Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

In AQA A-level Physics, case study questions go beyond simple recall. You are typically given an unfamiliar scenario and must apply your knowledge of mechanics, fields, nuclear physics, waves or materials to analyse data, make predictions and evaluate conclusions. This article walks you through eight extended case studies that mirror the style of synoptic exam questions, each targeting key Year 13 content. You will see how to break down a problem, identify the relevant principles, carry out calculations and comment on uncertainties. Working through these examples will strengthen your ability to tackle the most demanding sections of Paper 2 and Paper 3.

在 AQA A-level 物理中，案例分析题远超简单的回忆。你会被给到一个陌生情境，并需要运用力学、场论、核物理、波或材料的知识来分析数据、做出预测和评价结论。本文带领你走过八个拓展案例，每个都针对关键的 Year 13 内容，并模拟综合性真题风格。你将学会如何拆解问题、确定相关原理、进行计算并评论不确定度。吃透这些实例，能极大提升你应对卷2和卷3高难度部分的能力。

1. Understanding the Case Study Approach | 理解案例分析题型

Case studies in the AQA specification are designed to assess your ability to think like a physicist. They often combine two or more topic areas, require you to extract information from text, graphs or tables, and expect you to justify every step. The first action is always to list the quantities given and the quantity you need to find, converting any prefixed units to SI. Next, identify the underlying physical model, for example a uniform field, ideal gas, simple harmonic oscillator or standard model interaction. Write down the relevant equation(s) in symbolic form before substituting numbers. Finally, check that your final answer has the correct units, a sensible number of significant figures and fits the physical context.

AQA 考试大纲中的案例分析旨在评估你像物理学家一样思考的能力。它常结合两个或多个主题领域，要求你从文字、图表或表格中提取信息，并期望你每步都能论证。第一步永远是列出已知量和待求量，将任何带词头的单位换算为国际单位制。接着，确定背后的物理模型，比如匀强场、理想气体、简谐振子或标准模型相互作用。写出符号形式的方程再进行数值代入。最后，核验最终答案是否单位正确、有效数字合理且符合物理情境。

2. Case Study 1: Satellite Orbits and Gravitational Fields | 案例1：卫星轨道与引力场

A telecommunications company plans to place a satellite in a geostationary orbit above the equator. The data sheet gives the mass of Earth M = 5.97 × 10²⁴ kg, the universal gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻² and the length of a sidereal day T = 8.64 × 10⁴ s. The task is to show that the required orbital radius is about 4.2 × 10⁷ m and to calculate the gravitational potential at this altitude.

某通信公司计划将一颗卫星送入赤道上空的地球静止轨道。数据表给出地球质量 M = 5.97 × 10²⁴ kg，万有引力常数 G = 6.67 × 10⁻¹¹ N m² kg⁻²，恒星日时长 T = 8.64 × 10⁴ s。任务是证明所需轨道半径约为 4.2 × 10⁷ m，并计算该高度处的引力势。

For a circular orbit, the centripetal force is provided by gravity: GmM/r² = mω²r. Since ω = 2π/T, substituting and rearranging gives r³ = GMT² / 4π².

对圆轨道，向心力由引力提供：GmM/r² = mω²r。利用 ω = 2π/T 代入整理得 r³ = GMT² / 4π²。

r³ = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (8.64 × 10⁴)²) / (4π²) ≈ 7.54 × 10²² m³

r = ∛(7.54 × 10²²) ≈ 4.22 × 10⁷ m

This confirms the stated radius. Now gravitational potential V = -GM/r. Using r = 4.22 × 10⁷ m gives V ≈ -9.44 × 10⁶ J kg⁻¹. Because the potential is negative, work must be done to move the satellite to a higher orbit. The company must also consider that a small error in injection speed would cause the satellite to drift, showing the importance of sensitivity analysis in the case study.

这确认了给出的半径。现在引力势 V = -GM/r，代入 r = 4.22 × 10⁷ m 得 V ≈ -9.44 × 10⁶ J kg⁻¹。因势能是负的，要将卫星送入更高轨道需做功。公司还需考虑注入速度上微小的误差就会导致卫星漂移，显示了案例分析中灵敏度分析的重要性。

3. Case Study 2: Charged Particle in Electric and Magnetic Fields | 案例2：电场与磁场中的带电粒子

A mass spectrometer uses a velocity selector with crossed electric and magnetic fields. Protons enter a region where E = 2.5 × 10⁴ V m⁻¹ and B = 0.15 T. Only particles with speed v = E/B pass undeflected. These then enter a semicircular deflection chamber with the same magnetic field B and follow a path of radius r. The charge-to-mass ratio e/m for a proton is 9.58 × 10⁷ C kg⁻¹.

一台质谱仪使用交叉电场和磁场的速度选择器。质子进入一个 E = 2.5 × 10⁴ V m⁻¹ 且 B = 0.15 T 的区域。只有速度 v = E/B 的粒子不偏转。然后它们进入具有相同磁场 B 的半圆形偏转室，运行半径为 r。质子的荷质比 e/m = 9.58 × 10⁷ C kg⁻¹。

First calculate the speed: v = 2.5 × 10⁴ / 0.15 = 1.67 × 10⁵ m s⁻¹. In the deflection chamber the magnetic force provides the centripetal force: Bev = mv²/r, which rearranges to r = mv / (Be). Since e/m is known, r = v / (B × (e/m)).

首先计算速率：v = 2.5 × 10⁴ / 0.15 = 1.67 × 10⁵ m s⁻¹。在偏转室中磁力提供向心力：Bev = mv²/r，整理得 r = mv/(Be)。因已知 e/m，r = v / (B × (e/m))。

r = 1.67 × 10⁵ / (0.15 × 9.58 × 10⁷) ≈ 1.16 × 10⁻² m (1.16 cm)

Next, evaluate what would happen if the electric field fluctuated by ±2%. The exit speed would vary, causing a spread in radius and reducing mass resolution. A table of v and r for E at 2.45, 2.50 and 2.55 × 10⁴ V m⁻¹ shows r changing by about 0.5 mm. This highlights that temperature stability in the magnet coils is crucial. The examiner expects you to discuss systematic uncertainty and suggest improvements such as using a Hall probe to monitor B.

接下来，评估如果电场波动 ±2% 会发生什么。出射速度会变化，导致半径分散并降低质量分辨率。一个显示 E 为 2.45、2.50 和 2.55 × 10⁴ V m⁻¹ 时 v 和 r 的表格表明 r 变化约 0.5 mm。这凸显了磁体线圈温度稳定的关键性。考官希望你讨论系统不确定度并提出改进，例如用霍尔探头监测 B。

4. Case Study 3: Radioactive Decay and Half-life | 案例3：放射性衰变与半衰期

A hospital receives a sample of technetium-99m, which decays by gamma emission with a half-life of 6.01 hours. The activity at 8:00 am is 800 MBq. A scan requires an activity of at least 150 MBq. The medical physicist must decide the latest time the sample can be used.

一家医院收到一批锝-99m 样品，它通过伽马辐射衰变，半衰期为 6.01 小时。早 8:00 的活度为 800 MBq。一次扫描需要至少 150 MBq 的活度。医学物理师必须决定样品可被使用的最迟时间。

Use the decay law A = A₀ e^{-λt}. The decay constant λ = ln2 / T_½ = 0.693 / 6.01 = 0.1153 h⁻¹. Set A = 150 = 800 e^{-0.1153 t}. Taking natural logs: ln(150/800) = -0.1153 t → t = ln(800/150) / 0.1153 = ln(5.333) / 0.1153 ≈ 1.674 / 0.1153 ≈ 14.5 hours. So the deadline is 10:30 pm.

使用衰变定律 A = A₀ e^{-λt}。衰变常数 λ = ln2 / T_½ = 0.693 / 6.01 = 0.1153 h⁻¹。设 A = 150 = 800 e^{-0.1153 t}。两边取自然对数：ln(150/800) = -0.1153 t → t = ln(800/150) / 0.1153 = ln(5.333) / 0.1153 ≈ 1.674 / 0.1153 ≈ 14.5 小时。因此截止时间是晚 10:30。

But the decay is a random process; the measured activity follows a Poisson distribution. With an initial count rate of, say, 800 counts per second, the standard uncertainty is √800 ≈ 28 counts s⁻¹. Propagating this through the exponential gives an uncertainty of roughly ±0.3 hours. Therefore, to guarantee a 95% confidence level the sample should be used by 10:00 pm. The case study illustrates why hospitals build in a safety margin and routinely recalibrate.

然而衰变是随机过程；测得的活度遵循泊松分布。假设计数率初始为每秒 800 次，标准不确定度为 √800 ≈ 28 s⁻¹。将其传播到指数函数中给出约 ±0.3 小时的不确定度。因此为确保 95% 置信水平，样品应在晚 10:00 前使用。该案例说明医院为何要设置安全余量并定期重新校准。

5. Case Study 4: Simple Harmonic Motion and Pendulum | 案例4：简谐运动与摆

A student investigates a simple pendulum of length L = 0.980 m and claims its period T is exactly 2.00 s. She uses a light gate to record the time for ten oscillations, finding 10T = 19.87 s, 19.92 s and 19.90 s. The local gravitational acceleration g is to be determined and compared with the theoretical value 9.81 m s⁻².

一位学生研究一个长 L = 0.980 m 的单摆，声称其周期 T 恰为 2.00 s。她用光闸记录十次振荡的时间，得到 10T = 19.87 s, 19.92 s 和 19.90 s。要求测定当地重力加速度 g 并与理论值 9.81 m s⁻² 比较。

The mean period T = (1.987 + 1.992 + 1.990)/3 = 1.9897 s. The uncertainty in the mean can be estimated by half the range: (1.992 – 1.987)/2 = 0.0025 s. For small angles, T = 2π √(L/g), so g = 4π²L / T².

平均周期 T = (1.987 + 1.992 + 1.990)/3 = 1.9897 s。平均值的不确定度可用半极差估计：(1.992 – 1.987)/2 = 0.0025 s。对小角度，T = 2π √(L/g)，故 g = 4π²L / T²。

g = 4π² × 0.980 / (1.9897²) ≈ 39.48 / 3.959 ≈ 9.97 m s⁻²

The value is higher than 9.81, possibly due to the approximation sinθ ≈ θ failing if the initial amplitude exceeds about 10°. Indeed, if the amplitude was 15°, a correction factor of (1 + θ²/16) increases the period, making g appear smaller if uncorrected; but here g is larger, so systematic error likely came from a short ruler used to measure L, which added a constant 2 mm positive offset. The student should measure L from the pivot to the centre of mass with a metre ruler holding a set square and repeat with a longer pendulum to reduce percentage uncertainty.

该值高于 9.81，可能原因是如果初始摆幅超过约 10°，近似 sinθ ≈ θ 不成立。实际上，若振幅为 15°，修正因子 (1 + θ²/16) 会使周期变大，若不修正则 g 会被低估；但这里 g 更大，因此系统误差很可能来自测量 L 的短尺，它引入了 2 mm 的正向偏移。学生应用米尺和三角板测量从悬点到质心的距离，并用更长的摆重复实验以减小百分比不确定度。

6. Case Study 5: Capacitor Discharge and Time Constant | 案例5：电容器放电与时间常数

A 2200 μF capacitor is charged to 6.00 V and discharged through a 4.7 kΩ resistor. The voltage across the capacitor is logged every 5 seconds. The technician needs to know the time constant τ and verify if the capacitance is within its ±10% tolerance.

一只 2200 μF 电容器充电至 6.00 V，并通过 4.7 kΩ 电阻放电。电容器两端电压每 5 秒记录一次。技术员需知道时间常数 τ 并验证电容是否在 ±10% 容差内。

The theoretical time constant τ = RC = 4700 × 2200 × 10⁻⁶ = 10.34 s. The discharge equation is V = V₀ e^{-t/RC}. Taking natural logs gives ln V = ln V₀ – t/RC. Plotting ln V against t yields a straight line with gradient -1/RC.

理论时间常数 τ = RC = 4700 × 2200 × 10⁻⁶ = 10.34 s。放电方程为 V = V₀ e^{-t/RC}。取自然对数得 ln V = ln V₀ – t/RC。绘出 ln V 对 t 的图线，可得一条斜率为 -1/RC 的直线。

t / s	V / V	ln(V)
0	6.00	1.792
5	3.63	1.289
10	2.20	0.788
15	1.33	0.285
20	0.81	-0.211

The gradient from the first and last points is (−0.211 − 1.792) / (20 − 0) = −0.1002 s⁻¹. Hence RC = 1/0.1002 = 9.98 s. The experimental capacitance C_exp = τ / R = 9.98 / 4700 = 2.12 × 10⁻³ F = 2120 μF. The difference from the labelled value is (2120 − 2200)/2200 = −3.6%, well within tolerance. However, the voltmeter might have a non-negligible internal resistance, which would make the measured τ smaller. To improve, the student should use a high-impedance data logger and include the ESR of the capacitor in the model.

从首末两点求斜率：(−0.211 − 1.792) / (20 − 0) = −0.1002 s⁻¹。因此 RC = 1/0.1002 = 9.98 s。实验电容值 C_exp = τ / R = 9.98 / 4700 = 2.12 × 10⁻³ F = 2120 μF。与标称值的差异为 (2120 − 2200)/2200 = −3.6%，在容差内。但电压表可能有不可忽略的内阻，会使测得的 τ 偏小。要改善，学生应使用高阻抗数据记录仪，并在模型中纳入电容的等效串联电阻（ESR）。

7. Case Study 6: Nuclear Fission and Binding Energy | 案例6：核裂变与结合能

One possible fission reaction of uranium-235 is: n + ²³⁵U → ¹⁴¹Ba + ⁹²Kr + 3n. The masses are: m(²³⁵U) = 235.0439 u, m(¹⁴¹Ba) = 140.9144 u, m(⁹²Kr) = 91.9262 u, m(n) = 1.00866 u. 1 u is equivalent to 931.5 MeV. The case study asks you to calculate the energy released per fission and to suggest why the fragments are radioactive.

铀-235 的一种可能裂变反应为：n + ²³⁵U → ¹⁴¹Ba + ⁹²Kr + 3n。质量分别为：m(²³⁵U) = 235.0439 u，m(¹⁴¹Ba) = 140.9144 u，m(⁹²Kr) = 91.9262 u，m(n) = 1.00866 u。1 u 相当于 931.5 MeV。该案例要求计算每次裂变释放的能量，并解释为何碎块具有放射性。

Left-hand side mass = 235.0439 + 1.00866 = 236.05256 u. Right-hand side mass = 140.9144 + 91.9262 + 3×1.00866 = 140.9144 + 91.9262 + 3.02598 = 235.86658 u. Mass defect Δm = 236.05256 − 235.86658 = 0.18598 u. Energy released = 0.18598 × 931.5 ≈ 173.2 MeV.

左边质量 = 235.0439 + 1.00866 = 236.05256 u。右边质量 = 140.9144 + 91.9262 + 3 × 1.00866 = 235.86658 u。质量亏损 Δm = 236.05256 − 235.86658 = 0.18598 u。释放能量 = 0.18598 × 931.5 ≈ 173.2 MeV。

This matches typical values. The barium and krypton isotopes are neutron-rich and lie above the stability belt, so they undergo beta-minus decay. A follow-up calculation can find the total power if the sample undergoes 2.5 × 10¹⁰ fissions per second: P = 173.2 MeV × 1.602 × 10⁻¹³ J/MeV × 2.5 × 10¹⁰ ≈ 693 W. In a reactor, energy is absorbed by a moderator and coolant, emphasising the need for thermal management. The case study therefore bridges nuclear physics and engineering.

这与典型值相符。钡和氪同位素富含中子且处于稳定带上方，因此发生β⁻衰变。后续计算可求得若样品每秒发生 2.5 × 10¹⁰ 次裂变，总功率 P = 173.2 MeV × 1.602 × 10⁻¹³ J/MeV × 2.5 × 10¹⁰ ≈ 693 W。在反应堆中，能量被慢化剂和冷却剂吸收，凸显热管理的必要性。该案例因此搭接了核物理与工程学。

8. Case Study 7: Thermistor and Resistivity | 案例7：热敏电阻与电阻率

A thermistor is calibrated to monitor the temperature in a solar panel. The relationship is R = R₀ exp(B/T), where B = 3500 K and R₀ = 0.025 Ω. The resistance is measured with a digital multimeter giving values of 85.0 Ω, 84.7 Ω and 85.3 Ω over 30 seconds. The ambient temperature needs to be determined.

一只热敏电阻被校准以监测太阳能板中的温度。其关系为 R = R₀ exp(B/T)，其中 B = 3500 K，R₀ = 0.025 Ω。用数字万用表在 30 秒内测得电阻值为 85.0 Ω，84.7 Ω 和 85.3 Ω。需确定环境温度。

First find mean resistance: R_mean = 85.0 Ω. Then T = B / ln(R/R₀). Substituting: T = 3500 / ln(85.0 / 0.025) = 3500 / ln(3400) = 3500 / 8.132 ≈ 430 K. That is 157 °C, too high for ambient air, suggesting the solar panel is under intense illumination and the thermistor is self-heating. The case prompts an evaluation of measurement uncertainty: with a standard deviation of only 0.3 Ω, random errors are small, but systematic error from self-heating dominates. The cure is to use a lower current or to measure the resistance with a pulsed method.

先求平均电阻：R_mean = 85.0 Ω。然后 T = B / ln(R/R₀)。代入：T = 3500 / ln(85.0 / 0.025) = 3500 / ln(3400) = 3500 / 8.132 ≈ 430 K。这是 157 °C，对周围空气而言过高，表明太阳能板正受强光照射且热敏电阻自身加热。该案例引发对测量不确定度的评价：标准差仅为 0.3 Ω，随机误差很小，但自热导致的系统误差占主导。解决方法是用更小电流或用脉冲法测电阻。

9. Case Study 8: Diffraction Grating and Wavelength Measurement | 案例8：衍射光栅与波长测定

A laser of unknown wavelength illuminates a diffraction grating with 300 lines per mm. A screen is placed 1.50 m away. The distance between the two second-order bright spots is measured as 84.6 cm ± 0.4 cm. Determine the wavelength, estimate its uncertainty and comment on the suitability of the grating.

一未知波长的激光照射每毫米 300 条线的衍射光栅。屏距 1.50 m。测得两第二级亮斑之间的距离为 84.6 cm ± 0.4 cm。求波长，估算不确定度并评论光栅的适用性。

Grating spacing d = 1 × 10⁻³ m / 300 = 3.333 × 10⁻⁶ m. For second order (n=2), the angle θ satisfies d sinθ = nλ. The distance from central maximum to one second-order spot is y = 42.3 cm = 0.423 m. tanθ = y / D = 0.423 / 1.50 = 0.282, so θ = arctan(0.282) ≈ 15.8°. Then sinθ ≈ 0.272. Hence λ = d sinθ / 2 = (3.333 × 10⁻⁶ × 0.272) / 2 = 4.53 × 10⁻⁷ m = 453 nm.

光栅间距 d = 1 × 10⁻³ m / 300 = 3.333 × 10⁻⁶ m。对第二级 (n=2)，角 θ 满足 d sinθ = nλ。从中央极大到一个第二级光斑的距离 y = 42.3 cm = 0.423 m。tanθ = y / D = 0.423 / 1.50 = 0.282，故 θ = arctan(0.282) ≈ 15.8°。则 sinθ ≈ 0.272。于是 λ = d sinθ / 2 = (3.333 × 10⁻⁶ × 0.272) / 2 = 4.53 × 10⁻⁷ m = 453 nm。

The absolute uncertainty in y is 0.2 cm, so the percentage uncertainty is 0.2/42.3 ≈ 0.47%. That translates to approximately ±2 nm in λ. However, the grating equation assumes normal incidence; a slight tilt could shift the pattern. Moreover, with only 300 lines mm⁻¹, higher orders may be dim, limiting precision. The student could improve by using a finer grating and measuring several orders to plot sinθ against n, where the gradient is λ/d.

y 的绝对不确定度为 0.2 cm，百分比不确定度为 0.2/42.3 ≈ 0.47%。这造成 λ 约 ±2 nm 的不确定度。但光栅方程假设垂直入射；轻微倾斜会移动图样。此外，仅 300 条 mm⁻¹，高级次可能很暗，限制了精度。学生可改用更密的光栅，并测量多级次，作 sinθ 对 n 的图线，斜率为 λ/d。

10. Bringing It All Together | 融会贯通

These eight case studies demonstrate a pattern: read, list, model, calculate, evaluate. In every case, the physics stays firmly rooted in the AQA data sheet equations, but the context demands you think about assumptions, limitations and experimental design. Practise by rewriting each study as a series of exam-style bullet points and then explaining your reasoning to a peer. Remember that quality of written communication matters; use correct terminology and structure your answers logically. When you encounter a novel situation in the exam, stay calm and follow the same methodical approach. The ability to link concepts across forces, fields, waves and particles is exactly what the AQA examiners are testing.

以上八个案例展示了一个模式：读题、列出已知量、建立模型、计算、评估。每种情况下，物理都牢牢扎根于 AQA 数据表里的公式，但情景要求你思考假设、限制和实验设计。通过将每个研究改写为一串考试风格的要点，并向同伴解释推理来练习。记住书面表达的质量很重要；使用正确的术语并在逻辑上组织答案。当在考试中遇到新情境时，保持冷静，按照同样的系统方法处理。横跨力、场、波和粒子的概念衔接能力，正是 AQA 考官要检验的。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Case Study Practical Exercises | 案例分析实战演练