📚 A2 Physics: Past Paper Analysis | A2 物理：历年真题解析
Mastering A2 Physics requires more than understanding concepts; it demands familiarity with how exam questions are structured and marked. This article analyses typical past paper questions across core A2 topics, offering step-by-step solutions and highlighting common pitfalls. By working through these examples, you will develop the problem-solving skills needed to excel in your exams.
掌握 A2 物理不仅要理解概念，更需要熟悉考题的结构和评分标准。本文分析 A2 核心主题中的典型历年真题，提供分步解答并指出常见的错误。通过练习这些例题，你将培养出在考试中取得高分所需的解题技巧。
1. Circular Motion | 圆周运动
A favourite question involves a mass moving in a vertical circle. 'A particle of mass m is attached to a light string of length r and swung in a vertical circle. Find the minimum speed at the top of the circle so that the string remains taut.'
一个常见考题是：一个质量为 m 的物体系在长度为 r 的轻绳一端，在竖直平面内做圆周运动。求物体在最高点的最小速度，使绳子保持绷紧。
on T and weight mg. For minimum speed, tension T = 0. Thus, mg = m v²/r. Solving gives v = √g r. This result appears frequently and must be justified, not just quoted.
在最高点，向心力由绳的拉力 T 和重力 mg 提供。当速度最小时，拉力 T = 0，因此 mg = m v²/r。解得 v = √g r。这一结果经常出现，必须给出推导过程，不能直接引用。
2. Gravitational Fields | 引力场
Past papers often test the application of Newton's law of gravitation to satellite motion. 'A satellite orbits Earth at radius R. Derive an expression for its orbital speed v in terms of G, M (Earth's mass) and R.'
真题经常考察如何将万有引力定律应用于卫星运动。“一颗卫星在半径为 R 的轨道上绕地球运行。用 G、地球质量 M 和 R 推导其轨道速度 v 的表达式。”
force provides centripetal force: GMm/R² = m v²/R. Cancel m and R to get v² = GM/R, so v = √GM/R speed is independent of satellite mass. Many students forget to cancel mass and obtain v = √GMm/R, losing marks.
由于万有引力提供向心力：GMm/R² = m v²/R。消去 m 和一个 R 得到 v² = GM/R，因此 v = √GM/R。注意速度与卫星质量无关。很多学生忘记消去质量而得到 v = √GMm/R，导致失分。
3. Simple Harmonic Motion | 简谐运动
r amplitude, period, frequency, and maximum acceleration. For a mass-spring system, T = 2π√m/k, and amax = ω² xmax, with ω = 2π/T. Often, energy calculations are required: total energy Etotal = ½ k A².
典型的真题会给出一张位移-时间图，要求写出振幅、周期、频率和最大加速度。对于弹簧–质量系统，T = 2π√m/k，且 amax = ω² xmax，其中 ω = 2π/T。题目经常要求计算能量：总能量 Etotal = ½ k A²。
restoring force is proportional to displacement and acts towards equilibrium. Remember to use F = -k x, linking to Newton's second law.
在一套考题中，学生需要证明漂浮在水面上的金属托盘被按下时做简谐运动，并说明条件。关键是回复力与位移成正比且指向平衡位置。要记得利用 F = -k x，并与牛顿第二定律联系起来。
4. Ideal Gas & Thermodynamics | 理想气体与热力学
Questions frequently combine the ideal gas equation pV = nRT with kinetic theory. ' ture 300 K has volume 0.020 m³. Calculate the number of moles and the number of molecules.' Use n = pV/(RT) and number = n × Nₐ. Remember that temperature must be in kelvin.
考题经常将理想气体状态方程 pV = nRT 与分子动理论结合。“一种理想气体在压强 1.0×10⁵ Pa、温度 300 K 时体积为 0.020 m³。计算摩尔数和分子数。”使用 n = pV/(RT)，分子数 = n × Nₐ。注意温度必须使用开尔文。
rage kinetic energy of a molecule is 3/2 kT. Another common task is to derive pV = 1/3 N m
分子的平均动能为 3/2 kT。另一个常见任务是推导 pV = 1/3 N m
5. Electric Fields & Capacitance | 电场与电容
Capacitor charging and discharging curves appear almost every year. ' rom a 6.0 V battery. Calculate the time constant and the charge stored when fully charged.' τ = RC = 470×10⁻⁶ × 10×10³ = 4.7 s. Q = C V = 470×10⁻⁶ × 6.0 = 2.82×10⁻³ C.
电容器的充放电曲线几乎每年都考。“一个 470 μF 的电容器通过一个 10 kΩ 的电阻从 6.0 V 电池充电。计算时间常数和充满电时的电荷。” τ = RC = 4.7 s，Q = CV = 2.82×10⁻³ C。
For discharging, V = V₀ e-t/RC y stored in a capacitor is E = ½ C V², which must be derived by considering work done charging the capacitor.
对于放电，V = V₀ e-t/RC。ln V 对 t 的图是一条直线，斜率为 -1/RC。要能利用此方法从实验数据求出电容。此外，电容器储存的能量为 E = ½ C V²，需要通过考虑充电过程的做功来推导。
6. Magnetic Fields & Electromagnetic Induction | 磁场与电磁感应
Faraday's law ε = - dΦ/dt and Lenz' ng through a coil. The induced emf opposes the change in flux linkage. A typical question: ' pendicular to the coil increases from 0 to 0.80 T in 0.40 s. Find the average induced emf.'
法拉第定律 ε = - dΦ/dt 和楞次定律的考查常以磁铁通过线圈等情景出现。感应电动势反抗磁通链的变化。典型考题：“一个 200 匝的线圈面积为 0.050 m²，垂直于线圈的匀强磁场在 0.40 s 内从 0 增加到 0.80 T。求平均感应电动势。”
B = 200 × 0.050 × 0.80 = 8.0 Wb. Average emf magnitude = ΔΦ/Δt = 8.0/0.40 = 20 V. Use Lenz' urrent that creates a field opposing the increase). Many students forget to multiply by N.
Φ = N B A，因此 ΔΦ = N A ΔB = 200 × 0.050 × 0.80 = 8.0 Wb。平均电动势的大小 = ΔΦ/Δt = 8.0/0.40 = 20 V。用楞次定律说明方向（例如感应电流产生的磁场阻碍磁场的增加）。许多学生忘记乘以匝数 N。
7. Alternating Current | 交流电
Questions on transformers and rectification are common. 'An ideal transformer has 500 primary turns and 50 secondary turns. The primary is connected to 240 V rms AC. Calculate the secondary rms voltage.' Vs / Vp = Ns / Np, so Vs r full-wave rectification, the output waveform and smoothing capacitor are often sketched.
关于变压器和整流的题目很常见。“一个理想变压器初级线圈 500 匝，次级 50 匝。初级接在 240 V 均方根交流电上。计算次级均方根电压。” Vs / Vp = Ns / Np，故 Vs = 24 V。全波整流的输出波形和滤波电容常需要画图。
RMS values: for sinusoidal AC, Vrms = V0/√2. A heating effect calculation: compare power dissipated by the same resistor connected to DC and AC. Make sure to use average power
= Irms² R. Also, the mean power in a purely resistive circuit is half the peak power.
均方根值：对于正弦交流电，Vrms = V0/√2。热效应计算：比较同一电阻在直流和交流下的功率。要确保使用平均功率
= Irms² R。纯电阻
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