📚 Case Study Practical Exercises for Year 13 OCR Physics | Year 13 OCR 物理：案例分析实战演练
Case studies in OCR A Level Physics challenge you to apply core principles from Modules 5 and 6 to unfamiliar real‑world scenarios. This article walks through ten high‑frequency case types, modelling how to identify the relevant physics, select the right equations, and produce rigorous quantitative analyses. Use these worked outlines to build confidence for the synoptic elements of your exam.
OCR A Level 物理的案例分析题要求你将模块5和模块6的核心原理运用到陌生的真实情境中。本文精选十种高频案例类型，示范如何识别涉及的物理知识、选择合适的方程并进行严谨的定量分析。通过这些示范来建立对考试综合题的信心。
1. Satellite Orbital Motion | 案例1：卫星轨道运动
A common scenario gives the satellite's altitude above a planet and asks for orbital speed, period, or the mass of the central body. The physics engine is the balance between gravitational force and centripetal force.
常见情境是给出卫星在行星上方的高度，要求计算轨道速率、周期或中心天体的质量。背后的物理是引力与向心力的平衡。
The centripetal force is supplied by gravity: GMm/r² = mv²/r, which simplifies to v = √GM/r. The orbital period follows from T = 2πr/v, giving Kepler's third law T² = (4π²/GM) r³.
向心力由引力提供：GMm/r² = mv²/r，化简得 v = √GM/r。轨道周期根据 T = 2πr/v 得出，即开普勒第三定律 T² = (4π²/GM) r³。
When the altitude is h and the planet's radius is R, the orbital radius is r = R + h. If the problem provides surface gravity g instead of G and M, use GM = gR². Then v = √(gR²/(R+h)). Always convert units to SI before substituting.
当高度为 h、行星半径为 R 时，轨道半径 r = R + h。如果题目给出表面重力加速度 g 而非 G 和 M，利用 GM = gR²。于是 v = √(gR²/(R+h))。代入数据前务必将所有单位转换为国际单位制。
Example: A reconnaissance satellite orbits Earth at 300 km altitude. Earth's radius is 6370 km and GM = 3.986 × 10¹⁴ m³ s⁻². r = 6.67 × 10⁶ m. Orbital speed v = √3.986 × 10¹⁴ / 6.67 × 10⁶ ≈ 7.73 × 10³ m s⁻¹. Period T = 2π × 6.67 × 10⁶ / 7.73 × 10³ ≈ 5.42 × 10³ s (≈ 90 min).
示例：一颗侦察卫星在300 km高度绕地球运转。地球半径6370 km，GM = 3.986 × 10¹⁴ m³ s⁻²。r = 6.67 × 10⁶ m。轨道速率 v ≈ 7.73 × 10³ m s⁻¹，周期 T ≈ 5.42 × 10³ s（约90分钟）。
2. Mass Spectrometer | 案例2：质谱仪
A sample is ionised, accelerated through a potential difference V, and then deflected in a uniform magnetic field B. The radius of the semicircular path enables mass identification.
样品被离子化，经过电势差 V 加速，再进入匀强磁场 B 中偏转。半圆形路径的半径可用于确定质量。
Acceleration stage: ½mv² = qV → v = √2qV/m. In the magnetic field, the magnetic force provides the centripetal force: qvB = mv²/r → r = mv/(qB). Combine to eliminate v: m = qB²r²/(2V). This expression is the workhorse for mass spectrometry.
加速阶段：½mv² = qV → v = √2qV/m。在磁场中，洛伦兹力提供向心力：qvB = mv²/r → r = mv/(qB)。消去 v 得 m = qB²r²/(2V)。这是质谱分析的核心表达式。
When analysing ions of the same charge state, mass is proportional to r². For isotopes, the separation Δr is related to Δm. Often the question will ask you to determine V or B from a measured radius for a known mass, or to find the mass of an unknown isotope given a calibration point.
当分析带相同电荷的离子时，质量与 r² 成正比。对于同位素，半径差 Δr 与质量差 Δm 有关。题目常要求根据已知质量的测量半径求出 V 或 B，或利用校准点推算未知同位素的质量。
Example: Singly‑charged carbon‑12 ions (m = 1.99 × 10⁻²⁶ kg) follow a radius of 0.152 m when B = 0.500 T. Unknown singly‑charged ions of the same element produce a radius of 0.162 m. Their mass is m₂ = m₁ × (r₂/r₁)² = 1.99 × 10⁻²⁶ × (0.162/0.152)² ≈ 2.26 × 10⁻²⁶ kg, corresponding to carbon‑14.
示例：单电荷碳‑12 离子（m = 1.99 × 10⁻²⁶ kg）在 B = 0.500 T 时的轨迹半径为0.152 m。同一元素的未知单电荷离子半径为0.162 m。其质量 m₂ = m₁ × (r₂/r₁)² ≈ 2.26 × 10⁻²⁶ kg，对应碳‑14。
3. Capacitor Discharge in a Defibrillator | 案例3：除颤器电容放电
Defibrillators store energy in a capacitor and deliver a controlled pulse to the patient through an RC discharge. The analysis links exponential decay, time constant, and energy delivery.
除颤器将能量储存在电容器中，通过 RC 放电为患者提供受控脉冲。分析涉及指数衰减、时间常数与能量释放。
During discharge, the voltage obeys V = V₀ exp(−t/RC) same exponential law. The time constant τ = RC is the time for the voltage to fall to ≈ 37% of its initial value. The half‑life t₁/₂ = RC ln 2 ≈ 0.693 RC.
放电过程中电压遵循 V = V₀ exp(−t/RC)。电流和电荷也遵循相同的指数规律。时间常数 τ = RC 是电压降至初始值约37%所需的时间。半衰期 t₁/₂ = RC ln 2 ≈ 0.693 RC。
The energy initially stored is E₀ = ½CV₀². After a time t, the energy remaining is E = ½CV₀² exp(−2t/RC). A common task is to determine the resistance required to deliver a certain fraction of stored energy within a specified time window.
初始储存能量 E₀ = ½CV₀²。经过时间 t，剩余能量 E = ½CV₀² exp(−2t/RC)。常见任务是确定在规定时间窗口内输送特定能量比例所需的电阻值。
.0 kV is discharged through a 10 Ω resistor. τ = (50 × 10⁻⁶) × 10 = 5.0 × 10⁻⁴ s. The half‑life is 3.5 × 10⁻⁴ s. If the pulse must deliver 90% of stored energy, solve 0.10 = exp(−2t/RC) → t = ½ RC ln(10) ≈ 5.8 × 10⁻⁴ s. The designer selects R to match this timing.
示例：一只 50 μF 电容充电至 5.0 kV，通过 10 Ω 电阻放电。τ = 5.0 × 10⁻⁴ s，半衰期 3.5 × 10⁻⁴ s。若脉冲需释放 90% 储存能量，解 0.10 = exp(−2t/RC) 得 t = ½ RC ln(10) ≈ 5.8 × 10⁻⁴ s。设计者据此选择 R。
4. Radioactive Dating with Carbon‑14 | 案例4：碳‑14 放射性定年
Archaeological specimens containing organic material can be dated by measuring the present activity of ¹⁴C and comparing it with the activity of a modern reference sample.
含有机物质的考古样本可通过测量当前碳‑14 的活度，并与现代参考样品的活度比较来进行年代测定。
The decay law: A = A₀ exp(−λt), where λ = ln 2 / T₁/₂. For ¹⁴C, T₁/₂ = 5730 years. The time since the organism died is t = (1/λ) ln(A₀/A). A₀ is the activity of an equivalent living sample, typically 0.23 Bq per gram of carbon.
衰变规律：A = A₀ exp(−λt)，其中 λ = ln 2 / T₁/₂。对于碳‑14，T₁/₂ = 5730 年。生物死亡至今的时间 t = (1/λ) ln(A₀/A)。A₀ 是等量活体样本的活度，通常为每克碳 0.23 Bq。
Questions often provide the count rate from a sample of known mass after correcting for background. Remember that activity is proportional to the number of ¹⁴C nuclei present. Always correct for background radiation and express activities in Bq; if mass is given, use specific activity.
题目常给出已知质量样本修正背景后的计数率。记住活度与存在的碳‑14 核数目成正比。务必扣除背景辐射，并以 Bq 为单位表示活度；若给出质量，则使用比活度。
Example: A wooden tool fragment of mass 50 g yields a net count rate of 0.30 Bq. Modern wood of the same mass gives 0.92 Bq. A₀/A = 0.92/0.30 = 3.07. λ = ln 2 / 5730 = 1.21 × 10⁻⁴ year⁻¹. t = ln(3.07) / 1.21 × 10⁻⁴ ≈ 9.3 × 10³ years. The tool dates to about 7300 BCE.
示例：一块 50 g 的木工具碎片净计数率为 0.30 Bq。同质量的现代木头为 0.92 Bq。A₀/A = 3.07。λ = 1.21 × 10⁻⁴ year⁻¹。t = ln(3.07) / 1.21 × 10⁻⁴ ≈ 9.3 × 10³ 年，工具可追溯至约公元前7300年。
5. Simple Harmonic Motion of a Mass–Spring System | 案例5：质量–弹簧系统的简谐运动
A mass attached to an ideal spring is a building-block case for SHM. Given data on extension or oscillation period, you can determine spring constant, maximum velocity, or acceleration.
连接在理想弹簧上的质量块是简谐运动的基本案例。根据伸长量或振动周期数据，可以确定劲度系数、最大速度或加速度。
The defining relation is F = −kx, leading to ω = √k/m and T = 2π √m/k. Displacement is x = A cos(ωt), velocity v = −ωA sin(ωt) (max |v| = ωA), and acceleration a = −ω²x (max |a| = ω²A).
基本关系是 F = −kx，由此导出 ω = √k/m 和 T = 2π √m/k。位移 x = A cos(ωt)，速度 v = −ωA sin(ωt)（最大速率 ωA），加速度 a = −ω²x（最大加速度 ω²A）。
The total mechanical energy is constant: E = ½kA² = ½mv² + ½kx². This can be used to find speed at a given displacement: v = ± ω √A² − x². In a vertical spring system, the equilibrium position shifts by mg/k, but the dynamics about that point are identical.
总机械能守恒：E = ½kA² = ½mv² + ½kx²。可用于求任一位置的速度：v = ± ω √A² − x²。在竖直弹簧系统中，平衡位置下移 mg/k，但绕该平衡点的动力学完全一样。
30.8 N m⁻¹. Maximum speed = ωA = 7.85 × 0.120 = 0.942 m s⁻¹. When x = 0.080 m, v = 7.85 × √0.120² − 0.080² ≈ 0.70 m s⁻¹.
示例：一个 0.500 kg 的质量块以振幅 0.120 m、周期 0.800 s 振动。ω = 7.85 rad s⁻¹，k = 30.8 N m⁻¹。最大速度 0.942 m s⁻¹。当 x = 0.080 m 时，v ≈ 0.70 m s⁻¹。
6. Young's Double‑Slit Interference | 案例6：杨氏双缝干涉
Measuring the fringe spacing in a double‑slit setup allows the wavelength of light to be determined, or conversely the slit separation to be found using a known laser.
通过测量双缝装置中的条纹间距，可以确定光的波长，或反过来利用已知激光求出缝间距。
The key formula is λ = a x / D, where a is the slit separation, x is the fringe separation (distance between consecutive bright fringes), and D is the distance from slits to screen. This approximation holds when D ≫ a and the angles are small.
关键公式为 λ = a x / D，其中 a 为缝间距，x 为条纹间距（相邻亮纹中心间的距离），D 为双缝到屏幕的距离。该近似在 D ≫ a 且角度很小时成立。
In a typical practical, a laser of known wavelength illuminates the slits, and x is obtained by measuring across multiple fringes. The separation a can then be calculated: a = λ D / x. For white‑light fringes, the central fringe is white and the outer fringes show spectra with blue on the inner edge.
在典型实验中，已知波长的激光照射双缝，通过测量多条条纹的跨度得出 x。然后计算 a = λ D / x。对于白光条纹，中央条纹为白色，外侧条纹呈现光谱，且内侧为蓝色。
Example: A He–Ne laser (λ = 633 nm) produces fringes separated by 0.450 cm on a screen 2.00 m away. a = (633 × 10⁻⁹ × 2.00) / (0.450 × 10⁻²) = 2.81 × 10⁻⁴ m. The slit separation is 0.281 mm. When repeating with a different laser, the fringe spacing changes to 0.375 cm, so the new wavelength is λ' = (2.81 × 10⁻⁴ × 0.375 × 10⁻²) / 2.00 = 5.27 × 10⁻⁷ m (527 nm), green light.
示例：He–Ne 激光（λ = 633 nm）在 2.00 m 外的屏上产生的条纹间距为 0.450 cm。a = 2.81 × 10⁻⁴ m。换用另一激光，条纹间距变为 0.375 cm，新的波长 λ' = 5.27 × 10⁻⁷ m（527 nm），为绿光。
7. Ideal Gas Law and Kinetic Theory | 案例7：理想气体状态方程与分子动理论
Problems may link macroscopic gas laws to the microscopic kinetic model, requiring conversion between pressure, volume, temperature and root‑mean‑square speed.
题目可能将宏观气体定律与微观动理论模型联系起来，要求在压强、体积、温度与方均根速率之间进行换算。
The equation of state: pV = nRT and pV = NkT. The pressure from kinetic theory: p = ⅓ ρ
状态方程：pV = nRT 以及 pV = NkT。动理论给出的压强：p = ⅓ ρ ⟨c²⟩，其中 ρ 为密度，⟨c²⟩ 为方均速率。方均根速率 cᵣₘₛ = √3RT/M，M 以 kg mol⁻¹ 为单位。
Average translational kinetic energy per molecule: ½ m ⟨c²⟩ = (3/2) kT. This link is often used to compare gases at different temperatures or to find temperature from measured r.m.s. speed.
每个分子的平均平动动能：½ m ⟨c²⟩ = (3/2) kT。这一联系常用于比较不同温度下的气体，或从测得的方均根速率求温度。
Example: At 300 K, the r.m.s. speed of helium atoms (M = 4.0 × 10⁻³ kg mol⁻¹) is cᵣₘₛ = √3 × 8.31 × 300 / 4.0 × 10⁻³ ≈ 1.37 × 10³ m s⁻¹. If the temperature doubles to 600 K, cᵣₘₛ becomes √2 times the original, ≈ 1.93 × 10³ m s⁻¹. This demonstrates the √T dependence.
示例：在 300 K 时，氦原子 (M = 4.0 × 10⁻³ kg mol⁻¹) 的 cᵣₘₛ ≈ 1.37 × 10³ m s⁻¹。温度升至 600 K 时，cᵣₘₛ 变为原来的 √2 倍，约 1.93 × 10³ m s⁻¹。这体现了 √T 的依赖关系。
8. Cyclotron and Circular Motion in a Magnetic Field | 案例8：回旋加速器与磁场中的圆周运动
A charged particle moving perpendicular to a uniform magnetic field travels in a circle. The cyclotron exploits the fact that the period is independent of speed, enabling repeated acceleration.
带电粒子垂直进入匀强磁场后做圆周运动。回旋加速器利用周期与速率无关的特点，实现反复加速。
Circular motion: qvB = mv²/r → r = mv/(qB). The angular speed ω = qB/m and the period T = 2πm/(qB). Since T does not depend on speed, the synchronised alternating voltage always reverses when the particle reaches the gap.
圆周运动：qvB = mv²/r → r = mv/(qB)。角速度 ω = qB/m，周期 T = 2πm/(qB)。由于T与速率无关，同步的交变电压总能在粒子到达间隙时换向。
The maximum kinetic energy is fixed by the dee radius R: KE_max = ½mv_max² = q²B²R²/(2m). This shows that to attain higher energies, one needs stronger magnetic fields or larger radius. In practice, relativistic mass increase limits the cyclotron, which motivates the synchrotron.
最大动能由D形盒半径 R 决定：KE_max = ½mv_max² = q²B²R²/(2m)。可见要获得更高能量，需要更强的磁场或更大的半径。实际中相对论质量增大会限制回旋加速器，由此催生了同步加速器。
proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) in a 1.20 T field and D radius 0.90 m has _max = (1.60 × 10⁻¹⁹)² × (1.20)² × (0.90)² / (2 × 1.67 × 10⁻²⁷) ≈ 1.49 × 10⁻¹² J ≈ 9.3 MeV.
示例：一个质子（m = 1.67 × 10⁻²⁷ kg，q = 1.60 × 10⁻¹⁹ C）在 1.20 T 的磁场和半径 0.90 m 的D形盒中，KE_max ≈ 1.49 × 10⁻¹² J，约 9.3 MeV。
9. Photoelectric Effect and Stopping Potential | 案例9：光电效应与遏止电势
Data from a phototube experiment—variation of stopping potential Vₛ with light frequency f—enable determination of Planck's constant and the work function of the cathode material.
来自光电池实验的数据——遏止电势 Vₛ 随光频率 f 的变化——可用于测定普朗克常数和阴极材料的功函数。
Einstein's equation: hf = φ + Kmax. With the stopping potential, e Vₛ = Kmax, so Vₛ = (h/e) f − φ/e. A graph of Vₛ against f is a straight line with slope h/e and intercept −φ/e.
爱因斯坦方程：hf = φ + Kmax。利用遏止电势 e Vₛ = Kmax，得 Vₛ = (h/e) f − φ/e。Vₛ 对 f 作图是一条直线，斜率为 h/e，截
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