📚 Case Study Practice in Cambridge IGCSE Physics | 剑桥IGCSE物理案例分析实战演练
Case studies are a powerful way to apply physics concepts to real-world situations. In the Cambridge IGCSE Physics exam, you may be asked to analyse unfamiliar scenarios, interpret data, and solve problems by combining multiple topics. This article provides a series of practice case studies covering mechanics, waves, electricity, thermal physics, and nuclear physics. Each case includes a description, key questions, step-by-step reasoning, and model answers, helping you build confidence and exam technique.
案例分析是将物理概念应用于真实情境的有效方式。在剑桥IGCSE物理考试中，你可能需要分析不熟悉的场景、解读数据，并结合多个知识点解决问题。本文提供了一系列涵盖力学、波动、电学、热物理和核物理的练习案例。每个案例包括描述、关键问题、逐步推理和模型答案，帮助你建立信心并掌握应试技巧。
1. Car Braking Distance | 汽车制动距离
A driver travelling at 20 m/s sees a hazard and applies the brakes. The car decelerates uniformly at 5 m/s² until it stops. The driver's reaction time is 0.7 s before the brakes are applied.
一位司机以20 m/s的速度行驶，看到危险后踩下刹车。汽车以5 m/s²的匀减速直到停止。司机的反应时间（从看到危险到踩下刹车）为0.7秒。
(a) Calculate the thinking distance. (b) Calculate the braking distance. (c) Determine the total stopping distance. (d) Explain how the stopping distance would change if the road is wet.
(a) 计算反应距离。(b) 计算制动距离。(c) 求总停车距离。(d) 如果路面潮湿，说明停车距离会如何变化。
Step 1: Thinking distance dr = v × t = 20 m/s × 0.7 s = 14 m.
步骤1：反应距离 dr = v × t = 20 m/s × 0.7 s = 14 m。
speed u = 20 m/s, final speed v = 0, a = -5 m/s². 0 = 20² + 2×(-5)×s → s = 400/10 = 40 m.
步骤2：制动距离使用 v² = u² + 2as。初速度 u = 20 m/s，末速度 v = 0，加速度 a = -5 m/s²。0 = 20² + 2×(-5)×s → s = 400/10 = 40 m。
Step 3: Total stopping distance = 14 + 40 = 54 m.
步骤3：总停车距离 = 14 + 40 = 54 m。
Step 4: On wet roads, friction is reduced, so deceleration is smaller. This increases braking distance, therefore total stopping distance increases. Thinking distance unaffected.
步骤4：湿滑路面摩擦力减小，因此减速度变小，制动距离增加，总停车距离增加。反应距离不受影响。
2. Fuse Selection in a Household Circuit | 家用电路中的保险丝选择
A 230 V mains circuit supplies a 2 kW electric heater and a 100 W lamp in parallel. The fuse is rated at 13 A. Analyse whether the fuse is suitable.
一个230 V的市电电路，并联连接一台2 kW的电暖器和一盏100 W的灯泡。保险丝的额定电流为13 A。请分析该保险丝是否合适。
culate current for each appliance. For heater: P = IV → I = P/V = 2000 W / 230 V ≈ 8.70 A.
步骤1：计算每个电器的电流。电暖器：P = IV → I = P/V = 2000 W / 230 V ≈ 8.70 A。
Step 2: For lamp: I = 100 W / 230 V ≈ 0.43 A.
步骤2：灯泡：I = 100 W / 230 V ≈ 0.43 A。
13 A. This is less than 13 A, so fuse will not blow under normal operation. However, a fault causing short circuit could draw high current. The 13 A fuse is appropriate for normal load but provides protection against overload.
步骤3：并联电路中，总电流 = 各支路电流之和 = 8.70 + 0.43 = 9.13 A。该值小于13 A，正常工作下保险丝不会熔断。但发生短路的故障可能会产生大电流。13 A保险丝适合正常负载，并能提供过载保护。
3. Water Waves and Refraction | 水波与折射
Plane water waves travel from deep water into shallow water at an angle. The wave speed in deep water is 0.8 m/s, and in shallow water is 0.5 m/s. The angle of incidence is 45°.
平面水波从深水区域以一定角度进入浅水区域。深水波速为0.8 m/s，浅水波速为0.5 m/s，入射角为45°。
Use Snell's law for waves: sin
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