📚 Year 10 CIE Physics Case Study Practice | 10年级CIE物理案例分析实战演练
Case studies are an integral part of CIE IGCSE Physics, testing your ability to apply theoretical knowledge to real-world situations. This article provides hands-on practice with nine carefully selected scenarios covering mechanics, energy, thermal physics, waves, electricity, and nuclear physics. Each case study breaks down the problem into clear steps, helping you build confidence in quantitative analysis and logical reasoning.
案例分析是CIE IGCSE物理的重要组成部分，考察你将理论知识应用于实际情境的能力。本文提供了九个精心挑选的实战演练，涵盖力学、能量、热物理、波、电学和核物理。每个案例将问题分解为清晰的步骤，帮助你在定量分析和逻辑推理方面建立信心。
1. Case Study: Stopping Distance | 案例研究：刹车距离
A car is travelling at a constant 72 km/h (which is 20 m/s) on a dry road. The driver notices an obstacle ahead and takes 0.7 s to react before hitting the brakes. The brakes produce a steady deceleration of 6 m/s². Find the total distance the car travels from the moment the driver sees the obstacle until the car comes to rest.
一辆汽车以72 km/h（即20 m/s）的恒定速度在干燥路面上行驶。驾驶员发现前方有障碍物，经过0.7秒的反应时间后才踩下刹车。刹车产生6 m/s²的恒定减速度。求从驾驶员看到障碍物到汽车静止所经过的总距离。
Total stopping distance is made up of thinking distance and braking distance. During the reaction time, the car continues at constant speed, so thinking distance (dthink) = speed × reaction time = 20 m/s × 0.7 s = 14 m.
总停车距离由反应距离和刹车距离组成。在反应时间内，汽车以原速度匀速行驶，因此反应距离(dthink) = 速度 × 反应时间 = 20 m/s × 0.7 s = 14 m。
m/s to v = 0 with a = -6 m/s². Using the equation of motion v² = u² + 2 a s, we can write:
0² = 20² + 2 × (-6) × s → 0 = 400 - 12 s → s = 400/12 ≈ 33.3 m
在刹车阶段，汽车从初速度u = 20 m/s匀减速到v = 0，加速度a = -6 m/s²。使用运动学方程v² = u² + 2 a s，可得：
0² = 20² + 2 × (-6) × s → 0 = 400 - 12 s → s = 400/12 ≈ 33.3 m
alved to 3 m/s², the braking distance would become 400/(2×3) = 66.7 m, increasing the total distance to 80.7 m. This highlights how critical speed and road conditions are for safety.
因此刹车距离约为33.3 m，总停车距离为14 m + 33.3 m = 47.3 m。如果路面湿滑，减速度降低一半至3 m/s²，刹车距离将变为400/(2×3) = 66.7 m，总距离增至80.7 m。这凸显了速度和路况对安全的关键影响。
2. Case Study: Solar Panel Efficiency | 案例研究：太阳能电池板效率
A solar panel with an area of 2 m² receives solar radiation of 1000 W/m². Under these conditions, the panel delivers an electrical output of 300 W. Calculate the efficiency of the panel. If the radiation intensity drops to 800 W/m² but the efficiency remains unchanged, what would the new output power be?
一块面积为2 m²的太阳能电池板接收1000 W/m²的太阳辐射。在这种条件下，电池板输出300 W的电力。计算电池板的效率。如果辐射强度降至800 W/m²但效率保持不变，新的输出功率为多少？
Efficiency = (useful power output) / (total power input) × 100% Total input power = 1000 W/m² × 2 m² = 2000 W. So efficiency = 300 W / 2000 W = 0.15 = 15%.
效率 = (有用输出功率) / (总输入功率) × 100%。总输入功率 = 1000 W/m² × 2 m² = 2000 W。因此效率 = 300 W / 2000 W = 0.15 = 15%。
rops to 800 W/m², the new input power = 800 × 2 = 1600 W. Since the efficiency stays at 15% temperature and angle of incidence, but this linear scaling is a good first approximation.
如果辐射强度降至800 W/m²，新的输入功率 = 800 × 2 = 1600 W。由于效率保持15%，输出功率 = 0.15 × 1600 = 240 W。实际上效率会随温度和入射角变化，但这种线性比例是一个很好的初步近似。
3. Case Study: Heating Water with an Electric Kettle | 案例研究：电热水壶加热水
An electric kettle rated at 2000 W is used to heat 0.5 kg of water from an initial temperature of 20 °C to the boiling point (100 °C). The specific heat capacity of water is 4200 J/(kg °C). Assuming no heat losses, calculate the energy required and the minimum time to bring the water to a boil.
一个额定功率为2000 W的电热水壶用来加热0.5 kg的水，使其从20 °C升至沸点100 °C。水的比热容为4200 J/(kg °C)。假设无热量损失，计算所需能量及将水加热至沸腾的最短时间。
where Δθ = 100 °C - 20 °C = 80 °C. Thus ΔE = 0.5 kg × 4200 J/(kg °C) × 80 °C = 168,000 J.
所需热量ΔE = m c Δθ，其中Δθ = 100 °C - 20 °C = 80 °C。因此ΔE = 0.5 kg × 4200 J/(kg °C) × 80 °C = 168,000 J。
some energy is lost to the surroundings, so the actual time will be longer. If the kettle's efficiency is only 80%, the useful power is 1600 W and the time becomes 168,000 / 1600 = 105 s.
由于电壶每秒提供2000 J的能量，理想加热时间t = ΔE / P = 168,000 J / 2000 W = 84 s。实际中部分能量散失到环境中，实际时间会更长。如果电壶效率仅为80%，有用功率为1600 W，则时间变为168,000 / 1600 = 105 s。
4. Case Study: Hydraulic Jack | 案例研究：液压千斤顶
A small piston of cross-sectional area 0.02 m² is pushed down with a force of 50 N. It is connected to a larger piston of area 0.5 m² by an incompressible fluid. Determine the upward force exerted by the large piston. The small piston moves down by 0.1 m; how far does the large piston rise? Published by TutorHao | Year 10 Physics Revision Series | aleveler.com
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