📚 Case Study Practical Drill for Year 13 Cambridge Physics | Year 13 剑桥物理案例分析实战演练
Mastering physics at A Level means moving beyond memorising formulas to applying them in unfamiliar, real-world scenarios. In this article, we work through nine detailed case studies that span mechanics, fields, thermodynamics, quantum physics, nuclear physics, oscillations, circuits and waves. Each case is broken down step by step, with paired English–Chinese explanations, so you can train your problem-solving muscles exactly as required in Cambridge Year 13 examinations.
在A Level阶段掌握物理意味着不再死记公式，而是将其应用到陌生的真实情境中。本文将通过九个详尽的案例分析，涵盖力学、场、热力学、量子物理、核物理、振动、电路和波动。每个案例都配有步骤拆解和中英对照讲解，帮助你像剑桥Year 13考试所要求的那样训练解题思维。
1. Satellite Orbits and Gravitational Fields | 卫星轨道与引力场
A communications satellite must remain fixed above a point on Earth’s equator. For this geostationary orbit, the orbital period T must match Earth’s rotation – exactly 86 400 s.
一颗通信卫星必须固定在地球赤道上空某点上方。对于地球同步轨道，轨道周期 T 必须等于地球自转周期——精确为 86 400 s。
Using Newton’s law of gravitation and the condition for circular motion, the centripetal force mv²/r equals the gravitational force GMm/r². Replacing the speed v with the circumference divided by period (2πr/T) gives the relation between radius and period.
利用牛顿万有引力定律和圆周运动条件，向心力 mv²/r 等于万有引力 GMm/r²。用周长除以周期（2πr/T）替换线速度 v，可得出轨道半径与周期的关系。
GMm/r² = m(2πr/T)²/r ⇒ r³ = GMT²/(4π²)
Substituting the Earth’s mass M = 5.97×10²⁴ kg and the gravitational constant G = 6.67×10⁻¹¹ N m² kg⁻² yields an orbital radius r ≈ 4.22×10⁷ m, which is about 35 800 km above Earth’s surface.
代入地球质量 M = 5.97×10²⁴ kg 和引力常量 G = 6.67×10⁻¹¹ N m² kg⁻²，可求得轨道半径 r ≈ 4.22×10⁷ m，即距地面约 35 800 km。
2. Electromagnetic Induction in Generators | 发电机中的电磁感应
A simple AC generator consists of a rectangular coil of N turns, each of area A, rotating at a steady angular speed ω inside a uniform magnetic field B. The flux linkage through the coil varies as N B A cos(ωt).
一台简易交流发电机由一个 N 匝矩形线圈构成，每匝面积为 A，在匀强磁场 B 中以恒定角速度 ω 旋转。线圈的磁链按 N B A cos(ωt) 变化。
Faraday’s law states that the induced emf is the negative rate of change of flux linkage. Differentiating the cosine gives a sinusoidal output.
法拉第定律指出，感应电动势等于磁链的负变化率。对余弦函数求导得到正弦输出。
ε = -dΦ/dt = N B A ω sin(ωt) ⇒ ε₀ = N B A ω
For a coil with 200 turns, an area of 0.05 m², B = 0.8 T and a rotation frequency of 50 Hz (ω = 100π rad/s), the peak emf ε₀ reaches about 2.5 kV. This illustrates how the design parameters directly control the generator’s output.
对于一个 200 匝、面积为 0.05 m² 的线圈，B = 0.8 T、旋转频率 50 Hz（ω = 100π rad/s），峰值电动势 ε₀ 约达 2.5 kV。这表明设计参数如何直接控制发电机的输出。
3. Thermodynamics of an Ideal Gas Engine | 理想气体热机热力学
A car engine can be modelled by the ideal Otto cycle. Air (γ = 1.4) is compressed adiabatically with a compression ratio r = V_max / V_min = 8. The theoretical efficiency depends only on r and γ.
汽车发动机可以用理想奥托循环建模。空气（γ = 1.4）按压缩比 r = V_max / V_min = 8 进行绝热压缩。理论效率仅取决于 r 和 γ。
η = 1 − 1/r^(γ−1) = 1 − 1/8^(0.4) ≈ 0.565 (56.5%)
Real engines fall short of this value because of friction, incomplete combustion and heat losses to the cooling system. Nevertheless, the formula shows that increasing the compression ratio can boost efficiency – a core design principle in modern engines.
实际发动机由于摩擦、不完全燃烧和冷却系统散热，效率低于此值。不过该公式表明，提高压缩比可提升效率——这是现代发动机设计的核心原则。
4. Photoelectric Effect and Quantum Efficiency | 光电效应与量子效率
A photosensitive metal has a work function φ = 2.3 eV. When light of wavelength 350 nm strikes its surface, photon energy can be calculated using the Planck equation E = hc/λ.
一种光敏金属的功函数为 φ = 2.3 eV。当波长 350 nm 的光照射其表面时，可利用普朗克公式 E = hc/λ 计算光子能量。
Using the convenient value hc = 1240 eV·nm, the incident photon energy is E = 1240/350 ≈ 3.54 eV. The maximum kinetic energy of the emitted electrons is simply the difference.
利用常用值 hc = 1240 eV·nm，入射光子能量为 E = 1240/350 ≈ 3.54 eV。发射电子的最大动能就是两者的差值。
K_max = E − φ = 3.54 − 2.3 = 1.24 eV
The threshold frequency below which no electrons are emitted is f₀ = φ/h. With h = 4.14×10⁻¹⁵ eV·s, f₀ ≈ 5.56×10¹⁴ Hz. This case links photon energy to measurable stopping potentials in a photocell.
截止频率 f₀ = φ/h，即低于此频率不会发射电子。取 h = 4.14×10⁻¹⁵ eV·s，f₀ ≈ 5.56×10¹⁴ Hz。该案例将光子能量与光电池中可测的遏止电压联系起来。
5. Radioactive Decay and Carbon Dating | 放射性衰变与碳定年法
An archaeological wooden artefact has a carbon-14 activity of 0.12 Bq per gram of carbon, while a living sample of the same material records 0.23 Bq/g. The half-life of ¹⁴C is 5730 years.
一件考古木制文物的每克碳的碳-14 活度为 0.12 Bq，而同类活体样本为 0.23 Bq/g。¹⁴C 的半衰期为 5730 年。
The decay constant λ = ln2 / T₁/₂ = 0.693 / 5730 yr⁻¹. The activity follows A = A₀ e^(−λt), so solving for the age t gives:
衰变常量 λ = ln2 / T₁/₂ = 0.693 / 5730 yr⁻¹。活度遵循 A = A₀ e^(−λt)，解出年代 t：
t = (T₁/₂ / ln2) × ln(A₀/A) = (5730 yr / 0.693) × ln(0.23/0.12) ≈ 8260 years
This result places the artefact in the early Neolithic period. Carbon dating remains reliable only for organic materials less than about 50 000 years old, because the remaining ¹⁴C activity becomes too low to measure accurately.
这一结果将该文物定位于新石器时代早期。碳定年法仅对不超过约 50 000 年的有机材料可靠，因为残留的 ¹⁴C 活度太低而无法准确测量。
6. Damped Simple Harmonic Motion in Shock Absorbers | 减震器中的阻尼简谐运动
A car’s shock absorber is engineered to damp out oscillations quickly after a bump. The system can be modelled as a mass–spring–damper with mass m = 1000 kg and spring constant k = 40 000 N/m.
汽车减震器被设计用来在颠簸后迅速衰减振动。该系统可建模为质量–弹簧–阻尼系统，质量 m = 1000 kg，劲度系数 k = 40 000 N/m。
The natural angular frequency is ω₀ = √(k/m) = √(40 000/1000) ≈ 6.32 rad/s. Critical damping occurs when the damping coefficient b satisfies b² = 4mk.
固有角频率 ω₀ = √(k/m) = √(40 000/1000) ≈ 6.32 rad/s。当阻尼系数 b 满足 b² = 4mk 时出现临界阻尼。
b_crit = 2√(mk) = 2 × √(1000 × 40 000) ≈ 12 650 N s/m
For underdamped motion, the amplitude decays as A(t) = A₀ e^(−bt/2m). Designers choose b slightly lower than critical to avoid a harsh ride while still suppressing prolonged bouncing. The quality factor Q = ω₀ m / b helps quantify the damping effectiveness.
对于欠阻尼运动，振幅按 A(t) = A₀ e^(−bt/2m) 衰减。设计者会选择略低于临界值的阻尼系数，以避免驾乘过硬，同时抑制长时间跳动。品质因数 Q = ω₀ m / b 可用于量化阻尼效果。
7. Charging and Discharging Capacitors in Timing Circuits | 定时电路中的电容充放电
A common 555 timer relies on the exponential charge and discharge of a capacitor. Consider a 10 μF capacitor charged from a 9 V battery through a 1 MΩ resistor.
常见的 555 定时器依靠电容器的指数充放电工作。考虑一个 10 μF 电容器通过 1 MΩ 电阻由 9 V 电池充电的情形。
The time constant τ = RC = (1×10⁶ Ω) × (10×10⁻⁶ F) = 10 s. After one time constant, the voltage across the capacitor reaches about 63% of the supply.
时间常数 τ = RC = (1×10⁶ Ω) × (10×10⁻⁶ F) = 10 s。经过一个时间常数后，电容器两端电压约达到电源电压的 63%。
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