📚 A2 Physics: Worked Examples Step-by-Step | A2 物理：典型例题详解
Mastering A2 Physics requires more than just memorising formulas — it demands the ability to apply concepts to unfamiliar scenarios. This article walks through carefully selected worked examples from key A2 topics, showing you exactly how to break down problems, choose the right equations, and avoid common pitfalls. Each example is paired with detailed reasoning so you can build confidence for your examinations.
掌握 A2 物理不仅需要记忆公式，更需要将概念应用于陌生情境的能力。本文精选了 A2 阶段核心主题的典型例题，一步步展示如何拆解问题、选择正确的方程并避开常见陷阱。每道例题都配有详细的解题思路，帮助你建立考试信心。
1. Circular Motion: Calculating Centripetal Force | 圆周运动：计算向心力
A car of mass 1200 kg travels around a roundabout of radius 15 m at a constant speed of 8.0 m/s. Calculate the centripetal force required to keep the car moving in a circle. Identify what provides this force.
一辆质量为 1200 kg 的汽车以 8.0 m/s 的恒定速度绕行一个半径为 15 m 的环形交叉口。计算维持汽车圆周运动所需的向心力，并指出该力由什么提供。
We use the formula for centripetal force: F = mv²/r. Substituting the values: F = (1200 kg) × (8.0 m/s)² / (15 m) = 1200 × 64 / 15 = 5120 N. The force is provided by friction between the tyres and the road surface. Without sufficient friction, the car would skid outwards tangentially.
我们使用向心力公式：F = mv²/r。代入数值：F = (1200 kg) × (8.0 m/s)² / (15 m) = 1200 × 64 / 15 = 5120 N。该力由轮胎与路面之间的摩擦力提供。如果摩擦力不足，汽车将沿切线方向向外侧滑。
2. Gravitational Fields: Orbital Period Derivation | 引力场：轨道周期推导
A satellite orbits Earth at an altitude where the gravitational field strength is 2.5 N/kg. Given that the radius of the orbit is 9.0 × 10⁶ m, calculate its orbital period. (Mass of Earth = 6.0 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻².)
一颗卫星在引力场强度为 2.5 N/kg 的高度绕地球运行。已知轨道半径为 9.0 × 10⁶ m，计算其轨道周期。（地球质量 = 6.0 × 10²⁴ kg，G = 6.67 × 10⁻¹¹ N m² kg⁻²。）
Gravitational field strength g = GM/r², so 2.5 = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (9.0 × 10⁶)². Wait — we already have r, so we don’t need to solve for it. Instead, use the relationship between centripetal acceleration and gravitational acceleration: v²/r = GM/r². The orbital period T = 2πr/v. From v² = GM/r, we get v = √(GM/r). Then T = 2πr / √(GM/r) = 2π √(r³/GM). Substituting: T = 2π × √((9.0 × 10⁶)³ / (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)) = 2π × √(7.29 × 10²⁰ / 4.0 × 10¹⁴) = 2π × √(1.82 × 10⁶) = 2π × 1350 = 8480 s ≈ 2.36 hours.
引力场强度 g = GM/r²，因此 2.5 = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (9.0 × 10⁶)²。等一等——我们已经知道 r，不需要再求解它。改用向心加速度与引力加速度的关系：v²/r = GM/r²。轨道周期 T = 2πr/v。由 v² = GM/r 可得 v = √(GM/r)。那么 T = 2πr / √(GM/r) = 2π √(r³/GM)。代入得：T = 2π × √((9.0 × 10⁶)³ / (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)) = 2π × √(7.29 × 10²⁰ / 4.0 × 10¹⁴) = 2π × √(1.82 × 10⁶) = 2π × 1350 = 8480 s ≈ 2.36 小时。
3. Simple Harmonic Motion: Energy Transformations | 简谐运动：能量转换
A mass of 0.50 kg attached to a spring oscillates with amplitude 0.040 m. The spring constant is 200 N/m. Determine the maximum speed of the mass and its speed when the displacement is 0.025 m.
一个 0.50 kg 的质量块连接在弹簧上，以振幅 0.040 m 振动。弹簧常数为 200 N/m。求质量块的最大速度以及位移为 0.025 m 时的速度。
In SHM, total energy E = ½kA². Maximum kinetic energy occurs at equilibrium, so ½mv_max² = ½kA². Thus v_max = A √(k/m) = 0.040 × √(200 / 0.50) = 0.040 × √400 = 0.040 × 20 = 0.80 m/s. For displacement x = 0.025 m, use energy conservation: ½mv² = ½kA² − ½kx². So v = √(k/m) × √(A² − x²) = 20 × √(0.040² − 0.025²) = 20 × √(0.0016 − 0.000625) = 20 × √0.000975 = 20 × 0.0312 = 0.624 m/s.
在简谐运动中，总能量 E = ½kA²。最大动能出现在平衡位置，因此 ½mv_max² = ½kA²。于是 v_max = A √(k/m) = 0.040 × √(200 / 0.50) = 0.040 × √400 = 0.040 × 20 = 0.80 m/s。对于位移 x = 0.025 m，使用能量守恒：½mv² = ½kA² − ½kx²。因此 v = √(k/m) × √(A² − x²) = 20 × √(0.040² − 0.025²) = 20 × √(0.0016 − 0.000625) = 20 × √0.000975 = 20 × 0.0312 = 0.624 m/s。
4. Thermodynamics: First Law Application | 热力学：第一定律应用
A gas expands at constant pressure of 1.0 × 10⁵ Pa, doing 150 J of work on its surroundings. During this process, 400 J of heat is supplied to the gas. Calculate the change in internal energy of the gas.
气体在 1.0 × 10⁵ Pa 的恒定压强下膨胀，对外做功 150 J。在此过程中，气体吸收了 400 J 的热量。计算气体内能的变化。
First law of thermodynamics: ΔU = Q − W, where Q is heat added to the system and W is work done BY the system. Here Q = +400 J (heat into gas), W = +150 J (work done by gas). So ΔU = 400 − 150 = +250 J. The internal energy increases by 250 J. Note the sign convention carefully — many students lose marks by confusing the direction of energy transfer.
热力学第一定律：ΔU = Q − W，其中 Q 是系统吸收的热量，W 是系统对外做的功。本题中 Q = +400 J（气体吸热），W = +150 J（气体对外做功）。因此 ΔU = 400 − 150 = +250 J。内能增加了 250 J。注意符号规则——很多学生因混淆能量传递方向而丢分。
5. Electric Fields: Trajectory of a Charged Particle | 电场：带电粒子的轨迹
An electron enters a uniform electric field of strength 5000 V/m at right angles to the field lines, with an initial speed of 2.0 × 10⁶ m/s. The field plates are 0.020 m long. Calculate the angular deflection of the electron as it leaves the field. (Electron charge = 1.60 × 10⁻¹⁹ C, mass = 9.11 × 10⁻³¹ kg.)
一个电子以 2.0 × 10⁶ m/s 的初速度垂直于电场线进入强度为 5000 V/m 的匀强电场。电场板长度为 0.020 m。计算电子离开电场时的偏转角。（电子电荷 = 1.60 × 10⁻¹⁹ C，质量 = 9.11 × 10⁻³¹ kg。）
The electric force F = eE gives vertical acceleration a = eE/m = (1.60 × 10⁻¹⁹ × 5000) / (9.11 × 10⁻³¹) = 8.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 8.78 × 10¹⁴ m/s². Time in field t = length / horizontal speed = 0.020 / (2.0 × 10⁶) = 1.0 × 10⁻⁸ s. Vertical velocity gained: v_y = a × t = 8.78 × 10¹⁴ × 1.0 × 10⁻⁸ = 8.78 × 10⁶ m/s. Deflection angle θ: tan θ = v_y / v_x = 8.78 × 10⁶ / 2.0 × 10⁶ = 4.39. So θ = tan⁻¹(4.39) ≈ 77° from original direction.
电场力 F = eE 给出竖直加速度 a = eE/m = (1.60 × 10⁻¹⁹ × 5000) / (9.11 × 10⁻³¹) = 8.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 8.78 × 10¹⁴ m/s²。在电场中的时间 t = 板长 / 水平速度 = 0.020 / (2.0 × 10⁶) = 1.0 × 10⁻⁸ s。获得的竖直速度：v_y = a × t = 8.78 × 10¹⁴ × 1.0 × 10⁻⁸ = 8.78 × 10⁶ m/s。偏转角 θ：tan θ = v_y / v_x = 8.78 × 10⁶ / 2.0 × 10⁶ = 4.39。因此 θ = tan⁻¹(4.39) ≈ 77°（相对于原方向）。
6. Capacitance: RC Time Constant | 电容：RC 时间常数
A 100 μF capacitor is charged through a 50 kΩ resistor from a 12 V battery. Calculate the time taken for the potential difference across the capacitor to reach 8.0 V. (Assume initially uncharged.)
一个 100 μF 的电容器通过一个 50 kΩ 的电阻从 12 V 电池充电。计算电容器两端电势差达到 8.0 V 所需的时间。（假设初始未充电。）
Time constant τ = RC = (50 × 10³) × (100 × 10⁻⁶) = 5.0 s. For charging: V = V₀(1 − e^(−t/τ)). Rearranging: e^(−t/τ) = 1 − V/V₀ = 1 − 8.0/12 = 1 − 0.667 = 0.333. Taking natural log: −t/τ = ln(0.333) = −1.099. So t = 1.099 × 5.0 = 5.5 s. Always check your ratio V/V₀ is less than 1 for charging — if you get a negative time, you’ve made an algebra error.
时间常数 τ = RC = (50 × 10³) × (100 × 10⁻⁶) = 5.0 s。充电公式：V = V₀(1 − e^(−t/τ))。整理得：e^(−t/τ) = 1 − V/V₀ = 1 − 8.0/12 = 1 − 0.667 = 0.333。取自然对数：−t/τ = ln(0.333) = −1.099。因此 t = 1.099 × 5.0 = 5.5 s。务必检查充电过程中 V/V₀ 的比值小于 1——如果算出负的时间，说明代数处理有误。
7. Magnetic Fields: Force on a Current-Carrying Conductor | 磁场：载流导体受力
A straight wire of length 0.30 m carries a current of 4.0 A and is placed at 30° to a uniform magnetic field of flux density 0.25 T. Calculate the magnitude of the force experienced by the wire. Describe how the direction of the force is determined.
一根长 0.30 m 的直导线通有 4.0 A 的电流，与磁通密度为 0.25 T 的匀强磁场成 30° 角放置。计算导线受力的大小，并说明如何判定力的方向。
Force F = BIL sin θ, where θ is the angle between current and field lines. F = 0.25 × 4.0 × 0.30 × sin 30° = 0.25 × 4.0 × 0.30 × 0.5 = 0.15 N. The direction is given by Fleming’s left-hand rule: first finger = field (N to S), second finger = current (conventional, + to −), thumb = force. With 30° between I and B, the force is perpendicular to both, reduced by sin θ compared to maximum at 90°.
力 F = BIL sin θ，其中 θ 是电流与磁力线之间的夹角。F = 0.25 × 4.0 × 0.30 × sin 30° = 0.25 × 4.0 × 0.30 × 0.5 = 0.15 N。方向由弗莱明左手定则判定：食指 = 磁场（N 到 S），中指 = 电流（常规方向，+ 到 −），拇指 = 力。当电流与磁场夹角为 30° 时，力垂直于两者，与 90° 时的最大值相比缩小了 sin θ 倍。
8. Electromagnetic Induction: Faraday’s Law with Flux Linkage | 电磁感应：法拉第定律与磁链
A coil of 200 turns encloses an area of 0.025 m². It is placed in a magnetic field of 0.40 T, with its plane perpendicular to the field. The field is reduced uniformly to zero in 0.50 s. Calculate the average induced emf in the coil.
一个 200 匝的线圈包围面积为 0.025 m²，置于 0.40 T 的磁场中，线圈平面与磁场垂直。磁场在 0.50 s 内均匀减小为零。计算线圈中的平均感应电动势。
Initial flux linkage NΦ = N × B × A = 200 × 0.40 × 0.025 = 2.0 Wb-turns. Final flux linkage = 0. Change in flux linkage Δ(NΦ) = 0 − 2.0 = −2.0 Wb-turns. Average induced emf ε = −Δ(NΦ) / Δt = −(−2.0) / 0.50 = 4.0 V. The negative sign in Faraday’s law relates to Lenz’s law — the induced current opposes the change, but magnitude is what’s often required.
初始磁链 NΦ = N × B × A = 200 × 0.40 × 0.025 = 2.0 Wb-匝。最终磁链 = 0。磁链变化量 Δ(NΦ) = 0 − 2.0 = −2.0 Wb-匝。平均感应电动势 ε = −Δ(NΦ) / Δt = −(−2.0) / 0.50 = 4.0 V。法拉第定律中的负号与楞次定律有关——感应电流反抗磁通量的变化，但通常题目只要求大小。
9. Nuclear Physics: Binding Energy per Nucleon | 核物理：每个核子的结合能
Calculate the binding energy per nucleon for an iron-56 nucleus, given the following masses: proton mass = 1.00728 u, neutron mass = 1.00867 u, iron-56 nuclear mass = 55.93494 u, and 1 u = 931.5 MeV/c².
计算铁-56 原子核的每个核子结合能，已知以下质量：质子质量 = 1.00728 u，中子质量 = 1.00867 u，铁-56 核质量 = 55.93494 u，1 u = 931.5 MeV/c²。
Iron-56 has 26 protons and 30 neutrons. Total mass of separate nucleons = 26 × 1.00728 + 30 × 1.00867 = 26.18928 + 30.26010 = 56.44938 u. Mass defect Δm = 56.44938 − 55.93494 = 0.51444 u. Binding energy = Δm × 931.5 = 0.51444 × 931.5 = 479.2 MeV. Binding energy per nucleon = 479.2 / 56 = 8.56 MeV. This high value explains iron’s exceptional stability — it lies near the peak of the binding energy curve.
铁-56 有 26 个质子和 30 个中子。分离核子的总质量 = 26 × 1.00728 + 30 × 1.00867 = 26.18928 + 30.26010 = 56.44938 u。质量亏损 Δm = 56.44938 − 55.93494 = 0.51444 u。结合能 = Δm × 931.5 = 0.51444 × 931.5 = 479.2 MeV。每个核子的结合能 = 479.2 / 56 = 8.56 MeV。这一高值解释了铁的特殊稳定性——它位于结合能曲线的峰值附近。
10. Quantum Physics: Photoelectric Effect | 量子物理：光电效应
Ultraviolet light of wavelength 200 nm strikes a clean metal surface with work function 4.5 eV. Determine whether electrons are emitted and, if so, calculate their maximum kinetic energy in joules. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10⁻¹⁹ J.)
波长为 200 nm 的紫外光照射在功函数为 4.5 eV 的洁净金属表面上。判断是否会有电子逸出，如果有，计算其最大动能（以焦耳为单位）。（h = 6.63 × 10⁻³⁴ J s，c = 3.00 × 10⁸ m/s，1 eV = 1.60 × 10⁻¹⁹ J。）
Photon energy E = hf = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 2.00 × 10⁻⁷ = 9.945 × 10⁻¹⁹ J. Convert to eV: 9.945 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 6.22 eV. Since 6.22 eV > work function 4.5 eV, emission occurs. Maximum kinetic energy K_max = E_photon − Φ = 6.22 − 4.5 = 1.72 eV. In joules: 1.72 × 1.60 × 10⁻¹⁹ = 2.75 × 10⁻¹⁹ J. The threshold wavelength would be λ₀ = hc/Φ = 1240/4.5 ≈ 276 nm, so 200 nm is indeed above threshold frequency.
光子能量 E = hf = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (200 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 2.00 × 10⁻⁷ = 9.945 × 10⁻¹⁹ J。转换为 eV：9.945 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 6.22 eV。由于 6.22 eV > 功函数 4.5 eV，逸出发生。最大动能 K_max = E_光子 − Φ = 6.22 − 4.5 = 1.72 eV。以焦耳为单位：1.72 × 1.60 × 10⁻¹⁹ = 2.75 × 10⁻¹⁹ J。阈值波长为 λ₀ = hc/Φ = 1240/4.5 ≈ 276 nm，因此 200 nm 确实高于阈值频率。
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