Mock Test Walkthrough: Year 13 Physics Unit Test | 物理单元测试模拟卷解析

📚 Mock Test Walkthrough: Year 13 Physics Unit Test | 物理单元测试模拟卷解析

This comprehensive walkthrough analyses a mock unit test designed for Year 13 Cambridge A2 Physics. The paper samples questions from the core topics of the syllabus, including circular motion and gravitation, simple harmonic motion, ideal gases, electric fields, magnetic fields and charged particles, quantum physics, nuclear decay, and alternating currents. Each section presents a typical exam question, proceeds through a detailed step‑by‑step solution, and highlights the key physical principles and common pitfalls. The model answers are written to the standard expected in the Cambridge examination, with all equations rendered using Unicode characters and standard notation.

这份详尽的解析稿针对一份为剑桥A2物理Year 13设计的单元模拟卷。试卷从课程核心主题中精选典型问题，涵盖圆周运动与万有引力、简谐运动、理想气体、电场、磁场与带电粒子、量子物理、核衰变以及交流电。每个部分先呈现一道典型考题，再逐步给出详细解答，并强调关键物理原理与常见易错点。所有模型答案均符合剑桥考试评分标准，方程均以Unicode字符和标准记号表示。

1. Circular Motion and Gravitation | 圆周运动与万有引力

A satellite of mass m orbits the Earth in a circular path of radius r. The Earth has mass M and the universal gravitational constant is G. (a) By equating the gravitational force to the centripetal force, show that the orbital period T is given by T² = (4π²/GM) r³. (b) Given GM = 4.0 × 10¹⁴ m³ s⁻² for Earth, calculate the orbital radius of a geostationary satellite and explain why it must be in the equatorial plane.

一颗质量为m的卫星在半径为r的圆形轨道上绕地球运行。地球质量为M，万有引力常数为G。(a) 令万有引力等于向心力，证明轨道周期T满足 T² = (4π²/GM) r³。(b) 已知地球的GM = 4.0 × 10¹⁴ m³ s⁻²，计算地球同步卫星的轨道半径，并解释为何它必须在赤道平面内。

Solution: For part (a), the gravitational force provides the centripetal acceleration: GMm/r² = m v²/r. Cancelling m and multiplying by r gives v² = GM/r. The orbital speed is also the circumference divided by the period: v = 2πr/T. Substituting yields (2πr/T)² = GM/r → 4π²r²/T² = GM/r → T² = (4π²/GM) r³. For part (b), a geostationary satellite has a period of one sidereal day, T = 24 × 3600 = 86 400 s. Substituting into the derived relation: r³ = (GM × T²)/(4π²) = (4.0×10¹⁴ × (86400)²) / (4π²). First compute T² = 7.46496×10⁹ s², then numerator = 4.0×10¹⁴ × 7.46496×10⁹ = 2.985984×10²⁴. Dividing by 4π² ≈ 39.478 gives r³ ≈ 7.56×10²² m³, so r = ∛(7.56×10²²) ≈ 4.23×10⁷ m. This is about 42 300 km from the Earth’s centre, or approximately 35 900 km above the surface. The satellite must orbit in the equatorial plane and in the same direction as the Earth’s rotation so that its angular speed matches that of the Earth and it appears stationary relative to a point on the equator.

解答：对于(a)部分，万有引力提供向心加速度：GMm/r² = m v²/r。消去m并两边乘r得 v² = GM/r。轨道速率也可用周长除以周期表示：v = 2πr/T。代入得 (2πr/T)² = GM/r → 4π²r²/T² = GM/r → T² = (4π²/GM) r³。对于(b)部分，地球同步卫星的周期为一个恒星日，T = 24 × 3600 = 86 400 s。代入推导出的关系式：r³ = (GM × T²)/(4π²) = (4.0×10¹⁴ × (86400)²) / (4π²)。先计算 T² = 7.46496×10⁹ s²，分子 = 4.0×10¹⁴ × 7.46496×10⁹ = 2.985984×10²⁴。除以 4π² ≈ 39.478 得到 r³ ≈ 7.56×10²² m³，因此 r = ∛(7.56×10²²) ≈ 4.23×10⁷ m。这距离地心约42 300 km，或距地面约35 900 km。卫星必须在赤道平面内且与地球自转同向运行，使其角速度与地球匹配，从而相对于赤道上一点保持静止。

2. Simple Harmonic Motion | 简谐运动

A glider of mass 0.50 kg on a frictionless horizontal air track is attached to a spring of force constant 18 N m⁻¹. The glider is displaced 0.040 m from equilibrium and released from rest. (a) Determine the angular frequency and period of the motion. (b) Write expressions for the displacement, velocity and acceleration as functions of time, taking the moment of release as t = 0. (c) Calculate the maximum kinetic energy and the position where it occurs.

一个质量0.50 kg的气垫导轨滑车与一根劲度系数为18 N m⁻¹的弹簧相连，置于无摩擦的水平气垫上。滑车被拉离平衡位置0.040 m并从静止释放。(a) 求运动的角频率和周期。(b) 以释放时刻为t = 0，写出位移、速度和加速度随时间变化的表达式。(c) 计算最大动能及其出现的位置。

Solution: (a) The angular frequency ω = √(k/m) = √(18 / 0.50) = √36 = 6.0 rad s⁻¹. The period T = 2π/ω = 2π/6.0 ≈ 1.05 s. (b) The motion starts at maximum displacement, so we use a cosine function for displacement: x(t) = A cos(ωt) with A = 0.040 m. Thus x(t) = 0.040 cos(6.0 t). Velocity is v = –Aω sin(ωt) = –0.040 × 6.0 sin(6.0 t) = –0.24 sin(6.0 t) m s⁻¹. Acceleration a = –Aω² cos(ωt) = –0.040 × 36 cos(6.0 t) = –1.44 cos(6.0 t) m s⁻². (c) The total mechanical energy is constant and equal to the initial elastic potential energy: E_total = ½ k A² = ½ × 18 × (0.040)² = 9 × 0.0016 = 0.0144 J. At the equilibrium position (x = 0), all energy is kinetic, so the maximum kinetic energy is 0.0144 J. This occurs as the glider passes through the equilibrium point where speed is maximum.

解答：(a) 角频率 ω = √(k/m) = √(18 / 0.50) = √36 = 6.0 rad s⁻¹。周期 T = 2π/ω = 2π/6.0 ≈ 1.05 s。(b) 运动从最大位移处开始，因此位移用余弦函数表示：x(t) = A cos(ωt)，其中 A = 0.040 m。故 x(t) = 0.040 cos(6.0 t)。速度 v = –Aω sin(ωt) = –0.040 × 6.0 sin(6.0 t) = –0.24 sin(6.0 t) m s⁻¹。加速度 a = –Aω² cos(ωt) = –0.040 × 36 cos(6.0 t) = –1.44 cos(6.0 t) m s⁻²。(c) 总机械能守恒且等于初始弹性势能：E_总 = ½ k A² = ½ × 18 × (0.040)² = 9 × 0.0016 = 0.0144 J。在平衡位置 (x = 0) 处，所有能量为动能，故最大动能为0.0144 J。这发生在滑车通过平衡点速率最大时。

3. Ideal Gases and Kinetic Theory | 理想气体与动力学理论

A sealed cylinder of volume 2.0 × 10⁻³ m³ contains helium gas at a pressure of 1.2 × 10⁵ Pa and temperature 300 K. The molar mass of helium is 4.0 g mol⁻¹. (a) Calculate the number of moles of helium and the total mass. (b) The gas is heated to 450 K while the volume is kept constant. Find the new pressure. (c) Using the kinetic theory equation pV = ⅓ N m c²_rms, estimate the root‑mean‑square speed of the helium atoms at 450 K. (Take R = 8.31 J K⁻¹ mol⁻¹, N_A = 6.02 × 10²³ mol⁻¹)

一个体积为2.0 × 10⁻³ m³的密封气缸装有压强为1.2 × 10⁵ Pa、温度为300 K的氦气。氦的摩尔质量为4.0 g mol⁻¹。(a) 计算氦气的摩尔数和总质量。(b) 保持体积不变，将气体加热到450 K，求新的压强。(c) 利用动力学理论方程 pV = ⅓ N m c²_rms，估算氦原子在450 K时的方均根速率。（取 R = 8.31 J K⁻¹ mol⁻¹，N_A = 6.02 × 10²³ mol⁻¹）

Solution: (a) Using the ideal gas equation pV = nRT → n = pV/(RT) = (1.2×10⁵ × 2.0×10⁻³) / (8.31 × 300) = 240 / 2493 ≈ 0.0963 mol. The mass is m = n × M = 0.0963 × 4.0 = 0.385 g, or 3.85×10⁻⁴ kg. (b) For constant volume, p₁/T₁ = p₂/T₂, so p₂ = p₁ × (T₂/T₁) = 1.2×10⁵ × (450/300) = 1.2×10⁵ × 1.5 = 1.8×10⁵ Pa. (c) First find the number of molecules N = n × N_A = 0.0963 × 6.02×10²³ = 5.80×10²². The total mass of the gas is N m (with m as atomic mass), but we can use the equation: pV = ⅓ N m c²_rms = ⅓ M_total c²_rms, where M_total = 3.85×10⁻⁴ kg. Rearranging: c_rms = √(3pV / M_total) = √(3 × 1.8×10⁵ × 2.0×10⁻³ / 3.85×10⁻⁴). Compute numerator: 3 × 1.8×10⁵ × 2.0×10⁻³ = 1080. Then c_rms = √(1080 / 3.85×10⁻⁴) = √(2.805×10⁶) ≈ 1.67×10³ m s⁻¹. Alternatively, using c_rms = √(3RT/M_molar) = √(3×8.31×450 / 0.0040) ≈ √(11218.5 / 0.004) = √(2.805×10⁶) gives the same result. This matches the kinetic theory prediction that c_rms ∝ √T.

解答：(a) 由理想气体方程 pV = nRT 得 n = pV/(RT) = (1.2×10⁵ × 2.0×10⁻³) / (8.31 × 300) = 240 / 2493 ≈ 0.0963 mol。质量 m = n × M = 0.0963 × 4.0 = 0.385 g，即 3.85×10⁻⁴ kg。(b) 体积不变时，p₁/T₁ = p₂/T₂，故 p₂ = p₁ × (T₂/T₁) = 1.2×10⁵ × (450/300) = 1.2×10⁵ × 1.5 = 1.8×10⁵ Pa。(c) 先求分子数 N = n × N_A = 0.0963 × 6.02×10²³ = 5.80×10²²。气体总质量为 N m，其中 m 为原子质量，但可直接利用公式：pV = ⅓ M_total c²_rms，其中 M_total = 3.85×10⁻⁴ kg。整理得 c_rms = √(3pV / M_total) = √(3 × 1.8×10⁵ × 2.0×10⁻³ / 3.85×10⁻⁴)。计算分子：3 × 1.8×10⁵ × 2.0×10⁻³ = 1080。则 c_rms = √(1080 / 3.85×10⁻⁴) = √(2.805×10⁶) ≈ 1.67×10³ m s⁻¹。另一种方法：c_rms = √(3RT/M_molar) = √(3×8.31×450 / 0.0040) ≈ √(2.805×10⁶)，结果一致。这符合动力学理论预测，即 c_rms ∝ √T。

4. Electric Fields and Coulomb’s Law | 电场与库仑定律

Two point charges, +8.0 μC and –2.0 μC, are placed 0.60 m apart in a vacuum. (a) Determine the magnitude and direction of the electrostatic force on the negative charge. (b) Find the electric field strength at the midpoint between the charges and state its direction. (c) Show that the electric potential is zero at a point 0.20 m from the negative charge along the line joining the two charges, closer to the negative charge. (Use k = 8.99 × 10⁹ N m² C⁻²)

两个点电荷 +8.0 μC 和 –2.0 μC 在真空中相距 0.60 m。(a) 求负电荷所受静电力的大小和方向。(b) 求两点电荷连线中点处的电场强度并标明方向。(c) 证明在与负电荷相距 0.20 m、且更靠近负电荷的连线上，电势为零。（使用 k = 8.99 × 10⁹ N m² C⁻²）

Solution: (a) The magnitude of the force is given by Coulomb’s law: F = k |Q₁ Q₂| / r² = (8.99×10⁹) × (8.0×10⁻⁶ × 2.0×10⁻⁶) / (0.60)² = (8.99×10⁹ × 1.6×10⁻¹¹) / 0.36 ≈ (0.14384) / 0.36 ≈ 0.40 N. Since the charges are opposite, the force is attractive. Thus, the negative charge experiences a force of 0.40 N directed towards the positive charge. (b) At the midpoint, the distance from each charge is 0.30 m. The electric field due to a point charge is E = k |Q| / r². E₊ = (8.99×10⁹ × 8.0×10⁻⁶) / (0.30)² = 7.192×10⁴ / 0.09 ≈ 7.99×10⁵ N C⁻¹, pointing away from the positive charge (to the right if + is left). E₋ = (8.99×10⁹ × 2.0×10⁻⁶) / 0.09 ≈ 1.998×10⁵ N C⁻¹, pointing towards the negative charge (also to the right, because field points towards negative charge). Therefore, the fields are in the same direction, so the resultant E = 7.99×10⁵ + 2.00×10⁵ = 1.0×10⁶ N C⁻¹ towards the negative charge. (c) Let the required point be at a distance x from the negative charge and (0.60 – x) from the positive charge, with x = 0.20 m. The total potential V = k (+8.0×10⁻⁶) / (0.40) + k (–2.0×10⁻⁶) / (0.20). The positive term = (8.99×10⁹ × 8.0×10⁻⁶) / 0.40 = (7.192×10⁴) / 0.40 = 1.798×10⁵ V. The negative term = (8.99×10⁹ × –2.0×10⁻⁶) / 0.20 = (–1.798×10⁴) / 0.20 = –8.99×10⁴ V. The sum is 1.798×10⁵ – 8.99×10⁴ ≈ 8.99×10⁴ V? Wait, check: 1.798×10⁵ – 0.899×10⁵ = 0.899×10⁵ V, not zero. Recalculate: actually the question expects a point where the repulsive potential from the positive charge cancels the attractive potential from the negative charge. Solve for x such that k(8.0×10⁻⁶)/(0.60 – x) + k(–2.0×10⁻⁶)/x = 0 → 8.0/(0.60 – x) = 2.0/x → 8x = 2(0.60 – x) → 8x = 1.2 – 2x → 10x = 1.2 → x = 0.12 m. So the potential is zero at 0.12 m from the negative charge, not 0.20 m. The problem statement likely contains a typo; but the proof would show that at x = 0.12 m the potentials cancel. Nevertheless, following the given distance 0.20 m: V = 8.99×10⁹ × [8.0×10⁻⁶/0.40 – 2.0×10⁻⁶/0.20] = 8.99×10⁹ × (2.0×10⁻⁵ – 1.0×10⁻⁵) = 8.99×10⁹ × 1.0×10⁻⁵ = 8.99×10⁴ V, not zero. So the correct zero‑potential point is at x = 0.12 m. This illustrates the importance of verifying conditions.

解答：(a) 力的大小由库仑定律给出：F = k |Q₁ Q₂| / r² = (8.99×10⁹) × (8.0×10⁻⁶ × 2.0×10⁻⁶) / (0.60)² = (8.99×10⁹ × 1.6×10⁻¹¹) / 0.36 ≈ 0.14384 / 0.36 ≈ 0.40 N。由于电荷异号，力为吸引力，因此负电荷受到大小为0.40 N、指向正电荷的力。(b) 在中点，距每个电荷的距离均为0.30 m。点电荷的电场强度 E = k |Q| / r²。E₊ = (8.99×10⁹ × 8.0×10⁻⁶) / 0.30² ≈ 7.99×10⁵ N C⁻¹，方向远离正电荷（若正电荷在左，则向右）。E₋ = (8.99×10⁹ × 2.0×10⁻⁶) / 0.30² ≈ 2.00×10⁵ N C⁻¹，方向指向负电荷（同样向右）。两个场强方向相同，因此合场强 E = 7.99×10⁵ + 2.00×10⁵ = 1.0×10⁶ N C⁻¹，指向负电荷。(c) 设所求点距负电荷为 x，距正电荷为 (0.60 – x)。题目要求验证 x = 0.20 m 处电势为零。总电势 V = k(8.0×10⁻⁶)/(0.40) + k(–2.0×10⁻⁶)/(0.20) = 8.99×10⁴ V，并不为零。实际上，令 V = 0 可解得 x = 0.12 m 处电势为零。因此题目所给0.20 m并非零点；这表明解题时需仔细验证条件。若设方程 8.0/(0.60–x) = 2.0/x，得 x = 0.12 m，此即为电势为零的位置。

5. Magnetic Fields and Charged Particles | 磁场与带电粒子

An electron (mass 9.11 × 10⁻³¹ kg, charge –1.60 × 10⁻¹⁹ C) is accelerated from rest through a potential difference of 2.0 kV and then enters a uniform magnetic field of flux density 5.0 mT perpendicular to its velocity. (a) Calculate the speed of the electron before entering the field. (b) Determine the radius of the circular path it follows in the magnetic field. (c) Explain how the time taken to complete one full circle depends on the speed of the particle.

一个电子（质量 9.11 × 10⁻³¹ kg，电荷 –1.60 × 10⁻¹⁹ C）从静止经 2.0 kV 的电压加速，然后垂直进入磁感应强度为 5.0 mT 的匀强磁场。(a) 计算电子进入磁场前的速率。(b) 求电子在磁场中做圆周运动的轨道半径。(c) 解释电子绕行一整圈所需的时间与粒子速率的关系。

Solution: (a) Kinetic energy gained equals work done by electric field: ½ m v² = e V, where V = 2000 V. Thus v = √(2eV/m) = √(2 × 1.60×10⁻¹⁹ × 2000 / 9.11×10⁻³¹) = √(6.4×10⁻¹⁶ / 9.11×10⁻³¹) = √(7.025×10¹⁴) ≈ 2.65×10⁷ m s⁻¹. (b) In the magnetic field, the magnetic force provides the centripetal force: B e v = m v²/r, so r = m v/(B e). Substituting: r = (9.11×10⁻³¹ × 2.65×10⁷) / (5.0×10⁻³ × 1.60×10⁻¹⁹) = (2.414×10⁻²³) / (8.0×10⁻²²) = 0.0302 m = 3.02 cm. (c) The period of circular motion is T = 2πr/v = 2π (m v/(B e)) / v = 2π m/(B e). Notice that v cancels out, so the period is independent of the speed. The time for one revolution, called the cyclotron period, depends only on the charge-to-mass ratio and the magnetic field strength. Thus, all particles with the same e/m complete a circle in the same time, regardless of their initial speeds (provided v ≪ c). This principle is fundamental in cyclotron design.

解答：(a) 获得的动能等于电场力所做的功：½ m v² = e V，其中 V = 2000 V。所以 v = √(2eV/m) = √(2 × 1.60×10⁻¹⁹ × 2000 / 9.11×10⁻³¹) = √(6.4×10⁻¹⁶ / 9.11×10⁻³¹) = √(7.025×10¹⁴) ≈ 2.65×10⁷ m s⁻¹。(b) 在磁场中，磁力提供向心力：B e v = m v²/r，得 r = m v/(B e)。代入数据：r = (9.11×10⁻³¹ × 2.65×10⁷) / (5.0×10⁻³ × 1.60×10⁻¹⁹

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