\\theta$ both remain unchanged over a long period of time, meaning they have the same value in the initial and final states, which is called an adiabatic invariant. Find the final state radius $r^{\\prime}$ (express it as a power of $\\lambda$ to indicate the growth factor).", "solution": "", "answer": "" }, { "id": 348, "tag": "MECHANICS", "content": "There is now an elastic beam AB fixed to a vertical wall, where at the contact point A with the wall, it is tangentially horizontal. The length of the elastic beam is $L$, and the bending stiffness is $EI$ (meaning that the restoring moment of the elastic beam is $M=EI \\frac{d^2y}{dx^2}$). The coefficient of friction between the block and the beam is $\\mu$. There is now a slider C, whose dimensions relative to $L$ can be neglected, with a weight of $G$. Assume the block is at point D on the beam, and set the initial sliding position of the block to be $s$, where it just begins to slide. Given this setup, find the velocity $v$ at the end point B as the block slides along the beam and falls off.\n", "solution": "", "answer": "" }, { "id": 214, "tag": "MODERN", "content": "The uncertainty principle in quantum mechanics refers to the fact that two non-commuting quantities cannot be simultaneously measured with perfect accuracy. Suppose we observe two mechanical operators $A$ and $B$, which satisfy the commutation relation:\n$$\n[A,B]=A B-B A=C\n$$\nThen, when we simultaneously measure $A$ and $B$, their measurement uncertainties satisfy:\n$$\n\\langle(\\Delta A)^{2}\\rangle\\langle(\\Delta B)^{2}\\rangle\\geq\\frac{1}{4}\\langle-i C\\rangle^{2}\n$$\nIn classical mechanics, the ground state energy of a harmonic oscillator is zero. However, in quantum mechanics, due to the influence of the uncertainty principle, the harmonic oscillator’s ground state energy is not zero but rather $\\begin{array}{r}{\\cdot\\frac{1}{2}\\hbar\\omega=\\frac{1}{2}h\\nu.}\\end{array}$ The ground state energy of a harmonic oscillator, $\\cdot{\\frac{1}{2}}\\hbar\\omega$, is also called the vacuum zero-point energy. Consider the electromagnetic field can also be quantized as infinite harmonic modes. The electric field strength of the standing wave field between two parallel conducting plates can be decomposed as:\n$$\n\\overrightarrow{E}_{(z,t)}=\\sum_{j}\\widehat{\\epsilon}_{j}A_{j}q_{j}(t)\\sin{(k_{j}z)}\n$$\nSubstituting each harmonic oscillator into the equation satisfied by electromagnetic waves in a vacuum:\n$$\n\\nabla^{2}\\vec{E}-\\frac{1}{c^{2}}\\frac{\\partial^{2}\\vec{E}}{\\partial t^{2}}=0\n$$\nwhere $c$ is the speed of light, and $\\nabla^{2}=\\frac{\\partial^{2}}{\\partial x^{2}}+\\frac{\\partial^{2}}{\\partial y^{2}}+\\frac{\\partial^{2}}{\\partial z^{2}}$ is the scalar operator. We obtain:\n$$\n\\ddot{q}_{j}+c^{2}k_{j}^{2}q_{j}=0\n$$\nwhich is the harmonic oscillator equation, where $k_{j}$ represents the wavevector of the propagating electromagnetic wave. Therefore, the vacuum also possesses the ground state energy of the electromagnetic field (for each harmonic oscillator):\n$$\nE_{0}=\\frac{1}{2}h\\nu\n$$\nThis energy is referred to as the vacuum zero-point energy. As a result, even in the absence of photons, there exists a Casimir force between two parallel conducting plates in a vacuum.\n\nNext, we use a simplified model to theoretically calculate the magnitude of the force arising from the Casimir effect. Assume our universe is a one-dimensional ring of length $L$, and the two conducting plates are placed parallel to each other at a distance $x$. In this system, a series of electromagnetic wave modes satisfying the boundary conditions contribute to the vacuum zero-point energy of the system. In reality, metallic plates cannot reflect arbitrarily high-frequency radiation: the highest energy modes will escape. To account for this effect, an exponential term is introduced to truncate the high-energy modes. The truncation position is arbitrary, so we expect it will not affect any measurable quantities. Assume that when summing over different electromagnetic wave modes to obtain the total energy, a Boltzmann factor is introduced, i.e.,\n$$\nE=\\sum_{j}e_{j}^{-\\frac{E_{j}}{\\Lambda}}E_{j}\n$$\nwhere $E_{j}$ is the zero-point energy for each mode, and $\\Lambda$ is a constant.\n\nFind the Casimir force $F_{(x)}$ in the lowest-order approximation, assuming $L \\gg x$ and $h c \\ll x\\Lambda$.", "solution": "", "answer": "" }, { "id": 602, "tag": "ELECTRICITY", "content": "In a vertical plane, there is a circular coil with a radius of $a$ carrying a current $I$. OP is perpendicular to the plane of the circle and has a length of $h$. Point O is the center of the circular coil. At point P, a very small metal ring is placed, with its center at P and a radius of $b$, where $b$ is much smaller than $a$ and $h$. At $t=0$, the plane of the small ring is also perpendicular to OP. The current in the large ring remains constant, and the small ring rotates uniformly with an angular velocity $\\omega$. The small ring has a resistance of $R$, with its self-inductance neglected. The angular velocity direction of the small ring is vertically upward, and the direction of the current $I$ in the large coil, when viewed from along the PO direction, is counterclockwise. To maintain uniform angular velocity of the small ring, find the magnitude of the torque required on the small ring at time $t$. Given the vacuum permeability $\\mu_0$.", "solution": "", "answer": "" }, { "id": 638, "tag": "OPTICS", "content": "Laser has been widely used due to its high power; however, various dissipations in reality can reduce the laser power. It is known that due to the gain of the medium, light satisfies $$ \\frac{dA}{dx}=\\alpha A $$ where \\(A\\) is the light intensity, and \\(\\alpha\\) is a constant. It is also known that the length of the laser \\(L\\). Assume that the light will not be completely reflected at the front and back surfaces of the laser but is reflected in the opposite direction with a light intensity reflection rate \\( r < 1 \\). In fact, the medium not only brings gain but also causes some reflection. Define \\(k\\) as the ratio of the reflected light intensity to the original light intensity when advancing a unit distance in the medium. In order to enhance the laser, try to express the minimum value of \\(L\\) using \\(r\\), \\(k\\), and \\(\\alpha\\). Known \\(\\alpha > k\\), directly perform light intensity superposition without considering the phase.", "solution": "", "answer": "" }, { "id": 210, "tag": "OPTICS", "content": "Let a beam of monochromatic parallel light with wavelength $\\lambda$ and amplitude $A$ be irradiated from left to right onto a biconvex lens with a radius $R$, curvature radii of the left and right surfaces $\\rho_{1}, \\rho_{2}(\\rho_{1,2} \\gg R)$, and refractive index $n$. The direction of the light beam is parallel to the optical axis of the lens.\n\nLet the position on the axis and the optical center distance be $z$. Calculate the intensity distribution $I(z)$ on the optical axis behind the lens.", "solution": "", "answer": "" }, { "id": 642, "tag": "MECHANICS", "content": "On a free homogeneous disk A with a radius $R$, there is an inextensible light string of length $L$ wound around it (which does not affect the disk's diameter). The free end of the string is connected to object B, and then passes over a frictionless fixed pulley to connect to object C. Initially, A and B are at rest, with disk A exactly situated in the vertical plane, and C is stationary on the ground. The segments of the string between A and B, between B and the fixed pulley, and between the fixed pulley and C all remain vertical. It is known that the string length $L$ is much greater than the radius $R$ of the disk, and that after the entire string on disk A is unwound, disk A will fall off the string; during motion, the string remains in the vertical direction, and disk A will not collide with the ground. The masses of objects A, B, and C are $M_A$, $M_B$, and $M_C$, respectively. If the pulley is smooth, and disk $A$ is released from rest, $A, B, C$ all start moving simultaneously. Try to find the acceleration of $B$ at the moment of release.\n", "solution": "", "answer": "" }, { "id": 589, "tag": "MECHANICS", "content": "A ladder of length \\(2l\\) and mass \\(m\\) is placed against a vertical wall, initially making an angle \\(\\alpha\\) with the horizontal. It starts to slide under frictionless conditions. While the ladder is still in contact with the wall, its total mechanical energy (taking potential energy at \\(y=0\\) as zero) can be expressed as \\[ E(\\theta,\\dot{\\theta})=\\frac{2}{3}ml^2\\dot{\\theta}^2+mgl\\sin\\theta, \\] where \\(\\theta(t)\\) is the angle between the ladder and the horizontal. Derive the relationship between the critical angle \\(\\theta_c\\), at which the ladder loses contact with the wall, and the initial angle \\(\\alpha\\). Provide the final answer in a clear expression.\",\n", "solution": "", "answer": "" }, { "id": 489, "tag": "MODERN", "content": "The Bohr model holds historical significance as the first semi-classical quantum model of the hydrogen atom. It explained the challenges of classical electrodynamics regarding electron radiation and atomic stable states, as well as the results of spectroscopic observations at the time. Due to the limitations of quantum theory during its early development, it underwent modifications by Sommerfeld and others, eventually leading directly to the establishment of the Schrödinger equation by Schrödinger and others, which serves as the fundamental equation of quantum mechanics.\n\nIn the Bohr model, the principal quantum number $n$ simultaneously determines the energy and angular momentum of the electron state, which is actually incorrect. Through observation of hydrogen atoms in electric fields and magnetic fields, we point out that degenerate states under the same principal quantum number $n$ with different angular quantum numbers $l$, magnetic quantum numbers $m$, and even spin quantum numbers $s$ will experience energy level splitting. Specifically, some states' energy levels increase, while others decrease, and spectroscopic observations can reveal distinct spectral lines resulting from electronic level transitions between different states. The Stark effect is a typical effect when the hydrogen atom couples with an electric field. This problem attempts to estimate of the magnitude of energy level splitting due to the Stark effect using a semi-classical approach.\n\nConsider the slow evolution of the electron's orbital in space under the influence of a weak uniform electric field $\\vec{E}$. First, consider a simpler situation where at time $t=0$, the normal vector $\\vec{n}$ of the plane of the electron's orbit and the electric field strength $\\vec{E}$, along with the axis pointing from the nucleus to the pericenter of the orbit (the $x$-axis), are all situated in the $xOz$ plane, with an angle $\\theta$ between the latter two. Use the perturbation method to prove that the magnitude of the angular momentum of the electron's orbit does not change over time, and the orbit merely precesses around the direction of the electric field $\\vec{E}$ (that is, rotates around the axis). Determine this angular velocity $\\Omega$. Given the orbital energy $-U$, eccentricity $e$, electron mass $m$, and the magnitude of charge for both the atomic nucleus and electron as $q$.\n\nHint: The so-called perturbation method refers to calculating the instantaneous derivative of the dynamical conserved quantity of the original system under the action of the newly added perturbing force at each moment. Substitute the position and velocity of the periodic motion of the original system into this derivative and calculate the periodic average. Finally, this averaged quantity is interpreted as the time derivative of a dynamical conserved quantity along the instantaneous orbit.\n", "solution": "", "answer": "" }, { "id": 775, "tag": "THERMODYNAMICS", "content": "We study a specific thermodynamic system, often exploring it from two perspectives: constant volume and constant pressure.\n\nConsider a system describable by pressure p and volume V.\n\nDefine its constant-pressure temperature scale t such that, when pressure is held constant, the temperature scale t has a linear relationship with the volume V.\n\nDefine its constant-volume temperature scale t \n′\n such that, when volume V is held constant, the temperature scale t \n′\n has a linear relationship with the pressure p.\n\nIt is also known that when the pressure is constant at p \n0\n​\n , the temperature scale t is directly proportional to the volume V, with a proportionality constant of \nV \n0\n​\n \n1\n​\n . When the volume is constant at V \n0\n​\n , the temperature scale t \n′\n is directly proportional to the pressure p, with a proportionality constant of \np \n0\n​\n \n1\n​\n .\n\nIf, for a thermodynamic system, the values given by both temperature scales are always equal (i.e., t=t \n′\n , denoted as t \nsys\n​\n ), and it satisfies:\n\nWhen the pressure p is constantly p \n0\n​\n ,\nt \nsys\n​\n = \nV \n0\n​\n \nV\n​\n \n.\nWhen the volume V is constantly V \n0\n​\n ,\nt \nsys\n​\n = \np \n0\n​\n \np\n​\n \n.\nWhen both pressure p and volume V are sufficiently large, the temperature\nt \nsys\n​\n \nsatisfies the asymptotic relation\nt \nsys\n​\n +α∝pV\n, where\nα\nis a known parameter (easily determined experimentally).\nDerive the equation of state for this thermodynamic system, expressed as a relationship \nt \nsys\n​\n =t(p,V)\n involving p,V, and \nt \nsys\n​\n", "solution": "", "answer": "" }, { "id": 79, "tag": "ELECTRICITY", "content": "In vacuum, there is a plane wave with angular frequency $\\omega$ propagating in the $\\pmb{x}$ direction, given by $\\begin{array}{r}{\\vec{E}(x,y,z,t)=E_{0}\\cos(\\omega t-k x)\\hat{y},}\\end{array}$ $\\begin{array}{r}{\\vec{B}(x,y,z,t)=\\frac{E_{0}}{c}\\cos(\\omega t-k x)\\hat{z},k=\\frac{\\omega}{c}.}\\end{array}$ \nThe electromagnetic wave propagates in the $\\pmb{x}$ direction with amplitude $E_0$. An ideal mirror has an angle $\\pmb{\\theta}$ with the direction $-\\hat{\\pmb{x}}$, and has velocity $\\pmb{v}$, such that the wave is incident on the mirror in its reference frame. Find the radiation pressure $_{p_{2}}$, expressing the answer using $I_{0}={\\textstyle\\frac{1}{2}}\\varepsilon_{0}E_{0}^{2}c$. The permittivity in vacuum is $\\epsilon_0$, the speed of light in vacuum is $c$, and consider special relativity.", "solution": "", "answer": "" }, { "id": 520, "tag": "ELECTRICITY", "content": "\n\n---\n\nA ring with mass $m$, charge $q$, and radius $R$, which is uniformly charged, homogeneous, and made of insulating material, is located on a smooth horizontal surface. The plane of the ring coincides with the surface of the ground. Friction between the ground and the ring is negligible. In the space above, there exists a non-uniform magnetic field directed vertically upward (in the $-z$ direction), given by ${{B}}={{B_{0}}}+k{{x}}$. Initially, the center of the ring is located at the origin. Analyze the motion of the ring after being subjected to a small perturbation along the positive $x$ direction. \n\nThe ring will undergo small oscillations in the $x$ direction. Ignoring higher-order small terms of $x$, find the period of small oscillations in the $x$ direction.\n\n---", "solution": "", "answer": "" }, { "id": 447, "tag": "ELECTRICITY", "content": "A small permanent magnetic needle can be considered as a current loop with a very small radius, and its magnetic moment $\\mu$ is given by $\\mu=I S$, where $I$ is the constant current strength of the loop, $S=\\pi R^{2}$, and $R$ is the radius of the current loop. The direction of the magnetic moment is perpendicular to the plane of the current loop and follows the right-hand rule with the direction of the current. Two small magnetic needles, A and B, have a constant magnetic moment of size $\\mu$ and a mass of $m$. We establish a coordinate system by taking the downward vertical direction as the positive $z$ axis. Magnetic needle A is fixed at the origin $\\mathrm{o}$ of this coordinate system, with its magnetic moment pointing in the positive $x$ direction; magnetic needle B is placed directly below A. The magnetic permeability of vacuum is given as $\\mu_{\\mathrm{0}}$, and the gravitational acceleration is $g$. Assuming the magnetic moment direction of B makes an angle $\\theta$ with the $x$ axis, and its center of mass coordinates are $(0,0,z)$ ($z >> R$), find the angle $\\theta_{\\oplus}$ that B can maintain stable equilibrium with the $x$ axis. Stable equilibrium means that if B's magnetic moment direction deviates slightly from $\\theta_{\\oplus_{\\tt G}^{\\tt G}}$, there will still be a tendency to return to the original direction.", "solution": "", "answer": "" }, { "id": 604, "tag": "THERMODYNAMICS", "content": "Consider a material with emissivity $e$, which consists of two large parallel plates, $A$ and $B$, with constant temperatures $T_{A}$ and $T_{B}$, where $T_{A} > T_{B}$. The emissivities of $\\boldsymbol{A}$ and $\\boldsymbol{B}$ are $e_{1}$ and $e_{2}$, respectively. Find the net heat flow $J$ from $A$ to $B$.", "solution": "", "answer": "" }, { "id": 620, "tag": "OPTICS", "content": "As is well known, the human eye is most sensitive to green light around $532 \\mathrm{nm}$, so a laser at this wavelength can cause the most damage if it directly hits the eye. Consider an optical system where the refractive index on the left side is $n_{1}$, and the refractive index on the right side is $n_{2}$. A point $A$ on the optical axis is perfectly imaged at point $A'$. It is known that the angle between the light emitted from $A$ and the optical axis is $u$, and the angle between the light received at $A'$ and the optical axis is $u'$. For another point $B$, which is very close to the optical axis and at a perpendicular distance $\\delta y$ from $A$, it can also be perfectly imaged at point $B'$ on the image plane. Place a green laser pointer at point $O$, located at a distance $R$ from point $A$ along the line of the light emitted from $A$ at an angle $u$ to the optical axis, but extending in the opposite direction. The laser beam has a divergence angle of $\\theta \\ll 1$ and is incident at a small angle $u+\\theta/2$ with the optical axis. Find the minimum distance that the eye needs to maintain from the optical axis in the original $A'B'$ image plane region below the optical axis to avoid being damaged by the green laser.", "solution": "", "answer": "" }, { "id": 218, "tag": "OPTICS", "content": "Humans utilized the reflection of the ionosphere to achieve the first transatlantic wireless communication, and the ionosphere is also the most crucial auxiliary weapon in modern electronic warfare. Imagine ultra-high precision directional signal transmission (emitting signals only in one precise direction), using extremely high-frequency signals to eliminate diffraction, striving to achieve \"imaging communication\" (where both parties are precisely positioned at each other's image points), and converging the transmitted electromagnetic waves at the receiving site (thus achieving fixed-point signal reception). Here, \"imaging communication\" means that the two points of the object can achieve convergence under the third-order small quantity approximation. Since an ellipse can achieve ideal imaging, if the reflective surface is similar to an ellipse within certain precision and range, then ideal imaging can be realized. Only consider signal propagation within a plane. The angular separation between the signal source and the receiving station with respect to the Earth's center is $\\varphi$. Consider ground-to-ground communication with a single reflection; the base height of the ionosphere is $z$, and the Earth's radius is $R$. $\\varphi=2\\operatorname{arccos}R/(R+z)$, if the maximum deviation angle of the transmitted signal from the set value is $\\alpha(\\alpha\\ll1)$, calculate how far from the ground the receiving station can be to receive the signal, and assume $z\\ll R$ in this question.", "solution": "", "answer": "" }, { "id": 641, "tag": "MECHANICS", "content": "A particle \\( P \\) with a mass of \\( m \\) is subjected to a constant force of magnitude \\( F \\) that always points towards point \\( O \\). Initially, \\( P \\) is at a distance \\( r_{0} \\) from \\( O \\) and has a velocity perpendicular to \\( OP \\) with a magnitude of \\( u_{0} \\) (\\( u_{0} > \\sqrt{\\frac{F r_{0}}{m}}\\)). At the moment when the particle \\( P \\) reaches its farthest point from \\( O \\), an impulse is applied to it so that \\( P \\) can perform uniform circular motion. At some point during its subsequent movement, a perturbation is applied to \\( P \\) along the radial direction. Find the period \\( T \\) of the radial oscillation of particle \\( P \\).\n", "solution": "", "answer": "" }, { "id": 753, "tag": "ELECTRICITY", "content": "In a certain space, there is a uniform magnetic field $ \\vec{B} = B_0 \\hat{z}(B_0 > 0)$ in the region where $y > 0$. A homogeneous ring with a mass $M$, radius $R$, insulated and with a charge per unit length of $ \\lambda$, moves with a velocity $ \\vec{v} = v_0 \\hat{y}(v_0 > 0)$. At time $t = 0$, the ring is located in the $xy$ plane, and the center of the ring is at $(0, -R, 0)$. Ignore the effects of all non-electromagnetic forces. Find the minimum initial velocity $v_{1}$ required for the ring to completely enter the magnetic field.", "solution": "", "answer": "" }, { "id": 727, "tag": "MECHANICS", "content": "A bomb explodes at a height $H$, splitting into many small fragments. After the explosion, each fragment is projected outward with the same velocity $u$ in all directions (with uniform angular distribution). Afterwards, all fragments fall to the ground, experiencing completely inelastic collisions upon landing. Find the distribution radius \\(R\\) of the bomb debris.\n", "solution": "", "answer": "" }, { "id": 781, "tag": "ELECTRICITY", "content": "In a vacuum, there is an infinitely long thin-walled conductive cylinder with a cross-sectional radius of $R$, carrying a uniform and constant azimuthal current $I$. The cylinder is cut along the axial direction into two halves, separated by a very short distance, assuming the current distribution remains unchanged. Given that the permeability of the vacuum is $ \\mu_0$, find the force per unit length between the two halves.", "solution": "", "answer": "" }, { "id": 623, "tag": "MODERN", "content": "Consider the decay reaction $A \\rightarrow B+C$, where the rest masses of the particles are $m_{B}=m_{C}=m$ and $m_{A}=\\alpha m(\\alpha >2)$. In the ground frame, the initial velocity of $A$ is $v=\\beta c$. Let $\\beta_{0}=\\sqrt{1-\\frac{4}{\\alpha^{2}}}$, and it is known that $\\beta_{0}<\\beta<\\frac{\\beta_{0}}{\\sqrt{1-\\beta_{0}^{2}}}$. Find the maximum value of the angle between the velocity directions of $B$ and $C$ in the ground frame.", "solution": "", "answer": "" }, { "id": 433, "tag": "ELECTRICITY", "content": "When a metal conductor is placed into a plasma, changes in the electron and ion number densities and potential will occur near the surface. There is a presheath layer where the electron and ion number densities are equal, and a sheath layer where they are unequal, creating a current between them. By measuring the $U-I$ characteristic curve, certain characteristics of the plasma can be obtained, known as a plasma probe.\n\nThe problem is simplified to a one-dimensional problem, with the one-dimensional coordinate $x$. For $x>0$, it's the conductor, and for $x<0$, it's the plasma. From right to left, there are the metal conductor, the sheath, and the presheath.\n\nAt $x \\rightarrow \\infty$, the potential is set as the reference potential $V \\approx 0$, and the electron and ion number densities are $n_0$.\n\nAssume the ion mass is $m$, and the kinetic energy at the top of the presheath moving towards the conductor is $W_0$. It is assumed that ions do not experience collisions during their motion.\n\nAssume the potential at the interface between the presheath and sheath is 0, the electric field intensity is $E_d$, and the potential at the conductor's surface is $V_f < 0$. It is assumed that the electron number density follows the Boltzmann distribution, $n_e = n_0 \\exp\\left(\\frac{eV(x)}{kT}\\right)$, where $T$ is the plasma temperature.\n\nThe electron charge is known as $e$, and the vacuum permittivity is $\\epsilon_0$.\n\nAssume the surface area is $A$. If a small alternating voltage is superimposed on the conductor on the basis of the potential $V_f$, an alternating current will be generated between them. Find the dynamic capacitance $C$ of the conductor. (The dynamic capacitance is defined as $C = \\frac{dQ}{dV}$, which is the derivative of charge with respect to potential.)", "solution": "", "answer": "" }, { "id": 351, "tag": "MECHANICS", "content": "The acceleration due to gravity $g$ is directed vertically downward. A light rope of length $2c$ symmetrically passes through a smooth fixed hole $O$ and suspends a rectangular thin plate with length $2a$ and width $2b$. The suspension points are the two vertices on one edge of length $2a$. The plate has a mass $m$ uniformly distributed over its surface. Initially, the plate is in a state of symmetric suspended equilibrium. Suitable choices of $a,b,c$ ensure that the plate is in stable equilibrium.\n\nWe restrict disturbances and motion to occur only within a plane parallel to the plate, so the number of possible vibration modes is $2$. Let $a=3l$, $b=4l$, and $c=5l$.\n\nPlease output the larger angular frequency, requiring an exact solution of the larger angular frequency rather than a numerical solution, using $g$ and $l$ for expression.", "solution": "", "answer": "" }, { "id": 611, "tag": "MECHANICS", "content": "In a viscous fluid, a slender rod experiences two types of resistance. In this problem, we refer to the resistance acting in the normal direction to the rod as resistance, and the resistance acting along the rod as friction. The resistance per unit length is approximately $-\\mu v_n$, and the friction per unit length is $-\\mu v_t /2$, where $v_n$ and $v_t$ are the velocity components perpendicular and parallel to the rod axis, respectively, and $\\mu$ is a constant related to the fluid viscosity and the rod's diameter. Consider a slender rod of length $l$. The instantaneous center of the rod is point P (which may not necessarily be on the rod), and the radial and tangential distances from point P to the center of the rod are $a$ and $b$, respectively. The rod rotates in the plane around the instantaneous center P with an angular velocity $\\omega$, and point P is instantaneously at rest. The resistance and friction will exert forces and moments on the rod. The model described above can be used to calculate the resistance and its moment acting on the rod. Derive the moment of fluid resistance (including normal resistance and tangential friction) about the instantaneous center P. Express the result using the symbols defined in this problem.\n", "solution": "", "answer": "" }, { "id": 785, "tag": "MECHANICS", "content": "A thin cylinder with a radius of \\( R \\) and a height of \\( H \\) is situated on the horizontal surface, with its axis parallel to the vertical direction. Inside, there is a solid small ball with a radius of \\( r \\) and a mass of \\( m \\) that is in **pure rolling** motion while remaining in contact with the cylinder wall. At \\( t = 0 \\), the center of the ball is at a height of \\( H/2 \\) from the horizontal ground. The velocity of the ball in the vertical direction is upward with a magnitude of \\( v_0 \\), and the vertical acceleration is 0 (note that the **angular velocity perpendicular to the cylinder wall direction is non-zero**, but it is not required to be given here). The horizontal velocity is \\( v_1 \\). During the subsequent motion, the ball **does not collide with the ground**, **does not leave the cylinder**, and the cylinder itself **does not slip or topple over**. The gravitational acceleration is given as \\( g \\). If the thin cylinder is fixed on the ground, find the **minimum value** of the coefficient of friction between the ball and the cylinder.", "solution": "", "answer": "" }, { "id": 733, "tag": "MODERN", "content": "When a photon is scattered by an electron, if the initial electron has sufficient kinetic energy such that energy is transferred from the electron to the photon during the scattering process, this scattering is called inverse Compton scattering. Inverse Compton scattering occurs when a low-energy photon and a high-energy electron collide head-on. The rest mass of an electron is known to be $m_{e}$, and the speed of light in vacuum is $c$. If an electron with energy $E_{e}$ and a photon with energy $E_{y}$ collide head-on, If the incident photon energy is $2.00\\mathrm{eV}$ and the electron energy is $1.00\\times10^{9}~\\mathrm{eV}$, find the energy of the photon after scattering. It is known that $m_{e}=0.511\\times10^{6}~\\mathrm{eV}/c^{2}$. When necessary in calculations, approximations such as $\\sqrt{1-x} \\approx 1 - \\frac{1}{2}x$ can be used, and relativistic effects should be considered.", "solution": "", "answer": "" }, { "id": 298, "tag": "MODERN", "content": "People discovered that Mercury's orbit is constantly precessing, rather than following the fixed orbit predicted by Newton's laws. This problem was discovered in 1859, and for 50 years people were unable to provide a satisfactory solution. At the beginning of the 20th century, Einstein's special theory of relativity emerged, offering new insights for explaining Mercury's precession. M represents the mass of the sun (considered stationary), $\\mathrm{m}$ represents the rest mass of Mercury, E represents the total energy, L represents Mercury's angular momentum, G is the gravitational constant, and c is the speed of light. Under special relativity, inertial mass and gravitational mass are not equivalent; gravitational mass equals rest mass. \n\nPlease derive the precession angle $\\Delta\\theta_{t1}$ for one period based on special relativity.", "solution": "", "answer": "" }, { "id": 558, "tag": "ELECTRICITY", "content": "In 1917, Stewa and Tolman discovered that for a closed coil wound around a cylinder, when the cylinder rotates about its central axis with a constant angular acceleration, a current will flow through the coil. \n\nThere is now a coil with many turns, each turn having a radius of \\( r \\), and each turn is wound with a thin metal wire having a resistance \\( R \\). The coil is uniformly wound around a very long hollow glass cylinder, and the inside of the cylinder is a vacuum. Each turn of the coil is fixed on the cylinder with adhesive, with \\( n \\) turns per unit length, and the plane of the coil is perpendicular to the central axis of the cylinder. Starting from a certain moment, the cylinder and coil rotate about the central axis of the cylinder with a constant angular acceleration \\( \\beta \\). Given the electron mass \\( m \\) and elementary charge \\( e \\). We assume that only electrons can move freely, while positive ions cannot.\n\nFind the magnetic induction \\( B \\) at the central axis of the cylinder after a sufficiently long time.", "solution": "", "answer": "" }, { "id": 146, "tag": "MECHANICS", "content": "Three perfectly elastic balls are initially arranged in a straight line at rest. Striking the first ball with mass $m_1$ gives it a velocity of $v_{1}$ along the line, causing it to collide with the second ball of mass $m_2$, which in turn collides with the third ball of mass $m_3$. To maximize the velocity of the third ball after the collisions, given $m_{2} \\neq m_{3}$, what should the mass of the second ball be?", "solution": "", "answer": "" }, { "id": 283, "tag": "OPTICS", "content": "When a strong laser beam shines on a semi-transparent plate, due to uneven heating of the material, the transmitted light can self-focus to a point behind the plate. This effect is known as the thermal lens effect. In materials where the refractive index increases with temperature, this effect is characterized by a positive thermo-optic coefficient $\\gamma = \\mathrm{d}n/\\mathrm{d}T$. \n\nA semi-transparent disk, with radius $a$, thickness $b$, and light absorption rate $A$, is composed of material with thermal conductivity $k$ and thermo-optic coefficient $\\gamma$. The outer edge of the disk is in thermal contact with a circular metal frame (not shown in the figure), which is maintained at a constant temperature $T_{h}$. A parallel laser beam with radius $\\sigma$ and power $P_{L}$ vertically irradiates the center of the disk. The intensity distribution of the beam across its cross-section is uniform.\n\nUnder steady-state conditions, neglecting thermal convection and thermal radiation, the laser beam will focus at a point. Find the distance $f$ from this point to the disk.", "solution": "", "answer": "" }, { "id": 690, "tag": "MODERN", "content": "In the rotational spectroscopy analysis of diatomic molecules, we usually consider the two constituent atoms of the molecule as point masses, assuming their masses are respectively ${m}_{1}, m_{2}$, and that the connection between the two atoms is rigid, assumed to be at a distance $r$ apart. The quantization condition for angular momentum is given by: $$ \\mathrm{L}^{2}=l(l+1)\\hbar^{2} $$ where $L$ is the angular momentum, and $l$ is any positive integer. Knowing that transitions occur only between two adjacent energy levels, try to find the frequency of the photon absorbed during the transition from the $l$-th level to a higher energy level in the rotational spectrum of this molecule.", "solution": "", "answer": "" }, { "id": 509, "tag": "MECHANICS", "content": "A cylindrical rod with a radius of $R$ is fixed to a vertical wall (its base is fixed to the wall, and the axis of the cylinder is perpendicular to the vertical wall). The length of the cylinder is $L$ ($R \\ll L$). The cylinder has a mass $m$, Young's modulus $E$, and the effect of gravity is neglected. A torque $M$ is applied at the right end of the cylinder, causing the right end to experience a very small displacement $d$ ($d \\ll L$) downward.\n\nAfter removing $M$, assume that during vibration, the shape of the cylinder satisfies $y=\\frac{\\theta x^2}{2L}$, where $x$ represents the distance from a point on the axis of the undeformed cylinder to the wall, $y$ represents the vertical displacement of the point at position $x$ from its equilibrium position, and $\\theta$ is the central angle corresponding to the approximate circular arc formed by the deformation of the cylinder. Determine the period of small vibrations of the cylinder.", "solution": "", "answer": "" }, { "id": 435, "tag": "MECHANICS", "content": "When studying the motion of viscous fluids, when the flow velocity is not large, the motion of viscous fluids can be divided into a layer-by-layer motion. There is relative sliding between these layers, with viscous resistance acting upon each other. This type of motion is called laminar flow. Due to the viscous resistance between the layers, the distribution of flow velocity and the magnitude of the flow rate will be affected. As shown in the figure, there is a horizontally placed cylindrical tube with uniform diameter. The radius of the cylindrical tube is $r$, and the length of the tube is $L$. The tube undergoes uniformly accelerated linear motion to the right, with an acceleration of $a$. In the tube, there is an incompressible viscous fluid undergoing laminar flow. The fluid density is $\\rho$. The left end of the fluid block is subjected to a pressure $P_{1}$, and the right end to a pressure $P_{2}$ ($P_{1}>P_{2}$). The velocity is greatest at the center of the tube, and as the distance from the central axis increases, the flow velocity gradually decreases. At the point adjacent to the tube wall, the fluid velocity is zero due to adherence to the wall surface.\n\nSupplementary Knowledge:\n\nNewton's law of viscosity states that the magnitude of the viscous force between two adjacent layers within a fluid is proportional to the contact area $A$ between the two layers and is also proportional to the velocity gradient at the contact surface of the two layers $\\frac{\\mathrm{d}\\nu}{\\mathrm{d}r}$, namely\n\n$$\nf=\\eta A{\\frac{\\mathrm{d}\\nu}{\\mathrm{d}r}}\n$$\n\nUsing the horizontal cylindrical tube as the reference frame, calculate the flow velocity $\\nu(y)$ at a distance $y$ from the axis for the incompressible viscous fluid.", "solution": "", "answer": "" }, { "id": 309, "tag": "MECHANICS", "content": "Precession refers to the phenomenon where the axis of rotation of a spinning rigid body rotates around a certain center due to the influence of external forces. A round coin rolls purely on a horizontal surface, and under certain conditions, the homogeneous coin will move with an inclined state in a looping motion, maintaining an angle $\\theta$ between its plane and the horizontal plane, and precess uniformly around a vertical axis. The center of the coin is O.Assume the coin precesses uniformly around a vertical axis passing through point $S$ (with $S$ at the same height as point $O$), the gravitational acceleration is $g$, the coin's mass is $m$, and the radius is $r$. The radius of the coin's looping motion is $R$. Answer the following question. \nFind the expression for the precession angular velocity of the coin, $\\omega_{pr}$.", "solution": "", "answer": "" }, { "id": 474, "tag": "MECHANICS", "content": "Please solve the following physics problem. Use \\boxed{} to enclose the final answer: From the ceiling, a uniform rod is hinged vertically downward. The rod has a length of $l$ and a mass of $m$. A second identical rod is hinged to the end of the first rod. The entire system is subject to vertical downward gravity with an acceleration of $g$. The double-rod system is evidently stable, so in the case of small oscillations, it can perform simple harmonic motion near the equilibrium position with a certain natural frequency. In the eigenmode, the angles $\\theta_{1}$ and $\\theta_{2}$ between the rods and the vertical direction have a common angular frequency. Find the larger angular frequency of such a system.", "solution": "", "answer": "" }, { "id": 236, "tag": "MECHANICS", "content": "As one of the most common means of transportation, bicycles are renowned for their excellent stability and portability and have long been widely popular among the public. Even today, with transportation methods and technology rapidly changing, bicycles are still considered the most labor-saving and superior non-motorized means of transportation and are highly regarded for their outstanding environmental protection benefits. Early research on bicycles focused on gear coupling, pedal acceleration, and similar processes, but the study of bicycle control stability has been challenging. Why does a bicycle with two wheels remain stable while moving and not topple over? For over a hundred years, this question has intrigued many famous mechanicians, physicists, and mathematicians, resulting in more than a hundred renowned papers published in various languages such as English, German, French, Russian, and Italian. In 1897, the French Academy of Sciences even offered a reward for this research. Today, the study of the control and riding stability of bicycles with different structures and parameters continues to be a hot topic in the engineering field.\n(In this problem, the rider does not hold the handlebar and imparts no force on it)\nA ridden bicycle can ensure its own stability and not fall over. Based on our own experience, the faster the speed, the better the stability. However, is it possible to attach some devices to the bicycle to make it stable and upright even when stationary? Below, we provide a schematic of a bicycle structure with an attached momentum wheel, where a brushless motor is fixed to the bicycle body, connected to a momentum wheel, and can be driven by the motor to achieve any desired angular velocity.\n\nNext, we simplify this system. Since we are now studying the upright stability of the bicycle in a non-moving state, we can consider the front and rear wheels, body, and motor as a rigid body, abstracted as a homogeneous rigid rod that can only rotate in the plane around the contact point with the ground. The rod has a length $L$, mass $M$, and the momentum wheel connected to the rod's end can be driven by the motor to achieve any angular velocity. The momentum wheel is viewed as a disk with mass $m$ and radius $r$. (Ignore lateral slip at point $O$)\n\nIf the bicycle body obtains a clockwise rotation angle $\\theta$ , the angular velocity that the previously stationary momentum wheel subsequently needs to acquire is \n\nIn fact, when the bicycle tilt angle changes, the angular velocity and angular acceleration of the momentum wheel should also change correspondingly. We assume that sensors installed on the bicycle body can record the tilt angle $\\theta$ and feedback to the motor, allowing the momentum wheel to acquire an angular acceleration of size $\\dot{\\omega}=K\\theta$. To bring the bicycle back to a balanced position, solve for the dependency of $\\omega$ on time $t$. ( $\\theta<<1$, ignore any resistance, at $t=0$, $\\theta=\\theta_{0},\\dot{\\theta}=0$, and the angular velocity of the momentum wheel is 0)", "solution": "", "answer": "" }, { "id": 557, "tag": "ELECTRICITY", "content": "Two conductor spheres with equal radius \\( R \\) come into contact to form an isolated conductor. The distance between the centers of the two spheres is exactly \\( 2R \\). Find their capacitance \\( C \\).", "solution": "", "answer": "" }, { "id": 722, "tag": "MODERN", "content": "The two identical long straight wires 1 and 2, which are placed vertically and fixed parallel, are spaced apart by a distance of a ($0 << a$, much smaller than the length of the wires). Both wires carry a steady current of the same direction and magnitude, with the current flowing upwards. The positive ions in the wires are stationary, and the charge of the positive ions per unit length of the wire is $\\lambda$; the conduction electrons forming the current move uniformly downwards along the wire at a speed of $\\boldsymbol{\\mathrm{\\Delta}}v_{0}$, with the charge of the conduction electrons per unit length of the wire being $-\\lambda$. It is known that an infinitely long uniformly charged straight wire with a charge per unit length of $\\eta$ produces an electric field of magnitude $E=k_{e}\\frac{2\\eta}{r}$ at a distance $r$ from it, where $k_{e}$ is a constant; and when an infinitely long straight wire carries a steady current $I$, the magnetic field strength generated at a distance $r$ from the wire is $B=k_{m}\\frac{2I}{r}$, where $k_{m}$ is a constant. Use the length contraction formula from special relativity to find the ratio of the constants $k_{e}$ and $k_{m}$. Hint: Ignore gravity; the charge of positive ions and electrons is independent of the choice of inertial reference frame; the speed of light in a vacuum is $c$.", "solution": "", "answer": "" }, { "id": 569, "tag": "ELECTRICITY", "content": "On an infinitely large superconducting plate with a friction coefficient of $\\mu$ placed horizontally on the ground, there is a uniformly charged insulating spherical shell with a total charge of $Q$, radius $R$, and mass $M$. Fixed smooth panels are positioned on both sides of the spherical shell, ensuring that its center of mass does not move in the horizontal direction. It is known that during subsequent motion, the sphere will not leave the superconducting plate. Initially, the sphere is given an angular velocity of $\\omega_{0}$.\n\nDetermine the relationship between the angular displacement and angular velocity of the sphere following this setup. Express the angular displacement $\\theta$ in terms of the instantaneous angular velocity $\\omega$, initial angular velocity $\\omega_{0}$, and physical quantities such as $M$, $Q$, $R$, $\\mu$, and $g$.\n\n(Hint: Consider the electric and magnetic images created by the superconducting plate.)", "solution": "", "answer": "" }, { "id": 779, "tag": "MECHANICS", "content": "In this problem, we are given a zero-natural-length elastic rope with a constant elastic coefficient \\( k \\), a natural length of 0, and uniformly distributed mass (it can be considered that the natural length is almost 0). The total mass is \\( m \\), and it is suspended in Earth's gravitational field. We are to determine its shape under the following conditions. \n\nShape of a Zero-Natural-Length Elastic Rope in a Gravitational Field:\nIf the rope is close to the Earth's surface and its stretched length is much smaller than the Earth's size, the gravitational acceleration is \\( g \\), and the suspension points are at the same height with a spacing of \\( D \\), determine the shape of the rope. \n(For this question, it is recommended to set up a Cartesian coordinate system where the rope passes through the points (0, 0) and (D, 0).) \n", "solution": "", "answer": "" }, { "id": 418, "tag": "ELECTRICITY", "content": "An infinitely long cylindrical conductor shell with a radius of $R$ carries a positive charge of $+\\lambda$ per unit length. It is placed in front of an infinitely large grounded conducting plate, with the distance from the axis to the conductor plate being $d$. Given the vacuum permittivity constant as $\\varepsilon_{0}$, we establish a rectangular coordinate system with the foot of the perpendicular $O$ as the origin. The direction opposite to the normal vector of the infinite plane points inward to the plane, defined as the z-direction. The direction parallel to the cross-section of the cylinder on the infinite plane is the y-direction, and the direction perpendicular to the cross-section is the x-direction. In the $xOy$ plane, find the distribution of the charge density $\\sigma(y)$ on the conducting plate.", "solution": "", "answer": "" }, { "id": 247, "tag": "MECHANICS", "content": "A uniform rectangular thin plate oscillates around one of its edges, where this edge makes an angle $\\theta$ with the vertical direction, and the length of the other side is $b$. Given the gravitational acceleration $g$, find the period of small oscillations near the equilibrium position.", "solution": "", "answer": "" }, { "id": 299, "tag": "MECHANICS", "content": "Two infinitely long smooth rails intersect at point $O$, and there are two blocks with mass $m$ each on the rails. When passing through point $O$, the blocks do not collide with the other rail and pass directly through. One block on a rail is initially at a distance $b$ from point $O$ and remains stationary. The block on the other rail is initially far enough from point $O$ and slides towards $O$ with an initial velocity magnitude $v_{0}=\\lambda\\sqrt{G m/b}$. Consider the gravitational force between the two blocks, but ignore the gravitational force from the rails and gravity. Question: What is the final velocity magnitude of the block that initially moves with velocity $v_{0}$ after a sufficiently long time?", "solution": "", "answer": "" }, { "id": 539, "tag": "MECHANICS", "content": "**Transverse Waves on a String**\n\nIn this problem, we describe waves using the negative phase convention. In this convention, the most general form of a traveling wave is expressed as \n\n$$\n\\psi(x,t)=A e^{i(\\pm k x-\\omega t)}\n$$ \n\nwhere $\\omega$ is the angular frequency, $k$ is the magnitude of the wave vector, the $\"+\"$ sign represents a wave traveling to the right, and the $\"-\"$ sign represents a wave traveling to the left. $A$ is the complex amplitude (its magnitude corresponds to the amplitude, while its phase angle contains the information about the phase). Next, we will consider transverse waves on a string, whose wave equation is \n\n$$\n\\frac{\\partial^{2}\\psi}{\\partial x^{2}}-\\frac{1}{u^{2}}\\frac{\\partial^{2}\\psi}{\\partial t^{2}}=0\n$$ \n\nwhere $u$ is the wave velocity. When $k$ and $\\omega$ in the traveling wave expression satisfy $\\omega/k=u$, it is a solution to the wave equation above. Due to the linearity of the wave equation, the superposition of solutions to the traveling wave is also a solution to this equation. \n\nBoth ends A and B of a string are fixed to supports. We establish a coordinate axis with the midpoint (point O) as the origin and the positive direction of the $x$-axis pointing to the right. Thus, the coordinates of endpoints A and B are $±l/2$, where $l$ is the length of the string. At a point with a coordinate of $a$ on the string, a driver C with an angular frequency of $\\omega$ is installed, causing all mass elements on the string to oscillate with an angular frequency of $\\omega$. \n\nLet the amplitude reflection coefficient at the string's connection point with the supports be $\\boldsymbol{r}$, which is defined as the ratio of the complex amplitude of the reflected wave to that of the incident wave at the reflection point. Assume that the incident wave between C and B is \n\n$$\nA e^{i(k x-\\omega t)}\n$$ \n\nWe can introduce an inhomogeneous term into the wave equation to model the effect of the driver $C$: \n\n$$\n\\frac{\\partial^{2}\\psi}{\\partial x^{2}}-\\frac{1}{u^{2}}\\frac{\\partial^{2}\\psi}{\\partial t^{2}}=F\\delta(x-a)e^{-i\\omega t}\n$$ \n\nHere, $F$ characterizes the amplitude of the driving force. The function $\\delta(x-a)$ is a delta function with the following properties: it exhibits a very narrow, tall pulse centered near ${\\boldsymbol{x}}={\\boldsymbol{a}}$, with the area under the pulse curve equal to 1. Using this property, we can write the relationship satisfied by the wave functions $\\psi_{1}$ and $\\psi_{2}$ to the left and right of C. This relationship is also known as the continuity condition. \n\nFinally, express $A$ in terms of $F$, $k$, $l$, and $a$.", "solution": "", "answer": "" }, { "id": 652, "tag": "MECHANICS", "content": "A bottle with a height of $H$ is filled with water. One side of the bottle is uniformly distributed with small holes, with a number of holes per unit length being $n$. The area of each hole is $s$ (but its dimension is much smaller than $H$). Each hole sprays water outward, and it is assumed that the direction of water sprayed from the holes is along the horizontal direction. The decrease in water surface height is not considered during the spraying process. Find the amount of water hitting the ground per unit time and per unit length at a distance $x$ from the bottle.\n", "solution": "", "answer": "" }, { "id": 436, "tag": "ADVANCED", "content": "There is a mechanism that simulates a helicopter rotor, but it has only one blade. One end of the blade is hinged at a fixed rod $O$ and can rotate around the fixed point $O$ in the plane $β$ formed by the rod and the blade's center of mass. The center of mass is denoted as point $C$. The mass of this rigid body is $m$, and the distance $OC$ is $r$. A three-dimensional Cartesian coordinate system is established at the center of mass $C$. The $OC$ direction is designated as the $y$-axis, the axis perpendicular to the $y$-axis and within plane $β$ is the $z$-axis, and the axial direction is denoted as the $x$-axis. The principal moments of inertia of the rigid body around the three axes are $J_x, J_y, J_z$. Assuming the fixed axis rotates with a constant angular velocity $\\Omega$, determine the angle between $OC$ and the vertical rod at stable equilibrium (by default, take the acute angle).", "solution": "", "answer": "" }, { "id": 160, "tag": "MECHANICS", "content": "A smooth horizontal plane rotates uniformly with angular velocity $\\omega$ around a fixed axis passing through point $O$, and the axis is perpendicular to this plane. There is a particle $P$ with mass $m$ on the plane which is attracted by point $A$. The distance between $A$ and $O$ is $c$, and the gravitational force is proportional to the distance from the particle to point $A$, with the proportional constant being $4\\omega^{2}m$. At $t=0$, the particle is launched from $\\left({\\frac{8}{3}}c,0\\right)$ with velocity ${\\frac{4}{3}}c\\omega$ perpendicular to the $OA$ axis. Determine the trajectory equation $r(\\theta)$ of the particle relative to the fixed coordinate system.", "solution": "", "answer": "" }, { "id": 732, "tag": "THERMODYNAMICS", "content": "Consider a thermodynamic cycle composed of a monoatomic ideal gas, which undergoes the following four processes on the \\(p\\)–\\(V\\) diagram: \n- **Process A→B:** Isobaric expansion at pressure \\(rp\\) (where \\(r>1\\)), with temperature rising from \\(T_C\\) to \\(T_H\\); \n- **Process B→C:** Isothermal expansion at temperature \\(T_H\\); \n- **Process C→D:** Isobaric compression at pressure \\(p\\); \n- **Process D→A:** Isothermal compression at temperature \\(T_C\\). It is known that when the gas absorbs heat, it is always in contact with the heat reservoir at temperature \\(T_H\\), and when it releases heat, it is always in contact with the heat reservoir at temperature \\(T_C\\). In the above thermodynamic cycle, it is known that the efficiency of an ideal Carnot engine working between a high-temperature reservoir \\(T_H\\) and a low-temperature reservoir \\(T_C\\) is \\[ e_C=1-\\frac{T_C}{T_H}. \\] Please write the expression for the ratio \\(\\frac{e}{e_C}\\), where \\(e\\) is the efficiency of the cycle engine, using \\(T_C, T_H, r\\) to express your answer.", "solution": "", "answer": "" }, { "id": 486, "tag": "OPTICS", "content": "A perfect plano-convex thin lens is placed in a vacuum, with a refractive index of \\(n\\). A broad parallel light beam is incident perpendicularly on the flat side of the lens, and the transmitted light rays are strictly focused at point \\(F\\), where the focal length is \\(f\\). A parallel light beam with a cross-sectional diameter of \\(D\\) is incident perpendicularly on one side of the lens, with its axis aligned with the principal axis of the lens. Assume the incident light intensity is \\(I\\), and neglect the reflection and absorption of light by the lens. Calculate the force exerted by the light field on the lens.", "solution": "", "answer": "" }, { "id": 366, "tag": "ELECTRICITY", "content": "In a uniform magnetic field $\\vec{B}=B_0 \\hat{z}$, an ideal conductive spherical shell with a radius of $a$ rotates around the z-axis at an angular velocity $\\omega$. Find the electromotive force between its north pole and equator.", "solution": "", "answer": "" }, { "id": 276, "tag": "ELECTRICITY", "content": "The strict formulation of Green's reciprocity theorem is as follows: When the charges on $n$ conductors are $Q_{1}, Q_{2}, \\ldots, Q_{n}$, the potentials on each conductor are $\\varphi_{1}, \\varphi_{2}, \\ldots, \\varphi_{n}$; if the charges on the conductors are reset to $q_{1}, q_{2}, \\ldots, q_{n}$, the corresponding potentials change to $\\varphi_{1}^{\\prime}, \\varphi_{2}^{\\prime}, \\ldots, \\varphi_{n}^{\\prime}$, then the following equation holds:\n\n$$\n\\sum_{i=1}^{n} Q_{i} \\varphi_{i}^{\\prime} = \\sum_{i=1}^{n} q_{i} \\varphi_{i}\n$$\n\nFor an ellipsoidal conductor rotating around the minor axis, take the minor axis as the polar axis. At a distance from the center $r, r > a$, at an azimuthal angle $\\theta$, a charge with an amount of $q$ is placed, and the conductor is grounded. Try to solve for the induced charge $q_{a}^{\\prime}$ on the conductor in the stable state (to simplify the answer, introduce $\\begin{array}{r} {\\tan \\alpha_{0} = \\frac{2r c \\cos \\theta}{r^{2} - c^{2}}} \\end{array}$ to replace $r^{2} - c^{2}$).", "solution": "", "answer": "" }, { "id": 296, "tag": "MECHANICS", "content": "A smooth steel wire is bent into a semicircle with a radius of $R$ and placed vertically. Two particles are placed on the wire, with weights $P$ and $Q$ respectively. The two particles are connected by a lightweight, inextensible rope of length $2L$. Find the angle $\\theta$ that the rope makes with the horizontal plane at equilibrium.", "solution": "", "answer": "" }, { "id": 531, "tag": "THERMODYNAMICS", "content": "# 3. Asymmetric Adiabatic Process\n\nInside a horizontal cylinder, there is an enclosed piston. The piston is connected to a handle, allowing us to control the volume of the container. Initially, saturated water vapor at temperature $T_{0}$ is contained within the cylinder, with no liquid water present.\n\nAssume that the water vapor can be regarded as an ideal gas made of polyatomic molecules. At temperature $\\scriptstyle{T_{0}}$, the heat of vaporization of water is $L$, and for the purposes of this problem, it is assumed that $L$ is independent of temperature. The universal gas constant is $R$. The molar mass of water is $\\mu$.\n\nIt is well known that when there is a slight change in temperature around $\\scriptstyle{T_{0}}$, the relative change in saturated vapor pressure and the relative change in temperature satisfy the following relationship:\n\n$$\n\\varepsilon_{p}=\\frac{\\Delta p}{p_{0}}=\\alpha\\varepsilon_{T}=\\alpha\\frac{\\Delta T}{T_{0}}\n$$\n\nNext, assume the container and piston are adiabatic. When the volume of the container slowly increases so that its relative change reaches $\\beta$, find the change in temperature $\\Delta T_{2}$.", "solution": "", "answer": "" }, { "id": 286, "tag": "MECHANICS", "content": "Two concentric homogeneous steel wire loops are held by a soap film in the middle. The mass, radius, and surface tension coefficient of the soap film are $m_{1}$, $m_{2}$, $r_{1}$, $r_{2}$, and $\\sigma$. There is no gravity in space. The problem is:\n\nIf the ambient temperature changes, one physical fact is that the surface tension coefficient of the soap film will not remain constant. Assume that the two loops are initially given very small velocities along the axial direction, $v_1$ and $v_2$. During the subsequent motion of the system, the surface tension coefficient of the soap film very gradually increases from $\\sigma$ to $2\\sigma$ and then remains unchanged. Calculate the maximum kinetic energy of the system during the final stable motion. Retain terms up to the second order of $(v_2 - v_1)$.", "solution": "", "answer": "" }, { "id": 408, "tag": "MECHANICS", "content": "Sometimes, a golf ball approaches the edge of the hole, rolls along the wall of the hole while moving downward, and then bounces out of the hole. We examine this absurd motion.\n\nConsider a sufficiently rough ball rolling without slipping on the inner wall of an infinitely deep cylindrical hole. The mass of the ball is $m$, its moment of inertia is $I$, its radius is $a$, and the radius of the hole is $b > a$. The point of contact $P$ between the ball and the wall has an angular coordinate $\\theta$ in cylindrical coordinates. The unit normal vector at point $P$ on the wall is denoted by $\\hat{n}$, and gravity $g$ is directed along the $-\\hat{z}$ axis, which is also the axis of the hole. The friction between the ball and the wall is sufficient to ensure rolling without slipping, meaning the velocity at the point of contact is $0$, but it is not sufficient to eliminate the rotation perpendicular to the wall, i.e., $w_{n} = \\vec{\\omega} \\cdot \\hat{n}$ does not vanish.\n\nAssuming the initial precession angular velocity of the ball is $\\omega_{0}$, i.e., $\\dot{\\theta}(0) = \\omega_{0}$, determine the period $T$ of the ball's simple harmonic motion in the vertical direction.", "solution": "", "answer": "" }, { "id": 382, "tag": "MECHANICS", "content": "Uranus is a spherically symmetric planet with uniform mass distribution, density $\\rho_{1}$, and radius $R$. In the cosmic space between distances $R$ to $2R$ from the planet's center, cosmic dust with a uniform density of $\\rho_{2}$ is distributed.\n\nBefore leaving Uranus, a probe adjusts its orbit and performs uniform circular motion at a distance of $\\frac{3}{2}R$ from the center. Ignoring all drag caused by collisions with the dust, if the probe experiences a small radial impulse, determine the precession angular velocity of its orbit.", "solution": "", "answer": "" }, { "id": 715, "tag": "ELECTRICITY", "content": "In the $xy$ plane, there is a fixed charged ring with radius $R$, on which the charge $Q > 0$ is uniformly distributed. At the center of the ring, a point charge with mass $m$ and charge $q > 0$ is placed. The point charge is constrained to move along a smooth tube that passes through the center of the ring and forms an angle $\\theta$ with the $z$-axis. Now, move the point charge along the tube to a distance $A$ away from the center ($A \\ll R$). Neglect the effects of gravity, and let the electrostatic constant be $k$. Given that $\\theta > \\arctan{\\sqrt{2}}$, find the oscillation period of the point charge (expressed in terms of $Q$, $q$, $m$, $k$, $R$, and $\\theta$).", "solution": "", "answer": "" }, { "id": 692, "tag": "ELECTRICITY", "content": "There is an ideal electret dielectric with intrinsic polarization strength $P_0$, which can also be polarized by an internal electric field. The equation of state is $\\vec P=\\varepsilon _0\\chi\\vec E+\\vec P_0$. Place this dielectric sphere in a vacuum, with an external uniform electric field $E_0$ (which can be oriented differently from $P_0$). Find the distribution of bound surface charge. Use $\\cos\\theta = \\cos<\\vec r, \\vec E_0>, \\cos \\phi = \\cos<\\vec r, \\vec P_0>$ and the symbols provided in the problem to express the answer.", "solution": "", "answer": "" }, { "id": 322, "tag": "ELECTRICITY", "content": "This problem does not consider relativistic effects.\n\nAn infinitely long leaky dielectric cylinder, with a radius of $R$ and an absolute permittivity of $\\varepsilon$, initially stationary and overall uncharged. The carriers have a very small mass $m \\to 0$, a charge of $q$, and a mobility of $\\mu$ (i.e., the drift velocity of the carriers generated under a uniform electric field $E$ is $v=\\mu E$). The number denstiy of the carriers is $n$, Within the cylinder, a cylindrical coordinate system $(z,r,\\theta)$ is established, with the origin at the center of the circular top surface of the cylinder, the z-direction pointing axially outwards from the cylinder, and r pointing radially outwards. Throughout this process, the volume charge density $\\rho$, and the current densities $j_{r}, j_{\\theta}$ within the cylinder will vary with time $t$ and are functions of the spatial coordinates $r, \\theta$. Due to translational symmetry in the z-axis direction, the current $j_{z}$ in the z-direction is always zero, and all non-zero physical quantities are independent of the coordinate $z$.\n\nFirst, keep the cylinder stationary and slowly apply a uniform magnetic field $B$ along the cylinder's axis. Next, maintain the uniform magnetic field unchanged and suddenly apply a resistive torque to bring the cylinder to a complete stop, and marking this moment as $t=0$. Note that during this process, while the charges do not have time to transfer, the carriers (still due to the mass $m\\rightarrow0$) will change their drift velocity under the new electromagnetic forces, thereby generating a new current distribution $j_{r}(t=0)$, $j_{\\theta}(t=0)$. Find the current vector $\\vec{j}$ within the cylinder at $t=0$.", "solution": "", "answer": "" }, { "id": 191, "tag": "MECHANICS", "content": "In 1975, in the graduate admissions exam of the Department of Physics at the University of Wisconsin, USA, there was a seemingly simple hard ball collision problem: Two solid spheres made of the same material, the radius of the lower one is $2a$, and the radius of the upper one is $a$. They fall from a height of $h$ above the ground (measured from the center of the larger sphere). Assuming the centers of both spheres are always on a vertical line and all collisions are elastic, what is the maximum height that the center of the upper sphere can reach? (Hint: assume the larger sphere first collides with the ground and bounces up before colliding with the smaller sphere.)\n\nWe do not know what the answer from the examiner was. However, when this question was included in a certain book, its solution assumed that after the larger sphere falls and hits the ground, it rebounds and collides with the smaller sphere that is still in a downward trajectory. The maximum height that the smaller sphere can reach upon rebounding and rising is the maximum height achievable by the smaller sphere. Request to solve for the maximum height $h_{1}$ of the center of the smaller sphere during its first rebound and ascent.", "solution": "", "answer": "" }, { "id": 749, "tag": "ELECTRICITY", "content": "In space, there are four infinitely large charged planes located at $x = a$, $x = -a$, $y = a$, and $y = -a$. The intersection of each plane with the $x$-axis or the $y$-axis serves as the center point of the plane. The charge distribution on all four planes is identical: at a distance $ \\rho$ from the plane's center point, the surface charge density is $ \\sigma = \\frac{ \\sigma_{0}}{(1+ \\alpha^{2} \\rho^{2})^{ \\frac{3}{2}}}$, where $ \\alpha > 0$ and $ \\sigma_{0} > 0$. At the origin of coordinates, a particle with charge $q$ (where $q < 0$) and mass $m$ is placed. If the particle is constrained to move along the $z$-axis, determine the frequency of small oscillations $ \\omega$ of the particle near the origin.", "solution": "", "answer": "" }, { "id": 290, "tag": "THERMODYNAMICS", "content": "The Boltzmann constant is $k$. \n\nImagine a general potential energy existing in space, which consists of positive power laws $r$ for three directions:\n\n$$\nE_{p}=\\alpha(|x|^{r}+|y|^{r}+|z|^{r})\n$$ \n\nA classical ideal gas with temperature $T$ exists in this potential energy. Find the average potential energy $\\langle E_{p}\\rangle$ of each molecule.", "solution": "", "answer": "" }, { "id": 187, "tag": "MECHANICS", "content": " Under the gravitational influence of the moon and sun, the phenomenon of the sea's periodic rise and fall twice a day is called tides. To simplify the model, we will ignore Earth's rotation, and this question only discusses the tidal phenomenon in the Earth-Moon system. Let the mass of the moon be $m$, the gravitational acceleration on the Earth's surface be $g$, the Earth's radius be $R$, the distance from the Earth-Moon barycenter be $r_{\\mathrm{m}}$, and the gravitational constant be $G$. Assume that the Earth can be modeled as a rigid sphere with uniformly distributed mass, covered by a layer of seawater. The height of this seawater layer is negligible compared to the mass of Earth and the seawater, in comparison to the mass of the rigid Earth. The seawater covering the Earth's surface will become ellipsoidal. We establish a coordinate system $O x y z$ with the Earth's center of mass as the origin, where the $z$-axis is along the Earth-Moon line. Let $\\theta$ be the angle with the $z$-axis. Provide an expression for the maximum amplitude of the tidal rise and fall.\n", "solution": "", "answer": "" }, { "id": 315, "tag": "ELECTRICITY", "content": "In a vacuum, two parallel conductor plates with an area of $S$ are separated by a distance of $l_{0}$ and face each other to form a capacitor. The two plates are connected by a spring with an unstretched length of $l_{0}$, a cross-sectional area of $A$, and $N$ turns. The spring has a stiffness coefficient $k$. The spring is conductive and can also be treated as an inductive component. Neglect edge effects $\\langle\\sqrt{S}\\gg l_{0}\\gg\\sqrt{A}\\rangle$ and radiation effects, and assume that the stress and strain in the spring are always uniformly distributed. \nIf the two plates are initially charged at $t=0$ with $\\pm Q_{0}$, and the current through the spring is zero, then if an external force $F(t)$ (positive outward) is applied to keep the plates stationary in their initial position, the charge on the plates will vary periodically over time. \nNow, the external force is very slowly reduced to zero. During this process, the plate separation can be considered fixed over short periods of time but will change over a long period from the initial length $l_{0}$ to a final length $l$. Find the value of $l$.", "solution": "", "answer": "" }, { "id": 624, "tag": "THERMODYNAMICS", "content": "Given that the ambient temperature is $T$, the atmospheric pressure is $P_0$, and the initial pressure inside a tire is $P_i$ (which is greater than the atmospheric pressure), with the temperature remaining at the ambient temperature. The maximum pressure in the tire right after inflation is $P_{max}$, with the temperature at that moment being $T_{max}$ (both $P_{max}$ and $T_{max}$ are unknown). After cooling slowly back to the ambient temperature, the measured pressure is $P_f$. Assumptions: 1. Air is an ideal gas with a heat capacity ratio $\\gamma$. 2. The volume of the tire remains constant. 3. The inflation process is adiabatic. A small hand pump is used to inflate the tire. The pump's volume is much smaller than that of the tire, and the initial pressure of the gas inside is the atmospheric pressure. Due to the small volume of the pump, many strokes are required for inflation. We assume the inflation process is as follows: the air in the pump is first adiabatically compressed, increasing the pressure from the atmospheric pressure $P_0$ to the current pressure in the tire, and then the air is introduced into the tire at constant pressure. Replace discrete summation with integration to derive the expression for the tire pressure $P_{max}$ right after inflation using the parameters given in this problem: $P_0, P_i, P_f, T, \\gamma$.", "solution": "", "answer": "" }, { "id": 190, "tag": "MECHANICS", "content": "The Blue Star people have always fantasized about one day being able to abandon the inefficient transportation method of rockets and directly establish a \"sky ladder\" from the ground to the heavens. Some call it a \"space elevator,\" while others refer to it as an \"orbital lift.\" However, they do not realize that due to the limitations imposed by physical laws, building a space elevator on the Blue Star is unrealistic. In contrast, several years later, on Mars, which is farther from the sun, space elevators can \"stand tall.\"\n\nAs early as the Book of Genesis in the Bible, humans hoped to jointly construct a towering tower that reaches from the ground to the sky—the Tower of Babel—to spread their fame. The person who first proposed the concept of a space elevator should be Russia's \"father of rockets\"—Tsiolkovsky. His envisioned space elevator extends from the ground to the geostationary orbit, tens of thousands of kilometers high. As the elevator rises, the gravity inside gradually decreases. When cargo reaches the endpoint with the elevator, its speed is enough to maintain synchronous orbital movement around the Earth. Therefore, the station at the geostationary orbit is in a completely weightless state.\n\nIn 1979, Arthur C. Clarke's science fiction novel, \"The Fountains of Paradise,\" first brought this advanced concept of the space elevator into the public's view. Since then, space elevators have widely existed in various science fiction works. Whether it's the space elevator made with advanced nanomaterials in \"The Three-Body Problem\" or the three orbital lifts during the three-way division among Union, AEU, and Human Reform League in \"Mobile Suit Gundam 00,\" their basic concepts and principles are the same.\n\nNowadays, we can occasionally hear some related news reports, whether it's the Japanese company \"Obayashi Corporation\" planning to complete the first space elevator by 2050, or a Canadian company proposing a new space elevator plan. Regardless of whether these are just commercial companies engaging in hype, such news does indeed cause more people to wonder—are we really close to the era of abandoning rockets and embracing space elevators?\n\nIs this really the case? How far away are we from the technology required to build space elevators? We will conduct a detailed analysis in this question (the space elevators mentioned below are all built at the equator, and any bending is not considered).\n\nIn fact, whether a material is destroyed or not depends not only on the magnitude of the force but also on the area over which the force acts. This is easy to understand; under the same amount of tension, thinner pipes are more easily pulled apart. Therefore, we need to use \"stress (force divided by the area of action)\" to measure whether a space elevator will be damaged. If the cross-sectional area of the space elevator is uniform, then obviously the place where the force is greatest, in other words, the geostationary orbit area, is the most prone to breaking.\n\nDo we have any way to make the stress inside the entire space elevator equal? Of course we do! We know that the tension increases with height, so the cross-sectional area of an equal strength (internally equal tensile stress) space elevator should also increase with height. Therefore, the optimized space elevator design changes from a uniformly thick \"catenary\" to a funnel-shaped \"catenary\" with a thick top and thin bottom.\n\nIn this question, we use the following model: A mass at the end of the space elevator is considered infinitely large to ensure that only tensile forces act inside the elevator, so there is no force acting at the contact point with the ground. The elevator is thick at the top and thin at the bottom, with each cross-section having an equal tensile stress of σ. The material density is ρ, and the material elongation is not considered. The cross-sectional area at the ground level is \\(A_{0}\\). Solve the following problem:\n\nFind the expression for the cross-sectional area \\(A_{x}\\) at a distance \\(X\\) from the center of the Earth, expressed in terms of \\(g\\), \\(R\\), \\(\\omega_{0}\\) (Earth's rotational angular velocity), \\(\\sigma\\), \\(\\rho\\), and \\(A_{0}\\).", "solution": "", "answer": "" }, { "id": 597, "tag": "ELECTRICITY", "content": "According to the principles of electrochemistry, the spontaneity of redox reactions in aqueous solutions can be determined using the standard electrode potential $E_0$. The standard electrode potential $E_0$ is related to the change in Gibbs free energy of the reaction. Under conditions of constant temperature and pressure, the decrease in the Gibbs free energy of the system is equal to the maximum work $W_{max}$ done by the system. During the course of a battery reaction, the decrease in Gibbs free energy is equal to the maximum electrical work done by the battery. At standard conditions, $-\\Delta G_0 = W_{max}$, where $\\Delta G_0$ represents the change in Gibbs free energy of the chemical reaction under standard conditions. The reaction equation for a silver-zinc battery is $\\mathrm{Ag_2O+Zn=2Ag+ZnO}$, with a Gibbs free energy change set as $\\Delta G_0 = -309.7 \\text{ kJ/mol}$. Since both reactants and products are solids, near standard conditions, the variation of $\\Delta G$ with temperature and pressure can be ignored. There is a type of silver-zinc button cell where the silver positive electrode and the zinc negative electrode are separated by a membrane material soaked with a KOH solution. The concentration of the KOH solution is $c=14 \\text{ mol/L}$, the thickness is $h=0.5 \\text{ mm}$, and the resistivity is $\\rho=2.37 \\times 10^{-4} \\, \\Omega \\cdot \\text{m}$. The cross-sectional area of the battery is $S=0.465 \\text{ cm}^2$. The battery electrodes can be considered as a pair of parallel plate electrodes with a small gap $h$, neglecting edge effects. Consider only the resistance of the solution, ignoring the resistance of the materials of the battery's positive and negative electrodes. It is known that Avogadro's constant is $N_A=6.022 \\times 10^{23} \\text{ mol}^{-1}$, and the elementary charge is $e=1.602 \\times 10^{-19} \\text{ C}$. Short-circuit the battery, and determine the drift velocity $v$ of the charge carriers ($OH^{-}$ ions) in the electrolyte, expressed in terms of the physical quantities given in the problem, without requiring a numerical solution.", "solution": "", "answer": "" }, { "id": 114, "tag": "ELECTRICITY", "content": "Place an electric dipole $p$ at the origin and establish a polar coordinate system $(r,\\theta)$ with its direction as the polar axis. A negatively charged particle with charge $-q$ and mass $m$ starts from the point in polar coordinates $r=R,$ $\\theta=\\pi/2$ with initial velocities ${V}_{r}$ and $V_{\\theta}$. The particle then moves to a known point with polar coordinates $r_0,\\theta_0$. Find the angular momentum ${L}$ of the particle at this point. The permittivity of free space is $\\varepsilon_0$.", "solution": "", "answer": "" }, { "id": 503, "tag": "MODERN", "content": "Suppose the total mass of a galaxy is the sum of visible mass and dark matter mass, with only gravitational interaction between dark matter and visible matter. Consider a young galaxy, whose mass is primarily composed of visible interstellar gas and invisible dark matter (ignoring the mass of stars). The interstellar gas is composed of identical particles with mass \\(m_p\\), and the number density \\(n(r)\\) of these particles depends on the distance \\(r\\) from the galaxy center. The temperature \\(T(r)\\) of the interstellar gas is also a function of \\(r\\). We can assume that the interstellar gas is always in a state of hydrostatic equilibrium, with the pressure gradient force balanced by the gravitational force in the galaxy. It is known that all mass distribution in the galaxy is spherically symmetric. Given the gravitational constant \\(G\\), the pressure gradient \\(\\frac{dp}{dr}\\) of the interstellar gas balances the gravitational force (pressure is generated only by interstellar gas, ignoring the pressure contribution from dark matter). Assume the interstellar gas is an ideal gas, and for simplification, consider the temperature of the interstellar gas to be uniform everywhere at \\(T_0\\), and the number density of interstellar gas particles satisfies \\(n(r)=\\frac{\\alpha}{r(\\beta + r)^2}\\), where \\(\\alpha,\\beta\\) are known constants. Find the mass density \\(\\rho_d(r)\\) of dark matter at a distance \\(r\\) from the galaxy center.", "solution": "", "answer": "" }, { "id": 748, "tag": "ELECTRICITY", "content": "In an LRC circuit, the change in current in the inductor coil leads to a change in the magnetic field, thereby affecting the voltage and current in the circuit. Inserting a non-magnetic metal rod or a ferromagnetic rod will change the magnetic field distribution of the inductor coil, subsequently influencing the circuit's response. For a non-magnetic rod, due to electromagnetic induction, the changing magnetic field will produce an induced current within the metal rod, accompanied by Joule heating, thus generating an equivalent non-contact resistance $R$. If edge effects are ignored, the magnetic field inside the coil is uniform, and the number of turns per unit length of the coil is $n$. Assume the conductivity of the metal rod is $\\sigma$, the angular frequency of the power source (sine signal) is $\\omega$, and the lengths of both the coil and the metal rod are $l_{0}$, with their axes coincident, the radius of the metal rod is $r_{1}$, and the radius of the coil is $r_{0}$. For the non-magnetic metal rod, derive the expression for the non-contact resistance (assuming $\\mu_{0}\\omega\\sigma r_{0}^{2}\\ll1$, take the lowest order approximation).", "solution": "", "answer": "" }, { "id": 563, "tag": "MECHANICS", "content": "When a child was bathing, they accidentally slipped in the tub. Next, we will analyze the possible motion of the child in the tub.We model the tub as a smooth semicircular groove with a radius of $R$. We also model the child as a rigid hemispherical body with a mass of $m$, a radius of $R$, and a uniformly distributed mass . During the motion, the hemispherical child coincides with the center of the bathtub sphere and does not leave the bathtub.\n\nNow we assume that the hemisphere does not leave the bathtub during its motion. The child rotates about a fixed point through the sphere's center. The child's back remains at an angle $ \\alpha $ (i.e., the angle between the normal vector of the semicircular cross-section and the vertical direction is $ \\alpha $). During the motion of the hemisphere, the magnitude of the angular velocity remains constant, but its direction changes continuously. Determine the minimum angular velocity magnitude that allows the child's back to always remain at an angle $\\alpha$.", "solution": "", "answer": "" }, { "id": 424, "tag": "MECHANICS", "content": "There is water vapor with density $ \\rho_0 $ suspended in the air, and the density of water is $ \\rho $. When raindrops fall, they are approximated as spherical and grow larger by absorbing water vapor (water vapor becomes water upon absorption). Please prove that the raindrops eventually tend toward uniformly accelerated linear motion and find their acceleration. The acceleration due to gravity is $ g $.\n\nBased on the above model, we should first consider the resistance of dry air (excluding forces from interactions with water vapor):\n$ f = k A v^2 $\nwhere $ k $ is to be determined, $ A $ is the maximum cross-sectional area in the direction of the raindrop's motion, and $ v $ is the raindrop's speed. Additionally, the falling raindrops are not spherical, and assume $ A $ and the raindrop's mass $ m $ satisfy:\n$ A = a m^\\alpha $\nwhere $ a $ and $ \\alpha $ are known constants.\n\nUnder this model, find the raindrop's acceleration in the final state.\nNote: Consider the effect produced by water vapor and treat $ k $ directly as a known parameter.", "solution": "", "answer": "" }, { "id": 394, "tag": "MECHANICS", "content": "A smooth bowl with a radius of $R$ is fixed, and the plane at the mouth of the bowl is horizontal. A smooth, homogeneous, thin rod $AB$ with length $L = \\frac{4\\sqrt{3}R}{3}$ Translation:\nEnd B is located outside the bowl, while end A presses against a point inside the bowl. The rod achieves static equilibrium in a plane passing through the center of the sphere $O$. Points $D$ and $D^{\\prime}$ on the rod are nearly coincident with the point of contact at the rim of the bowl, but $D$ is slightly lower-left, and $D^{\\prime}$ is slightly upper-right. Let the angle between the rod and the horizontal plane be $\\theta$.\n\nThe rod is suddenly cut at point $D$. Note that after being cut, point $D$ will gently rest on the inner surface of the bowl. Find the angular acceleration $\\beta = {\\ddot{\\theta}}$ of the rod at this instant.", "solution": "", "answer": "" }, { "id": 158, "tag": "ADVANCED", "content": "## **Theory of Surface States of Topological Insulators**\n\nThe surface states of topological insulators have unique electromagnetic wave propagation characteristics, involving **Berry phase**, **Chern number**, and **nonlinear effects**. These concepts can be understood through step-by-step derivation and analysis.\n\nConsider the surface states of a three-dimensional topological insulator, whose low-energy effective Hamiltonian matrix is:\n$$\nH = v_F (\\sigma_x k_y - \\sigma_y k_x) + m \\sigma_z\n$$\n\nWhere $v_F$ is the Fermi velocity, $\\sigma_x, \\sigma_y, \\sigma_z$ are the Pauli matrices, $k_x$ and $k_y$ are the wave vectors in momentum space, and $m$ is the mass term.\n\nLet's define $\\vert \\psi(\\mathbf{k})\\rangle$ as the normalized eigenvector. Since there are two eigenvectors, for convenience we take the one with the negative eigenvalue (although the final results are clearly the same).\n\nBerry connection is defined as:\n$$\n\\vert A\\rangle =\\begin{pmatrix}\n\\langle \\psi(\\mathbf{k}) | \\frac{\\partial }{\\partial k_x} | \\psi(\\mathbf{k})\\rangle\\\\\n\\langle \\psi(\\mathbf{k}) | \\frac{\\partial }{\\partial k_y} | \\psi(\\mathbf{k})\\rangle\n\\end{pmatrix}\n$$\n\nDefinition of Chern number**: Chern number C is the integral of the Berry curvature $\\Omega(\\mathbf{k})$ over the Brillouin zone, where the integration region can be considered as the entire plane:\n$$\nC = \\frac{1}{2\\pi} \\int_{\\text{BZ}} \\Omega(\\mathbf{k}) \\, d^2 k\n$$\n\nWhere Berry curvature $\\Omega(\\mathbf{k})$ is defined as:\n$$\n\\Omega(\\mathbf{k}) = \\nabla_{\\mathbf{k}} \\times \\langle \\psi(\\mathbf{k}) | i \\nabla_{\\mathbf{k}} | \\psi(\\mathbf{k}) \\rangle\n$$\n\nCalculate the Chern number when $m>0$.", "solution": "", "answer": "" }, { "id": 519, "tag": "ELECTRICITY", "content": "Thunderstorm clouds are the source of chaotic and intense atmospheric electric fields. Under the combined influences of turbulent disturbances, precipitation, horizontal winds, and more complex convergence and uplift factors, their distribution becomes highly irregular. However, as an approximate model, we can consider only the vertical influences. This is because for the electric field of thunderstorm clouds, what truly concerns everyday life is essentially just the \"lightning that reaches the ground.\" This portion of lightning can be understood by the vertical electric field surpassing the atmospheric breakdown limit, and the accumulation of the vertical electric field is of paramount importance.\n\nWithin thunderstorm clouds, negative ions are more likely to dissolve into raindrops and fall, resulting in a general pattern of positive charges above and negative charges below. As precipitation occurs, charge separation will continue to expand until the electric field becomes sufficiently large for lightning to occur.\n\nLet us consider two cloud layers of equal thickness, with their thickness being much smaller than their vertical separation. Precipitation exists between them. Each raindrop has a radius of $r$, density $\\rho$, carries negative charge, and the charge of each raindrop is proportional to its radius: $q = br$. The number density of raindrops is $n$. The friction force acting on a falling raindrop is proportional to the first power of its velocity: $f = -kv$, where $k$ is a constant. The resistance experienced by ions moving through the air can be expressed using the modified Stokes' formula: $F = -Ku$, where $K$ is a constant, and $u$ is the magnitude of the ion's velocity. Suppose all positive ions carry a single positive charge $e$, and compared to the electric field force, the gravitational force acting on the ions can be ignored. The gravitational acceleration is $g$, and the vacuum permittivity is $\\epsilon_0$.\n\nIf lightning has not occurred, and the electric field between the cloud layers is considered uniform, with the field strength being zero at time $t = 0$, find the function of the electric field $E(t)$ between the cloud layers as a function of time.", "solution": "", "answer": "" }, { "id": 275, "tag": "ELECTRICITY", "content": "As one of the four fundamental interactions, electromagnetic interactions play an important role in determining the internal properties of matter in everyday life. In particular, the electromagnetic interactions become more complex when studying the interactions between molecules. Below, we will simply calculate a few types of intermolecular interactions. \n\nBetween two molecules, their charges polarize one another, causing a shift in the centers of positive and negative charges. Suppose the electron distribution of the molecules is approximately uniformly distributed within a sphere with radius $a$, and the electron cloud does not deform under the influence of an external electric field. The atomic polarizability $\\alpha$ is defined as the ratio of the electric dipole moment induced by polarization in the molecule to the external polarizing electric field.\n\n**Problem:** Calculate the case in which a nonpolar molecule $A$ is polarized by a molecule $B$ that possesses a permanent electric dipole moment $p_{B}$. Molecule $B$ is located at the origin of the coordinate system, with the dipole moment direction pointing in the positive direction of the polar axis. The polar coordinates of the nonpolar molecule $A$ are $(r, \\theta)$. Calculate the force between the two molecules.", "solution": "", "answer": "" }, { "id": 501, "tag": "THERMODYNAMICS", "content": "The atmosphere is divided from low to high into the troposphere, the stratosphere... The temperature in the troposphere and the stratosphere varies with height according to the following formulas, respectively: $T_t(h) = T_0(1-\\alpha h)$, $T_s(h) = T_t(h_t)[1+\\beta (h-h_t)]$, where $h$ is the height above the ground, $h_t$ is the distance from the ground to the top of the troposphere (known), and $\\alpha, \\beta, T_0$ are known constants. To simplify the problem, we assume that the air behaves as an ideal gas and that no convection occurs, with the gas distributed in a stable and equilibrium state. Derive the relationship for the variation of air density in the stratosphere with height, $\\rho_s(h)$. It is known that the atmospheric pressure at the ground is $p_0$, the molar mass of air is $\\mu$, the ideal gas constant is $R$, and the gravitational acceleration is a constant $g$.", "solution": "", "answer": "" }, { "id": 663, "tag": "THERMODYNAMICS", "content": "The equation of state for a certain gas can be expressed within a certain range as: \\( p V = n R(T + a T^{2}) \\). Here, \\( a \\) is a constant, \\( R \\) is the universal gas constant, and \\( n \\) is the number of moles. The molar mass of this gas is \\(\\mu\\). The relationship between internal energy and state is known to be: \\( U = \\int_{V_{0}}^{V}\\left(T{\\left(\\frac{\\partial p}{\\partial T}\\right)}_{V}-p\\right)\\mathrm{d}V+f(T) \\). In this problem, \\( f(T) \\) is zero. Assume that the surface of a certain planet consists entirely of this gas, the surface gravitational acceleration of the planet is \\( g \\), and the temperature changes with altitude according to the relation \\( T = T_{0}(1+\\alpha h) \\), where \\(\\alpha\\) is a constant and \\( h \\) is the altitude. The gas density at an altitude of zero is \\(\\rho_{0}\\). Find the expression for air density as a function of altitude, the expression must not contain the variable \\( T \\).", "solution": "", "answer": "" }, { "id": 667, "tag": "ELECTRICITY", "content": "According to the theory of special relativity, we can derive that if the electric field strength in the $S^{\\prime}$ frame is $\\vec{E}^{\\prime}$ and the magnetic induction strength is ${\\vec{B}}^{\\prime}$, then in the $S$ frame (where the $S^{\\prime}$ frame moves along the positive $x$-axis of the $S$ frame with a velocity $ u$), the electric field strength is $\\vec{E}$ and the magnetic induction strength is $\\vec{B}$, and there are the following formulas: $$ \\left\\{\\begin{array}{l l}{E_{x}=E_{x}^{'}}\\ {E_{y}=\\gamma\\left(E_{y}^{'}+ u B_{z}^{'}\\right),}\\ {E_{z}=\\gamma\\left(E_{z}^{'}- u B_{y}^{'}\\right)}\\ {B_{z}=\\gamma\\left(B_{z}^{'}+\\frac{ u}{c^{2}}E_{y}^{'}\\right)}\\end{array}\\right.\\left\\{\\begin{array}{l l}{\\displaystyle B_{x}=B_{x}^{'}}\\ {\\displaystyle B_{y}=\\gamma\\left(B_{y}^{'}-\\frac{ u}{c^{2}}E_{z}^{'}\\right)}\\ {\\displaystyle B_{z}=\\gamma\\left(B_{z}^{'}+\\frac{ u}{c^{2}}E_{y}^{'}\\right)}\\end{array}\\right. $$ Now consider a point charge moving along the $x$-axis direction with a velocity $ u=\\beta c$. Find the electric field strength $\\vec{E}$ at point $\\boldsymbol{A}$ located at a vector position $\\vec{r}$ from the charge, as observed in the $S$ frame.The result is indicated by $\\beta$, not $u$ (the angle between $\\vec{r}$ and $\\vec{ u}$ is denoted by $\\theta$).", "solution": "", "answer": "" }, { "id": 183, "tag": "ADVANCED", "content": "A cylindrical insulating container of mass 𝑀 is stationary in a vacuum, with one end sealed. Initially, an insulating piston of mass 𝑚 and negligible width divides the container into two equal parts. The sealed part contains 𝑛 moles of a monoatomic ideal gas with a temperature 𝑇 and molar mass $𝑀_0$. Assume the container is smooth. It is assumed that during the expansion process, the state of the gas can be approximated under thermal equilibrium conditions. At the moment the piston leaves the container, the gas and the container will move at a velocity 𝑣, while the piston moves at a velocity 𝑢. After all the gas has left the container, the container's final velocity further changes from 𝑣 to 𝑣 + 𝑣′. Use the kinetic theory of gases to estimate 𝑣′. Assume the container's final velocity is much smaller than the thermal velocity of the molecules. The gas constant is 𝑅. There is no heat exchange among the gas, container, and piston. Temperature changes of the gas after leaving the container can be considered negligible. Earth's gravitational force can be ignored.", "solution": "", "answer": "" }, { "id": 700, "tag": "ELECTRICITY", "content": "Coaxial thin-walled cylinders with radii of $R$ and $2R$ extend infinitely in the direction perpendicular to the plane of the paper, maintaining a potential difference of $U$ between the inner and outer cylinders (the inner cylinder is at a higher potential than the outer cylinder). The region between the inner and outer cylinders is vacuum and has a constant uniform magnetic field $B$ directed into the plane of the paper. Inside the region of the inner cylinder, there is a variable uniform magnetic field $B_{v} = kt$ directed into the plane of the paper, where $k$ is a positive constant, and $t$ is time. Considering a charged particle with mass $m$ and charge $+q$ moving in the region between the inner and outer cylinders, ignore the effect of gravity and the influence of the charged particle and its induced charge on the potential of the inner and outer cylinders. At $t=0$, the charged particle is released from rest from the outer surface of the inner cylinder. It is known that the charged particle first reaches the inner surface of the outer cylinder after rotating $180^{\\circ}$ around the axis of the cylinder, with its velocity direction exactly tangent to the cylinder at that time. Determine the time $t$ when the particle first reaches the inner surface of the outer cylinder after being released (just provide the expression, it is not necessary to discuss the existence of a positive real solution).", "solution": "", "answer": "" }, { "id": 485, "tag": "MECHANICS", "content": "Consider the problem of the trajectory under the influence of an inverse-square force. It is well known that an electron, under the Coulomb force exerted by a near-static atomic nucleus, follows a classical elliptical trajectory:\n\n$$\nr = \\frac{p}{1 + e \\cos \\theta}\n$$\n\nwhere $r$ is the radial distance and $\\theta$ is the angle of the radial vector with respect to the $x$-axis. We aim to calculate the \"average position\" of the electron over one period $T$, defined as:\n\n$$\n\\overrightarrow{O P} = \\frac{1}{T} \\int_{0}^{T} \\overrightarrow{r} \\, \\mathrm{d}t\n$$\n\nPlease compute the length $l$ of $OP$ (express the answer using the quantities provided in the problem).", "solution": "", "answer": "" }, { "id": 376, "tag": "MODERN", "content": "The mass of the star Proxima Centauri is \\( M \\), its surface temperature is \\( T_s \\), and its radius is \\( R \\). Planet B orbits Proxima Centauri in a circular orbit, with its orbital period given as \\( T \\). Assuming both the star and the planet are black bodies and perfect heat conductors, estimate the steady-state temperature of the planet. Known constants include the gravitational constant \\( G \\) and the Stefan-Boltzmann constant \\( \\kappa \\).", "solution": "", "answer": "" }, { "id": 364, "tag": "OPTICS", "content": "Using illumination light with wavelength $\\lambda$, perform Fraunhofer diffraction. The diffraction screen is composed of two overlapping square apertures, each with a side length of \\(a\\). Both square apertures have sides that are either horizontal or vertical. One square is located at the upper left position, and the other is at the lower right position. The overlapping part is exactly a square with a side length of \\(\\frac{a}{2}\\). Only the square apertures and their overlapping section are transparent on the plane. A plane wave is incident perpendicular to the screen. Determine the light intensity distribution $\\widetilde{I}(\\theta_{1},\\theta_{2})$ of the Fraunhofer diffraction field on the observation screen. Express the result using $\\alpha=\\frac{\\pi a\\mathrm{sin}\\theta_{1}}{\\lambda}$ and $\\beta=\\frac{\\pi a\\mathrm{sin}\\theta_{2}}{\\lambda}$. Ensure that the intensity is $I_0$ when $\\theta_{1}\\to 0, \\theta_{2}\\to 0$.", "solution": "", "answer": "" }, { "id": 521, "tag": "MECHANICS", "content": "Assume there exists a substance that always satisfies $p=\\frac{1}{2}k\\rho^2$, where $p$ is the pressure, $\\rho$ is the density of the substance, and $k$ is a constant. In the universe, there is a spherically symmetric celestial body composed of this substance, and the body remains stable (every mass element experiences a net force of zero). Find the radius $R$ of this celestial body in its stable state. Assume that the law of universal gravitation always holds, with the gravitational constant being $G$. Hint: Derive the differential equation that $\\rho(r)$ satisfies, where $r$ is the distance from a point to the center of the sphere, and use the substitution $u=r\\rho$ to solve.", "solution": "", "answer": "" }, { "id": 291, "tag": "MECHANICS", "content": "Internal combustion engine and crankshaft connecting rod mechanism. The connecting rod $AB$ has a length of $l$; one endpoint $A$ rotates on a circle with radius $r$, with $O$ as the center of the circle. The piston is constrained to move along the line $OB$. The gas pressure in the cylinder is $P$, the atmospheric pressure is $p$, and the piston is circular with a radius of $R$. When the angle between $OA$ and $OB$ is $\\psi$, with what force does the connecting rod rotate the crank?", "solution": "", "answer": "" }, { "id": 596, "tag": "ELECTRICITY", "content": "In recent years, some have proposed using plasma lenses to focus high-energy charged particles. A plasma lens can be considered as a long cylindrical good conductor with a radius of \\( R \\), carrying a uniform axial current \\( I \\) (which can be treated as infinitely long), with vacuum outside the cylinder. The origin is set at the middle point of the cylinder on the axis, and the axial direction is the \\( z \\)-axis (the direction of current is the +z direction), with the radial coordinate denoted as \\( r \\). At a certain moment, a relativistic particle with charge \\( +q \\) and momentum magnitude \\( p_0 \\) (the magnitude to be determined) is emitted from a particle source at \\( z=0, r=r_0 (r_0>R) \\), in the direction along the +z axis. It is assumed that the plasma is sufficiently sparse, so the particle only experiences the magnetic field generated by the plasma lens, and even if it enters within the cylindrical plasma lens, collisions between the particle and the plasma can be completely ignored. Since the plasma is considered a good conductor, the electric field within it has negligible effect on the charged particle. It is assumed that the magnetic permeability of the plasma equals the vacuum permeability \\( \\mu_0 \\). During the focusing process, particles with smaller momentum may reverse direction within the plasma lens (i.e., have a momentum z-component opposite to the initial direction), a situation known as over-focusing of the plasma lens. To allow the charged particle to transmit in the axial direction without reversing, find the particle's minimum initial momentum \\( p_0 \\). Note: For convenience, you may need to assume the particle's rest mass or speed, but surprisingly, the final result depends only on momentum. This actually allows us to conveniently use readily available particles to simulate less common situations—for example, using a large and stable number of high-energy protons produced by an accelerator to simulate the extremely rare muons in cosmic rays, as long as their momentum is equal.", "solution": "", "answer": "" }, { "id": 723, "tag": "ELECTRICITY", "content": "Consider a regular polygon with 2017 sides, where a point charge \\(q\\) is placed at each of the 2016 vertices, and a point charge \\(Q\\) is placed at the center of the polygon. The distance from the center of the polygon to each vertex is \\(a\\). Utilizing symmetry and the principle of superposition of electric fields, determine the magnitude of the net electrostatic force on the point charge \\(Q\\). The answer should be expressed as a clear formula.", "solution": "", "answer": "" }, { "id": 698, "tag": "ADVANCED", "content": "To solve the surface wave equation for a viscous fluid under the approximation of small movements, we take the bottom boundary to be horizontal at $z=0$, with liquid depth $h$, kinematic viscosity $ u$, density $\\rho$, and surface tension coefficient $\\tau$. There exists only a uniform downward gravitational field $-g \\vec e_z$. The surface equation corresponding to the displacement at $z=h$ is $\\eta(\\vec{x},t)$, where $\\vec{x}$ represents the coordinates on the surface. Given the Laplace transform $\\bar{\\eta}_{k}(p)$ of the $k$ wavevector component $\\eta_{k}(t)$ of the surface displacement $\\eta(\\vec{x},t)$. Initial conditions $\\eta_{k}(0)$ and $\\eta'_{k}(0)$ are provided. Hint: Solve the linearized Navier-Stokes equation $$ \\partial_t \\vec v = -\\frac{1}{\\rho} abla p - g \\vec e_z + u abla^2 \\vec v $$ Decompose the velocity field as $$ \\vec v(x,z,t) = abla\\times abla\\times \\vec e_z S(x,z,t) + abla\\times \\vec e_z Z(x,z,t) $$ You need to take the curl of the N-S equation twice to obtain a partial differential equation for $S$, requiring the no-stress boundary condition at the bottom boundary (this aids in smoothly transitioning to the ideal fluid case) $$ S(x,z,t)=\\partial_z^2 S(z,x,t)=0|_{z=0} $$", "solution": "", "answer": "" }, { "id": 138, "tag": "ELECTRICITY", "content": "Maxwell's electromagnetic theory tells the world that the electromagnetic field has gauge symmetry. Under the gauge transformation $$ \\vec{E} \\equiv- abla \\varPhi- \\frac{ \\partial \\vec{A}}{ \\partial t}, \\quad \\vec{B} \\equiv abla \\times \\vec{A}. $$ $$ \\vec{A^{ \\prime}}= \\vec{A}- abla \\varPsi, \\quad \\varPhi^{'}= \\varPhi+ \\frac{ \\partial \\varPsi}{ \\partial t} $$ the electromagnetic field $ \\vec{E}$ and $ \\vec{B}$ are unchanged. In the Lorentz gauge $$ abla \\cdot{ \\vec{A}}+{ \\frac{1}{c^{2}}}{ \\frac{ \\partial \\phi}{ \\partial t}}=0 $$ the electromagnetic potentials satisfy the source-included d'Alembert equation $$ abla^{2} \\varPhi- \\frac{1}{c^{2}} \\frac{ \\partial^{2} \\varPhi}{ \\partial t^{2}}=- \\frac{ \\rho}{ \\epsilon_{0}}, \\quad abla^{2} \\vec{A}- \\frac{1}{c^{2}} \\frac{ \\partial^{2} \\vec{A}}{ \\partial t^{2}}=- \\mu_{0} \\vec{J}. $$ However, if the gauge symmetry is broken, the equation satisfied by the electromagnetic potential takes the following form $$ abla^{2} \\varPhi- \\frac{1}{c^{2}} \\frac{ \\partial^{2} \\varPhi}{ \\partial t^{2}}- \\mu^{2} \\varPhi=- \\frac{ \\rho}{ \\epsilon_{0}}, \\quad abla^{2} \\vec{A}- \\frac{1}{c^{2}} \\frac{ \\partial^{2} \\vec{A}}{ \\partial t^{2}}- \\mu^{2} \\vec{A}=- \\mu_{0} \\vec{J}. $$ Or, using four-vectors, it can be written in the following form $$ \\left( \\boldsymbol{ \\Pi}+ \\mu^{2} \\right) \\boldsymbol{A}^{ u}= \\mu_{0} \\boldsymbol{J}^{ u}, $$ where $ \\begin{array}{r}{ \\exists \\equiv{ \\frac{1}{c^{2}}}{ \\frac{ \\partial^{2}}{ \\partial t^{2}}}- abla^{2}} \\end{array}$ In quantum mechanics, an equation in this form is called the Klein-Gordon equation. Study Coulomb's law in the case of $ \\mu eq0$. The electric potential $ \\varPhi$ satisfies $$ \\left( abla^{2}- \\mu^{2} \\right) \\varPhi \\left( \\vec{r} \\right)=- \\frac{q}{ \\epsilon_{0}} \\delta^{3} \\left( \\vec{r} \\right). $$ It is known that the general solution of the inhomogeneous Helmholtz equation in the form $$ \\left( abla^{2}+k^{2} \\right)G \\left({ \\vec{r}} \\right)=-4 \\pi \\delta^{3} \\left({ \\vec{r}} \\right) $$ is $$ G \\left( \\vec{r} \\right)=A \\frac{ \\mathrm{e}^{ \\mathrm{i}k r}}{r}+B \\frac{ \\mathrm{e}^{- \\mathrm{i}k r}}{r}+g, $$ where $A+B=1$, $g$ is a solution to the homogeneous Helmholtz equation. Thus, the modified formula for the electrostatic field of a point charge in Coulomb's law is $$ \\varPhi \\left( \\vec{r} \\right)= \\frac{q}{4 \\pi \\epsilon_{0}} \\frac{ \\mathrm{e}^{- \\mu r}}{r}. $$ Maxwell used the following experiment to measure deviations in the forces between charges relative to Coulomb's law: Place two concentric thin spherical conducting shells with radii $a$ and $b(a>b)$, and connect them with a thin wire. Charge the outer shell to a potential $U$, remove the power source, then remove the wire connecting the two shells, and finally ground the outer shell. At this point, the potential measured at the inner shell is no greater than $u$. Estimate the upper limit of the photon's static mass from this, retaining the leading order contribution.", "solution": "", "answer": "" }, { "id": 619, "tag": "THERMODYNAMICS", "content": "Heat engines and heat pumps utilize thermodynamic cycles of substances to achieve opposite functions: the former absorbs heat from a high-temperature source, converts part of the heat into work for external output, and releases the remaining heat to a low-temperature sink; the latter takes in external work to absorb heat from a low-temperature source and discharges it to a high-temperature sink, together with the heat converted from the external work. According to the second law of thermodynamics, whether for a heat engine or a heat pump, if the working substance during its cycle only interacts with two heat reservoirs at temperatures $T_{1}$ and $T_{2}$, the absorbed heat $Q_{1}$ and $Q_{2}$ satisfies the inequality $$ \\frac{Q_{1}}{T_{1}}+ \\frac{Q_{2}}{T_{2}} \\leq0 $$ Here, the heat can be positive or negative, indicating absorbing heat from or releasing heat to a heat source, respectively. A certain heating device originally uses heat released by a boiler at temperature $T_{0}$ to directly heat a room, maintaining the room temperature at a constant $T_{1}$, which is higher than the outdoor temperature $T_{2}$. To improve energy efficiency, it is proposed to use the aforementioned machines based on existing energy to improve the heating scheme. Compared to direct heating, what is the theoretical limit of the reduction rate $ \\eta$ in boiler energy consumption?", "solution": "", "answer": "" }, { "id": 550, "tag": "OPTICS", "content": "Solve the following physics problem, and frame the final answer in \\boxed{}: General relativity is the astounding theory Einstein provided to theoretical physics in the early 20th century. It is profound for its unique concepts of spacetime and matter, elegant for its geometric mathematical structure, and distinguished for its successful predictions of physical phenomena, garnering the attention of later theoretical physicists. Mr. Yang Zhenning once said that the Yang-Mills theory (the main theory governing our understanding of the other three fundamental interactions) is actually similar in mathematical structure to general relativity.\n\nAs a crucial part of verifying general relativity, in 1919, Sir Arthur Eddington embarked on an expedition to Brazil and the Gulf of Guinea, leading an experiment observing the deflection of starlight during a solar eclipse, which directly announced the validity of general relativity to the world. Let's use a non-relativistic equivalent method to derive the classical results of this experiment. It can be proven that if a light ray originating from infinity behaves like a particle in general, incident with an impact parameter $b$ and its propagation direction being deflected under the gravitational field of a centrally fixed celestial body $M$, then its trajectory will be identical to that in a refractive index field $n(r)$. This refractive index field is:\n\n$$\nn^{2}=n_{0}^{2}\\left(1+{\\frac{2G M b^{2}}{c^{2}r^{3}}}\\right)\n$$\n\nwhere $n_{0}$ is the refractive index when $r\\rightarrow\\infty$. Consider the following problem: \n\nLet us study the influence of different $b$ on the shape of the orbit. Even if the radius of the central body $M$ can be neglected, it can be proven that when $b$ is smaller than a certain value, the radius of the light ray will decrease infinitely during its propagation and will not be able to escape the central body $M$. This phenomenon is called the capture of photons by the celestial body. Please strictly calculate, for a parallel beam of light originating from infinity, within what area range the light will be captured by the central body (capture cross-section)?", "solution": "", "answer": "" }, { "id": 442, "tag": "MODERN", "content": "A uniformly positively charged infinite straight conductor has a linear charge density $\\lambda$. A particle with a static mass $m$ and negative charge $-q$ can, under the appropriate initial velocity, move in a uniformly spiraling motion on a cylindrical surface with radius $R$ due to the electrostatic force between it and the straight conductor. Based on this, consider solving for the angular frequency $\\omega$ of small radial oscillations when the initial velocity of the negative point charge changes to the composition of a very small radial velocity $v_r \\ll v_z$ in the radial direction. The result must be expressed using $$ k = \\frac{\\lambda q}{2\\pi\\varepsilon_0 m c^2} $$ and $v_z$. Both $k$ and $v_z$ are considered known.", "solution": "", "answer": "" }, { "id": 588, "tag": "ELECTRICITY", "content": "In a three-dimensional space without gravity, let us consider a thin metal spherical shell moving in a uniform magnetic field. The thickness of the metal spherical shell is \\(h\\), the radius of the metal spherical shell is \\(a\\), the density is \\(\\rho_m\\), and the conductivity is \\(\\sigma\\), where \\(a \\gg h\\). The angular velocity \\(\\omega\\) of the metal spherical shell is along the positive \\(x\\)-axis. There is a uniform magnetic field in the space along the positive \\(z\\)-axis, with magnitude \\(B\\). We assume that the current distribution is established instantaneously, and there is no transient process (i.e., when the angular velocity changes, the current at each moment is stable, satisfying \\( abla\\cdot \\vec J=0\\)). If the initial angular velocity of the spherical shell is \\(\\omega_0\\), determine the relationship of the angular velocity with time \\(t\\). We assume that the spherical shell is not subject to any forces and torques other than electromagnetic forces, and we ignore the magnetic field generated by the shell's own current. Hint: Since the shell is very thin, the current only has a tangential component, so the problem can be transformed into a two-dimensional problem on the sphere's surface. You can use \\(\\vec J=\\sigma(\\vec E+\\vec v\\times B)\\) to solve for the tangential electric field, back-calculate the current, and then complete the mechanical part of the calculation.", "solution": "", "answer": "" }, { "id": 768, "tag": "ELECTRICITY", "content": "A semicylindrical dielectric structure of length $l$ with inner and outer radii $a$ and $b$ respectively is composed of two different lossy dielectrics. Their relative permittivities and conductivities are $ \\pmb{ \\varepsilon}_{1}$ and $ \\pmb{ \\sigma}_{1}$ (in the region $0 \\leq \\varphi < \\theta_{0}$), and $ \\pmb{ \\varepsilon}_{2}$ and $ \\pmb{ \\sigma}_{2}$ (in the region $ \\theta_0 < \\varphi \\leq \\pi$). The bottom sides of the semicylinder ($ \\theta = 0$, $ \\theta = \\pi$) are coated with metallic films, between which a DC voltage $V_0$ is applied, reaching a steady state. The vacuum permittivity is given as $ \\varepsilon_0$, and the relative permittivities of the two media are large, while their conductivities are small. Edge effects are neglected. Find the total charge at the interface $ \\varphi = \\theta_0$. The answer should not contain integral expressions.", "solution": "", "answer": "" }, { "id": 473, "tag": "MECHANICS", "content": "Please solve the following physics problem and put the final answer in \\boxed{}: Consider a statics version of Buffon's needle problem: A steel needle of length $l^{\\prime}=x l$ is randomly placed on infinitely thin horizontal iron racks that are evenly spaced at distance $l$, where $x\\geq2$. Find the probability $p$ that the steel needle can rest stably on the racks without falling off.", "solution": "", "answer": "" }, { "id": 448, "tag": "MODERN", "content": "In the cosmic space, there is a spaceship that is approximately spherical in shape, initially at rest, with a radius of $R$ and a mass of $M$. The spaceship is propelled by radiation. Assume that the spaceship's speed is very high, so we must consider relativistic effects.\n\nAfter a period of travel, the spaceship enters a region of uniform, stationary space dust, where the mass of dust per unit volume in the cosmic space reference frame is $p$. Upon contact with the spaceship, the dust adheres to the surface of the spaceship. For simplicity, let the spaceship's velocity at this moment be $v_{0}$ and its rest mass be $M_{1}$. If after entering the dust region, the spaceship shuts off its engines and moves without propulsion, assume that the mass of the dust attached to the spaceship is very small relative to the spaceship's own mass. Just upon entering the dust region, set the clock on the spaceship to $t^{\\prime}=0$. Determine the relationship between the spaceship's speed $v$ and the time $t'$ on the spaceship.", "solution": "", "answer": "" }, { "id": 165, "tag": "MECHANICS", "content": "The plane of the uniform right-angle side $AOB$ is vertical, with side $AO$ having a length of $l_1$ and side $BO$ having a length of $l_2$. It can swing left and right around a horizontal axis passing through point $O$ and perpendicular to the plane of $AOB$, and at the same time, it can rotate around a vertical axis passing through point $O$. Assume that during the uniform rotation around the vertical axis, the angle between side $OA$ and the vertical line stabilizes at angle $\\alpha$, and the gravitational acceleration is $g$. Find the angular velocity of the rotation.", "solution": "", "answer": "" }, { "id": 566, "tag": "ADVANCED", "content": "Alice and Bob are playing a game, roughly described as follows. Charlie is the host of the game. In each round, Charlie gives each of Alice and Bob a bit, \\(x, y\\), which means that each of them gets a number, 0 or 1. After receiving the bits, Alice and Bob each reply to Charlie with a bit, \\(a, b\\). To win the game, it is required that \\(x\\,\\text{and}\\, y = a\\,\\text{xor}\\,b\\). Alice and Bob can devise a strategy before the game, but they cannot communicate during the game.\n\nWe consider the following strategy:\nA pair of entangled quantum bits can be described as\n\\[\\ket{\\phi}=\\frac{1}{\\sqrt{2}}\\left(\\ket{00}+\\ket{11}\\right)\\]\nHere, \\(\\ket{mn}\\) refers to the state of the system composed of two quantum bits, meaning the quantum state of the system where the quantum bit in Alice's hand is in state \\(\\ket{m}\\) and the quantum bit in Bob's hand is in state \\(\\ket{n}\\). The game process is that Alice first performs a measurement based on the \\(x\\) she received, and then Bob performs a measurement based on the \\(y\\) he received. They determine their reply bits \\(a, b\\) based on the measurement results.\n\nLet's define the following states:\n\\[\\begin{align*}\n \\ket +&=\\frac{1}{\\sqrt2}\\ket0+\\frac{1}{\\sqrt{2}}\\ket 1&\n \\ket -&=\\frac{1}{\\sqrt2}\\ket0-\\frac{1}{\\sqrt{2}}\\ket 1\\\\\n \\ket{a_0}&=\\cos\\theta\\ket0+\\sin\\theta\\ket1&\\ket{a_1}&=-\\sin\\theta\\ket0+\\cos\\theta\\ket1\\\\\n \\ket{b_0}&=\\cos\\theta\\ket 0-\\sin\\theta\\ket1&\\ket{b_1}&=\\sin\\theta\\ket 0+\\cos\\theta\\ket 1\n \\end{align*}\\]\nwhere \\(\\theta=\\pi/8\\).\n\nThe specific measurement strategy is as follows:\n\n**Alice's Measurement:**\n\n* When Alice's random number \\(x\\) is 0, she uses the measurement basis \\(\\{\\ket{1}, \\ket{0}\\}\\).\n * If she measures \\(\\ket{1}\\), the value she sends is 1.\n * If she measures \\(\\ket{0}\\), the value she sends is 0.\n* When Alice's random number \\(x\\) is 1, she uses the measurement basis \\(\\{\\ket{+}, \\ket{-}\\}\\).\n * If she measures \\(\\ket{+}\\), the value she sends is 0.\n * If she measures \\(\\ket{-}\\), the value she sends is 1.\n \n**Bob's Measurement:**\n\n* When Bob's random number \\(y\\) is 1, he uses the measurement basis \\(\\{\\ket{b_1}, \\ket{b_0}\\}\\).\n * If he measures \\(\\ket{b_1}\\), the value he sends is 1.\n * If he measures \\(\\ket{b_0}\\), the value he sends is 0.\n* When Bob's random number \\(y\\) is 0, he uses the measurement basis \\(\\{\\ket{a_0}, \\ket{a_1}\\}\\).\n * If he measures \\(\\ket{a_0}\\), the value he sends is 0.\n * If he measures \\(\\ket{a_1}\\), the value he sends is 1.\nCalculate the average probability of Alice and Bob's winning game with this quantum strategy.", "solution": "", "answer": "" }, { "id": 362, "tag": "MECHANICS", "content": "A cylinder with radius $r$ rolls purely on a horizontal surface. A rod $AB$ leans against it, with one point of the rod resting on the cylinder and the other end $A$ on the ground. There is no relative sliding between the rod and the cylinder, and the end $A$ of the rod does not leave the ground. Given that the velocity $v_0$ of the cylinder's center $O$ is a constant, and the angle between the rod and the ground is $\\varphi$, find the angular acceleration $\\varepsilon$ of the rod $AB$.\n", "solution": "", "answer": "" }, { "id": 560, "tag": "MECHANICS", "content": "There is a railway curve with a curvature radius of $r$, and the distance between the two rails is $L$. There is a height difference between the inner and outer tracks. When the train passes through this curve at its rated speed $v_{0}$, the tracks do not experience lateral thrust. When the train passes through the curve at a speed $v(v>v_{0})$, to prevent the train from overturning, how high can its center of gravity be above its chassis?", "solution": "", "answer": "" }, { "id": 499, "tag": "MECHANICS", "content": "A rigid metal wire in the shape of a parabola is fixed in a vertical plane, with the parabola's equation given by \\( y = ax^2 \\) (the \\( y \\)-axis is oriented vertically upwards, and \\( a \\) is a constant to be determined). A uniform rigid thin rod of length \\( 2l \\) has small holes at each of its ends, \\( A \\) and \\( B \\), which just fit onto the metal wire. The contact between the holes and the wire is very smooth, with negligible friction.\n\nIf an impulse is given to the rod causing it to start moving, after a sufficient amount of time, it comes to rest at an equilibrium position, where the angle between the rod and the horizontal direction is \\( \\theta = 30^\\circ \\). The magnitude of the gravitational acceleration is known to be \\( g \\).\n\nThe rod remains stationary in this equilibrium position. Now, a small mouse starts to climb from the bottom end of the rod upwards. During its ascent, the rod remains stationary. Assume the mouse can be regarded as a point mass, and it does not contact the metal wire at the ends of the rod. Find the displacement of the mouse along the rod at time \\( t \\) (consider the time when the mouse starts climbing the rod as time zero) expressed as \\( s(t) \\).", "solution": "", "answer": "" }, { "id": 481, "tag": "MECHANICS", "content": "Two small steel balls, each with mass $m$, are connected by a string of length $2L$ and placed on a smooth horizontal surface. A constant pulling force $F$ is applied at the midpoint $O$ of the string. The direction of this force is horizontal and perpendicular to the initial direction of the connection. The string is very flexible and non-stretchable, and its mass is negligible. After several collisions, the two steel balls eventually continue to move while remaining in contact. Find the total energy lost due to collisions.", "solution": "", "answer": "" }, { "id": 579, "tag": "THERMODYNAMICS", "content": "Percolation is a geometric phase transition with profound significance in the study of localization and related topics. We use a model here to understand it. On a 2D lattice, lattice points can be occupied by black particles or be empty (white particles). The probability of placing a black particle is $p$, and the probability of being empty is $q=1-p$, where $N$ is the total number of particles.\n\nIt is clear that the probability of large clusters $\\mathrm{\\Delta}s$ appearing, where $n_{s}(p)$ at low values of $p$ is significant, decreases rapidly with $s$. However, when $p$ is large, one observes the emergence of macroscale $s$, which is the localization phenomenon. The transition occurring with $p$ is referred to as percolation. To study this problem, consider the influence between lattice points and transform it. We provide some approximate treatments: firstly, the number density of black particles is redefined as a dynamically changing $n$. We treat it as a function containing time $t$ and lattice coordinates $(x,y)$, and continuoize $(x,y)$ into a two-dimensional real number set instead of integer lattice coordinates. In our simple continuum model, the flux of number density between lattice points is exactly proportional to the gradient of the number density (consistent with the two-dimensional diffusion problem). However, in the actual discrete model, when localization occurs, a lattice point with certain probability is in a connected large cluster, with an internal number density always being 1; or the lattice point is located in an empty space or a small cluster with other probabilities, where the number density is less than 1. This links the continuous model with the discrete model: if the continuous model's $n(x,y)$ ultimately forms a stable distribution $n(r)$ rotationally symmetric around the origin and time-independent, where $r=\\sqrt{x^{2}+y^{2}}$, in this model, take a circle of radius $r_{0}$ such that the total number of particles inside is $N$, representing the characteristic particle number of a large cluster, and the edge particle number density $n(r_{0})$ should be related to the critical probability $p_{c}$ at which localization begins. We directly assume $n(r_{0})\\cdot\\pi r_{0}^{2}=p_{c}N$. Now, let $N=1$, and use $p_{c},r_{0}$ to represent $n(r)$.", "solution": "", "answer": "" }, { "id": 476, "tag": "THERMODYNAMICS", "content": "The metal wire, with a length of $2l$ and a cross-sectional area of $A$, is connected to two steel terminal posts, $C$ and $D$, within the circuit. The metal wire is surrounded by materials with excellent thermal insulation. It is known that the resistivity of the metal wire is $\\rho$ and its thermal conductivity is $\\kappa$. A constant current $I$ is applied. Assume the temperatures of the two steel terminal posts, $C$ and $D$, are always maintained at room temperature $T_{0}$. After the entire system reaches a steady state, answer the following questions:\n\nDescription of the figure: The metal wire is a long cylindrical rod, surrounded by materials with excellent thermal insulation, and is connected at both ends to the steel terminal posts $C$ and $D$.\n\nTake the midpoint of the metal wire as the origin, with the positive $x$-axis extending along the wire to the right. Derive the temperature distribution function along the metal wire.\n\n**Hint**:\n\n1. Heat conduction in the metal wire follows the linear heat conduction law: the heat flux through a cross-section of the wire in unit time is $\\dot{Q}=-\\kappa \\frac{\\mathrm{d}T}{\\mathrm{d}x}A.$\n\n2. You may need the integration formula: $\\int \\ln x \\, dx = x(\\ln x - 1) + C$", "solution": "", "answer": "" }, { "id": 199, "tag": "THERMODYNAMICS", "content": "The thermal decoupling process $DM_1+DM_2\\longleftrightarrow SM_1+SM_2$ has a dark matter annihilation cross-section $\\sigma$, and the relative velocity of dark matter particles is $v$. Given that dark matter is cold, the cross-section can be expanded in a wave-like series as follows:\n\n$$\n\\sigma v=a_s+a_p v^2+a_d v^4+\\cdots\n$$\n\nIt is known that the leading order is the $P$-wave, and we are given $\\left\\langle \\sigma v\\right\\rangle=\\frac{b_0 T}{m}$. Express $a_p$ in terms of $b_0$.\n\nHint: Under thermal equilibrium, the particle number density distribution function is $f_{eq}=\\frac{1}{e^{E/T}\\pm 1}\\approx e^{-\\frac{E}{T}}$, and the number density is $n_{eq}=\\int e^{-\\frac{E}{T}}\\frac{d^3 \\vec{p}}{(2\\pi)^3}$. In this problem, dark matter particles are non-relativistic.", "solution": "", "answer": "" }, { "id": 373, "tag": "THERMODYNAMICS", "content": "At time $t=0$, an adiabatic, thin and lightweight piston, which is thin and lightweight, divides an adiabatic cylinder with a cross-sectional area $S$ into two equal parts, each with a volume of $V_{0}$. On the left side of the piston, there is an ideal gas with an adiabatic index of $\\gamma=3/2$, and an initial pressure of $p_0$. \n\nOn the walls of the container, there is a certain viscous material characterized by a negligible heat capacity and volume. Its behavior is as follows: when the piston moves with a speed of magnitude $v$ in a certain direction, the material exerts a resistive force on the piston given by $F=-k v$. Moreover, the heat generated due to friction exhibits two possible behaviors: \n\n- For a forward-directed viscous material, the heat is conducted along the original direction of motion to the adjacent gas on that side (if there is no gas, it is conducted to the external environment).\n- For a backward-directed viscous material, the heat is conducted opposite to the original direction, transferring to the gas on the opposite side (if there is no gas, it is conducted to the external environment).\n\nConsider the following scenario (retain results in analytical form without converting to numerical values):\n\nThe right side of the container is vacuum, and the material used is forward-directed viscous material. Calculate the time $t$ required for the piston to move to the far right.", "solution": "", "answer": "" }, { "id": 198, "tag": "ADVANCED", "content": "There is now a semi-classical Fermi gas system. It is approximated that the interaction between particles can be simplified to a hard sphere potential, with a characteristic radius $\\lambda_{th}=\\frac{h}{\\sqrt{2\\pi mkT}}$, and the system's free energy function can be written as:\n$$\nF=NkT\\ln\\frac{\\lambda_{th}^3}{\\frac{V}{N}-a\\lambda_{th}^3}\n$$\nGiven the definition of chemical potential $\\mu=\\frac{\\partial G}{\\partial N}$, the system's grand potential can be written as $\\phi=F-\\mu N$, and this system's grand potential has the following exact solution. Determine the constant $a$ in the free energy function. (You may use the approximation $\\mu\\ll kT$)\n$$\n\\phi=-\\frac{kTV(2\\pi m)^{3/2}}{h^3}\\cdot \\frac{2}{\\sqrt{\\pi}}\\int_{0}^{\\infty}\\ln\\left[1+e^{-\\frac{\\epsilon-\\mu}{kT}}\\right]\\epsilon^{\\frac{1}{2}}d \\epsilon\n$$", "solution": "", "answer": "" }, { "id": 761, "tag": "OPTICS", "content": "In a vacuum, there is a beam of circularly polarized light propagating along the $+z$ direction, represented by $\\begin{array}{r}{\\vec{E}=E_{0}\\cos(\\omega t-k z)\\hat{x}+E_{0}\\sin(\\omega t-k z)\\hat{y}}\\end{array}$. From a quantum perspective, circularly polarized light consists of photons that possess non-zero spin angular momentum. The energy density of the electromagnetic field in a vacuum is known to be $\\begin{array}{r}{w=\\frac{1}{2}\\varepsilon_{0}\\vec{E}^{2}+\\frac{1}{2\\mu_{0}}\\vec{B}^{2}}\\end{array}$, and the spin angular momentum density is given by $\\vec{s}=\\varepsilon_{0}\\vec{E}\\times\\vec{A}$, where $\\vec{A}$ is the magnetic vector potential. Considering only the time-oscillating term of the magnetic vector potential, for a monochromatic plane electromagnetic wave, we have $\\vec E=-\\frac{\\partial \\vec A}{\\partial t}$.\\n\\nNow, let this beam of light be incident perpendicularly on a wave plate with thickness $d$ and cross-sectional area $S$. The refractive indices of the wave plate for linearly polarized light in the $x$ and $y$ directions are $n_{o}$ and $n_{e}$, respectively. In this case, the angular momentum of the photons is no longer an eigenstate in a fixed direction but is in a superposition state of two opposite spin directions. Calculate the magnitude of the torque required to stabilize this wave plate. In this problem, we do not consider the absorption and reflection of light, i.e., only the phase change after passing through the wave plate needs to be considered. The speed of light in a vacuum is $c$.", "solution": "", "answer": "" }, { "id": 391, "tag": "THERMODYNAMICS", "content": "An adiabatic container is connected to a vacuum through a small hole. Initially, the container holds an ideal gas with a molar amount of $n_{0}$. The gas leaks very slowly, and the gas in the container is always in equilibrium. It is known that the adiabatic equation for an ideal gas is $p V^{\\gamma} = \\text{constant}$, where $\\gamma = \\frac{C_{v} + R}{C_{v}}$.\n\nAssuming the container holds a single type of ideal gas with a molar heat capacity at constant volume of $C_{\\scriptscriptstyle V} = \\frac{5}{2}R$, find the molar amount $n_{1/2}$ remaining in the container when the pressure in the container is halved.", "solution": "", "answer": "" }, { "id": 738, "tag": "MECHANICS", "content": "A water bottle with a height of $H$ is filled with water. One side of the bottle is uniformly distributed with several small holes, with the number of small holes per unit length being $n$. Each small hole has an area of $s$ (the dimensions of the small hole are much smaller than $H$). Each small hole sprays water outward, and it is assumed that the direction of water sprayed from the small holes is along the horizontal direction. During the spraying process, the decrease in water level is not considered. Find the amount of water, per unit time and per unit length, that hits the ground at a distance $x$ from the side of the water bottle that is spraying water. (Do not consider the process of water splashing back up after hitting the ground, just provide the expression for the area that can receive the water.\n", "solution": "", "answer": "" }, { "id": 705, "tag": "ELECTRICITY", "content": "Placed on the \\(xOy\\) plane is a fixed square wire loop carrying a constant current \\(I\\), with a side length of \\(2a\\). At its center, there is a small magnetic sphere, with a radius \\(r (r \\ll a)\\), mass \\(m\\), and relative permeability \\(\\mu_r\\). We assume the magnetization \\(\\vec{M}\\) of the sphere is nearly unchanged. (The magnetization \\(\\vec{M}\\) is not known.) The sphere's movement is constrained along the \\(z\\) axis, and it is subjected to a disturbance moving along the \\(z\\) axis. Find the period of small oscillations \\(T\\).", "solution": "", "answer": "" }, { "id": 402, "tag": "MECHANICS", "content": "In medical physics experiments, the capillary method is commonly used to measure the viscosity of low-viscosity liquids. The Ostwald viscometer, due to its ease of construction and simple operation, along with relatively high measurement accuracy, has become a common instrument for determining liquid viscosity using the capillary method. It is particularly suitable for studying liquids with small viscosity coefficients, such as water, gasoline, alcohol, plasma, or serum. It is widely used in clinical and pharmaceutical industries. The Ostwald viscometer consists of a $U$-shaped glass tube, with one side having a larger diameter and having a large glass bulb $A$, and the other side having a smaller diameter and having a small glass bulb $B$. The lower end of the small glass bulb $B$ has a cylindrical capillary tube with a length of $L$ and an inner radius of $R$. There are markings $m$ and $n$ (painted in red) at the upper and lower ends of the small glass bulb $B$. The height difference between $A$ and $B$ is $\\Delta h$. During the experiment, the volume of liquid flowing through the capillary is the volume inside the small glass bulb $B$, which is the volume as the liquid level drops from $m$ to $n$. The average liquid level point is $^b$, and the liquid level point $a$ inside the large glass bulb $A$ is in the middle of the large glass bulb and remains basically unchanged during the experiment.\n\nNeglecting capillary effects, the flow of the liquid can be considered equivalent to viscous fluid performing laminar flow in a horizontal fine circular tube with constant cross-section (cross-sectional radius is $R$) and a length of $L$, with the velocity at the contact position between the fluid and the tube wall being zero. Given that the Newtonian viscosity coefficient is $\\eta$ and the capillary tube is placed vertically, derive the expression for the flow rate $q_{V}$ of the liquid in the capillary.", "solution": "", "answer": "" }, { "id": 595, "tag": "ELECTRICITY", "content": "There is a planar resistor network. The circuit is composed of 2014 sectors, with each side of the sector being a resistor with resistance $r$. Obviously, there are 2014 radii, with one end of the resistors on these radii connected at the center to form the center point; the other end is connected using arc resistors. Try to determine the resistance between two adjacent points on the sector arc in the circuit. (Hint: You can consider 2014 as a very large number and approximate it as an infinite network.)", "solution": "", "answer": "" }, { "id": 46, "tag": "MECHANICS", "content": "Considering the relativistic case of a two-dimensional elastic collision, a particle $m_{1}$ collides with $m_{2}$, with an initial velocity of $v_{0}$. Given that $m_{1} > m_{2}$, discuss the maximum angle between the exit angle of particle $m_{1}$ after the collision and its initial incident angle.", "solution": "", "answer": "" }, { "id": 203, "tag": "ELECTRICITY", "content": "In the $x y$ plane, there are electric dipoles $P_{2}$ and $P_{1}$ located at $(a,0)$ and $(-a,0)$, respectively. Their magnitudes are both $P$, and they form angles $\\varphi_{2}(t)$ and $\\varphi_{1}(t)$ with the $x$-axis. The permittivity of vacuum is given as $\\varepsilon_{0}$. \n\nConsider a special case where $\\varphi_{1}(t) = \\omega t$ and $\\varphi_{2}(t) = \\pi - \\omega t$. A point charge with charge $-q$ and mass $m$ is located at the origin and experiences a small perturbation in the $x$ direction. It is known to undergo harmonic oscillation. Determine the angular frequency, under the condition that $\\omega$ is much larger than the angular frequency of the perturbation experienced by the point charge but much smaller than the frequency range where radiation needs to be considered.", "solution": "", "answer": "" }, { "id": 397, "tag": "MECHANICS", "content": "A uniform thin rod is placed on the track of a rotating hyperbolic surface. The rod has a known length $L$ and mass $M$. The equation of the hyperbolic curve in the plane is: \n\n$$\n{\\frac{y^{2}}{a^{2}}}-{\\frac{x^{2}}{b^{2}}}=1\n$$ \n\nThe gravitational acceleration $g$ is known. \n\nGiven that the rod is in a non-horizontal stable equilibrium position, determine the angular frequency of small oscillations of the rod within the plane defined by the rod and the origin.", "solution": "", "answer": "" }, { "id": 480, "tag": "OPTICS", "content": "In the natural light Young's double-slit interference experiment, the front slit is called $S_0$, and the back double slits are called $S_1$ and $S_2$. Initially, no polarizers are placed at any positions. Now, polarizers $P_1$ and $P_2$ are placed behind slits $S_1$ and $S_2$, respectively, so that the transmission direction of $P_1$ forms an angle $\\theta$ with $P_2$. Find the contrast of the interference fringes $\\gamma$, expressed in terms of $\\theta$.", "solution": "", "answer": "" }, { "id": 671, "tag": "ELECTRICITY", "content": "There is an isolated metal sphere with a charge of 0, having a radius of $r$. At point $A$, located at a distance $a$ from the sphere's center $O$, there is an electric dipole. The magnitude of the dipole moment is $p$, and the angle between the direction of the dipole and $\\vec{OA}$ is $\\theta$. Find the electric potential $\\varphi(x)$ produced by the conductor sphere at a distance $x$ from the center $O$, in the direction along the line OA from the center to the dipole.", "solution": "", "answer": "" }, { "id": 606, "tag": "MODERN", "content": "In the upper half-plane of the $x$-axis, there is a magnetic field with strength $B$ directed into the page. A charged particle with rest mass $m_{0}$ and charge $-q (q > 0)$ is launched from the origin at high speed in the positive $y$-axis direction. A resistance with constant magnitude $q c B$ is applied opposite to the direction of velocity. It is known that the charged particle comes to a stop after turning an angle $\\theta_{0}(\\theta_{0} > \\frac{\\pi}{2})$ in the direction of velocity, and at point $A$ on the trajectory, the direction of the particle's velocity is parallel to the $x$-axis ($\\theta_{A}=\\frac{\\pi}{2}$). Now, a smooth track is laid along the trajectory of the charged particle (taking $\\theta_{0}=\\pi$). Another charged particle, also with rest mass $m_{0}$ and charge $-q$, is launched from the origin along the inside of the track with the same initial momentum. At this time, a resistance proportional to velocity is applied, given by ${\\vec{f}}_{0}=-q B{\\vec{v}}$. The question is to determine the pressure $N_{A}$ exerted by the charged ion on the track at point $A$.", "solution": "", "answer": "" }, { "id": 162, "tag": "ADVANCED", "content": "选用单位使得 $\\hbar = c = 1$。考虑一个量子非相对论粒子，具有质量 $m = 1$（在某些单位中），电荷 $e = 1$，且没有自旋。粒子被约束在一个平面 $\\mathbb{R}^2$ 上运动（具有欧几里得度量 $ds^2 = dx^2 + dy^2$），存在一个*与平面垂直的均匀磁场*。系统的哈密顿量是存在磁势 $(A_x, A_y)$ 时的标准形式，即\n\n$$\nH = \\frac{1}{2} \\left( (p_x - A_x)^2 + (p_y - A_y)^2 \\right)。\n$$\n\n磁场的强度*未知*，但通过直接实验发现在相同强度的磁场作用下，对相同质量和电荷的粒子，该粒子在平坦的 2-环面 $T^2 = \\mathbb{R}^2 / \\mathbb{Z}^2$ 上运动，会产生一个具有 100 个基态的量子系统。\n\n为确定性，我们固定规范条件\n\n$$\nA_y(x, y) = 0 \\quad \\text{和} \\quad A_x(x, 0) = 0。\n$$\n\n某实验人员在时间 $t = 0$ 准备了这个系统的三个副本，处于状态下，具有波函数（在上述规范中）\n\n$$\n\\psi(x, y; 0) = \\frac{1}{\\sqrt{\\pi}} \\exp \\left[ ikx - \\frac{1}{2}(x^2 + y^2) \\right] \\in L^2(\\mathbb{R}^2)，\n$$\n\n然后检查系统在不同时间 $t$ 的演化。要求学生帮助预测他会发现什么。\n\n写出在时间 $t = 1$ 的时间演化波函数 $\\psi(x, y; 1)$；", "solution": "", "answer": "" }, { "id": 295, "tag": "MECHANICS", "content": "A particle with mass $m$ is released from rest at an unknown height $h$ and slides down a smooth track, colliding with a particle of mass $M$ located on a smooth horizontal surface. The coefficient of restitution for the collision is $e$. After the collision, $M$ will enter a vertical circular track with radius $R$ and a coefficient of friction $\\mu$. This rough vertical circular track is tangent to the original smooth horizontal surface. The particle $M$ moves along the inner side of the track and detaches at the appropriate position to pass through the center of the circular track. The gravitational acceleration is $g$. Find the release height $h$ for mass $m$.", "solution": "", "answer": "" }, { "id": 368, "tag": "OPTICS", "content": "A thermometer is cylindrical (which can be considered as infinitely long), with an outer radius of $R = 20.00\\mathrm{mm}$ and an inner radius of $r = 15.00\\mathrm{mm}$. The region between the inner and outer walls consists of a transparent medium with refractive index $n$ to be measured. Inside the inner radius is opaque mercury that is capable of completely reflecting light. **The result should be expressed with four significant figures**.\n\nThe refractive index of the medium is measured as follows. A distant object is placed far away, and the line of sight $P$ is fixed in the direction of that object without adjusting the viewing direction anymore. The thermometer is inserted between the object and the line of sight, and the perpendicular distance $h$ from the center of the thermometer to the line of sight is gradually reduced. When this distance is $19.00\\mathrm{mm}$, the human eye will first see the image of the object again along the line of sight. Based on this, the refractive index $n$ can be calculated. Furthermore, as $h$ continues to decrease, other possible images of the object can still be observed. Calculate the displacement $\\Delta h$ of the thermometer along the $h$ direction when the next image of the object is seen.", "solution": "", "answer": "" }, { "id": 292, "tag": "ELECTRICITY", "content": "In space, there are two homogeneous concentric spherical shells, each with its center of mass at rest. The inner shell has a mass of $m_1$, radius of $R_1$, and is uniformly charged with a charge of $q_1$. The outer shell has a mass of $m_2$, radius of $R_2$, and is uniformly charged with a charge of $q_2$. The relationship $\\frac{q_i^2}{4\\pi\\varepsilon_0 R_i c^2}\\ll m_i$ holds. After consideration, we find that the motion of the two shells should satisfy the following manner: both the inner and outer shells undergo precession around a common \"axis.\" Given that initially both centers of mass are at rest and the angular velocities are $\\vec{\\omega_1} = \\omega_1 \\hat{x}$ and $\\vec{\\omega_2} = \\omega_2 \\hat{y}$, we seek to determine the subsequent evolution of $\\vec{\\omega_1}$ for the inner shell.", "solution": "", "answer": "" }, { "id": 513, "tag": "MECHANICS", "content": "Consider an arrangement with $n$ ($n \\geq 3$) identical springs, where one end of each spring is attached to a small ball of mass $m$, and the other end is evenly distributed along the circumference of a rigid circular ring with radius $b$ (i.e., the angular spacing between adjacent spring endpoints is $\\frac{2\\pi}{n}$). The small ball is initially located at the center of the circular ring. The natural length of each spring is $a$ ($a \\ll b$), with a spring constant of $k$. Gravity is neglected, and the rigid circular ring remains stationary. When the small ball moves within the plane of the circular ring, determine the angular frequency $\\omega$ of the small ball’s small oscillations near its equilibrium position.", "solution": "", "answer": "" }, { "id": 156, "tag": "THERMODYNAMICS", "content": "When a disturbance propagates in a medium at a speed faster than the speed of sound in the medium, a shock wave forms. \nThe Mach number $M$ of a disturbance is defined as the ratio of the propagation speed of the disturbance in the medium to the speed of sound in the medium.\n\nDue to the motion of the shock wave, there is a discontinuity in physical quantities on either side of the shock wave front.\n\nChoose the shock wave reference frame. Let the upstream gas have a velocity magnitude $v_{1}$ (positive when directed opposite to the wave front), density $\\rho_{1}$, temperature $T_{1}$, and specific internal energy $\\boldsymbol{u}_{1}$. Let the downstream gas have a velocity magnitude $v_{2}$ (positive when directed opposite to the wave front), density $\\rho_{2}$, temperature $T_{2}$, and specific internal energy $u_{2}$. You can derive the conservation laws on both sides of the shock wave front.\n\nTreat the gas as an ideal gas and assume the adiabatic index is $\\gamma$. \n\nIn the ground reference frame, the upstream gas is stationary ($w_{1}=0$), the velocity magnitude of the shock wave front is $w$, and the downstream gas velocity magnitude is $w_{2}$ (positive in the same direction as the wave front's motion).\n\nThe speed of sound in an ideal gas is given as $v_{s} = \\sqrt{\\frac{\\gamma RT}{\\mu}}$, and the Mach number of the shock wave is $M$.\n\nFind $w_{2}$ in terms of $w$, $\\gamma$, and $M$.", "solution": "", "answer": "" }, { "id": 377, "tag": "THERMODYNAMICS", "content": "At $t=0$, an adiabatic ,thin and lightweight piston with a cross-sectional area of $S$ divides an adiabatic cylinder into two regions, each with an initial volume $V_0$. The left side contains an ideal gas with an adiabatic index $\\gamma=3/2$ and initial pressure $p_0$. The cylinder walls are coated with a specific optional viscous material of negligible heat capacity and volume, characterized as follows: when the piston moves in a direction at a speed $v$, the material exerts a resistive force $F=-k v$ on the piston. The heat generated due to friction can be transmitted in two distinct ways:\n\n- For forward-transmitting viscous material, the heat is conducted forward in the direction of the piston’s motion and absorbed by the gas on that side (or conducted externally if no gas is present).\n- For reverse-transmitting viscous material, the heat is conducted backward against the direction of the piston’s motion and absorbed by the gas on that side (or conducted externally if no gas is present).\n\nConsider the following problem (retain the results in analytic form, without converting to numerical values): \n\nThe region on the right side of the container is vacuum, and reverse-transmitting viscous material is used. Calculate the time $t$ required for the piston to move to the far-right position.", "solution": "", "answer": "" }, { "id": 441, "tag": "ELECTRICITY", "content": "In a vacuum, two infinitely long, uniformly charged thin-wall coaxial cylinders are placed. The radii of the cylinders are $r_{1}$ and $r_{2}$, respectively. In the direction parallel to the cylinder axis, the mass per unit length of both the inner and outer cylinders is $m$, and the charge per unit length of the inner and outer cylinders is $q$ $(q > 0)$ and $-q$, respectively. Both cylinders can freely rotate around their respective central axes. The permittivity and permeability of the vacuum are given as $\\varepsilon_{0}$ and $\\mu_{\\mathrm{0}}$, respectively. $r_1 < r_2$\n\nInitially, both the inner and outer cylinders are stationary. If a torque is applied to the outer cylinder, causing it to begin rotating, calculate the total mechanical angular momentum $L$ along the axis per unit length of the inner and outer cylinders when the angular velocity of the outer cylinder reaches $\\varOmega$.", "solution": "", "answer": "" }, { "id": 530, "tag": "MECHANICS", "content": "2. Frictional Transmission\n\nA large disk with a sufficient radius acts as the driving wheel and rotates counterclockwise around a fixed axis $o$ with a constant angular velocity $\\Omega$. There is another small disk (with a radius of $r$) that can freely rotate around a fixed axis $O_{1}$, which acts as the driven wheel. This small disk is pressed against the large disk with a force of magnitude $N$ by the axis $O_{1}$. If the large and small disks are coaxial, the small disk will be driven by the large disk and rotate with the same angular velocity. Now we consider the situation where the distance between the two axes is $d \\neq 0$. It is assumed that the pressure between the two disks is uniformly distributed. We can also add lubricant between the two disks, turning the friction between them into \"wet friction.\" The characteristic of this is: for two mutually contacting surface elements of size $\\Delta\\sigma$, the wet friction force on one of them can be expressed as\n\n$$\n\\Delta f = -\\gamma v_{r} \\Delta\\sigma\n$$\n\nwhere $v_{r}$ is the magnitude of the relative velocity between the two surface elements, $\\gamma$ is the viscosity coefficient, and the negative sign indicates that the force opposes the direction of relative motion. When the angular velocity of the small disk reaches $\\omega$ (but has not necessarily reached stable rotation yet), find the heating power at the lubricant $P_{Q}$.", "solution": "", "answer": "" }, { "id": 274, "tag": "MECHANICS", "content": "A sufficiently rough conical surface with a half-apex angle of $\\alpha$ is fixed vertically, with the cone's sharp angle pointing straight down. A homogeneous small ball with a radius of $R$ is placed on it, and the distance from the ball's center to point $O^{'}$ is denoted as $r$. The gravitational acceleration $g$ is known. It is assumed that the small ball will never leave the cone surface. A line parallel to the cone surface passes through the ball's center and intersects the central axis of the cone surface at point $O'$, with the spin angular velocity and the line connecting the ball's center and $O'$ forming an angle $\\phi$. It is known that $\\begin{array}{r}{\\alpha=\\frac{\\pi}{4}.}\\end{array}$ When the ball is in stable precession, it satisfies $r_{0}=4R$. Let the angle between the spin angular velocity and the cone surface be $\\varphi$. If a radial perturbation is applied to the ball, causing its center to oscillate back and forth near $r_{0}$, find the oscillation period and the range of values for $\\varphi$. Note: The angular velocity vector of the small ball in the ground frame is the sum of the precession angular velocity and the spin angular velocity.", "solution": "", "answer": "" }, { "id": 610, "tag": "THERMODYNAMICS", "content": "The Earth is round, and the street is also a circle with a length of $L$. There are 2024 point masses (completely elastic) on the street, and their tangential velocity distribution along the street follows the Maxwell velocity distribution. The probability that the velocity is between $v$ and $v+\\mathrm{d}v$ is: $$ \\mathrm{d}p={\\frac{1}{\\sqrt{2\\pi}\\sigma}}\\mathrm{e}^{-{\\frac{v^{2}}{2\\sigma^{2}}}}\\mathrm{d}v $$ Neglecting the possibility of three-body collisions, calculate the average frequency of elastic collisions between the point masses in the system.", "solution": "", "answer": "" }, { "id": 111, "tag": "MECHANICS", "content": "Under zero gravity, a soft rope with a mass line density of $\\lambda$ is used for rope skipping. The rope is in the same plane, and an orthogonal coordinate system $xOy$ is set up in this plane. The two endpoints of the rope are symmetrical about the $y$-axis. The ends of the rope are connected to handles, and the direction of the handles is perpendicular to the tangent direction of the rope ends. The angular velocity $\\omega$ of the rope rotation is along the positive direction of the ${\\pmb x}$-axis. The lowest point of the rope is at a distance $h$ from point $O$, and the tension at the lowest point is $T_{0}$. If $\\lambda\\omega^{2}h^{2}=2T_{0}$, find the angle between the handle on the left side and the positive direction of the $x$-axis.", "solution": "", "answer": "" }, { "id": 344, "tag": "MECHANICS", "content": "The dynamics of the following unique system are considered: A homogeneous cylindrical shell with a mass of $M$ and a radius of $R$ can freely rotate around its center $O$. Inside it, there is another smaller homogeneous cylinder with a mass of $m$ and an undetermined radius $r$. The two cylinders are completely rough, maintaining no relative sliding either during continuous motion or during short impact time intervals. The gravitational acceleration $g$ is vertically downward. Initially, the small cylinder is at the lowest point of the large cylinder, and both cylinders are at rest. Suddenly, a very brief impulse $I$ is applied at the center $O^{\\prime}$ of the small cylinder. The various angles of rotation during the impact process are negligible.\n\nWe find that if, at the initial moment, a point $P$ fixed to the small cylinder is located at a distance $l$ directly beneath $O^{\\prime}$ (possibly outside the cylinder), then, under appropriate initial conditions, this point will keep moving up and down vertically, and the small cylinder will never detach from the large cylinder. Find the value of impulse $I$, expressed solely in terms of $R, r, M$.", "solution": "", "answer": "" }, { "id": 593, "tag": "ELECTRICITY", "content": "A proton (charge \\(+e\\)) moves with a speed \\(v\\) parallel to an infinitely long, thin wire at a distance \\(r\\) from the wire's axis. The wire carries both a current \\(I\\) and a positive charge per unit length \\(\\lambda\\). If the net force on the proton due to the electric and magnetic fields is zero, allowing it to move in uniform linear motion parallel to the wire, find the relationship between the proton's speed \\(v\\) and the other parameters. Provide the final answer as a clear expression. The speed of light is given as \\(c\\).", "solution": "", "answer": "" }, { "id": 616, "tag": "OPTICS", "content": "In recent years, with the discovery of numerous planets orbiting stars, observing exoplanets at astronomical distances has become a challenge for scientists. Gravitational microlensing is one of the detection methods. It utilizes a principle discovered by Einstein in general relativity: when light passes by a spherically symmetric object with mass \\( M \\), its direction is deflected toward the object by a small angle \\(\\alpha = \\frac{4GM}{rc^2}\\), where \\( G \\) is the gravitational constant, \\( c \\) is the speed of light, and \\( r \\) is the shortest distance between the light and the object. In this problem, we will study the principle of detecting exoplanets through the microlensing effect. Consider a distant star \\( S \\), at a distance \\( D_s \\) from Earth \\( E \\), as the source star. Another star \\( L \\), with mass \\( M \\), is at a distance \\( D_l \\) from Earth (\\( D_l < D_s \\)), serving as the lens. Consider the situation where the lens star and the source star are aligned (the source star, lens star, and the observer on Earth are on the same straight line). To the observer on Earth, the image of \\( S \\) appears as a ring, known as the Einstein ring. Derive the expression for the angular radius of the Einstein ring using the parameters given in the problem.", "solution": "", "answer": "" }, { "id": 393, "tag": "OPTICS", "content": "A point light source is placed at the center of a spherical screen with a radius of $R$. A planar object is placed inside the sphere, and the distance from the light source to this plane is $d$. Establish a rectangular coordinate system $xOy$ on the plane, with the origin at the projection point of the light source on the plane. Calculate the relationship between the magnification of the image projected onto the spherical screen and the coordinates of the object plane. Here, magnification is defined as the ratio of the differential area of the projection on the screen to the differential area of the projection on the object plane before projection.", "solution": "", "answer": "" }, { "id": 109, "tag": "THERMODYNAMICS", "content": "A rigid adiabatic rectangular container with length $L$ (the thickness of the container walls can be ignored) is filled with an ideal gas composed of monoatomic molecules, where the mass of each gas molecule is $m$, and the initial temperature is $T_0$. Now, the container suddenly accelerates uniformly with an acceleration $a$ in the direction of $L$ ($maL \\ll kT_0$, where $k$ is the Boltzmann constant). Try to find the temperature $T$ of the system after reaching a stable state.", "solution": "", "answer": "" }, { "id": 784, "tag": "THERMODYNAMICS", "content": "Due to the sudden drop in temperature, several small ice crystals have appeared in the cylinder, severely affecting operational safety. Now, an external force $F$ is applied to slowly compress and melt all the ice into water, which is then expelled by centrifugal force. The atmospheric pressure is $p_0$. Given that the area of the piston in the cylinder is $S$ (the mass of the piston can be neglected), it can slide smoothly and freely within the cylinder. Initially, the cylinder is in equilibrium, the gas volume is $V_0$, the temperature is $T_0$, and the total mass of the ice crystals is $m$. The gas is known to be a diatomic ideal gas, the specific heat capacity of ice is $c$, the latent heat of fusion is $l$, and the melting point is $T_{ice}$ (approximately constant). Ignore the volume of the solid and liquid; the cylinder is considered adiabatic. By slowly compressing the substances in the cylinder, all the ice is melted. Find the final state gas pressure.", "solution": "", "answer": "" }, { "id": 209, "tag": "THERMODYNAMICS", "content": "Consider an adiabatic bubble wall with a constant surface tension coefficient $\\sigma$. The external pressure is much smaller than the additional pressure and can be regarded as 0. Let us examine the process of first injecting gas into the bubble and then expelling the gas from the bubble. This problem focuses on the gas injection process. Initially, the bubble's radius is denoted as 0, and the temperature of all the gas prior to injection is uniform (to be determined). After injection, a bubble with a radius of $r_{0}$ is formed, and the temperature of the gas is $T_{0}$. It is known that the adiabatic index of the gas is $\\gamma$. Neglecting heat exchange between the liquid and the gas, find the temperature of the gas before injection, denoted as $T_{1}$.", "solution": "", "answer": "" }, { "id": 201, "tag": "ELECTRICITY", "content": "Two smooth sliders with charges $Q$ and $4Q$, respectively, can freely slide on a horizontal straight insulated rail. The volume of the sliders can be ignored, and they are connected by two light, insulating strings of equal length $l$ to a massive particle with mass $m$. When the system is in static equilibrium, the angle between the strings and the horizontal direction is $ \\theta_{0}=60^{ \\circ}$. Now, introduce another slider with charge $2Q$ between the two existing sliders and keep it fixed in position. Due to factors such as air resistance, the system will settle into a new equilibrium configuration after a dynamic process. Calculate the angle $ \\theta$ between the strings and the horizontal direction at this new equilibrium.", "solution": "", "answer": "" }, { "id": 338, "tag": "ELECTRICITY", "content": "In the spatial rectangular coordinate system, point charges each with a charge amount of $+Q$ are fixed at $(0,0,-a)$ and $(0,0,a)$, and there is a free point charge with a charge amount of $+q$ at the origin. $(Q, q > 0)$. Electrostatic force alone cannot make the charge be in stable equilibrium. Therefore, to stabilize the charge (restrict the range of charge motion), a magnetic field will also be applied. If a uniform magnetic field is applied in the $z$ direction, with the magnitude $B$ (sufficiently large such that the charge is in stable equilibrium at the origin). If the $+q$ point charge leaves the origin with a very small initial velocity (the direction is arbitrary), try to discuss the condition under which the point charge can pass through the origin again in terms of the magnetic field's magnitude (limited to the meaning of small oscillations). Hint: You may use two unknown integers in your answer, denoted by $n_{1}$, $n_{2}$.", "solution": "", "answer": "" }, { "id": 411, "tag": "MODERN", "content": "A particle with rest mass $m$ is projected with an initial velocity of magnitude $v_0$ at a horizontal angle of elevation $\\theta$. During its motion, the particle is subject to a constant gravitational force $F$. Considering relativistic effects, strictly solve for its trajectory equation (express the result as $y(x)$, ensuring that the answer contains only the known quantities $u_{0}= \\frac {v_{0}}{\\sqrt {1- v_{0}^{2}/ c^{2}}} , a_{0}= \\frac Fm,\\theta$).", "solution": "", "answer": "" }, { "id": 787, "tag": "THERMODYNAMICS", "content": "The mechanical properties of the rubber membrane are similar to the surface layer in thermodynamics. In the first approximation, the tension within the rubber membrane can be described using the surface tension coefficient (a layer of rubber membrane is equivalent to a layer of surface layer). (The refractive index of the membrane is considered the same as that of water, and the membrane is very thin.) The tension formula is as follows: \\[ F = \\sigma l \\] --- A hollow cylinder with height \\(L \\) and radius \\(R \\) has its two ends covered with rubber membranes of the same surface tension coefficient. An ideal gas with pressure \\(p_0 \\) is sealed inside, exerting pressure when the membrane is taut. The cylinder is submerged in a pool to a depth \\(H \\) (with \\(L \\ll H \\) assumed, the liquid additional pressure on both membranes is the same), and when the cylinder is at a distance of \\(H/3 \\) from the water surface, the bottom of the pool is precisely imaged at the surface. The temperature of the water is independent of depth, the cylinder has good thermal conductivity, the refractive index of water is \\(n \\), its density is \\( \\rho \\), the gravitational acceleration is \\(g \\), and the atmospheric pressure is \\(p_0 \\). If the cylinder in water can be considered as a **thin lens**, and the pressure of the gas inside can be considered unchanged (i.e., the membrane is relatively rigid and the air volume is approximately unchanged). Continue to submerge the cylinder, and the pool bottom can still be precisely imaged at the water surface. Determine the distance from the water surface to the cylinder when it images again.", "solution": "", "answer": "" }, { "id": 240, "tag": "MECHANICS", "content": "It might seem that the gyroscopic effect is the reason why bicycles don't fall over, but in fact, the mass of the spinning wheels is not very large, and the gyroscopic effect it generates is not sufficient. Moreover, when people equipped bicycle wheels with counter-rotating wheels that provided equal but opposite angular momentum, it was found that bicycles still maintained stability. These facts are enough to show that the gyroscopic effect is not the absolute reason for a bicycle's stability. In fact, the decisive reason that bicycles don't fall is \"centrifugal force.\" (For ease of calculation, this question does not need to consider the influence of the gyroscopic effect or Coriolis torque in rotating systems.)\n\nA bicycle can be regarded as an inverted pendulum (unstable from side to side), requiring additional restorative forces to maintain balance, and the restorative force provided is precisely the \"centrifugal force\" when the bicycle is steering.\n\nCentrifugal force is a function of speed and handlebar steering angle. At a fixed speed, it can be considered that controlling the handlebar steering angle means controlling the restorative force.\n\nFirst, please imagine yourself riding a bicycle. At some moment, you notice that you have a tendency to tip over to the right. To prevent yourself from falling, which way will you turn the handlebars?\n\nThen please calculate the model below.\n\nThe bicycle is divided into four parts as rigid bodies: front wheel, rear wheel, rear frame and rider, front frame and handlebars. Both the front and rear wheels can be considered homogeneous rings with mass $m$ and radius $r$. The distance between the centers of the front and rear wheels is $L$. The mass of the rear frame and rider is $M_{1}$, with the center of mass at the coordinates $(x_{1},z_{1})$. The mass of the front frame and handlebars is $M_{2}$, with the center of mass at the coordinates $(x_{2},z_{2})$. \n\nTry to solve the minimum angle $\\alpha (\\ll 1)$ within the horizontal plane that must be rotated by the handlebars when the bicycle-rider system is moving forward at a constant speed $v$ and has a tilt angle $\\theta$ from the vertical balanced position.", "solution": "", "answer": "" }, { "id": 783, "tag": "ELECTRICITY", "content": "Consider a **lightweight solid ball** rotating around a fixed axis in a vacuum. The radius of the ball is \\( R \\), it is **uniformly charged**, and the volume charge density is \\( \\rho \\). Let the axis be the \\( z \\) axis, and establish a spherical coordinate system \\( (r, \\theta, \\varphi) \\), not considering relativistic effects. The ball is initially stationary, and a **positive external torque along the \\( z \\) axis** is applied to the ball, causing it to rotate with an **angular acceleration** \\( \\beta \\) along the \\( +z \\) direction, and at some moment the angular velocity is \\( \\omega \\). The vortex electric field in space is only in the \\( \\varphi \\) direction. Based on this, calculate the magnitude of the required external torque \\( M \\). Appendix: Conclusions that might be used in this problem: - The **energy density** of the electromagnetic field: \\[ w = \\frac{1}{2} \\varepsilon_0 E^2 + \\frac{1}{2 \\mu_0} B^2 \\] - Current density (energy flux density): \\[ \\vec{S} = \\vec{E} \\times \\vec{H} \\] - Momentum density: \\[ \\vec{g} = \\vec{D} \\times \\vec{B} \\] Where: - \\( \\vec{E} \\): Electric field strength - \\( \\vec{B} \\): Magnetic induction strength - \\( \\vec{D} \\): Electric displacement vector - \\( \\vec{H} \\): Magnetic field strength", "solution": "", "answer": "" }, { "id": 386, "tag": "THERMODYNAMICS", "content": "At time $t=0$, an adiabatic, thin and lightweight piston divides an adiabatic cylinder with a cross-sectional area $S$ into two equal parts, each having a volume $V_0$. On the left side, the cylinder contains an ideal gas with an adiabatic index $\\gamma = 3/2$ and an initial pressure $p_0$. The cylinder's walls are equipped with a type of viscous material possessing negligible heat capacity and volume. The viscous material has the following properties: when the piston moves at a velocity $v$ in a certain direction, it exerts a resistive force on the piston given by $F = -k v$. At the same time, the heat generated by friction has two possible behaviors. For **pro-directional viscous material**, the heat is conducted forward along the original velocity direction and absorbed by the gas on that side (or transferred out of the system if no gas is present). For **anti-directional viscous material**, the heat is conducted backward opposite to the original velocity direction and absorbed by the gas on that side (or transferred out of the system if no gas is present). Consider the following problem (retain results in analytical form without converting them to numerical values):\n\nThe right side of the container holds the same type of gas with a pressure of $2p_0$, and an anti-directional viscous material is used. Calculate the displacement $l$ of the piston after it comes to a stop.", "solution": "", "answer": "" }, { "id": 119, "tag": "ELECTRICITY", "content": "On an infinite superconductor plate placed horizontally on a surface with a friction coefficient of $\\mu$, there is a uniformly charged insulating spherical shell with total charge $Q$, radius $R$, and mass $M$. Fixed smooth baffles are placed on both sides of the spherical shell to prevent its center of mass from moving horizontally. It is known that in the subsequent motion, the sphere will not leave the superconducting plate. Now, a sphere is given an initial angular velocity of $\\pmb{\\omega}$. The magnetic permeability in a vacuum is $\\mu_0$, and the direction of $\\omega$ is vertical. Try to determine the relationship between the angular displacement and the angular velocity of the sphere afterward (considering the sphere will not leave the superconducting plate).", "solution": "", "answer": "" }, { "id": 780, "tag": "MECHANICS", "content": "In this problem, we are given a 0-length elastic string, with a spring constant \\(k \\) that is constant. The original length is 0, and the mass distribution is uniform (it can be considered that the original length is almost 0). The total mass is \\(m \\), and it is hanging in Earth's gravitational field. We need to find its shape under the following circumstances. If the string is far from the Earth's surface, and the stretched length is close to Earth's size, with Earth's mass being \\(M \\), the gravitational constant \\(G \\), the suspension points are at a distance \\(R \\) from Earth's center, and the angle between the line connecting to Earth's center is \\( \\varphi \\), we need to find the shape of the string. It is not necessary to consider the contact between the string and the Earth's surface. (This question uses polar coordinates \\((r , \\theta) \\), the string is symmetric about the polar axis, passing through \\((R, \\frac{ \\varphi}{2}) \\) and \\((R,- \\frac{ \\varphi}{2}) \\)). Due to the difficulty in providing internal force and deformation boundary conditions for this problem, we directly give an additional geometric boundary condition. That is, the angle between the string tangent at the suspension point and the line to Earth's center is \\( \\alpha \\). The final equation of the curve should be expressed in terms of \\( \\alpha \\), \\( \\varphi \\), and \\(R \\).", "solution": "", "answer": "" }, { "id": 438, "tag": "MECHANICS", "content": "There is a table with six legs placed on a horizontal surface. The length of the table is 2a, with three legs on the longer side: two located at the corners and one located at the midpoint of the long side. The width of the table is a. The tabletop is rigid, and the legs are elastic with very small elastic deformation. The self-weight of the table is neglected. Now, apply a concentrated force P at a certain point on the tabletop. Please try to calculate the total area on the table where the force P can be applied when only 4 out of the 6 table legs have internal pressure.", "solution": "", "answer": "" }, { "id": 461, "tag": "MECHANICS", "content": "On a rainy day, the process of raindrops falling is quite complex, involving knowledge of dynamics, fluid mechanics, phase transition, and other fields. By combining some actual phenomena and experience, we can establish a simple model to simulate the falling of droplets. Gravity causes a condensation nucleus to fall in uniform still water vapor, and it can absorb a mass of water vapor equal to \\( a \\) per unit distance fallen. The initial velocity and mass of the condensation nucleus, as well as the viscous force during the raindrop's fall, can be ignored. Find the speed \\( v \\) of the raindrop after falling vertically for \\( t \\) seconds (before reaching the ground). \\( g \\) is the constant gravitational acceleration.\n", "solution": "", "answer": "" }, { "id": 357, "tag": "MECHANICS", "content": "A rope with length $L$ and mass $m$ is lying on a rough horizontal surface (the rope can be considered as concentrated at one point), and the coefficient of friction between the rope and the horizontal surface is $\\mu$. Now, a horizontal constant force $F$ is applied to one end of the rope, gradually pulling the rope straight. Given that the acceleration due to gravity is $g$, find the velocity $v$ of the rope at the moment it is completely straightened.", "solution": "", "answer": "" }, { "id": 735, "tag": "OPTICS", "content": "The design of the laser targeting and strike system needs to consider changes in the air's refractive index. Due to the influence of various factors such as surface conditions, altitude, temperature, humidity, and air density, the distribution of the air's refractive index in the atmosphere is uneven, and hence the propagation path of the laser is not a straight line. For simplicity, assume that the variation of the air's refractive index with height \\( y \\) at a certain location is as follows: $$ n^2 = n_0^2 + \\alpha^2 y $$ Where: \\( n_0 \\) is the refractive index of air at \\( y = 0 \\) (ground level); \\( n_0 \\) and \\( \\alpha \\) are known constants greater than zero; The propagation time of the laser itself can be ignored; The laser emitter is located at the coordinate origin. If the emission direction of the laser forms an angle \\( \\theta_0 (0 \\leq \\theta_0 \\leq \\frac{\\pi}{2}) \\) with the vertical \\( y \\)-axis, find the expression for the height \\( y \\) when the horizontal propagation distance is \\( x \\). Express the result using the following physical quantities: Refractive index constant \\( n_0 \\), \\( \\alpha \\), emission angle \\( \\theta_0 \\), and horizontal propagation distance \\( x \\).", "solution": "", "answer": "" }, { "id": 358, "tag": "THERMODYNAMICS", "content": "A steel chain is composed of N oval-shaped steel rings connected together, with the rings having a long axis length of \\( l+a \\) and a short axis length of \\( l-a \\). The mass of the steel rings can be ignored. The upper end of the steel chain is fixed to the ceiling, and the lower end is connected to a weight with mass \\( m \\). It is known that the steel rings can only be in \"State 1\" (long axis oriented vertically) and \"State 2\" (long axis oriented horizontally). If there are \\( n \\) steel rings in State 1 and \\( N-n \\) steel rings in State 2, the energy of the system at this time is denoted as \\( E_{n} \\), with the potential energy corresponding to \\( n=0 \\) as the reference point for potential energy. Let \\( \\varOmega_{n} \\) be the number of states of the system with energy \\( E_{n} \\) (i.e. \\( E_{i}=E_{n} \\)). Calculate the partition function \\( Z=\\sum_{n}\\varOmega_{n}e^{-\\beta E_{n}} \\).", "solution": "", "answer": "" }, { "id": 390, "tag": "MECHANICS", "content": "A small ball with mass $m$ (considered as a point mass) is tied to one end of a very long but inextensible light string and placed on a smooth horizontal surface. The other end of the string is fixed to a vertical, stationary cylinder with radius $R$. Initially, the string wraps around the cylinder exactly once, and the ball is snug against the cylinder at point $P$. Let the center of the cylinder on this horizontal cross-section be O. At time $t=0$, an external force imparts an impulse to the ball in the radial direction, giving the ball an initial speed $v_0$, causing the ball and string to begin unwinding from the cylinder surface. Label the instant contact point where the string begins unwinding from the cylinder as $Q$. $\\theta$ denotes the angle between OP and OQ (i.e., the angle the segment OQ has rotated relative to OP on the cylinder). \n\nFor the case where the cylinder is completely free: Assume the cylinder has a mass $M$, radius $R$, and its mass is concentrated in a thin layer on the surface. Now, the cylinder is not constrained and can move freely on the smooth horizontal surface (the cylinder does not tip over). At time $t=0$, an external force imparts an impulse to the ball in the radial direction, giving the ball an initial speed $v_0$, causing the ball and string to begin unwinding from the cylinder surface. Use the angle $\\varphi$ to represent the angle PO has rotated relative to the horizontal plane. Determine $\\theta$ as a function of time under this condition.", "solution": "", "answer": "" }, { "id": 621, "tag": "THERMODYNAMICS", "content": "Consider a submarine that discharges a bubble of oxygen (molecular weight $32$, diatomic gas) at a temperature of ${T}_{1}$ and volume ${{V}}_{0}$ from a depth $h$ on the seabed, with an initial velocity of 0. The atmospheric pressure at the sea surface is ${\\mathrm{p}}_{0}$, and the temperature distribution below the sea surface is given by $\\begin{array}{r}{{T}_{{x}}={T}_{0}-\\frac{a}{h}x}\\end{array}$, where $x$ represents the depth at that point. The density of water is $\\rho$. Establish a simple model in which the underwater expansion and contraction of the gas, as well as heat equilibrium, occur quickly in a quasi-static process, and determine the amount of heat absorbed (released) throughout the process $\\Delta Q$. The acceleration due to gravity is a constant $g$, and the oxygen is assumed to be an ideal gas. Express your answer using $T_0,T_1,p,\\rho,g,h,a$.", "solution": "", "answer": "" }, { "id": 635, "tag": "MECHANICS", "content": "A person riding a bicycle can be simplified into the following model: A homogeneous rod of length 2L and mass 2m, with its two ends rigidly connected to two homogeneous rods of length L and mass m, respectively. The short rods are perpendicular to the long rod, and their ends are connected to wheels which have negligible volume and mass. The wheels are in contact with the ground. The person is considered as a point mass located at the center of the long rod, with a mass of m. If at a certain moment, the front wheel suddenly hits a small stone (with negligible size), causing the front wheel's speed to instantly become zero, what should be the constant speed V of the bicycle for the person and the bicycle to just avoid contact at this instant?", "solution": "", "answer": "" }, { "id": 459, "tag": "ELECTRICITY", "content": "The magnetoresistance effect refers to the phenomenon where the resistance value changes when a magnetic field is applied to a metal or semiconductor through which current is flowing. This phenomenon is called the magneto-resistance effect, also known as magnetoresistance (MR). Let the resistivity of the sample be $\\rho_0$ when no magnetic field is applied, and $\\rho_B$ when an external magnetic field is present. The magnitude of the magnetoresistance effect can be expressed as: $MR = \\frac{\\Delta\\rho}{\\rho_0}=\\frac{\\rho_B - \\rho_0}{\\rho_0}$.\n\nThe magnetoresistance effect in metals can usually be explained using a two-carrier model: In metals, there are two types of charge carriers, with mobilities $\\mu_1$ and $\\mu_2$ (mobility $=\\frac{\\text{drift velocity of the carrier}}{\\text{magnitude of the field}}$), and corresponding conductivities $\\sigma_1$ and $\\sigma_2$. Derive the relationship between the magnitude of the magnetoresistance effect and the external magnetic field strength $B$. It is known that the external magnetic field is perpendicular to the current direction, and the change in resistivity $\\Delta\\rho$ is much smaller than $\\rho_0$.", "solution": "", "answer": "" }, { "id": 34, "tag": "THERMODYNAMICS", "content": "Consider a cylinder with its axis along the vertical direction, with the base fixed on a horizontal surface and a piston sealing the top. The seal between the cylinder and the piston is perfect, without friction, and both are adiabatic. Inside the cylinder, there is a monoatomic gas composed of molecules with molecular mass $m$ and total molecule count $N_{0}$, at a temperature of $T_{0}$. The exterior of the cylinder is vacuum, and at this moment, the piston remains stationary. Let the gravitational acceleration be $g$, the bottom area of the cylinder be $S$, and the initial distance between the piston and the horizontal surface be $h$. The gas follows the Boltzmann distribution, where the probability distribution is proportional to $e^{\\frac{E_{p}}{k T}}$, with $E_{p}$ representing gravitational potential energy and $k$ being the Boltzmann constant. \n\nIn the initial state, a certain mass of sand is slowly placed on top of the piston. Determine the temperature $\\scriptstyle{T_{1}}$ inside the cylinder when the piston descends by $h/2$. For calculation purposes, assume $frac{mgh}{kT_{0}} = 1$, and express the result in terms of $T_{0}$, retaining four significant figures. \n\nPossible formula that may be used: \n\n$$\n\\mathsf{d}\\left(\\frac{x}{e^{x}-1}\\right)=x\\mathsf{d}\\left(\\frac{1}{e^{x}-1}\\right)+\\mathsf{d}\\left(\\ln\\left(1-e^{-x}\\right)\\right)\n$$", "solution": "", "answer": "" }, { "id": 51, "tag": "MECHANICS", "content": "$n$ uniform thin rods, each with a length of $a$ and mass $m$, are sequentially connected by lightweight smooth hinges. The gravitational acceleration is known to be $g$. The head of the first rod is suspended at a fixed point O (the origin in the diagram), and the end of the last rod is subjected to a horizontal force $F$. Find the tangent value tan $\\theta_{i}$ of the angle $\\theta_{i}(i=1,2,...,n)$ between each rod in the equilibrium state and the vertical direction (the $y$-axis direction in the diagram).", "solution": "", "answer": "" }, { "id": 157, "tag": "MECHANICS", "content": "In fact, resonance does not necessarily need to be caused by an external force. A system may exhibit approximate resonance even if its own parameters are stable. Below is the study of a spring pendulum.\n\nA light spring with original length $L$ and spring constant $k$ is suspended from the ceiling, with a mass $m$ hanging below. The gravitational acceleration is $g$. Assume the spring always remains as a straight line, and the system moves only within a two-dimensional plane. Initially, the spring is vertical, and the system reaches equilibrium.\n\nIntroduce the spring's inclination angle $\\theta$ and the extension amount $x$ relative to the vertically suspended equilibrium to describe the system's motion.\n\nHere we define $\\delta=\\frac{x}{L}$, $\\frac{mg}{kL}=\\eta$, $\\omega_0^2=\\frac{g}{l}$ to simplify your expression.\n\nFirst expand to first-order approximation, with initial conditions as\n$$\n\\delta\\vert_{t=0}=0\\\\\\partial_t\\delta\\vert_{t=0}=0\\\\\n\\theta\\vert_{t=0}=0\\\\\\partial_t\\theta\\vert_{t=0}=\\omega\n$$\nAfter calculating the result, consider the second-order approximation. Please provide the resonance condition by calculating the equation of motion of $\\delta(t)$ under second-order approximation, using only $\\eta$ to express the answer.", "solution": "", "answer": "" }, { "id": 494, "tag": "ELECTRICITY", "content": "Within an infinitely long cylindrical region centered at point $O$, there exists a uniform magnetic field that varies with time. The magnetic field is directed into the paper, and its magnetic flux is proportional to time, satisfying $\\varPhi=K t$. From the center $O$, a light, inextensible rod of length $l_{1}$ is extended, with its end connected to the center of another light rod of length $2l_{2}$ through a smooth hinge. At the two ends of the rod of length $2l_{2}$, there are fixed one positive and one negative point charge, each with mass $m$ and charge $\\pm q$. Considering only the motion of the entire system in the plane of the paper, the charges do not enter the magnetic field region and are only subject to the vortex electric field forces caused by the changing magnetic field.\n\nUnder suitable initial conditions, the two rods precisely rotate uniformly about $O$ with a constant angular velocity, while maintaining a perpendicular orientation throughout the motion, and the tension in the rod of length $l_{1}$ remains zero at all times.\n\nBased on the above motion, if the system is subjected to a small perturbation, the angle between $l_{1}$ and $l_{2}$ will no longer remain $90^{\\circ}$ but will instead become $90^{\\circ}+\\delta$. Under suitable conditions, $\\delta$ can oscillate harmonically with time. Determine the angular frequency $\\omega$ of this oscillation.", "solution": "", "answer": "" }, { "id": 694, "tag": "ELECTRICITY", "content": "In a certain medium, electrons can be viewed as particles with an effective mass of \\( m \\), subject to a force \\( F = -m\\omega_{0}^{2}x - m\\gamma\\dot{x} \\) from the surroundings. The first term is a linear restoring force where \\( \\omega_{0} \\) is the natural frequency of the medium, and the second term is the damping force. Assume the electron has a charge \\( q \\) and an electron number density of \\( N \\) in the medium. Ignoring the magnetic part of the incident light, the electric field is \\( E = E_{0} \\text{e}^{\\text{i} \\omega t} \\). The vacuum permittivity is \\( \\epsilon_0 \\). By solving the electron motion, calculate the equivalent refractive index squared (which can be a complex number) of the medium.", "solution": "", "answer": "" }, { "id": 379, "tag": "MECHANICS", "content": "Uranus is a spherical planet with a uniformly distributed mass density of $\\rho_{1}$ and a radius of $R$. In the cosmic space from a distance of $R$ to $2R$ from the center of the sphere, there is uniformly distributed cosmic dust with a density of $\\rho_{2}$.\n\nA space probe is launched from the surface of the planet at an angle of 45 degrees to the horizontal plane. Determine the critical velocity $v_{1}$ required for the probe to escape the range of the cosmic dust.", "solution": "", "answer": "" }, { "id": 706, "tag": "MECHANICS", "content": "Before a new model of car leaves the factory, it has to undergo destructive testing. During one such test, one of the three spokes of a car wheel was knocked off, leaving two spokes with an angle of 120° between them. The wheel can be considered as consisting of a disk with inner and outer radii of $R_{1}{=}4R_{0}/5$ and $R_{0}$, respectively, and two spokes (assuming the disk can be viewed as a uniform annular disk and the spokes as uniform thin rods). The angle between the two spokes is 120°, each spoke has a mass of $m$, and the disk has a mass of $M = 8m$. The wheel starts from rest and is released such that the bisector of the angle between the spokes is horizontal to the ground. Subsequently, the wheel undergoes pure rolling on a horizontal surface. Find: The angular acceleration of the wheel upon release.\n", "solution": "", "answer": "" }, { "id": 598, "tag": "ELECTRICITY", "content": "As shown in the figure, four smooth particles, each with mass $m$ and charge $+q$, are initially at rest on an infinitely large smooth table, forming a square with side length $L$. Ignore the electrostatic force between the charges. There is a uniform magnetic field $B$ perpendicular to the table surface and directed upwards. A lightweight elastic rope with a spring constant $k$ and original length zero, whose thickness can be neglected, wraps around the four particles to form a closed loop. The distance from the center $O$ to each particle is denoted as the radial distance $r$. Suddenly, each of the four particles is given an initial velocity $v_{0}$ in a clockwise direction perpendicular to the radial distance within the plane. Analyze the subsequent motion of the particles. At a certain initial velocity $v_{0}$, the particles will be able to reach the center point $O$ during their subsequent motion. By deriving the expression for $v_{0}$, ultimately find the relationship between the radial distance $r$ and the angle $\\theta$ through which the particles rotate around $O$.", "solution": "", "answer": "" }, { "id": 356, "tag": "MECHANICS", "content": "A fine circular ring with a circular cross-section is subject to a uniformly distributed couple (whose vectors are all along the tangential direction of the ring). Experiments tell us that when the couple density \\( m \\) increases, the rotation angle \\( \\phi \\) of the ring’s cross-section also increases; when \\( m \\) reaches a certain critical value, the ring will suddenly flip. Both the uniformly distributed couple and couple density \\( m \\) refer to the couple per unit length, \\( m = \\frac{dM}{ds} \\); the flipping of the ring refers to the part that was originally on the inner side of the ring suddenly flipping to the outer side on the wire forming the ring. The wire forming the ring (radius \\( R \\)) has an elastic modulus \\( E \\) and a radius \\( r \\). Determine the critical value of the couple intensity at which this flipping occurs.", "solution": "", "answer": "" }, { "id": 758, "tag": "MECHANICS", "content": "The relationship between the energy and momentum of a quasiparticle is expressed as $H=H(p_{x},p_{y})$, which is significantly meaningful. Additionally, the velocity of the quasiparticle is given by the following equations: $$ v_{x}=\\frac{\\partial H}{\\partial p_{x}}\\quad,\\quad v_{y}=\\frac{\\partial H}{\\partial p_{y}} $$ Suppose in a certain crystal, the energy-momentum relationship of the quasiparticle is: $$ H=\\frac{p_{x}^{2}+p_{y}^{2}}{2m}+\\frac{p_{x}p_{y}}{m^{\\prime}}\\quad,\\quad m 0 \\). Let \\( x(r_1)=0 \\) at radius \\( r_1 \\), meaning \\( x \\) represents the relative thickness of the lens with respect to \\( r_1 \\). All angles are assumed to be small angles. The refractive index of the simulated lens is \\( n \\). Try deriving the expression for the thickness of the lens \\( x(r) \\). The answer should be expressed in terms of \\( n \\), \\( R \\), \\( r \\), and \\( r_1 \\).", "solution": "", "answer": "" }, { "id": 622, "tag": "THERMODYNAMICS", "content": "In a cylindrical cylinder with a cross-sectional area of ${A}$, there is a certain special gas. This gas has a molar constant-volume heat capacity $c_{ u}=\\frac{3}{2}R$ when the temperature is below $2T_{0}$, and a molar constant-volume heat capacity $c_{ u}=\\frac{5}{2}R$ when the temperature is above $2T_{0}$. The top of the cylinder is a piston with a mass of $m$, which separates the internal gas from the external vacuum. There is a small hole with an area of $S(S\\ll A)$ on the piston. The piston can slide up and down in the cylinder without friction. The relaxation time of the gas is much shorter than the characteristic time of the piston movement. Initially, the distance from the piston to the bottom of the cylinder is $L$, the initial temperature of the gas inside is $T_{0}$, the gravitational acceleration is $g$, the Boltzmann constant is $k$, and the mass of the gas molecules is $\\mu$. Do not consider the effect of gravity on the distribution of the gas. Calculate the time $t$ it takes for the gas to completely leak out starting from the initial moment. Express the answer using $L$, $k$, $\\mu$, and $T_{0}$, and please provide a precise numerical expression without accepting decimal results.", "solution": "", "answer": "" }, { "id": 38, "tag": "ELECTRICITY", "content": "In this problem, we consider an anisotropic linear medium. Assume $D_{x}=\\varepsilon_{x}E_{x}$, $\\mathit{D_{y}}\\overset{\\widehat{}}{=}\\varepsilon_{y}E_{y}$, $D_{z}=\\varepsilon_{z}E_{z}$. Suppose there is a conducting sphere with a radius of $R$ in space, and find its capacitance. In this question, assume $\\varepsilon_{x}>\\varepsilon_{y}=\\varepsilon_{z}$.", "solution": "", "answer": "" }, { "id": 656, "tag": "ELECTRICITY", "content": "A magnetic field exists in space with central symmetry, perpendicular to the paper and directed inward. The magnetic field changes with distance from the center, given by the formula \\( B(r) = B_0 \\left( \\frac{r}{R} \\right)^n \\). A particle with charge \\( q \\) and mass \\( m \\) moves in uniform circular motion with a radius \\( R \\), velocity \\( v \\), and the center of the circle as the symmetry center. Now, if a small radial disturbance is applied to the particle, it will undergo small oscillations along the radial direction. Find the period of these small oscillations.", "solution": "", "answer": "" }, { "id": 747, "tag": "MECHANICS", "content": "Above a horizontal ground at a height of $h$, there is a ball launcher that can shoot elastic balls in various directions in the upper half-plane. It is known that the launching directions are uniformly distributed on the sphere (i.e., the probability of launching within each unit solid angle in various directions is the same). The initial velocity of the launched elastic balls is $v = \\sqrt{g h}$, and the launched balls do not affect each other (i.e., disregarding collisions between balls). A single ball can be regarded as a solid sphere with uniform mass distribution, with a known mass of $m$. The launcher is activated at $t = 0$, and within an extremely short time (i.e., the launch time can be neglected, and it can be considered that the balls are launched simultaneously at $t = 0$), $N$ balls are launched ($N \\gg 1$). If each elastic ball can be considered as a rigid sphere, undergoing completely inelastic collisions with the ground, and the friction coefficient between the elastic ball and the ground is $\\mu = 0$. After all the launched balls have landed, try to find the number density distribution of the balls at a radial distance $r$ from the point directly below the launcher on the ground, $n(r, t)$.", "solution": "", "answer": "" }, { "id": 416, "tag": "THERMODYNAMICS", "content": "The left end of an adiabatic cylinder is connected to a large atmospheric reservoir via a magical semipermeable membrane. The atmospheric reservoir contains a monoatomic gas with constant pressure and temperature \\(p_0\\) and \\(T_0\\). Assume this semipermeable membrane has the following magical properties: diatomic molecules cannot pass through at all; when the partial pressure of monoatomic molecules inside the cylinder exceeds the partial pressure of monoatomic gas in the reservoir, monoatomic molecules can freely enter the reservoir, but molecules inside the reservoir can never enter the cylinder under any circumstances. Apart from allowing monoatomic molecules to pass through unidirectionally, this semipermeable membrane is considered adiabatic (heat exchange of diatomic gas cannot occur through the membrane, and monoatomic gas undergoes isobaric release when passing through the membrane).\n\nThe right side of the cylinder is an adiabatic piston. Initially, the cylinder is filled with equal molar amounts of monoatomic and diatomic molecules, with a total pressure of \\(2p_0\\), a temperature of \\(T_0\\), and a volume of \\(V_0\\). The piston is now slowly compressed to the left. Using the cylinder's temperature \\(T\\), volume \\(V\\), and their differentials \\(dT\\) and \\(dV\\) as variables at a given moment, derive a differential equation and solve to express \\(V\\) as a function of \\(T\\), \\(V(T)\\).", "solution": "", "answer": "" }, { "id": 565, "tag": "ELECTRICITY", "content": "A regular octahedral frame, with each edge having a resistance of $R$, an edge length of $l$, and negligible mass. Metal spheres with a mass of $m$ and no resistance are welded at each vertex. The frame is now placed in a uniform magnetic field $\\boldsymbol{B}$, initially ($t=0$) rotating relative to its geometric center at an angular velocity of $\\boldsymbol{\\omega}_{0}$.\n\nFind the angular velocity $\\boldsymbol{\\omega}(t)$ of the frame at any subsequent time.", "solution": "", "answer": "" }, { "id": 659, "tag": "ELECTRICITY", "content": "There is a smooth elliptical track, and the equation of the track satisfies $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1 (a>b>0)$. On the track, there is a small charged object that can move freely along the track. The mass is $m$, and the charge is $q$, located at $(a,0)$. Point charge $Q$ is located at the left focus of the elliptical track. Find the period of the small charged object's small oscillations around its stable equilibrium position.", "solution": "", "answer": "" }, { "id": 320, "tag": "MECHANICS", "content": "Three homogeneous rods OA, AB, and AC, each with a length of \\( L \\) and mass \\( m \\), are hinged together at point A. Between point O and point B, as well as between point C and point O, there is a spring with natural length 0 and stiffness coefficient \\( \\mathrm{k} \\). The rod OA is hinged at point O on a smooth surface, and the entire system is placed on a smooth surface, allowing it to rotate around an instantaneous axis through point O without being subjected to any additional forces. Initially, the system rotates uniformly at a certain angular velocity, and the rods AB and AC are perpendicular to rod OA. Now, a symmetrical disturbance is given to rods AB and AC, causing a small deviation of the angle between each rod and OA from \\( \\frac{\\pi}{2} \\). The system will then undergo small vibrations near the equilibrium state of uniform rotation. Find the period \\( T \\) of these small vibrations.", "solution": "", "answer": "" }, { "id": 760, "tag": "ELECTRICITY", "content": "Given that the angular momentum density of the electromagnetic field is $ \\pmb{l}= \\pmb{r} \\times( \\varepsilon_{0} \\pmb{E} \\times \\pmb{B})$, there is a lightweight cylinder with a radius of $R$ and a length of $ L$($L\\gg R$), carrying a uniform charge density $ \\rho$, rotating about its vertical central axis with an angular velocity $ \\omega$. To stop the cylinder from rotating, we apply a resistive torque on it that is proportional to the angular velocity, $ \\pmb{M}=- \\alpha \\pmb{ \\omega}$. Given that at $ \\scriptstyle t=0$ the angular velocity of the cylinder is ${{ \\omega}_{0}}$, determine the relationship between the angular velocity of the cylinder and time $t$.", "solution": "", "answer": "" }, { "id": 352, "tag": "ADVANCED", "content": "Here is a structure composed of three elastic beams. The left and right beams are both vertically suspended relative to the horizontal plane (with connection points at both ends), with lengths of 2l each. The upper connection points are marked as F and S, and the lower connection points are marked as G and T. The connections are hinged, meaning no additional constraint moments are generated; now a horizontal beam with length l is rigidly connected at the midpoints of both vertical beams, meaning any deformation here is continuous, and the two beams always maintain a perpendicular relationship. The bending stiffness of each beam is EI, that is $M=EI\\frac{d^2y}{dx^2}$. Now, a concentrated downward vertical force P is applied at the midpoint of the horizontal beam AB, disregarding gravity. Calculate the downward displacement at the midpoint of the beam. ```markdown A structure is composed of three elastic beams. The left and right beams are vertically suspended relative to the horizontal plane with lengths 2l and connection points F, S on top and G, T at the bottom, respectively. The connections are hinged (without any additional constraint moments). A horizontal beam with length l is rigidly connected at the midpoints of these vertical beams, maintaining a perpendicular relationship under deformation and keeping it continuous. Bending stiffness of each beam is EI, expressed as $M=EI\\frac{d^2y}{dx^2}$. A concentrated downward force P is applied at the midpoint of the horizontal beam AB. Ignoring gravity, the task is to determine the downward displacement at this midpoint.", "solution": "", "answer": "" }, { "id": 712, "tag": "THERMODYNAMICS", "content": "There is a closed container filled with water vapor, and the pressure inside the container remains constant. When we continuously cool the water vapor to reach the critical temperature of phase change, many small water droplets will condense inside the water vapor and suspend within it. Assume that the total mass of these small water droplets is much less than the total mass of the water vapor. Find the speed of sound propagation within it. To simplify the problem, assume that the wavelength of the sound wave is much greater than the scale of the system's inhomogeneity, and we can use the ideal gas state equation to describe the water vapor. The molar mass of water is $M$, the latent heat of vaporization is $L$, the temperature is $T$, the ideal gas constant is $R$, and the specific heat capacity at constant pressure of the gas is $c_{p}$", "solution": "", "answer": "" }, { "id": 180, "tag": "ELECTRICITY", "content": "Place a thick conductive spherical shell (with inner and outer radii $R_{1}$ and $R_{2}$, respectively) in a uniform electric field (with a field strength of $E_{0}$), and allow it to reach electrostatic equilibrium. The dielectric constant of the conductive sphere is the vacuum permittivity $\\varepsilon_{0}$. At time $t = 0$, the external electric field is suddenly removed, ignoring any induced effects caused by the removal of the field. The resistivity of the conductor is given as $\\rho$. Determine how the charge distribution on the outer surface of the conductive spherical shell changes over time.", "solution": "", "answer": "" }, { "id": 502, "tag": "MECHANICS", "content": "The Magnus effect is a common fluid dynamics phenomenon in our daily lives. In real situations, the Magnus effect is very complex. This problem uses a simplified model without viscosity derived from the Kutta–Joukowski theorem to deduce a formula for the Magnus effect. The Kutta–Joukowski theorem describes a scenario where a cylinder and air are in relative motion, and the cylinder rotates around an axis perpendicular to its base. The cylinder will experience a force perpendicular to the angular velocity vector and the direction of relative motion velocity. The magnitude of the force per unit length along the cylinder's axis is $f=\\rho V_{air} \\Gamma$ (the direction of relative motion to air must be perpendicular to the angular velocity direction), where $\\rho$ is the air density (known), $V_{air}$ is the relative motion speed of air and the cylinder as a whole, unaffected by the cylinder’s movement at infinity, and $\\Gamma$ is the circulation satisfying the loop integral $\\Gamma=|\\int \\vec{v}\\cdot d\\vec{l}|$, where the integral path is the circumference of the base of the cylinder. In this problem, you can consider $\\vec{v}$ as the speed generated at the edge points due to the rotation of the cylinder. Air resistance is not considered in this problem, and the air is considered stationary relative to the ground. Now there is a sphere rotating around a horizontal axis passing through its center at an angular velocity $\\omega$, while moving horizontally at a speed of $V$ perpendicular to the rotation axis relative to the ground. The sphere has a radius of $R$. Try to find the lift force magnitude caused by this motion of the sphere (the problem assures that the directions of angular velocity and motion velocity result in an upward lift on the sphere).", "solution": "", "answer": "" }, { "id": 440, "tag": "MECHANICS", "content": "A homogeneous hypersphere of radius $r$ is launched between two fixed parallel hard plates above and below, colliding with both plates 3 times in succession, nearly returning to the original position. The $x$-axis is directed horizontally to the right, the $y$-axis vertically downward, and the positive $z$-axis is determined by the right-hand rule. Initially, the horizontal component of the sphere's velocity is $\\nu_{0x}$, the component in the $z$ direction is 0, and the angular velocity about the sphere's axis (parallel to the $z$-axis) is $\\omega_{0z}$ $(\\omega_{0z}<\\nu_{0x}/r)$. Gravity is neglected.\n\nFind the horizontal component of the velocity of the sphere’s center $\\nu_{3x}$ and the angular velocity of the sphere $\\omega_{3z}$ after the third collision with the plates.\n\nHint: It is known that the moment of inertia of a homogeneous sphere with mass $m$ and radius $r$ about its axis through the center is $J=2m r^{2}/5$. A hypersphere is a hard rubber ball, whose rebound on hard plate surfaces can be considered completely elastic; in other words, there is no sliding at the contact point. The tangential deformation and normal deformation that occur at the contact point when subjected to static friction and normal pressure are considered to be elastic. For simplification, assume these two types of deformation are independent of each other (hence the corresponding elastic forces are conservative forces).", "solution": "", "answer": "" }, { "id": 739, "tag": "ELECTRICITY", "content": "Given a circuit, a power source connects in series with an inductor $L_1$, followed by two parallel branches. Branch one consists of an inductor $L_2$ and a capacitor $C$ in series, and branch two contains inductor $L_3$, after which it returns to the power source. The power source is a constant source with an electromotive force of $E$. At the initial moment, there is no current anywhere in the circuit, and there is no charge on the capacitor plates. Determine the current $I(t)$ flowing through $L_1$ at any subsequent time $t$.", "solution": "", "answer": "" }, { "id": 581, "tag": "MECHANICS", "content": "In the $y O z$ plane, a thin steel wire is fixed, and the shape of the wire satisfies $y^{2}+z^{2}=a^{2}$. The wire is smooth. There is a uniform thin rod with mass $m$, one end of which is smoothly hinged at $A(a,0,0)$, where $a>0$. The length of the rod is $l>\\sqrt{3}a$. The thin rod rests on the wire, and the contact point with the wire is $B$. The gravitational acceleration is known to be $g$.\n\nAssume that initially, the contact point is $B(0,0,-a)$ and within the wire. After giving the rod a tangential perturbation, calculate the angular frequency of the rod's oscillatory motion.", "solution": "", "answer": "" }, { "id": 684, "tag": "MODERN", "content": "Atoms confined in a fixed-volume box have an average energy \\[\\epsilon\\]. In a short time period \\(\\Delta \\tau\\), a small number of atoms \\(\\Delta N\\) (where \\(\\Delta N<0\\)) leave the box with an average energy of \\((1+\\beta)\\epsilon\\) (where \\(\\beta>0\\)), resulting in a slight change in the average energy of the remaining atoms \\(\\Delta\\epsilon\\) (where \\(\\Delta\\epsilon<0\\)). Assume \\[ \\left|\\frac{\\Delta\\epsilon}{\\epsilon}\\right|\\ll 1,\\quad \\left|\\frac{\\Delta N}{N}\\right|\\ll 1, \\] and the term \\(\\frac{\\Delta\\epsilon}{\\epsilon}\\frac{\\Delta N}{N}\\) can be neglected. Please derive the relationship between the change in average energy of the atoms remaining in the box and the change in the number of atoms, and provide the formula expressing \\(\\frac{\\Delta\\epsilon}{\\epsilon}\\) in terms of \\(\\frac{\\Delta N}{N}\\) and \\(\\beta\\).", "solution": "", "answer": "" }, { "id": 174, "tag": "MECHANICS", "content": "A smooth insulated cone with a half-angle of $\\theta$ is placed vertically with its vertex pointing downwards (the axis is vertical). At the vertex 0 of the cone, there is a fixed electric dipole with a dipole moment $p$, and the dipole moment direction is vertically upward along the cone's axis. Now, a small charged object (charge amount q>0, mass m) is released inside the cone at a vertical distance $h_{0}$ from the vertex. The initial speed of the small object is $v_{0}$, along the horizontal direction and tangent to the cone surface. Find the minimum vertical height from the vertex that the object can reach while moving on the cone surface. (Ignore the volume of the object)\n", "solution": "", "answer": "" }, { "id": 306, "tag": "MECHANICS", "content": "In a certain rigid vertical plane, a non-elastic light string of length \\( l \\) has one end fixed at point \\( O \\) and the other end connected to a small ball with mass \\( m \\). Initially, the string is straightened, forming an angle \\(\\theta\\) (less than \\(\\frac{\\pi}{4}\\)) with the horizontal direction.And at this time, the position where the small ball is is higher than point O.\n. At a certain moment, the small ball is released freely. Once the string is taut again, the small ball immediately starts moving along an arc orbit. Eventually, the ball will leave the arc orbit. Relative to point \\( O \\), find the maximum height \\( h_m \\) that the ball can reach after leaving the arc orbit.", "solution": "", "answer": "" }, { "id": 672, "tag": "OPTICS", "content": "A circular plate with radius $R$ and thickness $b$ is made of a material whose refractive index $n$ varies radially. The refractive index at the center is $n_{0}$, and at the edge is $n_{R}$. The refractive index inside the circular plate follows a certain distribution with respect to the radius, such that rays parallel to the principal axis can be perfectly focused. Please provide the focal length $f$ of this circular plate lens.", "solution": "", "answer": "" }, { "id": 456, "tag": "THERMODYNAMICS", "content": "Scientists are attempting to eradicate cancer cells using thermal therapy. We will model this process with a simple model. Assume a cancer cell is approximately a perfect sphere with a radius $R_{T}$, sharing the same thermal conductivity $k$ as the surrounding human body environment (which can be considered as a constant). The ambient temperature of the body is constant at $T_{0}$. The cancer cell contains a specific protein that, when irradiated by a certain laser, generates heat at a rate of $p$ per unit volume per unit time within the cancer cell. The effect of the laser on the human body environment outside the cancer cell can be neglected. Determine the temperature difference $\\Delta T$ between the highest temperature point within the cancer cell in steady state and the body temperature $T_{0}$ (since the cancer cell is quite small and has a low number density, it can be considered isolated, and for simplification, thermal expansion of the cancer cell can be ignored, assuming that the temperature at $r=\\infty$ is $T_0$).", "solution": "", "answer": "" }, { "id": 773, "tag": "MECHANICS", "content": "A right-angled isosceles triangular plate with mass M and uniformly distributed mass hangs from a hinge. Let the three vertices of the triangular plate be A, B, and C, where ∠A = 90°. Vertex B of the triangular plate is fixed on the hinge. Assume the right-angled sides of the isosceles triangular plate have a length of 𝑎, and the hinge is smooth. The triangular plate is in a stable equilibrium state. A particle with mass m enters the sloping side of the triangular plate at a horizontal velocity v. After entering, the particle embeds into the sloping side of the triangular plate. If the entire incident process is very short and the axis of rotation does not experience a horizontal impulse, the point where the particle enters is called the strike center. Try to give the larger of the two solutions for the vertical distance between the strike center and the hinge. Only the case within the plane is considered in this problem. Do not consider situations where the strike center does not exist.\n", "solution": "", "answer": "" }, { "id": 213, "tag": "ELECTRICITY", "content": "\n\nA long straight coaxial insulating cylindrical shell is placed vertically, with inner and outer radii of $a$ and $b(b>a)$, respectively. Two sufficiently long ideal electrode sheets are fixed between the inner and outer shells, each with a width of $l=b-a$. The planes containing the two metal sheets both pass through the cylindrical axis, and the angle between the two planes is $\\theta_{0}{\\left(<2\\pi\\right)}$. A third metal sheet identical to the aforementioned metal sheets is inserted between the two metal sheets (within the dihedral angle of $\\theta_{0}$). The mass surface density of the metal sheet is $\\sigma$, and the angle between the plane of this sheet and one of the fixed sheets is $\\theta\\left(0\\le\\theta\\le\\theta_{0}\\right)$.\n\nA constant voltage DC power supply with an electromotive force of $V$ and negligible internal resistance is connected between the two electrodes (with the high-potential pole connected to the plate at $\\theta=0$). The plane of the movable sheet always passes through the cylindrical axis. The collision between the movable sheet and the fixed sheet is completely inelastic, and during separation after colliding with the fixed sheet, the movable sheet can acquire $\\chi(\\chi<1)$ times the initial charge of the fixed sheet. The vacuum permittivity is given as $\\varepsilon_{0}$.\n\nWhen the sheet becomes positively charged due to collision with the plates and is at an angular position $\\theta$, find the electric field torque $M_{+}(\\theta)$ on the sheet relative to the axis of the cylinder.", "solution": "", "answer": "" }, { "id": 591, "tag": "ELECTRICITY", "content": "There are $n$ terminals, and there is a resistor $R$ between any two terminals. Find the resistance between any two terminals.", "solution": "", "answer": "" }, { "id": 651, "tag": "ELECTRICITY", "content": "Consider a tightly wound solenoid of finite length. The radius of the solenoid is $R$, the number of turns per unit length is $n$, and the current $I$ is known. However, the solenoid is only located between $-l 0$. The region beyond the metal wire and within the cylindrical metal surface is filled with a certain dielectric medium. Due to the influence of the electric field, the atoms of the medium nearby the exterior of the wire are ionized into free electrons and cations, where the free electrons are adsorbed by the wire upon ionization, while the cations move radially away from the wire. Assume the radial mobility of the cations (defined as the ratio of radial speed to electric field intensity) is a constant $w$, and during the migration process, the cations always form a uniform cylindrical thin layer surrounding the wire. \n\nAssume the total electric charge of all the cations is $Q$. To keep the potential of the wire unchanged at the original $U_0$, it is necessary to supply an additional electric charge $Q^{*}$ to the wire. Determine the relationship between $Q^{*}$ and time $t$.", "solution": "", "answer": "" }, { "id": 680, "tag": "MECHANICS", "content": "Three homogeneous rods 1, 2, and 3, each with length \\(2r\\) and mass \\(m\\), are connected by smooth hinges \\(A\\) and \\(B\\). The bottom rod 1 is connected to a fixed point \\(O\\) on the ground through a smooth hinge. Initially, the lower two rods 1 and 2 are vertical, while the top rod 3 deviates to the right from the vertical direction by a small angle \\(\\varepsilon \\ll 1\\). The system is then released from rest. The gravitational acceleration is \\(g\\). Find the horizontal component \\(F_x\\) of the force \\(\\vec F\\) at support point \\(O\\) at this moment, where the \\(+x\\) direction is horizontally to the right.\n", "solution": "", "answer": "" }, { "id": 400, "tag": "MODERN", "content": "In the $S$ frame, there is a clock positioned in the $_{\\mathrm{X-Y}}$ coordinate system, with the 12 o'clock direction pointing along the positive $y$-axis and the 3 o'clock direction pointing along the positive $\\mathbf{X}$-axis. The $S_{1}$ frame moves relative to the $S$ frame with velocity $\\nu_{0}$ along the $\\mathbf{X}$-axis. The $S_{2}$ frame moves relative to the $S_{1}$ frame with velocity $\\nu_{0}$ along the $y$-axis of the $S_{1}$ frame. The $S_{3}$ frame moves relative to the $S_{2}$ frame with velocity $-\\nu_{0}$ along the $\\mathbf{X}$-axis of the $S_{2}$ frame. The $S_{4}$ frame moves relative to the $S_{3}$ frame with velocity $-\\nu_{0}$ along the $y$-axis of the $S_{3}$ frame. Assuming $\\nu_{_0}/c \\ll 1$, keep only the first non-zero term in the result.\n\nFind the angle of the clockwise deflection of the 12 o'clock direction of the clock as seen in the $S_{4}$ frame.", "solution": "", "answer": "" }, { "id": 696, "tag": "MECHANICS", "content": "Objects with masses $M_1$ and $M_2$ move in a circular orbit around the center of mass, with a distance $R$, angular velocity $\\Omega$, and gravitational constant $G$. The center of mass is taken as the origin $O$ of the coordinate system, and the direction from $M_2$ to $M_1$ is the positive direction of the $x$-axis. We establish a rectangular coordinate system $xOy$ in the plane of the orbit, with angular velocity $\\Omega$ in the $+z$ direction. A mass $m$ ($m\\ll M_1, M_2$) is placed near the Lagrange point $L_4$, which is not located on the $x$-axis and has $y>0$, to study its small amplitude motion deviating from the equilibrium position. The initial deviation is known to be $(\\delta x_0, \\delta y_0) = (\\epsilon, \\delta y_0)$, and the initial velocity is $(\\delta v_{x0}, \\delta v_{y0}) = (-\\frac{1}{2}\\Omega\\epsilon, \\delta v_{y0})$, where $\\delta y_0$ and $\\delta v_{y0}$ are unknowns. Given that the masses satisfy $\\frac{M_1 - M_2}{M_1 + M_2} = \\sqrt{\\frac{11}{27}}$, and it is required that the mass can perform stable motion near the equilibrium position, find $\\delta v_{y0}$, expressed solely in terms of $\\epsilon$ and $\\Omega$.\",\n", "solution": "", "answer": "" }, { "id": 452, "tag": "OPTICS", "content": "There is a hemispherical uniform isotropic medium, which is a light-emitting diode made of semiconductor material gallium arsenide (GaAs). Its core ($AB$) is the circular light-emitting region, with a diameter $d$, and is concentric with the circle on the ground. To avoid total internal reflection, the upper part of the light-emitting diode is polished into a hemispherical shape, so that the light emitted from the core has the maximum transmittance when emitted outward. If the light emitted from the edges of the emitting region at points $A$ and $B$ is required not to undergo total internal reflection, what should the radius $r$ of the hemisphere be? It is given that the refractive index of GaAs is $n$.", "solution": "", "answer": "" }, { "id": 660, "tag": "MECHANICS", "content": "A spring with an original length of $L_{0}$ and total mass $m$ has a total elastic coefficient of $k$ when it is uniformly stretched, and the mass distribution of the spring is also uniform at this time. Now, the spring is placed inside a horizontal, smooth-walled cylindrical tube and rotates with the tube around a vertical axis at a constant angular velocity $\\omega$. One end of the spring is fixed at point $P$, which is a distance $r_{0}$ from the axis of rotation, and the spring remains stationary relative to the cylindrical tube. Try to find the length $l$ of the spring (you may introduce parameters $\\omega_{0} = \\sqrt{\\frac{k}{m}}, \\alpha = \\frac{\\omega}{\\omega_{0}}$ to simplify the expression).\n", "solution": "", "answer": "" }, { "id": 767, "tag": "MECHANICS", "content": "A uniform cylindrical shell with an inner radius of $\\pmb{R}_{1}$, an outer radius of $\\pmb{R_{2}}$, a total length of $\\pmb{L}$, and a mass of $M$ is placed vertically on a frictionless horizontal surface. Inside, there are $\\pmb{n}$ turns of a screw with a single-layer width of $\\pmb{t}$ and a single-layer height of $\\pmb{d}$. Both the screw and the cylindrical shell are made of the same material with a density of $ρ$. A homogeneous small spherical shell with mass $m$ and radius $r$ is fixed on the screw by a light partition, and the small spherical shell is in close contact with the cylinder. We assume the following: $R_{1}, R_{2}, L \\gg R_{2}-R_{1}, t, d, r; R_{1} \\approx R_{2} \\approx R$. Now suppose the inner wall of the cylinder is smooth, and the screw is also smooth. Remove the partition and release the small ball from the top (where the contact point of the small ball with the screw is on the same horizontal plane as the upper surface of the cylinder). Initially, the small ball is at rest and does not rotate. Calculate the speed of the ball when it falls to the bottom (where the contact point of the small ball with the screw is on the same horizontal plane as the lower surface of the cylinder). Note: the acceleration due to gravity is $g$. You can simplify the final expression using $$ \\tan\\theta={\\frac{L}{2n\\pi R}} $$ to ensure the final expression does not include $n$ and $R$.\n", "solution": "", "answer": "" }, { "id": 279, "tag": "ELECTRICITY", "content": "Consider a simple circuit that contains two parallel plate capacitors connected in parallel, with capacitances $C_{1}$ and $C_{2}$, connected via wires and a switch. Capacitor $C_{1}$ is initially charged to a voltage $V_{0}$, while $C_{2}$ is completely uncharged. Throughout the problem, the circuit remains within a square loop with side length $l$, and the conductive wire has a diameter of $D$.\n\nFeynman made the following assumption: The missing energy is converted into the kinetic energy of charge carriers moving from $C_{1}$ to $C_{2}$. Assume the average free path $\\lambda>2l$ of the carriers between collisions. Find the total kinetic energy $\\Delta K$ acquired by the carriers during the transfer of total charge from $C_{1}$ to $C_{2}$.", "solution": "", "answer": "" }, { "id": 613, "tag": "MECHANICS", "content": "In this problem, we will explore how modifying the law of gravity changes the orbit. Suppose the gravitational potential energy $$ U=-k{\\frac{e^{-r/a}}{r}} $$ is modified from $U=-{\\frac{k}{r}}$, where $k=G M m$, and $a$ is a known constant with units of length. When $a\\rightarrow\\infty$, the closest and farthest distances of the Earth to the Sun are $r_{close}$ and $r_{far}$, respectively. For the modified potential energy $U=-k\\frac{e^{-r/a}}{r}\\Big(a\\gg r_{far}\\Big)$, if the Earth has the same energy and angular momentum and the orbit is still approximately elliptical, and the Earth's orbital equation is $$ r={\\frac{b}{1+e \\cos\\theta}} $$ where $e$ is the eccentricity of the orbit. Try to find the expression for the change in eccentricity $\\delta e$ of the Earth’s new eccentricity $e^{\\prime}$. The answer should be expressed in terms of $r_{far},\\quad r_{close}$, and $a$, retaining the first-order term of ${\\frac{1}{a}}$.\",\n", "solution": "", "answer": "" }, { "id": 585, "tag": "ELECTRICITY", "content": "This problem considers the role of symmetry and scaling transformations in physics for solving practical problems. Given the dielectric constant in vacuum being $\\varepsilon_{0}$.\n\nConsider a logarithmic spiral, whose curve equation is $r=a\\mathrm{e}^{k\\theta}$. Place $n$ equal point charges $q$ at $\\theta=0, \\alpha, 2\\alpha, \\cdots, (n-1)\\alpha$. Find the electric potential $\\varphi$ at the origin.", "solution": "", "answer": "" }, { "id": 395, "tag": "ELECTRICITY", "content": "There is a metallic sphere shell with a radius of $R$, and inside, there is a spherical metallic electrode with a radius of $r$. The space in between is filled with a medium with conductivity $\\rho$. Assume the electrode is displaced from the center by a distance $l$, and that $r << l$, $r << R$, and the electrode is not near the boundary. Estimate the resistance between the electrode and the metallic sphere shell, accurate to the first-order approximation.", "solution": "", "answer": "" }, { "id": 215, "tag": "MECHANICS", "content": "A smooth horizontal shaft at a height $h$ from the ground is smoothly connected to a lightweight board of length $2L$. The shaft is fixed at the center of the board. A long, thin string is placed on the board, with a mass per unit length of $\\lambda$. The lightweight board is sufficiently rough such that the string does not slip on its surface. The remaining part of the string is piled on the ground, and the suspended string does not experience any tension from the string pile. Initially, the system is in equilibrium.\n\nLet $\\theta$ be the angle of rotation counterclockwise from the equilibrium position. At time $t=0$, a small impulse is applied to the right side of the board, giving it an initial angular velocity $\\omega_{0} = \\sqrt{\\alpha g/L} \\left(\\alpha \\ll 1\\right)$ in the counterclockwise direction, after which the board begins to oscillate. Since $\\alpha \\ll 1$ and $|\\theta| \\ll 1$, it can be approximated that the strings on both the left and right sides remain vertical. Let the tensions exerted on the endpoints of the board by the hanging strings on the left and right be $T_{l}$ and $T_{r}$, respectively. The gravitational acceleration is given as $g$.\n\nGiven $h = L/3$, find the time $T_{1/2}$ required for the average amplitude of the board to decay to half of its original value.", "solution": "", "answer": "" }, { "id": 466, "tag": "ELECTRICITY", "content": "On a smooth horizontal plane, there is a fixed regular $n$-gon $A_{1}A_{2}\\dotsm A_{n}(n>3)$ with a center $O$. Each vertex has a point charge $Q(Q>0)$, and there is a point charge $q$ with mass $m$ at the center. Assume the side length of the regular $n$-gon is $a$. If the central charge is $-q(q>0)$, and the permittivity of vacuum is $\\varepsilon_{0}$, a uniform magnetic field perpendicular to the plane is needed to constrain the point charge so it moves within a small range. Find the minimum value of the magnetic field.", "solution": "", "answer": "" }, { "id": 528, "tag": "ADVANCED", "content": "Consider the two-mode squeezing operator:\n\\[\nS(\\lambda, \\theta) = \\exp\\left(\\frac{\\lambda}{2} \\Big[ (a_1^{\\dagger\\,2} - a_2^{\\dagger\\,2})\\sin\\theta + 2a_1^\\dagger a_2^\\dagger \\cos\\theta - (a_1^2 - a_2^2)\\sin\\theta - 2 a_1 a_2 \\cos\\theta \\Big] \\right).\n\\]\nwhere:\n- \\(\\lambda\\) and \\(\\theta\\) are parameters,\n- \\(a_1, a_2\\) are the annihilation operators for the two modes.\n\nThe task is to calculate:\n\\[\nF = |\\langle 00 | S^\\dagger(\\lambda, 0) S(\\lambda, \\theta) | 00 \\rangle|^2,\n\\]\nwhich is the fidelity between the squeezed states with different parameters, \\(S(\\lambda, \\theta)|00\\rangle\\) and \\(S(\\lambda, 0)|00\\rangle\\).", "solution": "", "answer": "" }, { "id": 728, "tag": "MECHANICS", "content": "A bomb explodes on the ground, and after the explosion, all fragments are ejected at the same speed $u$, with directions confined within a narrow angle not exceeding \\(\\alpha \\ll 1\\) from the vertical upward direction (angles are uniformly distributed within this range). All fragments eventually land, and upon landing, undergo a completely inelastic collision. Let the mass of the bomb be $M$. Find the maximum radius \\(R\\) of the fragment distribution (retaining the first-order term of \\(\\alpha\\)).\n", "solution": "", "answer": "" }, { "id": 405, "tag": "ADVANCED", "content": "The known form of the electromagnetic field momentum conservation theorem is expressed as:\n\\[\\vec{f_m}+\\frac{\\partial \\vec{g}_{em}}{\\partial t}+\\nabla \\bm{T}_{em}\\]\nwhere $\\vec{g}_{em}$ is the momentum density of the electromagnetic field, and $\\bm{T}_{em}$ is the momentum flux density of the electromagnetic field. \n\nNow, consider a given point charge $q$ moving with velocity $\\vec{v}$ in a constant external magnetic field $\\vec{B}_0$. By taking into account only the momentum density of the electromagnetic field while ignoring the momentum flux density, derive the force on the charge, i.e., provide the mechanical force density \\[\\vec{f_m}\\]. The trajectory of the particle $\\vec{r}_q(t)$ may be retained.", "solution": "", "answer": "" }, { "id": 220, "tag": "ELECTRICITY", "content": "There is a ring with radius \\( R \\) and uniform charge \\( Q \\), with the center of the ring at the origin. The ring is located in the \\( oxy \\) plane. Try to find the electric potential at the spherical coordinates \\( (r, \\theta) \\) where \\( r \\gg R \\) (require keeping the lowest order correction term up to \\(\\frac{R}{r}\\)).", "solution": "", "answer": "" }, { "id": 658, "tag": "ELECTRICITY", "content": "A uniform metal ring with radius $a$ has 3 metal spokes connecting the center of the ring to the ring itself. The spokes and the ring are made of the same type of metal with an identical cross-sectional area. Each spoke has a resistance of $R_{1}$. The angles between the spokes are equal. The connections at the center and between the spokes and the ring are well established. There is a magnetic field perpendicular to the plane of the ring, with the upper half having a magnetic flux density of $3 B_{0}$ and the lower half having a magnetic flux density of $B_{0}$, both in the same direction. Now the ring (and spokes) rotate uniformly around the center within the plane at an angular velocity of $\\omega$. Assume that at $t=0$, spoke 1 is exactly at the boundary between different magnetic fields and is about to enter the $3 B_{0}$ magnetic field area. Calculate the external torque required to maintain the uniform rotation of the ring.", "solution": "", "answer": "" }, { "id": 272, "tag": "MECHANICS", "content": "Three stars shining together means the sun is at its zenith. However, such a state does not have a large enough stability range. If considering external disturbances, during the long history of stellar evolution, there is a considerable probability that they will break apart into a binary star system and a solitary star, separated by cosmic space, like \"divided by a mere stream, silently they gaze but cannot speak.\" The only thing that can transcend time and space might be the radiation that doesn't seem dazzling from afar and the propagation of gravitational fields that dominate their motion and evolution.\n\nStill considering three celestial bodies, each having equal mass $m$. Take the reference frame in translational motion of the binary system’s center of mass, and use its center of mass as the coordinate origin. The orbit of each sub-star in the binary system is an ellipse with semi-major axis $a$ and eccentricity $e$. Orient the major axis direction as the pole direction of the planar polar coordinate system. Since the distance within the binary system is much smaller than the distance between them and the solitary star $a \\ll r$, the binary system's period is much smaller than the time scale of the movement of its center of mass revolving with the solitary star under their mutual gravitational influence. Hence, when calculating the gravitational force exerted by the binary system on the solitary star, we can use the concept of period averaging. Now, we study the situation when the solitary star is at polar coordinates $(r,\\theta_{0})$.\n\nNext, consider the reaction exerted by the solitary star on the binary system, and the overall dynamic effect will cause a gradual change in the eccentricity of the binary orbit over time (with the semi-major axis unchanged). Calculate the rate of change $\\partial e/\\partial t$. Only keep the leading term. It is known that the gravitational constant is $G$.", "solution": "", "answer": "" }, { "id": 104, "tag": "MODERN", "content": "The power of the atomic bomb is immense, but triggering the force of nuclear weapons is not easy. One type of atomic bomb uses plutonium nuclear fission, with the core being a high-purity plutonium sphere with a radius of ${r_{0}}$. The first surrounding layer is uranium, whose sole function is to reflect neutrons overflowing from the center, with a reflectivity of ${R}$. Assume the diameter of the neutron reaction cross-section for plutonium is $d$, and the direction of neutrons reflected by the uranium layer is randomly distributed within a solid angle of $2\\pi$. The total number of plutonium atoms is $N_{0}$. If a neutron is generated at the center of the plutonium sphere, find the probability $P$ that the neutron will react with plutonium, using expressions involving $ \\alpha=\\frac{3N d^{2}}{16r_{0}^{3}}, r_{0}, R$.", "solution": "", "answer": "" }, { "id": 607, "tag": "MECHANICS", "content": "This question discusses electric torpedoes. Since thermal engines generate a large amount of bubbles while operating underwater, which can reveal the target's position, modern naval warfare more often utilizes electric torpedoes. During navigation, a battery-capacitor combined power supply is used; currently, only the capacitor is considered: capacitance $C$, its internal resistance $R_{1}$, leakage resistance $R_{2}$, and rotor internal resistance $R_{3}$. The resistance of the ocean is very large, assuming the torpedo's front cross-sectional area is $A$, the rear end is equipped with a four-blade propeller, which has rectangular blades with length $a$ and width $b$, and the blades' normal forms an angle $\\theta$ with the direction of propulsion. The resistance of seawater on the front end and blades is given by $\\mathrm{d}F= k v\\cdot\\mathrm{d}S$, acting perpendicular to $\\mathrm{d}S$, where $v$ is the velocity component perpendicular to $\\mathrm{d}S$ at that location. It is known that the blades are subject to mechanical torque $\\tau$ from the motor. Determine the forward speed $u$ of the torpedo.\n", "solution": "", "answer": "" }, { "id": 599, "tag": "MECHANICS", "content": "A perfectly flexible rope with a constant linear mass density $\\lambda$ is wound into a semicircle of radius $R$ in the ground reference frame $S$, and it is strictly inextensible. An unknown downward impulse $I_0$ is applied tangentially at the left side at $\\theta = \\pi$. As a result, an upward tangential velocity $v_0$ is produced at the right free end of the rope (at $\\theta = 0$). Determine the magnitude of $I_0$, disregarding relativistic effects.\n", "solution": "", "answer": "" }, { "id": 482, "tag": "MECHANICS", "content": "A rigid cubic component is composed only of three rigid homogeneous thin plates on the top, bottom, and right side. Each plate has an edge length of $2l$ and a mass of $m$. Establish a Cartesian coordinate system with the center $O$ of the cube as the origin. The vertical direction has a gravitational acceleration of $g$, with the positive $z$ axis pointing upwards. The $x$ and $y$ axes are parallel to the edges of the horizontal plates. Currently, at points $A(0,-l,-l)$, $B(0,l,-l)$, and $C(-l,0,l)$ on the plates, vertical, inextensible light ropes of length $l$ are used to connect to fixed horizontal supports from the wall (located at $z=0, x\\leq 0$) at points $D(0,-l,0)$, $E(0,l,0)$, and $F(-l,0,0)$. Miraculously, the component is in a state of tensegral equilibrium. Based on this, calculate:\n\nIf the rope $B E$ is suddenly cut, find the tension$T_{CF}$ in the rope $C F$.", "solution": "", "answer": "" }, { "id": 353, "tag": "MECHANICS", "content": "There is now a circular ring with a radius of $R$, formed by a metal wire with a diameter of $d$. Two mutually perpendicular line segments, AB and CD, pass through the center of the circle. Points A, B, C, and D are located on the ring. An identical force $P$ is applied vertically downward on AB, while an identical force $P$ is applied vertically upward on CD. Find the relative displacement $\\Delta c$ between points B and C in the plane perpendicular to the original undeformed plane of the ring. Given: Young’s modulus $E$ and shear modulus $G$.", "solution": "", "answer": "" }, { "id": 576, "tag": "THERMODYNAMICS", "content": "In outer space, there is a magical gas planet that has a layer of solid on its surface but is gas internally. The molecular mass is $m$, and there are a total of $\\pmb{N}$ molecules. The planet has a radius of $R$. The solid layer on the surface and the internal gas rotate at the same angular velocity $\\omega$. The mass of the solid surface can be ignored, as well as any interactions (including gravitational) between the gas molecules. In this problem, assume the gas can instantaneously reach equilibrium, and the external temperature is $\\pmb{T}$ (where $\\pmb{T}$ is very small) and constant.\n\nFind the moment of inertia $I$ of the planet around its axis of rotation when the rotation speed is $\\omega_{0}$, approximated to the first order of small quantities (using $T, \\omega_{0}, N, m, R$).", "solution": "", "answer": "" }, { "id": 725, "tag": "ELECTRICITY", "content": "Between two infinitely large parallel grounded conducting plates (with a distance \\(l\\) between the plates), a point charge \\(q\\) is placed, which is at a distance \\(z_0\\) from the lower plate. Find the total induced charge \\(Q_u\\) on the upper plate, expressing the result in terms of \\(q\\), \\(z_0\\), and \\(l\\).", "solution": "", "answer": "" }, { "id": 314, "tag": "THERMODYNAMICS", "content": "Entropy is an important physical quantity for studying the properties of thermodynamic processes of ideal gases. The entropy formula for an ideal gas can be written as:\n\n$$\nS = \\nu C_{V} \\ln T + \\nu R \\ln V + S_{0}\n$$\n\nwhere \\( C_{V} \\) is the molar heat capacity at constant volume for the ideal gas, and \\( \\nu \\) is the number of moles of the gas. \\( S_{0} \\) is related to the choice of the zero point of entropy and is a constant that does not change with the state variables \\( T \\) and \\( V \\).\n\nAssume the atmospheric environment has a sufficiently large volume \\( \\textstyle V_{0} \\to \\infty \\) (but fixed), thus having a constant pressure \\( p_{0} \\) and temperature \\( T_{0} \\). There is an insulated container with a volume of \\( V \\) in the atmosphere. The container has a small hole that can be controlled to open or close. Inside the container is an ideal gas with an adiabatic index \\( \\gamma = C_{p}/C_{V} \\), with pressure \\( p = \\alpha p_{0} \\) less than the external pressure but with temperature consistent with the external temperature.\n\nWhen the small hole is opened, the external gas enters the container. This process is slow enough for the gas inside the container to reach equilibrium, but occurs rapidly enough in terms of heat transfer between the inside and outside of the container (thereby not allowing enough time for heat transfer). Determine the total entropy change \\( \\Delta S_{1a} \\) of the system (including the atmosphere) during the process.", "solution": "", "answer": "" }, { "id": 490, "tag": "THERMODYNAMICS", "content": "1 mol of ideal gas undergoes a certain quasi-static process:\n\n\\[ P = 6P_0 + \\frac{P_0}{V_0} V \\]\n\nwhere \\( P_0 \\) and \\( V_0 \\) are positive constants. For this 1 mol of ideal gas, find a new quasi-static process expression \\( P'(V') \\) so that its state curve passes through the state point \\( (4P_0, 4V_0) \\), and has the same heat capacity as the aforementioned process at each temperature.", "solution": "", "answer": "" }, { "id": 500, "tag": "MODERN", "content": "The following systems are located in a vacuum, with the known vacuum permittivity \\(\\varepsilon_{0}\\) and vacuum permeability \\(\\mu_{0}\\). Ignore relativistic effects and electromagnetic radiation. A large number of small magnetic needles are arranged in a row with a spacing of \\(r\\), and the length of the magnetic needles is much smaller than \\(r\\). The positions of each small magnetic needle are fixed, and the magnitude of the magnetic moments is \\(m\\), with the direction of the magnetic moments able to rotate within the plane. Each small magnetic needle has a moment of inertia \\(I\\) with respect to its own fixed axis. To simplify the analysis, we agree that only adjacent small magnetic needles interact with each other. Initially, all small magnetic needles are in a stable equilibrium state, and their magnetic moment directions are all parallel to each other. We apply a disturbance to the system, causing the direction of the magnetic moments of the small magnetic needles to slightly deviate from the direction at stable equilibrium. Under certain conditions, waves will form and propagate in the array of small magnetic needles (which can be regarded as harmonic waves). Let the angular frequency of the wave be \\(\\omega\\). If \\(\\omega\\) is known and satisfies the conditions for stable wave propagation, please provide the phase velocity \\(v(\\omega)\\) of the waves in the array of small magnetic needles.", "solution": "", "answer": "" }, { "id": 263, "tag": "MECHANICS", "content": "The geometric shape uses an equilateral triangle with a side length of $L$ as the main body, with three sides extending outward into curves. Using the bottom-left vertex of the main equilateral triangle as the pole, establish a polar axis $x$ vertically upward. The three curves form a continuous function defined on $[0, 2\\pi)$ in polar coordinates, with exactly three possible non-differentiable points, which are the three vertices of the original main triangle. In order for the extended shape to become a Reuleaux triangle, it must satisfy the following geometric properties: for any set of parallel lines, when at least one of the parallel lines is just tangentially \"clamping\" the geometric shape, the distance between the lines is the same. One special solution is, of course, a circular shape (that is, three 1/3 circle arcs joined together). In fact, there should be countless such curves of constant width, but we impose the following restrictions on a \"Reuleaux triangle\": ⃝1 The curves extending from each side should be congruent, and the shape should have rotational symmetry ⃝2 At the vertex positions of the main body, the curve is non-differentiable, meaning when drawing equidistant parallel lines, one parallel line passes through the vertex, and another is tangent to the opposite curve ⃝3 Composed of arc or straight line segments.\n\nA Reuleaux triangle, when translated perpendicular to the plane of the paper, forms a cylinder in space called a Reuleaux prism. Consider a homogeneous Reuleaux prism with mass $m$, placed on a horizontal surface with its lateral side down, with a known gravitational acceleration $g$. Due to the excellent geometric properties of the Reuleaux triangle, it has inspired attempts to build a \"cart\" using Reuleaux triangles as wheels.\n\n$N=10$ homogeneous Reuleaux prisms of mass $m$ are placed at equal intervals, with the horizontal distance between adjacent centroids being $L$. A large flat board with mass $m$ (long enough, not considering the prism detachment in this question) is placed horizontally as the cart body. The coefficient of friction at the contact surface is sufficient to ensure all contacts maintain pure rolling throughout the process. The initial configuration is a stable equilibrium. At this point, an initial velocity is provided to the system to the right, with the initial speed of the chassis being $v$. To ensure that the cart moves forward, solve for the minimum value of $v$, denoted as $v_{01}$. (Keep five significant digits in the decimal part.)", "solution": "", "answer": "" }, { "id": 491, "tag": "MODERN", "content": "In the inertial frame $S$, there is a point source of light with an equation of motion ${\\pmb r}=(v t,0,0)$. The emitted light is monochromatic with a frequency $f$, and the space is a vacuum. At time $t$, there is an observer $P$ on the $y$-axis located at $(0,y_{0},0)$ and moving with a velocity of $(0,\\dot{y}_{0},0)$. Calculate the signal frequency $f^{\\prime}(y_0, \\dot y_0, t)$ received by the observer $P$.", "solution": "", "answer": "" }, { "id": 632, "tag": "OPTICS", "content": "A simple telescope is composed of a thin convex objective lens $L_{o}$ and a thick convex eyepiece $L_{E}$ installed in a tube. The leftmost end of the tube overlaps with the left side of the eyepiece, and the rightmost end is the objective lens. On the right side of the eyepiece, a reference line is engraved to assist in measuring the angular size of the target object. The entire system has an angular magnification $M$. To ensure that all light that strikes the right side of the eyepiece eventually propagates to the left side without any loss of light energy, the telescope design requires the following: if a small paraxial object is placed close to the left side of $L_{o}$, it should just pass through the right side of the eyepiece and form an image on the left side. Determine the ratio of the absolute values of the radii of curvature for both sides of the eyepiece, expressed in terms of $M$.", "solution": "", "answer": "" }, { "id": 639, "tag": "THERMODYNAMICS", "content": "A thin circular disk with mass $M$, area $S$, and heat capacity $C$ is initially at a temperature $T_{1}$. It is placed in an environment filled with a monoatomic gas that maintains a constant temperature $T_{0}$. The gas molecules have a mass $m$ and a number density $n$. Gravity is neglected. The properties of the two surfaces of the disk are different. The upper surface is ideally adiabatic, meaning gas molecules reflect off this surface like balls colliding on a smooth plane. The lower surface is ideally conductive, meaning when gas molecules collide with this surface in the disk's frame, they are re-emitted as if leaking from a small hole at temperature $T$, which is the temperature of the disk, with the emission directions being randomly distributed. Note: The problem uses the following approximations: The relative temperature difference between the initial disk temperature and the gas temperature is very small, so the speed of the disk is always much less than the average speed of the gas; also, $M \\gg m$. The Maxwell speed distribution law of the gas is known: $$ {\\mathrm{f}}({\\vec{v}})={\\Big(}{\\frac{m}{2\\pi k T}}{\\Big)}^{\\frac{3}{2}}\\mathrm{exp}\\left(-{\\frac{m}{2k T}}v^{2}\\right) $$ Assume that at a certain moment, the temperature of the disk is $T$, and its speed perpendicular to the plane of the disk is $u$ (positive upwards, negative downwards). Neglect gravity. Determine the rate of change of the disk's speed $\\frac{\\mathrm{d} u}{\\mathrm{d} t}$.", "solution": "", "answer": "" }, { "id": 765, "tag": "ELECTRICITY", "content": "A lightweight insulated plastic thin disc with a radius of $R$ is placed horizontally and can freely rotate without friction around a vertical fixed axis through the center of the disc. A lightweight small circular coil with a radius of $a$ $(a \\ll R)$ is fixed on the disc's surface, coaxial with the disc. Four positively charged metal balls, each with mass $m$, are fixed at equal intervals on the edge of the disc, and each metal ball carries a charge of $q$. This device is in a uniform strong magnetic field with a magnetic induction intensity of $B_{0}$, directed vertically upward. Initially, the disc is at rest, and a constant current $I$ flows through the small circular coil in a clockwise direction (as viewed from above). If the current in the small circular coil is cut off, the disc will start to rotate. Find the magnitude of the force in the horizontal direction acting on each metal ball after the thin disc reaches stable rotation. Assume the metal balls can be considered as point masses, neglect the self-inductance of the small circular coil and the magnetic field generated by the motion of the charged metal balls. It is known that a circular coil with a radius of $a$ fixed on the disc's surface carrying a current $I$ generates a magnetic induction intensity of $B = k_{m} \\frac{2 \\pi a^{2}I}{r^{3}}$ at a distance $r$ $(r \\gg a)$ from the center of the coil. In this formula, $k_m$ is a known constant, and the direction of the magnetic field is vertical and upward when the current in the coil flows clockwise. The electrostatic force constant is $k_e$.", "solution": "", "answer": "" }, { "id": 325, "tag": "OPTICS", "content": "There are $N$ identical, uniform, transparent dielectric spheres arranged coaxially and equidistantly from left to right. Each sphere has a radius $r$ and a refractive index of $n=3/2$. The distance between the centers of two adjacent spheres is $8r$. To the left of the first sphere, at a distance $u_1$ from its center, there is a small luminous object (located near the axis). The light emitted by this object undergoes refraction successively through the $N$ dielectric spheres, forming images. The final image distance (distance from the last image to the right vertex of the $N$-th sphere) is $\\nu_{N}$. It can be proven that, as $N$ becomes very large, $\\nu_{N}$ approaches a constant value. Determine this constant value, $\\nu_{\\infty}$.Note that this constant value should be stable against small perturbations in the initial conditions.", "solution": "", "answer": "" }, { "id": 518, "tag": "THERMODYNAMICS", "content": "Before the discovery of radioactivity (and the mass-energy equation), Kelvin (1899) suggested estimating the age of the Earth based on the process of decay in the Earth's internal heat flow. He assumed that the Earth initially had a molten crust, and the exothermic process of solidification from the outside to the inside led to a gradual increase in the thickness of the solid crust. The thickness of the solid crust simultaneously determined the rate of heat dissipation outward (according to the classical heat conduction and heat radiation theorem). This allowed him to calculate the relationship between the rate of heat dissipation and time. Given the latent heat of solid-liquid phase transition per unit mass of the crust as $L$, the density of the solid as $\\rho$, the temperature of the molten material as $T$, and the current surface temperature approximately as $T_0$, as well as the Stefan constant $\\sigma$ and Fourier's thermal conductivity coefficient $\\kappa$, estimate the age of the Earth.", "solution": "", "answer": "" }, { "id": 422, "tag": "MECHANICS", "content": "A triangular wedge with an angle of inclination $\\theta$ and mass $M$ is placed on a rough horizontal surface. A homogeneous sphere with mass $m$ and radius $r$ is placed on the inclined plane of the wedge. The sphere begins to move freely downward from rest, and the wedge does not rotate during the entire process. The gravitational acceleration is $g$. It is assumed that the static and kinetic coefficients of friction between the sphere and the wedge are both $\\mu_{1}$, and the static and kinetic coefficients of friction between the wedge and the ground are both $\\mu_{2}$, where $(\\mu_{1},\\mu_{2}>0)$.\n\nIf the sphere rolls down without slipping and there is relative sliding between the wedge and the ground, find the minimum value of $\\mu_1$, expressed as a function of $\\mu_2$.", "solution": "", "answer": "" }, { "id": 82, "tag": "MECHANICS", "content": "Consider the problem of the motion of an ideal rigid body under the influence of gravitational force. A stationary rigid homogeneous sphere has a mass $M$, its surface is smooth, and a homogeneous rod of length $L$ has a mass $m$, with $M \\gg m$. The rod is tangent to the sphere, and at the initial moment, the point of tangency between the rod and the sphere is at a distance $\\pmb{\\mathcal{x}_0}$ from the center of the rod. The rod is released from rest. The gravitational constant is known to be $G$, and the radius of the sphere is $R$.\n\nIf the offset value ${\\mathcal{x_0}}$ is small, the rod will perform small harmonic oscillations. Find the angular frequency of the small oscillations.", "solution": "", "answer": "" }, { "id": 668, "tag": "ELECTRICITY", "content": "A conducting sphere will generate a current under the influence of an external alternating electric field, thereby producing Joule heat. As shown in the figure, a metal conducting sphere with a radius of $R$ is placed in an electric field $E=E_{0}\\cos\\omega t$. Given the vacuum permittivity $\\epsilon_{0}$ and the sphere's resistivity $\\rho$, ignore all electromagnetic effects other than the electrostatic effect, and find the average power of heat generation $P$ when $\\omega \\neq 0$.", "solution": "", "answer": "" }, { "id": 729, "tag": "MECHANICS", "content": "A bomb explodes on the ground, and after the explosion, all fragments are ejected at the same speed $u$, with their directions confined to a narrow cone with an angle not exceeding \\(\\alpha\\) to the vertical (the angles are uniformly distributed within this range). All fragments eventually land, and the landing is completely inelastic. Let the mass of the bomb be $M$. Define the radial density distribution \\(\\rho(r)\\) such that \\(\\rho(r)2\\pi r\\,dr\\) is the mass of debris within the range of radius \\(r\\) to \\(r+dr\\) from the center. Find \\(\\rho(r)\\) within the maximum range \\(R\\) (retaining terms up to \\(r^2\\) in the approximation.\n", "solution": "", "answer": "" }, { "id": 444, "tag": "THERMODYNAMICS", "content": "Given that the ambient temperature is $T_{a}$ and the atmospheric pressure is $p_{a}$. The initial temperature in a tire is also $T_{a}$, with the pressure being $p_{0}$. The maximum pressure in the tire right after inflation is noted as $p_{m}$, and the temperature at this time is denoted as $T_{m}$. Afterward, it slowly cools down to the ambient temperature $T_{a}$, and the pressure decreases to the final target pressure $p_{f}$. Assuming the air is an ideal gas with a constant heat capacity ratio $C_{p}/C_{\\nu} = \\gamma$, the tire volume $V_{0}$ is considered constant, and the inflation process is adiabatic with all frictional dissipation neglected. The universal gas constant is $R$.\n\nUsing a large air compressor to inflate the tire, assume the pressure of the air in the compressor remains constant during inflation, and its temperature is $T_{a}$. What is the minimum air pressure $p_{c}$ in the compressor needed to achieve the final pressure $p_{f}$? (The answer to the above question should only contain $\\gamma$, $p_{0}$, and $p_{f}$.)", "solution": "", "answer": "" }, { "id": 337, "tag": "MECHANICS", "content": "Small balls $a$ and $b$, with masses $m_{a}$ and $m_{b}$ respectively, are placed on a smooth insulated horizontal surface. The two balls are connected by an insulating light spring with an original length of $l_{0}$ and a spring constant of $k_{0}$. If the two small balls are given equal charges of the same sign, the spring's length is $L_0$ when the system is in equilibrium. Let $K$ be the electrostatic force constant. Find the frequency of small oscillations of the system comprising the two balls and the spring (neglect the effects of induced charges). The final answer should not include any physical quantities not provided in the problem statement.", "solution": "", "answer": "" }, { "id": 151, "tag": "MODERN", "content": "The so-called photon rocket refers to a spacecraft that can achieve propulsion by converting its own rest energy into photons for emission. Consider the following theoretical problem:\n\nA photon rocket undergoes two stages of acceleration in a vacuum from an initial state to a final state. It is known that the photon rocket is initially at rest with a rest mass of $m$. In the final state, the photon rocket reaches a velocity of $0.8c$, and its rest mass decreases to $0.25m$. Here, $c$ is the speed of light. It is also known that if we take the instantaneous inertial frame of the spacecraft at the end of the first stage of acceleration (i.e., an inertial frame in which the spacecraft is momentarily at rest), the direction of acceleration of the spacecraft remains fixed in both stages, but the magnitude of acceleration is unspecified. Additionally, in this frame, the angle between the directions of acceleration in the two stages is $\\alpha$. Find the minimum value of $\\alpha$.", "solution": "", "answer": "" }, { "id": 637, "tag": "MODERN", "content": "A railroad is moving at a high speed with velocity \\( v \\) perpendicular to the direction of the railroad relative to the observer. In the railroad's reference frame, there is a rectangular train with the same width as the railroad, moving with velocity \\( 2v \\) parallel to the railroad with respect to the railroad. Considering relativistic effects, calculate the angle between each of the four sides of the train and the railroad as seen in the observer's reference frame. It is known that all four angles between the sides and the railroad are equal. Please express the result in terms of \\( v \\) and \\( c \\), where \\( c \\) is the speed of light.", "solution": "", "answer": "" }, { "id": 468, "tag": "MECHANICS", "content": "You are requested to solve the following physics problem and present the final answer using \\boxed{}: Four uniform-density slender rods $M A, A P, B N, P B$ are smoothly hinged together to form an \"M\"-shaped structure, with $M$ and $N$ constrained to move along horizontal rails, and $P$ constrained to move along a vertical rail. The width of the rails and the diameter of the rods are negligible. The rails do not affect the rotation of the rods. An $O x y$ coordinate system is established, and at the initial moment, the coordinates of the endpoints of the rods are given as: $M(-2l,0)$, $N(2l,0)$, $P(0,l)$, $A(-l,2l)$, $B(l,2l)$. \n\nStarting from the initial moment, motors control the movement of rod ends $M$ and $N$ along the negative and positive directions of the $x$ axis, respectively, with velocity $v$ and acceleration $a$. Meanwhile, rod end $P$ moves along the positive direction of the $y$ axis with velocity $2v$ and acceleration $2a$. The mass per unit length of the rods is $\\lambda$, and the gravitational acceleration is $g$. Ignoring all losses, determine the power exerted by the motors at the initial moment.", "solution": "", "answer": "" }, { "id": 577, "tag": "THERMODYNAMICS", "content": "One mole of a certain substance is a simple $p,v,T$ system. The coefficient of volumetric expansion under any circumstance is \n\n$$\n\\alpha=3/T\n$$ \n\nThe adiabatic equation under any circumstance is \n\n$$\np v^{2}=C\n$$ \n\nAnd the constant pressure heat capacity of the system under any circumstance is exactly: \n\n$$\nc_{p}=\\lambda{\\frac{p v}{T}}\n$$ \n\nHere, $\\lambda$ is a dimensionless constant independent of the state. Find all possible values of $\\lambda$. \n\nHint 1: According to the third law of thermodynamics, the absolute entropy should be zero at a temperature of zero for any volume. \n\nHint 2: Entropy $S$ can be expressed as a function of $T,p$, i.e., $S=S(T,p)$. In this case, we have $\\begin{array}{r}{\\mathrm{d}S=\\left(\\frac{\\partial S}{\\partial p}\\right)_{T}\\mathrm{d}p+\\left(\\frac{\\partial S}{\\partial T}\\right)_{p}\\mathrm{d}T}\\end{array}$, where the subscript T indicates the partial derivative is taken at constant T, and $\\begin{array}{r}{\\left(\\frac{\\partial S}{\\partial p}\\right)_{T}=-\\left(\\frac{\\partial v}{\\partial T}\\right)_{p}}\\end{array}$. If $\\mathrm{d}f=A\\mathrm{d}x+$ $B\\mathrm{d}y$, then $\\begin{array}{r}{\\left(\\frac{\\partial A}{\\partial y}\\right)_{x}=\\left(\\frac{\\partial B}{\\partial x}\\right)_{y}.}\\end{array}$", "solution": "", "answer": "" }, { "id": 417, "tag": "MODERN", "content": "There is a rocket in space with an initial rest mass of $m_{1}$, moving at a very high velocity $\\mathbf{v}$ (v can be compared with the speed of light c). Starting from a certain moment, the rocket is ejecting gas at a constant high speed $\\mathbf{u}$ in the direction perpendicular to its own velocity as observed in its reference frame. The mass of the gas ejected per unit time in its reference frame is $\\lambda$. Considering relativistic effects, find the angle turned by the rocket in its direction of motion after time t in the ground reference frame.", "solution": "", "answer": "" }, { "id": 682, "tag": "MECHANICS", "content": "A homogeneous disk with mass \\(m\\) and radius \\(R\\) is suspended by three light strings of length \\(L\\). In the equilibrium state, the three strings are equally spaced and remain vertical. If the disk is rotated through a small angle \\(\\theta\\) around its axis and then released, find the period of the disk's motion.\",\n", "solution": "", "answer": "" }, { "id": 159, "tag": "MODERN", "content": "A sphere in vacuum has a radius of $R$ in its rest frame, and its surface is uniformly charged with a charge $Q$. It is moving with a constant velocity $v=\\beta c$ relative to the ground frame, where $c$ is the speed of light in vacuum. Given the vacuum permittivity $\\varepsilon_0$, find the electromagnetic field energy in all space in the ground frame.", "solution": "", "answer": "" }, { "id": 147, "tag": "MECHANICS", "content": "A light thin wire of length $L$ is wound around a rough cylinder with radius $R$ to form a coil. The coil is subjected to a pulling force at one point, parallel to the axis of the cylinder. Due to the friction between the coil and the cylinder, with the coefficient of friction $\\mu$, there exists a critical length $L_{0}$ beyond which the coil cannot remain stationary on the surface of the cylinder no matter what. Determine this critical length $L_{0}$.", "solution": "", "answer": "" }, { "id": 239, "tag": "MECHANICS", "content": "The influence of the gyro effect: In the 20th century, Arnold Sommerfeld, Felix Klein, and Fritz Noether jointly proposed an explanation—the gyro effect. The gyro effect is essentially a manifestation of the angular momentum theorem. Under the action of external torque, the angular momentum of a rotating object changes, leading to the generation of precession angular momentum. In other words, under the influence of the gyro effect, a rapidly rotating object will not fall down directly but will instead experience a change in its rotational direction.\n\n$AB$ is fixed to the ceiling. A lightweight rod $BO$ with length $L$ is rigidly connected to the center of a uniform disk with mass $m$ and radius $R$. The rod $BO$ is connected to $AB$ by a ball joint, allowing it to rotate freely in all directions. The disk rotates with an angular velocity $\\omega$ around $BO$. Solve for the precession angular velocity $\\Omega$ required to keep $BO$ horizontal.", "solution": "", "answer": "" }, { "id": 472, "tag": "ELECTRICITY", "content": "Consider a uniform thin spherical shell whose center of mass is fixed but is free to rotate about the center of mass. The shell is uniformly charged, with a total charge of $Q$, mass $m$, and radius $R$. Suppose that there are many such thin spherical shells distributed in space, with their centers of mass fixed but free to rotate about the center of mass. The radius of each shell is $R$, their number density is $n$, and it satisfies $n R^{3} \\ll 1$. Considering the difference between the macroscopic average magnetic field $\\vec{B}$ of the system and the effective external magnetic field experienced by each thin shell, and accounting for the effects of this difference, determine the relative magnetic permeability $\\mu_{r}$ of the system. The vacuum permeability $\\mu_0$ is given.", "solution": "", "answer": "" }, { "id": 170, "tag": "OPTICS", "content": "The Nobel Prize-winning \"optical tweezer\" technology actually uses the single-beam gradient force to form an energy potential well for controlling small molecules. We are attempting to use a similar principle to design a small robot driven by light. We assume this micro-robot can be considered as a disk with a radius $a$, which is much larger than the wavelength of light used. We use a beam of parallel light with a radiation energy density of $u_{0}$ to uniformly illuminate the surface of the robot perpendicularly. The entire disk is within the illumination range of the parallel light. Assume that the exterior protective layer of the micro-robot is a medium whose refractive index continuously changes according to $n=\\alpha/\\left(z+\\beta\\right)$. Here, $\\alpha$ and $\\beta$ are undetermined constants, and $z$ is the depth into the surface. The refractive index changes continuously from $n_{1}$ at the outer surface $(z=0)$ through a transition layer of thickness $l_{0}$ to $n_{2}$ at the inner surface. The radius of the robot is much larger than the wavelength $\\lambda$ of the incident wave. Calculate the reflection coefficient $R$ of this protective layer, assuming the layer is extremely thin. Express $R$ in terms of $n_1$ and $n_2$, without using any approximations not provided in the problem statement.", "solution": "", "answer": "" }, { "id": 471, "tag": "ELECTRICITY", "content": "A rigid insulating cubic frame with side length $a$, labeled $A B C D - E F G H$, has charges fixed at each vertex with a magnitude of $q$. The charges at vertices $B$ and $H$ are negative, while the charges at vertices $A$, $C$, $D$, $E$, $F$, and $G$ are positive. A rigid and smoothly lined fixed track (whose inner diameter is much smaller than $a$) passes through vertices $B$ and $H$. At the midpoint $O$ of track $\\vec{B H}$ (which is also the center of the cube), a point charge with a negative charge $-Q$ and mass $m$ is placed. Neglect relativistic effects, with vacuum permittivity $\\varepsilon_{0}$ and vacuum permeability $\\mu_{0}$, and disregard gravity. When the cubic frame is fixed, discuss the period $T_r$ of the small oscillations of the point charge near equilibrium at $O$ due to a slight disturbance.", "solution": "", "answer": "" }, { "id": 169, "tag": "MECHANICS", "content": "The inhabitants of Blue Planet have always fantasized about one day getting rid of the inefficient rocket as a means of transportation and directly creating a \"Sky Ladder\" from the Earth to the heavens. Some call it the \"Space Elevator,\" while others refer to it as the \"Orbital Lift.\" However, they do not realize that due to the limitations imposed by physical laws, building a space elevator on Blue Planet is unrealistic. In contrast, years later, on Mars, which is further from the Sun, space elevators can \"stand tall.\"\n\nAs early as in the Book of Genesis in the Bible, humans hoped to jointly construct a giant tower reaching from the ground to the sky—the Tower of Babel—to proclaim their fame. The person who first proposed the concept of a space elevator should be the Russian \"father of rockets,\" Tsiolkovsky. The space elevator he envisioned would extend from the ground to the geosynchronous orbit tens of thousands of kilometers high. As the elevator ascends, the internal gravity gradually decreases. When cargo reaches the endpoint via the elevator, its speed already allows it to maintain synchronous orbital motion around Earth. Therefore, the synchronous orbit station is in a state of complete weightlessness.\n\nIn 1979, Arthur C. Clarke's science fiction novel \"The Fountains of Paradise\" introduced the revolutionary concept of the space elevator to the public's attention. Since then, space elevators have been widely present in various sci-fi works. Whether it is the space elevator made of advanced nanomaterials in \"The Three-Body Problem\" or the three orbital lifts operated by Union, AEU, and Human Reform League in \"Mobile Suit Gundam 00,\" the basic concepts and principles are the same.\n\nNowadays, we occasionally hear some related news reports, either from Japan's company \"Obayashi Corporation\" planning to build the first space elevator by 2050, or from a Canadian company proposing new space elevator schemes. Regardless of whether these are commercial companies creating hype, such news indeed prompts more people to wonder—are we really close to the era of abandoning rockets and embracing space elevators?\n\nIs this really the case? How far away are we from the technology needed to build a space elevator? We will conduct a detailed analysis in this matter (the space elevators discussed here are all constructed at the equator, with no consideration for any curvature).\n\nIn Tsiolkovsky's vision, the space elevator following Earth's rotation would reach a certain height, after which a satellite released freely would automatically become a geostationary satellite. Calculate the height $l_{1}$ of the space elevator in this model. Given are: gravitational acceleration $g$, Earth's radius $R$, and Earth's rotation period $T_{day}$.", "solution": "", "answer": "" }, { "id": 678, "tag": "ELECTRICITY", "content": "Calculate the magnetic field produced in space by a moving electric dipole (with dipole moment \\(\\vec{p}\\) and velocity \\(\\vec{v}\\)) under the following conditions: the movement is parallel to the dipole moment direction (\\(\\vec{v} // \\vec{p}\\)), and the velocity \\(v\\) is much smaller than the speed of light \\(c\\). Use the dipole as the origin, designate the direction of \\(\\vec{v}\\) as the polar axis, and represent the system in spherical coordinates \\((r,\\theta,\\phi)\\). Only the \\(\\phi\\) component of the magnetic field \\(B\\), denoted as \\(B_{\\phi}\\), needs to be provided.", "solution": "", "answer": "" }, { "id": 16, "tag": "MECHANICS", "content": "Consider a child with mass \\( m \\) sitting on a swing, the child can be regarded as a point mass with the mass concentrated at the seat plank. Ignore the mass of the other parts of the system. The distance from the swing seat plank to the pivot is \\( l \\). At this time, consider the frictional torque \\( M_f = a \\) (where \\( a \\) is a constant) at the swing's suspension point. There is someone behind who applies an impulsive torque \\( J_0 \\) to the swing every time it reaches the furthest back position. Find the difference in speed rates of the child after passing the lowest point twice successively when the motion reaches a steady state (with gravitational acceleration \\( g \\) and assuming the swing angle is relatively small).", "solution": "", "answer": "" }, { "id": 515, "tag": "MECHANICS", "content": "Collisions are one of the simplest mechanical interactions between objects, and they are the process of energy and momentum exchange between objects. The cascade of one-dimensional collisions resulting in chain events provides a simple model for a series of natural phenomena dominated by continuous collisions. When studying such systems, the physical quantities we are interested in are the proportion of energy or momentum transmitted by the collision cascade and its dependence on the coefficient of restitution \\( e \\) and the collision mass ratio. We arrange \\( n(n\\gg1) \\) balls (with masses \\( m_{i}(i=1,2,...,n) \\), adjustable) between an initial ball with mass \\( M \\) (fixed) and an end ball with mass \\( m \\) (fixed). Only the first collision between adjacent balls is considered. All collisions are inelastic with a coefficient of restitution \\( e \\) (very close to 1).\n\nRecently, Ricardo and Lee proved that when the mass distribution of the cascade ball chain is such that the mass \\( m_{i} \\) of each ball is the geometric mean of the masses of the two adjacent balls \\( \\sqrt{m_{i-1}m_{i+1}} \\), the efficiency of kinetic energy and momentum transfer between the initial and final balls is maximized. For a system of \\( n \\) balls, the maximum transfer efficiencies of kinetic energy and momentum between the initial and final balls can be expressed as \\( r_{K n},r_{p n} \\).\n\nWe aim to optimize the efficiency of kinetic energy and momentum transfer between the initial and final balls in this chain event. Provide the number of balls \\( n \\) that should be arranged between the initial and final balls for maximum transfer efficiency. (Expand to the lowest order in \\(\\delta=1-e\\), with the answer expressed in terms of \\( m \\), \\( M \\), and \\(\\delta\\)).", "solution": "", "answer": "" }, { "id": 176, "tag": "MECHANICS", "content": "A cylindrical thin-walled water bucket is fixed on a horizontal surface, with a cross-sectional area of \\( S_0 \\). Inside, there is water at a height of \\( a + b \\). Now, a small hole is suddenly opened at a distance \\( b \\) from the bottom, extending in a horizontal direction (the hole's area \\( S_1 \\ll S_0 \\)), causing the water to start leaking. The height \\( h \\) of the water above the hole gradually decreases from \\( a \\) to 0. The density of the water is \\( \\rho \\), and the gravitational acceleration is \\( g \\). The process goes from the beginning until all the water has flowed out, treating the water as an ideal fluid and ignoring the higher-order effects of internal flow and the bucket's shape and size. Find the functional relationship between the linear density of water distribution on the ground \\(\\lambda\\) and the horizontal distance \\( x \\) from the bucket wall (approximating the water trajectory as a thin line).", "solution": "", "answer": "" }, { "id": 626, "tag": "OPTICS", "content": "After parallel light enters a Michelson interferometer, it is split into two beams by the beam splitter. These two beams travel different paths and interfere at the observation screen. Mirrors $M_{1}$ and $M_{2}$ are plane mirrors with different distances to the beam splitter, causing the two beams that eventually reach the observation screen to have an optical path difference of $\\Delta L$. If the incident light consists of two wavelengths that are very close together, with an average wavelength of $\\lambda$ and a wavelength difference of $\\Delta\\lambda{(\\Delta\\lambda\\ll\\lambda\\bigr)}$, and both beams have the same intensity upon reaching the observation screen, calculate the relationship of the interference contrast $M$ with the change in $\\Delta\\lambda$, $M(\\Delta\\lambda)$.", "solution": "", "answer": "" }, { "id": 108, "tag": "OPTICS", "content": "When electromagnetic waves undergo total internal reflection at an interface, it has been observed that under certain circumstances, the reflected wave experiences a lateral shift along the interface relative to the point of incidence. This phenomenon is historically known as the \"Goos-Hänchen shift.\" The core concept is that during total internal reflection, the reflected wave acquires an additional phase $Φ(\\theta)$, which depends on the angle of incidence $\\theta$ (typically decreasing as the angle of incidence increases, becoming zero at the critical angle $\\theta$ for total internal reflection, with its derivative with respect to the angle of incidence becoming infinite at the critical angle). When the incident wave is not a perfect plane wave, electromagnetic waves with closely spaced but different angles of incidence can interfere. The condition for this interference to occur is that the reflected waves from different angles of incidence must achieve equal phase by undergoing a lateral shift $d$ along the interface. Given that the wavelength of the electromagnetic wave is $\\lambda$, and the phase shift in the \"Goos-Hänchen shift\" is $Φ(\\theta)$, determine the lateral shift $d$.", "solution": "", "answer": "" }, { "id": 363, "tag": "THERMODYNAMICS", "content": "The reversible Carnot cycle is an ideal cycle process; actual processes, due to irreversible factors such as thermal resistance, heat leakage, and friction, are always irreversible. Now, by solely considering the presence of thermal resistance (ignoring factors such as heat leakage and friction), an ideal cycle process—the internally irreversible Carnot cycle—is introduced. The internally irreversible Carnot cycle satisfies the following three conditions:\n\n(i) The working substance undergoes a quasi-static Carnot cycle; \n\n(ii) Due to the thermal resistance between the working substance and the heat source, the temperatures $T_{1},T_{2}$ of the two isothermal processes that the working substance undergoes are different from the heat source temperatures $T_{H},T_{L}$, specifically: $T_{H} > T_{1} > T_{2} > T_{L}$. In this way, there is a finite temperature difference between the heat source and the working substance, allowing heat conduction to occur over a finite period, resulting in power output from the cycle; \n\n(iii) Heat conduction satisfies Fourier's law of heat conduction: $\\frac{\\mathrm{d}Q}{\\mathrm{d}t}=\\kappa\\Delta T$, where $\\pmb{\\kappa}$ is a known proportionality coefficient. It is assumed that the adiabatic processes proceed very quickly. \n\nCompared to the ideal Carnot cycle, the internally irreversible Carnot cycle is of more practical significance. Using this model:\n\nFind the total entropy increase of the system per unit time when the power output of the internally irreversible Carnot cycle is maximized, using $\\kappa,\\:T_{H},\\:T_{L}$ for representation, assuming the heat source temperatures are fixed.", "solution": "", "answer": "" }, { "id": 754, "tag": "THERMODYNAMICS", "content": "Consider a closed system with a constant volume, initially containing liquid and vapor phases of the same substance in equal molar amounts. The initial pressure and temperature of the vapor are $P_{1}$ and $T_{1}$, respectively. Then, under the condition of keeping the total volume of the system unchanged, the system is heated to $T_{2}$. Assuming that the vapor satisfies the ideal gas condition during this process, the molar volume of the liquid phase is $v_{l}$, the constant pressure molar heat capacities of the gas and liquid are $C_{p}^{g}$ and $C_{p}^{l}$ respectively, and the molar latent heat of phase transition for the liquid is $\\Delta h$, which is independent of temperature and pressure. Find the ratio $f_{l}$ of the molar number of the liquid phase to the total molar number at the temperature $T_{2}$.", "solution": "", "answer": "" }, { "id": 664, "tag": "OPTICS", "content": "The flow of the medium will affect the additional bending of the light. Below the $x$-axis is a vacuum, and above the $x$-axis is a uniform medium moving at a constant speed of $\\beta c$ in the positive $x$-axis direction, with a refractive index of $n$. Now consider a beam of light obliquely incident into the medium from below. The speed of light in a vacuum is known to be $c$, and the angle of incidence is $i$. Considering relativistic effects, find the angle of refraction $t$.", "solution": "", "answer": "" }, { "id": 782, "tag": "ELECTRICITY", "content": "To consider the motion of an electron in a hydrogen atom as uniform circular motion around the hydrogen nucleus, when studying the magnetic effect of the electron's motion, the electron's motion can be equivalently considered as a loop current, with the radius of the loop equal to the orbital radius of the electron $r$. Now, apply an external magnetic field to a hydrogen atom, with the magnetic field having a magnetic induction strength of $B$, and a direction perpendicular to the orbital plane of the electron. At this time, the equivalent current of the electron's motion is represented by $I_{1}$. Now reverse the external magnetic field, but the strength of the magnetic induction remains unchanged, still $B$. At this time, the equivalent current of the electron's motion is represented by $I_{2}$. Assume that when the magnetic field is applied and when it is reversed, the position of the hydrogen nucleus, the orbital plane of the electron's motion, and the orbital radius all remain unchanged. Calculate the difference in the equivalent current of the electron's motion before and after the external magnetic field is reversed, that is, how much is $ \\mid I_{1}-I_{2} \\mid$? Use $m$ and $e$ to represent the mass and charge of the electron, respectively.", "solution": "", "answer": "" }, { "id": 217, "tag": "MECHANICS", "content": "An infinitely long uniformly negatively charged line with a charge linear density of $-\\lambda$ is fixed in space and vertically connected through two light strings of length $b$ to the ends of an insulating rod of length $a$. The rod has a mass $m$, is uniform, and carries a uniformly distributed charge with a linear density of $+\\lambda$. The gravitational acceleration is $g$. The rod can be normally suspended below a critical mass $m_0$. Under the condition that $m > m_0$, if the ends of the rod are subjected to equal amplitude perturbations in the vertical direction perpendicular to the plane, the system can undergo small oscillations. Calculate the angular frequency $\\omega$ of these oscillations.", "solution": "", "answer": "" }, { "id": 300, "tag": "MECHANICS", "content": "On a horizontal surface stands a homogeneous ring with a radius of $R$ and a mass of $m$. The coefficient of friction between the ring and the ground is infinite. A disk with the same mass $m$, and mass evenly distributed, can roll without slipping inside the ring. The disk has a radius of $r$. The gravitational acceleration is $g$. Initially, the system is at rest, with the disk in contact with the inside of the ring, and the centers of the disk and the ring are at the same horizontal height. Release the system and determine: the speed $v$ of the ring when the disk reaches the lowest point.", "solution": "", "answer": "" }, { "id": 736, "tag": "THERMODYNAMICS", "content": "There is a cone-shaped cylinder with internal gas separated from external gas by a special piston. The mass of the piston is not considered, but its area can change. Its edge is always fixed on the inner surface of the cone, moving without sliding friction, and the piston remains in a single plane during motion (always a cross-section of the cone), without bulging or sinking. The piston can be viewed as a thin film with variable area stretched by a ring-shaped rubber band, with the film not providing any force. The original length of the rubber band is $0$, and the tension is proportional to its length, where: ${F}_{T}=k L$. The angle between the cone's slant height and the axis of symmetry is $\\theta$. Assume initially there are $n\\ \\mathrm{mol}$ of monoatomic ideal gas in the cylinder with temperature $T$, and a heat quantity $Q$ is given to these gases. Find the pressure of the gas at this time. The ideal gas constant $R$ is known. Atmospheric pressure is ignored.", "solution": "", "answer": "" }, { "id": 527, "tag": "MECHANICS", "content": "The surface tension coefficient of a soap bubble is $\\sigma$. A soap bubble is placed in a vacuum, and the molar specific heat at constant volume of the internal gas is $C_v = \\frac{3}{2}R$. Given the mass of the soap bubble $m$, and the equilibrium radius of the bubble after change as $R_{0}$, determine the vibration period $T$ of the soap bubble after applying a disturbance by adding a charge $Q$, where the bubble maintains its spherical shape. The vacuum permittivity is $\\epsilon_0$.", "solution": "", "answer": "" }, { "id": 746, "tag": "MECHANICS", "content": "There is a sufficiently high smooth cylinder with a radius of $R$, vertically fixed in place. There is a vertical smooth track on the surface of the cylinder. Now we have an elastic thick rope (it can bend, but it is straight when not under force) with a radius of $r$, smooth surface, Young's modulus of $E$, negligible mass, and an original length of $l_{0}$. It is assumed that the thick rope can only elongate in the length direction, while changes in thickness are to be ignored. Now, one end of the thick rope is fixed at the top of the track (it cannot slide up and down), and then the rope is wrapped around the cylinder in a counterclockwise direction from top to bottom, making a total of $N$ turns. Throughout the process, due to the smooth surface of the thick rope, its shear deformation can be ignored. Additionally, because $l_{0}$ is much larger than $r$, the complex situation at the ends of the rope can be ignored. Suppose the angle between the midpoint of the rope at any point and the vertical direction is $\\theta$. Find the elastic potential energy $V_{\\mathrm{elastic}}$ of the rope. (Note: Consider the energy change caused by the bending of the rope)\"\n", "solution": "", "answer": "" }, { "id": 537, "tag": "OPTICS", "content": "Sometimes we see a slightly upward-curved colorful halo in the sky. This is not a rainbow. This halo is caused by the refraction of light in ice crystals, so it is called an ice halo. There are many types of ice halos, and one of them is called the circumhorizontal arc. Let's explore this phenomenon.\n\nIn the phenomenon of the circumhorizontal arc, the ice crystals involved have flat hexagonal shapes. Due to the resistance encountered while falling, these ice crystals tend to align their top and bottom faces horizontally. When light enters from the side of a hexagonal ice crystal and refracts twice before exiting from the bottom face, a circumhorizontal arc is formed. Assume the altitude angle of the sun at this time is $\\theta$, and the refractive index of ice is denoted as $n$. \nFind the lower bound of the central angle of the circumhorizontal arc for different altitude angles $\\theta$, providing the result in an expression containing $n$.", "solution": "", "answer": "" }, { "id": 778, "tag": "OPTICS", "content": "", "solution": "", "answer": "" }, { "id": 679, "tag": "ELECTRICITY", "content": "A uniformly charged homogeneous rigid small sphere has a radius of $R$, electric charge $Q$, mass $m$, and a moment of inertia of $s m R^{2}$. Now, suppose the small sphere is placed on a sufficiently rough turntable which rotates at a constant angular velocity $\\Omega$. A uniform magnetic field $B$ is perpendicular to the plane of the turntable and is in the same direction as the turntable's angular velocity. Assuming there is no gravity in the space and the small sphere will not detach from the turntable. It is known that the differential equation for the motion of the small sphere's center of mass can be written in the form $$ m\\dot{\\vec{v}}=Q\\vec{v}\\times\\vec{B_{e}}-k_{e}\\vec{r} $$ where $k_{e}$ and $\\vec{B_{e}}$ are effective parameters to be determined. You only need to find the product $k_{e}|\\vec{B_{e}}|$.", "solution": "", "answer": "" }, { "id": 661, "tag": "MECHANICS", "content": "In movies, there are often scenes like this: diving underwater immediately when there's an explosion on water, or surfacing immediately when there's an explosion underwater, seemingly to avoid the damage caused by the explosion. Let's study this problem below. In space, below the $y$-axis is water, and above the $y$-axis is air. A sound (with small amplitude) is transmitted from the water, where the sound speed and density are $c_{1}, \\rho_{1}$ respectively, to the air, where the sound speed and density are $c_{2}, \\rho_{2}$ respectively. The incident angle is $\\theta_{1}$. Find the ratio of the transmitted sound power to the incident sound power (expressed in terms of $c_{1}, \\rho_{1}, c_{2}, \\rho_{2}, \\theta_{1}$) (Hint: Both water and air are fluids, and their shear modulus is 0).", "solution": "", "answer": "" }, { "id": 154, "tag": "MODERN", "content": "In approximately flat spacetime, far away from any celestial bodies, four identical spaceships, A, B, C, and D, each have a rest length of $l_0 = 1.25 c t_0$ (where $c$ is the speed of light in vacuum, and $t_0$ is a known constant) and move uniformly in a straight line. Each spaceship is equipped with two fixed positioning rods $x_i$ and $y_i$ (where $i = A, B, C, D$, hereafter referred to as \"positioning rods\"), with rod $x_i$ along the direction of the spaceship.\n\nInitially, the tails of spaceships A, B, C, and D (denoted as $a_i$) coincide.\n\nThe observed motion of the spaceships and directions of the positioning rods in different reference frames are:\n\n1. In the reference frame of spaceship A, rod $y_A$ is perpendicular to rod $x_A$.\n2. In the reference frame of spaceship A, spaceship B moves at a velocity $v$ along the direction of rod $x_A$, and the two positioning rods of B are parallel to those of A.\n3. In the reference frame of spaceship B, spaceship C moves at a velocity $v$ along the direction of rod $y_B$, and the two positioning rods of C are parallel to those of B.\n4. In the reference frame of spaceship C, spaceship D remains stationary relative to spaceship A, and the two positioning rods of D are parallel to those of C.\n\nThe spaceships communicate with each other via electromagnetic waves, with each spaceship receiving electromagnetic waves at its bow (denoted as $b_i$) and emitting electromagnetic waves from its tail. In the reference frame of spaceship A, the initial moment is marked as $t=0$, and it is known that $v=0.6c$. Starting from $t=0$, the spaceships take turns to emit contact signals with electromagnetic waves in the order A→B→C→D→A→⋯, and in the reference frames of each spaceship, receiving the signal emitted by the previous spaceship and starting to emit a signal to the next spaceship occurs simultaneously. Find the moment when spaceship A receives the signal emitted by D for the first time (keep 4 significant digits in the coefficients in the result).", "solution": "", "answer": "" }, { "id": 467, "tag": "MODERN", "content": "Solve the following physics problem, and frame the final answer using \\boxed{}:\n\nConsider the following problem, which requires taking relativistic effects into account. A particle with a rest mass of $m$ starts from the origin and undergoes one-dimensional motion under the influence of a linear restoring force $F = -k x$. Its initial kinetic energy is $E_{k}$, and the speed of light in a vacuum is $c$. Calculate the period of motion $T^{\\prime}$ when $E_{k}$ is maximized.", "solution": "", "answer": "" }, { "id": 770, "tag": "ELECTRICITY", "content": "Consider a photonic crystal in the shape of a cube with side length 𝑎 and a relative permittivity of $ \\varepsilon_{r1}$. Inside, there are impurities with a radius of 𝑏, a relative permittivity of $ \\varepsilon_{r2}$, and a number density of $n$. Industrially, sparse pores are dug vertically through the photonic crystal on one of its faces, with a number density of $N$ and a radius of 𝑐. During problem-solving, edge effects do not need to be considered, and the relative permittivity of air is 1. Furthermore, $a \\gg b, a \\gg c$. Determine the effective relative permittivity $ \\varepsilon_{3}$ obtained in the direction along the pores.", "solution": "", "answer": "" }, { "id": 186, "tag": "MECHANICS", "content": "In daily life, due to various influences, a chandelier will vibrate near its equilibrium position, and we establish a model to study this phenomenon. A homogeneous disk with mass $\\pmb{m}$ and radius $\\pmb R$ is suspended by three light strings of length $I$. The suspension points above form an equilateral triangle with side length $\\sqrt{3}R$, and the points where the strings attach on the lower disk also form an equilateral triangle with side length $\\sqrt{3}R$. In the equilibrium state, the three strings are equally spaced and remain vertical. A homogeneous disk with mass $\\pmb{m}$ and radius $2R$ is suspended in a similar manner using three light strings below the above disk. The suspension points form an equilateral triangle with side length $\\sqrt{3}R$, and the points of attachment on the lower disk form an equilateral triangle with side length $2\\sqrt{3}R$. The height difference between the two disks is $l$. Find the maximum angular frequency among the possible vibration modes of the two disks.\n", "solution": "", "answer": "" }, { "id": 534, "tag": "ELECTRICITY", "content": "A positive charge \\( Q \\) is fixed at the center of a uniform magnetic field, which is distributed in a circular area. The direction of the magnetic field is perpendicular to the plane of the paper, pointing inward. The magnetic induction intensity is \\( B \\). A particle with mass \\( m \\), carrying a negative charge (charge quantity \\( -q \\)), moves in a clockwise circular motion at a constant rate around the fixed positive charge \\( Q \\) within the plane of the paper. The radius of the circular motion is \\( R \\). Ignore the effects of gravity and friction. Find: \nIf the magnetic field suddenly vanishes and the time taken for the magnetic field to disappear is very short (much less than the period of the circular motion), what is the period of the particle's motion if it continues to move periodically after the magnetic field disappears?", "solution": "", "answer": "" }, { "id": 514, "tag": "MECHANICS", "content": "A solid sphere with mass \\(m\\) and radius \\(a\\) is placed on a spherical surface with radius \\(R\\), starting from the initial angle \\(\\theta_0\\) and undergoing pure rolling without slipping. When the sphere rolls to a certain angle \\(\\theta\\), find the expression for the frictional force \\(f\\) on the sphere. Only consider scenarios where separation has not occurred.", "solution": "", "answer": "" }, { "id": 40, "tag": "MECHANICS", "content": "A rectangular wooden block with height $H$ and density $\\rho_{1}$ is gently placed on the water surface, where the density of water is $\\rho_{2}$, and $\\rho_{1} {<} \\rho_{2}$. The magnitude of gravitational acceleration is $g$. Only consider the vertical translational motion of the wooden block, neglect all resistance in the direction of motion, and assume that the height of the water surface remains unchanged. When the block is slowly pressed down to a certain depth and then released, the block's top edge enters the water (i.e., the block is fully submerged). After this, if the block does not leap out of the water surface, find the maximum distance $h$ by which the top of the block descends from its original equilibrium position.", "solution": "", "answer": "" }, { "id": 645, "tag": "MECHANICS", "content": "This problem discusses the situation of a snowball rolling down a slope under pure rolling conditions. Suppose at a certain moment there is a spherical snowball rolling down an inclined slope with an angle of inclination $\\theta$. It is assumed that as the snowball rolls downward, the increase in the radius of the snowball is the same each time it completes a full rotation, and during this process, the density of the snowball remains constant. It is known that at the initial moment the radius of the snowball approaches $0$. Assuming the snowball and the snow are thermally insulated from the external environment, calculate how much the temperature of the snowball increases after sliding a distance $L$ along the slope. It is known that the specific heat capacity of the snow is $c$, the acceleration due to gravity is $g$, and no phase change occurs when the temperature of the snow rises.\n", "solution": "", "answer": "" }, { "id": 255, "tag": "OPTICS", "content": "A massless rod can rotate without friction about a pivot along the $z$-axis at its center. Plane light waves propagate along the $x$-axis from left to right. The light's electric field is given by the equation\n\n$$\n\\vec{E}(x,t)=E_{0}\\hat{y}\\cos(k x-\\omega t)\n$$ \n\nAt the ends of the rod are disks parallel to the rod and the $z$-axis. One side of each disk is a 100% reflective perfect mirror, while the other side is 100% absorptive. The initial orientation of the disks is such that light always hits the absorptive side at the top of the rod (above the pivot), and hits the reflective side at the bottom. The mass of each disk is $m$, and the radius is $r\\circ$. Assume the distance from the pivot to the center of each disk, $R$, satisfies $R\\gg r$. Find the average angular acceleration of the rod for one full rotation. (In the answer, represent physical constants using $\\epsilon_0$.)", "solution": "", "answer": "" }, { "id": 618, "tag": "ELECTRICITY", "content": "A uniform sphere with a mass of $m$ and a radius of $R$, uniformly carrying a charge $Q$, rotates about the $z$ axis (through the sphere center) with a constant angular velocity $\\omega$. The calculation formula for the magnetic moment is known to be $\\sum_{i}I_{i}S_{i}$, where $I_{i}$ is the current of the current loop and $S_{i}$ is the area vector of the current loop. The sphere is placed on an infinitely large superconducting plane with the $z$ axis vertical and perpendicular to the plane surface, with gravitational acceleration $g$. Assuming that the selection of parameters ensures such a balanced state exists (i.e., no need to discuss the reasonableness of the parameters), find the distance $h$ from the sphere center to the plane surface. The vacuum permittivity $\\varepsilon_0$ and the vacuum permeability $\\mu_0$ are known.", "solution": "", "answer": "" }, { "id": 328, "tag": "MECHANICS", "content": "Consider the process of a step ladder collapsing and rebounding after hitting a wall. The step ladder can be viewed as two uniform slender rods of length $L$, hinged together at the top $A$ with a smooth, lightweight hinge. The mass of the left rod is $m$, while the mass of the right rod can be ignored. The coefficient of static friction and kinetic friction between the bottom end $B$ of the left rod and the ground is $\\mu$. The bottom end $C$ of the right rod makes smooth contact with the ground. Initially, the two rods are almost aligned, forming an angle of zero. The ground is horizontal, and the end $C$ is at a distance of $\\sqrt{2}L$ from a vertical wall corner $O$. The entire rod is released from rest, with gravitational acceleration $g$. The coefficient of friction satisfies $15 \\sqrt{10}/128<\\mu<1$. Solve for the energy lost at the moment of collision.", "solution": "", "answer": "" }, { "id": 17, "tag": "MECHANICS", "content": "During winter break, a student named Xiaoliang discovered an interesting phenomenon while playing with a slingshot. He noticed that when the projectiles had almost the same volume, overly light projectiles and overly heavy projectiles both failed to travel very far. Being thoughtful, Xiaoliang decided to build a model to analyze this process.\n\nAssume the slingshot consists of a rubber band with a spring constant $k$, fixed at both ends, and has an original length of $d$. The fixed ends are also $d$ apart. When firing, a projectile is placed in the middle of the rubber band, and the rubber band is pulled back, generating a displacement $l$ for the projectile,and the direction of the displacement is perpendicular to the initial direction of the rubber band,letting go completes the firing. Assume that the elastic potential energy of the rubber band is fully converted into the kinetic energy of the projectile during firing. The projectile is a small sphere with radius $a$ and uniform density $\\rho$. During its flight, the projectile is subject to air resistance given by $f=-\\gamma\\vec{v}$ (where $\\gamma$ is very small, and the approximation is retained to the lowest order of $\\gamma$). Assume the projectile is launched at an angle $\\theta$ with respect to the horizontal ground. \n\nFind the density of the projectile $\\rho$ that maximizes its range, given that the angle $\\theta$ remains constant. The acceleration due to gravity is given as $g$.", "solution": "", "answer": "" }, { "id": 675, "tag": "THERMODYNAMICS", "content": "From a microscopic perspective, the surface of a liquid is not a geometric surface, but rather a thin layer with a certain thickness, known as the surface layer. If a certain mass of liquid is stored in a closed container that is adiabatic to the outside and has a volume slightly larger than the liquid's volume at a certain temperature, a unit liquid equilibrium system consisting of the liquid's internal region, liquid surface layer region, and saturated vapor region will form. During the process where liquid molecules transform into vapor molecules while passing through the liquid surface layer region, they must overcome the conservative forces pointing inward from the surface layer to perform work and increase the surface free energy. The density of a certain liquid is given by $\\rho$, and the molecular molar mass is $\\mu$. Avogadro's constant is denoted as $N_{A}$, and the Boltzmann constant as $k$. At a certain temperature, the particle number density of components within the surface layer, as a percentage of the particle number density within the liquid interior, is $\\beta$ (understood as the number of bonds formed by surface layer molecules is $\\beta$ times that of internal particles). Its molar enthalpy of vaporization $L_{\\mathrm{m}}$ can be expressed by the interaction potential energy $\\varepsilon$ between its component particles and the number of bonds $n$ between each particle and adjacent particles as: $$ L_{m}=N_{\\mathrm{A}}\\cdot\\frac{\\left|\\varepsilon\\right|}{2}n $$ From a macroscopic perspective, when considering the liquid surface free energy, under dynamic equilibrium conditions, the molecular density distribution of the liquid internal region and saturated vapor region can be regarded as uniformly distributed, while the molecules within the liquid surface layer region are in the conservative force field of the liquid surface, satisfying the Boltzmann distribution law. Given the temperature $T$ of a certain substance's liquid, the molar volume of the liquid is $V_{i m}$, and the corresponding saturated vapor pressure is $P_{g}$. Assuming the liquid can be approximated as a quasi-crystal lattice model, the density at the liquid-gas surface layer interface is $\\frac{\\rho}{2}$, and the ratio of the number of molecules with surface free energy $\\varepsilon$ per unit area at the liquid-gas interface to the total number of molecules is $\\delta$. Find the expression for the surface tension coefficient of this liquid. The answer should be expressed in terms of $\\delta$, $N_{A}$, $\\rho$, $\\mu$, $k$, $T$, and $P_{g}$", "solution": "", "answer": "" }, { "id": 718, "tag": "MODERN", "content": "Protons are composed of smaller entities called \\\"partons.\\\" The Large Hadron Collider (LHC) is a high-energy proton-proton collider, where the energy of a single proton in the proton beam is $E=7.0\\mathrm{TeV}$ ( $1\\mathrm{TeV}=10^{3}\\mathrm{GeV}{=}10^{12}\\mathrm{eV}$ ). Two proton beams with the same energy collide head-on, and the two colliding partons a and b interact and annihilate to produce a new particle. Let the ratio of the kinetic energy of partons a and b to the proton energy be $x_{\\mathfrak{a}}$ and $x_{\\mathfrak{b}}$ respectively, and assume these ratios are much greater than their rest energy. Assuming that the two partons a and b collide and annihilate to produce a new particle S with a rest mass of $m_{\\mathrm{s}}=1.0\\mathrm{TeV}/\\mathrm{c}^{2}$, find the product $x_{\\mathfrak{a}} x_{\\mathfrak{b}}$, and express the answer numerically.", "solution": "", "answer": "" }, { "id": 113, "tag": "THERMODYNAMICS", "content": "The slogan of condensed matter theory is \"More is different\". Physics is not only the study of the composition and forms of interaction of matter, but also how new physical concepts are constructed at different theoretical levels, and more remarkably, how profound connections between physical concepts at different levels are revealed. Based on this idea, let's look at the following questions concerning connections, which are themselves interconnected.\n\nIn fact, diffusion is also a typical transport process. We can consider a gaseous medium, where a dilute component (thus the mechanical effects caused by its diffusion can be neglected) is composed of (quasi)particles with effective charge and mass $q, m$. Before discussing diffusion, let's consider the following question: if the system maintains a temperature $T$, and a uniform magnetic field $\\pmb{B}$ is applied to the gas. Then, when scattering of particles is not considered $(\\gamma\\rightarrow0)$, the particles originally conforming to the Maxwell distribution in the medium will now rotate around the magnetic field. Next, we consider the evolution caused by the diffusion of particles in the external magnetic field. We let a large number of particles start at rest from the origin. Its dynamics equation can be phenomenologically written as:\n\n$$\nm{\\ddot{r}}=f-\\gamma{\\dot{r}}+q{\\dot{r}}\\times B\n$$\n\nwhere $\\pmb{f}=-\\gamma\\pmb{v}$ reflects the scattering (or damping) of the particles, and $q\\dot{\\pmb r}\\times\\pmb B$ is the force caused by the magnetic field, with the magnetic field taken as ${\\boldsymbol{B}}= B e_{z}$. Meanwhile, $\\pmb{f}$ reflects the forces caused by thermal factors, which is a function that randomly changes over time and is independent of the particle's location, with an average value of zero. Compute the average of the angular momentum in the $z$ direction $J=\\langle m(x{\\dot{y}}-y{\\dot{x}})\\rangle$.\n\nHint: use the position vector perpendicular to the $z$ direction ${\\pmb r}^{\\prime}=({ x},{ y})$ to dot the components of the dynamics equation perpendicular to the $z$ direction, and assume $\\dot{R}=0$ at $t=0$. Boltzmann constant is $k$.", "solution": "", "answer": "" }, { "id": 305, "tag": "MECHANICS", "content": "A uniform spring with a spring constant $k$, an original length $L$, and a mass $m$, is connected at one end to a wall and at the other end to a block with mass $M$. It undergoes simple harmonic motion on a smooth horizontal surface. It is known that $m << M$.It can be approximately considered that the velocity of the spring varies linearly with the position.\n\nThe spring undergoes corrosion over a long period of use, causing the spring constant to decrease and the mass to increase. It is approximately assumed that the spring constant and mass change uniformly throughout the spring.\n\nA new spring initially has an amplitude $A_{0}$. After a period of time, the mass of the spring becomes $\\gamma_{1}$ times the original, and the spring constant becomes $\\left(1-\\gamma_{2}\\frac{m}{M}\\right)$ times the original. Assuming the mass added during the corrosion process moves with the spring, that is, the added mass has the same velocity as the point where it is added, find the change in amplitude $\\Delta A$.", "solution": "", "answer": "" }, { "id": 573, "tag": "ELECTRICITY", "content": "As we all know, the electric field generated by a point charge \\(q\\) in a vacuum at any point is described as:\n\n$$\n\\overrightarrow{E} = \\frac{q}{4\\pi\\varepsilon_{0}r^{3}}\\overrightarrow{r}\n$$\n\nHowever, this form presents the difficulty of the point charge's self-energy being infinitely large. To address this issue, Born and Infeld proposed a hypothesis in 1934: the vacuum also exhibits \"polarization\" and \"magnetization\" effects. Specifically, the vacuum's permittivity \\(\\varepsilon\\) and permeability \\(\\mu\\) are dependent on the electric field strength \\(\\vec{E}\\) and magnetic flux density \\(\\vec{B}\\) at that point:\n\n$$\n\\frac{\\varepsilon}{\\varepsilon_{0}} = \\frac{\\mu_{0}}{\\mu} = \\left[1 + \\frac{1}{b^{2}}\\left(c^{2}\\overrightarrow{B}^{2} - \\overrightarrow{E}^{2}\\right)\\right]^{-\\frac{1}{2}}\n$$\n\nHere, \\(\\varepsilon_{0}\\) and \\(\\mu_{0}\\) represent the permittivity and permeability of free space when the electromagnetic field approaches 0, \\(b\\) is a known parameter, \\(c\\) is the speed of light in vacuum, and \\(c^{2} = 1 / {\\varepsilon_{0}\\mu_{0}}\\). This problem takes relativistic effects into account.\n\nIn the inertial reference frame \\(S\\), an infinitely long uniformly charged straight wire, fixed along the \\(z\\) axis, moves in the positive \\(z\\) axis direction with a velocity \\({v}\\). The charge per unit length of the wire is measured as \\(\\lambda\\), where \\(\\lambda > 0\\). At a distance \\(r\\) from the wire, a point charge with charge \\(-q\\) (\\(q > 0\\)) is released with an initial radial velocity \\(v_{0}\\) relative to the \\(S\\) frame. The rest mass of the point charge is \\(m\\). Under the conditions specified in the problem, derive the expression for the maximum distance \\(r_{0}\\) that the point charge can reach from the \\(z\\) axis.", "solution": "", "answer": "" }, { "id": 91, "tag": "MECHANICS", "content": "An ellipsoid rigid body has a semi-major axis \\(a\\), semi-minor axis \\(b\\), and is formed by rotating about the major axis. Its mass distribution is rotationally symmetric about the major axis, with a total mass \\(m\\). The center of mass \\(C\\) is located at a focal point. The moment of inertia about the principal axis perpendicular to the major axis passing through the center of mass is \\(mr_1^2\\), and the moment of inertia about the principal axis parallel to the major axis passing through the center of mass is \\(mr_2^2\\), where \\(r_1, r_2\\) are called radii of gyration. The ellipsoid is placed on a horizontal ground with its major axis aligned vertically and the center of mass located on the upper half of the major axis. The ellipsoid is released from rest in this position and begins to roll after a slight disturbance. It is known that the ellipsoid maintains pure rolling throughout. Find the magnitude of the angular acceleration \\(\\beta\\) after the ellipsoid has rotated by 270°.", "solution": "", "answer": "" }, { "id": 426, "tag": "MECHANICS", "content": "The mass of the homogeneous elastic rope is $m$, and its length is $l_{0}$ when freely placed horizontally. The stiffness coefficient of the rope is $k$. Now place the elastic rope on a rough horizontal surface, where the coefficient of friction between the rope and the ground is $\\mu$. \nFind the maximum length at which the elastic rope can lie flat and stationary on the horizontal surface.", "solution": "", "answer": "" }, { "id": 487, "tag": "MODERN", "content": "Mesons are hadronic particles consisting of a quark–antiquark pair, bound by the strong force. For light meson systems composed of light quark-antiquark pairs (with rest mass of several \\(\\mathrm{MeV}/c^{2}\\)), both experimental measurements and theoretical calculations indicate that: In the static approximation, the strong interaction potential energy between quark-antiquark pairs is proportional to their distance \\(r\\), i.e., \\(U(r)=\\sigma r\\), where \\(\\sigma\\) is a positive constant. The linear behavior of the potential energy distribution described above prompted physicists to construct a semi-classical model known as the “string model” to describe the energy spectrum distribution of light mesons observed in experiments. As shown in the figure, light mesons are envisioned as rapidly rotating strings of length \\(r\\), with rest mass \\(m_0=\\sigma r/c^{2}\\) uniformly distributed along the length, where the mass (energy) of light quarks is negligible. Their kinematic effects are equivalently replaced by boundary conditions such as \"the speeds at the ends of the string are equal to the speed of light in vacuum\" in the center of mass frame, i.e., \\(\\omega r/2 = c\\). The quantization condition applied to the above strings is:\n\\[\nL_n=\\frac{nh}{2\\pi}, \\quad n = 1,2,3,\\cdots\n\\]\nwhere \\(L_n\\) is the angular momentum corresponding to the \\(n\\)-th energy level of the system in the center of mass frame. Try to provide the expression for the energy level \\(E_n\\).", "solution": "", "answer": "" }, { "id": 506, "tag": "ELECTRICITY", "content": "There are two infinitely long cylindrical ideal conductors forming a parallel transmission line pair. The conductor radius is $a$, and the distance between the axes is $2d$. Assume the conductors are perpendicular to the plane of the paper, one on the left side with the current direction perpendicular to the paper coming out, and one on the right side with the current direction perpendicular to the paper going in. The vacuum permeability $\\mu_0$ is known. Find the inductance per unit length $L$ between the wires.", "solution": "", "answer": "" }, { "id": 270, "tag": "MECHANICS", "content": "Made of the same metal material, two hollow spheres \\( A \\) and \\( B \\) are constructed, with the density of the metal material being \\( \\rho_1 \\). The mass of sphere \\( A \\) is \\( m_A \\), and the mass of sphere \\( B \\) is \\( m_B \\). The outer diameters are \\( R_A \\) and \\( R_B \\), respectively. The average density of sphere \\( B \\) is equal to \\( \\rho_0 \\). The cavities of the two spheres are connected by a thin metallic conduit. Due to the small diameter of the conduit, applying an excessive tensile force can cause breakage. This setup can be used to monitor the density of a liquid.\n\nWater with a mass of \\( m_w \\) is injected into the cavity of sphere \\( A \\), and wires are connected to both \\( A \\) and \\( B \\). After integrating the setup into an alarm circuit, it is placed horizontally on the surface of the liquid to be monitored. The density detector requires the liquid's density to not be lower than \\( \\rho_0 \\). The circuit triggers an alarm when a significant reduction in current is detected.\n\nWhen the liquid's density is just slightly lower than \\( \\rho_0 \\), how much does the volume of sphere \\( A \\) above the liquid surface increase compared to before the alarm was triggered? (Assume that after the alarm is triggered, no water leaks out from the cavity, and the acceleration due to gravity is \\( g \\)).", "solution": "", "answer": "" }, { "id": 132, "tag": "MECHANICS", "content": "Three individuals, ABC, simultaneously start from the vertices $A_{0}, B_{0}, C_{0}$ of an equilateral triangle with side length $L$, each moving at the same constant speed $v$. Throughout the motion, A's velocity is always directed towards B, B's velocity towards C, and C's velocity towards A. Determine the time it takes for the three individuals to meet.", "solution": "", "answer": "" }, { "id": 208, "tag": "ELECTRICITY", "content": "In a zero-gravity space, infinitely far above a conductive plate with thickness $D$ and conductivity $\\sigma$, there is a small magnet with its pole facing the plate. The magnet is incident from infinity with an initial velocity $v_0$. The magnet has a magnetic moment $\\mu$, mass $m$, and the vacuum permeability is $\\mu_0$. Neglect the magnetic fields generated by changing electric fields and conductive currents. Find the final stop position of the small magnet and the distance from the plate $h_e (h_e \\gg D)$.", "solution": "", "answer": "" }, { "id": 163, "tag": "ADVANCED", "content": "Compute the commutator of the total angular momentum squared operator $L^2$ with the raising operator of the vector $V$, $V_{+} = V_x + i V_y$. Express the result using $V_z, V_{+}, L_z, L_{+}$, where $L_{+} = L_x + i L_y$. Use $\\hbar = 1$ for the answer. Moreover, if there is a product of $V$ and $L$, please place $L$ at the end.", "solution": "", "answer": "" }, { "id": 572, "tag": "ELECTRICITY", "content": "Inside an insulating pipe with a cross-sectional area $S$, there are two frictionless pistons, each with mass $m$. The pistons seal a certain amount of monatomic ideal gas with a total number of molecules $N$. The initial distance between the pistons is $L$, and the initial gas temperature is $T$. The two plates carry positive and negative charges $\\pm Q$ respectively. It is known that if the molecules are in an external electric field $E$, they will induce a dipole moment $p=\\alpha E$ in the direction of the external electric field $E$, where $\\alpha$ is a temperature-independent constant. The dielectric constant in vacuum is $\\varepsilon_{0}$, and the Boltzmann constant is $k$. For convenience in expressing the answer, introduce a dimensionless constant $x=N\\alpha/\\varepsilon_{0}S L$, so $\\alpha$ no longer appears in the answer. Assume that the piston area is very large, allowing the use of a capacitor model. If the pistons are initially in static equilibrium and the outside of the pistons is a vacuum, calculate the quantity of positive and negative charges $\\pm Q$.", "solution": "", "answer": "" }, { "id": 516, "tag": "ELECTRICITY", "content": "Consider an L-shaped grounded conductor formed by the \\(xz\\) and \\(yz\\) planes, with its cross-section located in the region \\(x>0\\) and \\(y>0\\). Outside this region, there is a line charge (with line charge density \\(\\lambda\\)) located at the point \\((a,b)\\), where \\(b>a>0\\). Find the expression for the electric potential \\(V(x,y,z)\\) in the region.", "solution": "", "answer": "" }, { "id": 68, "tag": "ELECTRICITY", "content": "Establish a rectangular coordinate system with an infinitely large insulating plate as the $xy$ plane. There is a charge distribution on the plane:\n\n$$\n\\sigma(x,y)=\\sigma_{0}\\cos\\frac{x}{d}\\quad,\\quad(\\sigma_{0}>0,d>0)\n$$ \n\nTaking infinity as the zero point of potential, find the potential distribution $\\varphi(x,y,z)$. The vacuum permittivity is known to be $\\epsilon_0$.\n\nHint: \n$$\n\\int_{-\\infty}^{\\infty}{\\frac{\\cos(\\beta\\zeta)}{\\zeta^{2}+a^{2}}}\\mathrm{d}\\zeta=\\frac{\\pi}{a}e^{-a b}\n$$ \n\n$$\n\\int_{-\\infty}^{\\infty}\\frac{\\zeta\\sin(b\\zeta)}{\\zeta^{2}+a^{2}}\\mathrm{d}\\zeta=-\\frac{\\partial(\\frac{\\pi}{a}e^{-a b})}{\\partial b}=\\pi e^{-a b}\n$$", "solution": "", "answer": "" }, { "id": 107, "tag": "MECHANICS", "content": "A child with a mass of \\( m \\) is swinging on a swing. When the swing is swinging stably, the distance from the child's center of mass to the swing's fixed point is \\( l \\). Then, each time the swing passes through the lowest point of oscillation, the child raises their center of mass by a small distance \\( b \\) (\\( b << l \\)). Thus, the distance from the center of mass to the fixed point becomes \\( l-b \\). And at each swing's highest point, the child lowers their center of mass by a small distance \\( b \\), and the distance from the center of mass to the fixed point becomes \\( l+b \\). 1) Assuming the oscillation amplitude is very small, calculate the work \\(\\Delta E\\) done by the child in each oscillation cycle. Express it in terms of the initial mechanical energy \\( E_{i} \\) at the beginning of one cycle, using the lowest point of the swing as the zero point for gravitational potential energy. \\(\\begin{array}{r}{(\\cos\\theta\\approx1-\\frac{\\theta^{2}}{2})}\\end{array}\\)", "solution": "", "answer": "" }, { "id": 347, "tag": "MECHANICS", "content": "There are two thin rods $AB$ and $CD$ near a vertical wall. The $A$ end of the $AB$ rod is in contact with the ground and is located on the right side of the vertical wall, while the $B$ end is in contact with the vertical wall. The $D$ end of the $CD$ rod is in contact with the vertical wall and is positioned above the $B$ end of the $AB$ rod, and the $C$ end is located on the $AB$ rod and slides along it. The $A$ end moves to the right along the ground at a constant speed $v$, and the $C$ end slides downward relative to the $AB$ rod with a relative velocity of $v'$. At a certain moment, the $C$ end is located at the midpoint of $AB$, the angle between the $AB$ rod and the ground is $\\alpha$, the angle between the $CD$ rod and the vertical wall is $\\beta$, and $\\alpha + \\beta < \\pi/2$. Find the velocity $\\vec{v}_E$ of the midpoint $E$ of the $CD$ rod at this moment. (Use basis vectors $\\hat{i}, \\hat{j}$.)", "solution": "", "answer": "" }, { "id": 31, "tag": "MECHANICS", "content": "In the study of classical mechanics, the research on the properties of central force fields is the most representative and common. To facilitate the study of central force fields, for the central potential field with magnitude ${V}({r})$, assuming the particle mass is ${m}$ and its angular momentum in motion is L, the effective potential energy ${U}_{eff}$ in which the particle moves can be defined as:\n\n$$ {{U}}_{{eff}} = V(r) + {\\frac{L^{2}}{2m r^{2}}} $$\n\nAssuming a particle with mass m is moving in a central force field, and the potential energy $V(r) = \\frac{kr^2}{2} - \\frac{\\alpha}{r^2}$. Under the condition that the particle can avoid falling into the center, find the period of the particle's radial motion.", "solution": "", "answer": "" }, { "id": 216, "tag": "ELECTRICITY", "content": "At the initial time $t=0$, there are 100 metal plates with an area of $S$ positioned parallel to each other, with an equal spacing of $D\\ll\\sqrt{S}$. These plates are charged sequentially with $Q$, $2Q$, $3Q\\cdots 100Q$, such that the metal plate numbered $n$ carries a charge of $nQ$. Between all metal plates, a lossy dielectric with an absolute permittivity of $\\varepsilon$ and a conductivity of $\\sigma$ is used to fill the space. Throughout the process, edge effects are neglected. Determine how the voltage between plates 1 and 100 changes with time across the entire system.", "solution": "", "answer": "" }, { "id": 317, "tag": "MODERN", "content": "In space, there is a coordinate axis. In the $S$ system, $N$ particles numbered $1, 2, 3 \\cdots N$ are static at positions $x_{1}, x_{2}, x_{3}\\cdots x_{N}$ to the right of the origin, with distances $x_{1}, x_{2}, x_{3} \\cdots x_{N}$, respectively. All particles have a mass of $m$. At time $t = -t_{n}$, a constant leftward force $F_{n} = n m a$ is suddenly applied to particle number $n$. At $t = 0$, all particles arrive at point $O$ simultaneously and undergo a completely inelastic collision. Furthermore, during the acceleration process, the proper time interval for each particle (i.e., the time interval measured by the clock carried by the particle; or, equivalently, the sum of the time intervals of infinitesimal sub-processes when each sub-process is approximately at the same speed as the particle) is $\\tau$. \n\nFind the instantaneous velocity $V$ of the resulting single particle after all particles have undergone a completely inelastic collision and combined into one particle in the final state.", "solution": "", "answer": "" }, { "id": 601, "tag": "MECHANICS", "content": "A semi-cylinder of radius R and mass M is placed on a horizontal table with the plane of the cylinder facing up. Initially the cylinder is at rest and then a small perturbation is given to the cylinder. Consider only the direction of the axis of the half-column under the perturbation, i.e., the angular velocity of the half-column is always in the direction of the axis.\n\nThere is friction on the table, and assuming that the initial inclination of the half-column (the angle between the rectangular surface and the horizontal) is $\\theta_0$ and it is at rest, find the minimum value of the coefficient of static friction of the table so that the half-column can do pure rolling.", "solution": "", "answer": "" }, { "id": 388, "tag": "MECHANICS", "content": "Two spacecraft will pass through a region densely packed with stationary dust, where it is known that there is no friction between the dust and the spacecraft, and the dust will scatter tangentially upon collision with the spacecraft. Spaceship 1 is cylindrical with a cross-sectional radius of $R$, with its base facing the direction of motion; spaceship 2 is spherical with a radius of $R$. What is the ratio of the resistance experienced by the two when they are moving at the same speed? Ignore the effects of gravity.\n", "solution": "", "answer": "" }, { "id": 319, "tag": "ELECTRICITY", "content": "This problem does not consider relativistic effects.\n\nAn infinitely long leaky dielectric cylindrical object, with a radius of $R$, an absolute permittivity $\\varepsilon$, initially stationary and uncharged as a whole. The charge carriers have an extremely small mass $m \\to 0$, charge magnitude $q$, and a mobility $\\mu$ (i.e., the drift velocity of the charge carriers under a uniform electric field $E$ is $v = \\mu E$). The number density of carriers is $n$. A cylindrical coordinate system $(z,r,\\theta)$ is established inside the cylinder, with the origin being the center of the circular top surface, the $z$ axis pointing axially outward along the cylinder, and $r$ pointing radially outward. During the process, the charge volume density $\\rho$ and current densities $j_{r}, j_{\\theta}$ inside the cylinder will vary with time $t$, and they are also functions of spatial coordinates $r$ and $\\theta$. Due to the translational symmetry along the $z$ axis, the current density in the $z$ direction, $j_{z}$, will always be zero, and all non-zero physical quantities are independent of the $z$ coordinate.\n\nFirstly, keep the cylinder stationary, and slowly apply a uniform magnetic field $B$ along the cylinder axis. Then, maintain the uniform magnetic field and slowly rotate the cylinder steadily around the axis in the positive $\\theta$ direction with an angular velocity $\\omega$. Ignore the centripetal force required for the circular motion of the charge carriers (due to $m \\to 0$), and also ignore the magnetic field generated by the rotation of the charge carriers. In the steady state, the charge carriers will redistribute in the radial direction, resulting in a non-zero charge and current distribution $\\rho, j_{r}, j_{\\theta}$ in the ground frame. Solve for the total current vector $\\vec{j}$. \n\nHint: Although $\\rho \\neq 0$, assume that the number density of charge carriers is approximately $n$.", "solution": "", "answer": "" }, { "id": 484, "tag": "MECHANICS", "content": "Two steel balls, each with a mass of $m$, are connected by a string of length $2L$ and placed on a smooth horizontal surface. A constant force $F$ is applied at point $O$, the midpoint of the string. The force acts horizontally and is perpendicular to the initial direction of the string. The string is very soft, non-stretchable, and has negligible mass. Assume the two balls are rigid small spheres, undergo elastic collisions, and the collision time is negligible. Taking the initial midpoint of the two balls as the origin and the direction of $F$ as the positive $x$-axis, derive the function for the $x$-component of the trajectory vector $\\mathbf{r}$ for the upper ball on the horizontal plane as a function of time $t$, assuming $t=0$ at the start.", "solution": "", "answer": "" }, { "id": 584, "tag": "MECHANICS", "content": "A special material elastic ring is placed on a frictionless horizontal table. The average value of the inner and outer radii is $R$, the difference between the inner and outer radii is $b\\left(b\\ll R\\right)$, and the height is $h$. Initially, it is a stress-free circular ring. Assume that the material's Young's modulus is $E$ and the density is $\\rho_{\\mathrm{0}}$, and there is no energy loss during the vibration.\n\nAs a common form of vibration for metallic rings, the ring vibrates between a circle and an ellipse with a particularly small eccentricity, with the centers of both coinciding. If the amplitude of the vibration is very small, taking the center of the ring as the origin, the polar equation of the ellipse with a particularly small eccentricity can be approximately written as: $r=R{\\left(1+\\varepsilon\\mathrm{cos}2\\theta\\right)}$, where the change in the shape of the ring can be described by the change in the parameter $\\varepsilon$ (where $\\varepsilon$ is very small). During vibration, it can be approximately assumed that each micro-segment vibrates only radially, being compressed or stretched tangentially, meaning only radial kinetic energy is considered. (In fact, the tangential velocity is even smaller by an order of magnitude than the radial one.)\n\nFind the angular frequency of this vibration.", "solution": "", "answer": "" }, { "id": 327, "tag": "MECHANICS", "content": "A uniform rod of length $l$ is suspended by a spring above a liquid surface. When the system is at equilibrium, the lower end of the rod just touches the liquid surface, and the spring's elongation is exactly $l$. Now, the rod is now pulled downward vertically into the liquid, so that the distance between its lower end and the liquid surface becomes $2l$, and then the rod is released from rest. It is known that the rod and the liquid have the same density $\\rho$, and it is assumed that the rod moves only along the vertical direction. Determine the time elapsed from when the rod is released to when it rises to its maximum height.", "solution": "", "answer": "" }, { "id": 245, "tag": "MECHANICS", "content": "A flat pivot can be regarded as a cylinder, and a bearing can be treated as a plane. They are in contact and exert pressure on each other. It is known that the pressure $F$ between the pivot and the bearing is uniformly distributed over a circular cross-section with a radius $R$, and the coefficient of friction is $\\mu$. Find the torque of the frictional force.", "solution": "", "answer": "" }, { "id": 269, "tag": "ADVANCED", "content": "Using the natural units system \\[c=\\hbar=1\\], consider a particle with mass \\(\\frac{1}{2}\\) moving in an external field with the potential energy function \\(x^2+\\alpha \\delta(x)\\). Use non-relativistic quantum mechanics to solve for the energy spectrum of the particle (\\(\\alpha\\) is a known parameter).", "solution": "", "answer": "" }, { "id": 316, "tag": "ELECTRICITY", "content": "An ion with mass $m$ and charge $Q$ is projected from a very distant point towards a neutral conducting sphere with mass $M$ and radius $a$. Due to the ion's electric field, the neutral conducting sphere becomes polarized. When the polarization reaches equilibrium, the electric dipole moment of the neutral conducting sphere is $\\vec{p} = \\gamma \\vec{E}_0$, where $\\vec{E}_0$ is the electric field strength created by the ion at the location of the neutral conducting sphere, and $\\gamma$ is a constant related to the structure of the neutral conducting sphere. Assume the initial speed of the ion is $v_0$, and the impact parameter (the perpendicular distance from the neutral conducting sphere to the direction of incidence) is $b$. Neglect the effect of gravity, do not consider relativistic effects, and assume that at every moment during the ion's movement, the neutral conducting sphere can reach polarization equilibrium. Find the minimum distance between the ion and the center of the conducting sphere (assuming the minimum distance is still much larger than the radius $a$ of the sphere).", "solution": "", "answer": "" }, { "id": 173, "tag": "THERMODYNAMICS", "content": "In a cylinder, a certain amount of water vapor is enclosed, and the water vapor can be considered an ideal gas. The heat capacities of the liquid phase and gas phase are constant. The constant-pressure molar heat capacities of 1 mol of vapor and liquid are $C_{g p}$ and $C_{l p}$, respectively. At ${T}{=}0{K}$, the heat of vaporization of 1 mol of water is L(0). The saturated vapor pressure of water satisfies the Clausius equation:\n\n$$\np=A{e}^{-L(0)/k T}T^{( C_{g p}-C_{l p})/R}\n$$ \n\nDue to the high fluidity and expansiveness, water and water vapor are excellent choices for thermodynamic working substances. If water and water vapor undergo the following cycle, A-B-F-C-D-G-A is a complete cycle process, A and B are pure water, F and G are the transition points from mixed water and vapor to pure vapor. The pressure at point D is $P_0$, and the volumes at points C and D are both $V_0$. Find the cycle efficiency of this cycle. Express the final result using $L(T1), L(T2), A, T_1, T_2, V_0, k, C_{gp}, C_{lp}$, where $C_{gp}$ is the molar heat capacity of water vapor and $C_{lp}$ is the molar heat capacity of water. (The volume of water is negligible, the volume at points AB is considered 0, and B-F-C and D-G-A are isothermal processes at temperatures T1 and T2, respectively).", "solution": "", "answer": "" }, { "id": 536, "tag": "MECHANICS", "content": "A carriage of length $L$ is at rest on a smooth horizontal track. A launcher $A$, fixed at the center of the carriage floor, fires a particle of mass $m_1$ with an initial velocity $u_0$ relative to the horizontal track from the middle of the carriage along the smooth floor inside the carriage. The particle $m_1$ then collides and instantly sticks with another particle of mass $m_2$ placed on the carriage. At this moment, $m_2$ just comes into contact but does not adhere to a lightweight spring fixed at one end to the carriage at a horizontal position. The spring has a stiffness constant $k$ and its natural length is $l$. The total mass of the carriage and the launcher $A$ is $M$.\n\nFind the expression for the displacement of the carriage from rest until the spring reaches its maximum compression.\n\n", "solution": "", "answer": "" }, { "id": 533, "tag": "ELECTRICITY", "content": "A particle with mass $\\mathfrak{m}$ and positive charge $q$ moves in an electromagnetic field. The electric field is a uniform field with field strength $\\pmb{E}$, directed along the positive y-axis. The magnetic field is directed along the z-axis (perpendicular to the xy-plane), and the magnetic flux density varies only with y, specifically given by $B=\\alpha\\sqrt{|y|}$. The particle initially starts at the origin and is released from rest. During the subsequent motion, the particle's gravitational force can be neglected, and it comes to a stop for the first time at $t=T$. How far is the particle from the starting point when $\\scriptstyle t={\\frac{3}{2}}T$?", "solution": "", "answer": "" }, { "id": 545, "tag": "THERMODYNAMICS", "content": "A hemisphere of ice with a radius $R$ and refractive index $n$ is situated on a warm, flat table and melts slowly. The rate of heat transfer between the table and the ice is proportional to their contact area. The coordinate system is defined with the center of the semicircle as the origin; the $x$ and $y$ axes are horizontal, and the $z$ axis is vertical. It is known that the ice hemisphere completely melts in time $T_{0}$. Throughout the process, a laser beam shines on the ice from above. The beam is incident perpendicular to the plane along the line $x=R/2$, $y=0$. \n\nAssume the temperatures of the ice and the surrounding atmosphere are both $0^\\circ C$, and remain unchanged during the melting process. The laser beam does not transfer energy to the ice. The melted water immediately flows off the table, and the ice does not move along the table.\n\nFor $t \\geq 0$, where is the location of the point $x(t)$ on the table hit by the beam? Express the answer in terms of $n, R, T_{0}, t$ (and should not include trigonometric functions).", "solution": "", "answer": "" }, { "id": 13, "tag": "MECHANICS", "content": "10 identical wooden blocks are placed one next to the other on a horizontal floor. Each wooden block has a mass of $m=0.40kg$ and a length of $l=0.50m$ . The static and kinetic friction coefficients between the block and the ground are both $\\mu_{2}=0.10$. Initially, the blocks are at rest. A small lead block with mass $M=1.0$ kg is placed above the left end of the first wooden block on the left. The static and kinetic friction coefficients between the lead block and the wooden block are both $\\mu_{\\mathrm{i}}=0.20$. Now, the lead block is suddenly given an initial velocity to the right of $V_{0}=4.3m/s$ , causing it to slide over the large wooden blocks. Determine on which block the lead block finally stops. Take gravity acceleration $g=10m/s^2$ . Assume the dimensions of the lead block can be ignored compared to $l$. You need to give an integer between 1-10 as the answer.", "solution": "", "answer": "" }, { "id": 324, "tag": "ELECTRICITY", "content": "Two coaxial thin cylinders have radii of $R$ and $2R$, respectively, with both having a length of $L (L >> R)$, and a mass of $m$, with uniform distribution. The inner and outer surfaces are uniformly charged, carrying charges of $Q$ and $-Q$, respectively, and the charges do not move across the surfaces. Both cylinders can rotate about the central axis and are initially at rest. Now, due to the action of an external force (assuming the action time is very short), the inner cylinder gains an initial angular speed of $\\omega_{10}$. Assume the outer cylinder experiences a small frictional torque, while the frictional torque acting on the inner cylinder can be neglected. Find the work done by the frictional torque from the initial state to the final state.", "solution": "", "answer": "" }, { "id": 406, "tag": "ADVANCED", "content": "Given the form of the electromagnetic field momentum conservation theorem:\n\\[\\vec{f_m} + \\frac{\\partial \\vec{g}_{em}}{\\partial t} + \\nabla \\bm{T}_{em}\\]\nwhere $\\vec{g}_{em}$ is the electromagnetic field momentum density, and $\\bm{T}_{em}$ is the electromagnetic field momentum flow density.\n\nNow, consider a given point charge $q$ moving at velocity $\\vec{v}$ in a constant external magnetic field $\\vec{B}_0$. Consider only the momentum flow density of the electromagnetic field and ignore the momentum density. Provide the force on the charge, that is, provide the mechanical force density \\[\\vec{f_m}\\] (do not calculate \\[\\vec{g}_{em}\\]). To simplify the expression, assume the particle starts from the origin and moves in uniform linear motion.", "solution": "", "answer": "" }, { "id": 762, "tag": "MECHANICS", "content": "The spinning top is a children's toy with a long history, and its principles can also be applied in various fields such as navigation. Now we consider a uniform mass distribution conical top, with a half-apex angle of $\\alpha$, height $h$, and mass $m$. The top spins around its apex on a horizontal surface, with its apex fixed at the origin $O$. Possible resistance is not considered. Now let the axis of the spinning top maintain a constant angle $\\beta$ with the vertical direction while spinning. Find the expression for the minimum total angular velocity magnitude $\\omega_{0}$ of the top's rotation.\n", "solution": "", "answer": "" }, { "id": 686, "tag": "MODERN", "content": "In a three-dimensional harmonic trap, atoms oscillate approximately with frequency \\(\\omega_i\\) in the \\(i\\) direction (\\(i=x,y,z\\)). According to the kinetic energy distribution of a classical gas, the characteristic kinetic energy in the \\(x\\) direction is \\[ \\frac{1}{2}k_BT, \\] with the corresponding harmonic potential energy \\[ \\frac{1}{2}m\\omega_x^2R_x^2, \\] When these two are equal, the maximum displacement \\(R_x\\) of the atom in the \\(x\\) direction can be obtained. Please write out the characteristic confined volume \\(V\\) of the atom (expressed in terms of temperature \\(T\\) and the harmonic frequencies \\(\\omega_x,\\omega_y,\\omega_z\\)).", "solution": "", "answer": "" }, { "id": 225, "tag": "MECHANICS", "content": "In the fifth century AD, the Saxons designed a timekeeping device. They placed a bowl with a hole in the bottom in water and timed it by observing the bowl sinking. This device was called the \"Saxon Bowl.\" The bowl is a hollow open cylinder with a mass of $M$, the thickness of the cylinder wall and bottom is $D$, the internal radius is $R$, the height is $H$, and there is a small hole at the center of the bottom with a radius of $r$. Assume that $H$ and $h$ are much larger than $D$, but $R$ is of the same order of magnitude as $D$. The liquid density is $\\rho$. The critical condition for sinking: When the bowl is gently placed on the water surface, it is found that when $r$ meets certain conditions, the bowl remains balanced on the water surface without water flowing into the bowl. Given the surface tension coefficient of water is $\\sigma$, and the acceleration due to gravity is $g$, find the critical radius $r_{m}$?", "solution": "", "answer": "" } ]