📚 Year 11 WJEC Physics: Case Study Practice Drill | WJEC 物理案例分析实战演练
In WJEC GCSE Physics, case study questions challenge you to apply scientific principles to unfamiliar, real-world situations. You must analyse data, perform calculations, evaluate evidence and communicate your reasoning clearly. This drill provides eight structured case studies covering forces, energy, electricity, waves, radioactivity and data handling, all designed to strengthen your skills for the examination.
在 WJEC GCSE 物理中，案例分析题考查你将科学原理应用于陌生的真实情境的能力。你需要分析数据、进行计算、评估证据，并清晰地表达推理过程。本次实战演练提供了八个结构化的案例，涵盖力、能量、电学、波、放射性以及数据处理，旨在加强你的应试技巧。
1. Decoding Case Study Questions | 解读案例分析题
Start by highlighting the command words such as 'calculate', 'explain' or 'evaluate'. Identify the key physical quantities given and the ones you need to find. Always link the scenario to the relevant physics topic – for example, a braking car relates to stopping distances and equations of motion.
首先圈出指令词，如“计算”、“解释”或“评估”。识别已知和待求的物理量。始终将情境与相关的物理主题联系起来——例如，汽车刹车涉及停车距离和运动学方程。
8 m/s² or the speed of sound. Show your working step by step, and check that units are consistent before substituting into formulae.
使用题目给出的数据，但也要准备回忆标准值，如 g = 9.8 米/秒² 或声速。逐步展示计算过程，并确保在代入公式前所有单位一致。
Finally, consider the limitations of the model or experiment. Is friction ignored? Could human reaction time vary? This evaluation is often required to reach the highest marks.
最后，考虑模型或实验的局限性。是否忽略了摩擦？人的反应时间可能变化吗？这类评估往往是获得高分的关键。
2. Case Study: Stopping Distance on a Wet Road | 案例：湿滑路面的刹车距离
A driver in a car travelling at 30 m/s sees a hazard. Her reaction time is 0.7 s. The braking distance on a wet road is measured as 75 m.
一位司机以 30 m/s 的速度行驶时发现危险。她的反应时间为 0.7 s，在湿滑路面上的刹车距离经测量为 75 m。
7 = 21 m. Total stopping distance = thinking distance + braking distance = 21 + 75 = 96 m.
计算反应距离：反应距离 = 速度 × 反应时间 = 30 × 0.7 = 21 m。总停车距离 = 反应距离 + 刹车距离 = 21 + 75 = 96 m。
king, use the equation v² = u² + 2as, where final velocity v = 0, u = 30 m/s and s = 75 m.
为求刹车过程中的减速度 a，使用公式 v² = u² + 2as，其中末速度 v = 0，u = 30 m/s，位移 s = 75 m。
0 = 30² + 2a × 75 → -900 = 150a → a = -6.0 m/s²
代入得 0 = 30² + 2a × 75，解得 a = -6.0 m/s²。负号表示减速。湿滑路面增大了刹车距离，因为轮胎与路面的摩擦力减小。
The negative sign indicates deceleration. A wet road increases braking distance because the friction between tyres and road surface is reduced. In dry conditions, the braking distance would be shorter and the deceleration larger in magnitude.
在干燥条件下，刹车距离会更短，减速度值更大。评估时还应提到反应时间会因疲劳或分心而变化，影响总停车距离。
3. Case Study: Skydiver's Terminal Velocity | 案例：跳伞者的终端速度
During the fall, air resistance increases with speed. At the instant her speed is 20 m/s, the air resistance acting on her is 200 N.
一位质量为 70 kg 的跳伞者从静止的飞机上跳下，取 g = 10 N/kg。下落过程中，空气阻力随速度增大。当她的速度为 20 m/s 时，空气阻力为 200 N。
70 × 10 = 700 N. Resultant force = weight - air resistance = 700 - 200 = 500 N downwards.
计算她的重力：重力 = m × g = 70 × 10 = 700 N。合力 = 重力 - 空气阻力 = 700 - 200 = 500 N，方向向下。
sultant force is still downward, she continues to accelerate, but the acceleration is less than g.
由 F = m × a 可得加速度 a = 500/70 ≈ 7.14 m/s²。由于合力仍然向下，她继续加速，但加速度小于 g。
Eventually, when air resistance equals her weight (700 N), the resultant force becomes zero and she reaches terminal velocity. At this stage, her acceleration is 0 m/s² and her velocity is constant. If her terminal velocity is 55 m/s, the air resistance at that speed is exactly 700 N.
最终，当空气阻力等于重力（700 N）时，合力为零，她达到终端速度。此时加速度为 0，速度恒定。若其终端速度为 55 m/s，那么该速度下的空气阻力恰好为 700 N。
4. Case Study: Fuse Selection for a Household Appliance | 案例：家用电器保险丝选择
An electric kettle is rated at 2.2 kW and is connected to the UK mains supply of 230 V. The kettle has a 3 A fuse fitted, but the manufacturer recommends checking the correct fuse rating.
一个电水壶额定功率为 2.2 kW，连接到 230 V 的英国家用电源。水壶原本装有 3 A 保险丝，但制造商建议检查正确的保险丝额定值。
er normal operation. Use the power equation P = I × V, so I = P / V = 2200 / 230 = 9.57 A.
首先计算水壶正常工作时的电流。使用功率公式 P = I × V，得 I = P/V = 2200/230 ≈ 9.57 A。
The value 9.57 A is significantly higher than the 3 A fuse rating, so the 3 A fuse would blow as soon as the kettle is switched on. The standard fuse ratings are 3 A, 5 A and 13 A. A 13 A fuse is the most appropriate choice here because it can safely carry the current without blowing under normal conditions, but will still melt if a fault causes a current much larger than 13 A.
9.57 A 远高于 3 A 保险丝的额定值，因此 3 A 保险丝在打开水壶时便会熔断。标准保险丝额定值包括 3 A、5 A 和 13 A。此处应选择 13 A 保险丝，因为它能安全承载正常工作电流，同时仍能在故障产生远大于 13 A 的电流时熔断。
Choosing a fuse with too low a rating interrupts normal function; too high a rating may fail to protect the appliance. Fuses are safety devices that prevent overheating and potential fires caused by excessive current.
选择额定值过低的保险丝会中断正常功能；额定值过高则可能无法保护电器。保险丝是安全装置，可防止因电流过大导致的过热和潜在的火灾。
5. Case Study: Radioactive Tracers in Medicine | 案例：医学中的放射性示踪剂
Iodine-131 is a radioactive isotope used to treat thyroid cancer. It decays by beta-minus emission with a half-life of 8 days. A patient is administered a sample with an initial activity of 40 MBq.
碘-131 是一种用于治疗甲状腺癌的放射性同位素，它通过 β⁻ 衰变释放射线，半衰期为 8 天。一名患者被给予初始活度为 40 MBq 的样品。
ter each half-life, the activity halves. Therefore activity = 40 × (½)² = 40 × ¼ = 10 MBq.
计算 16 天后的剩余活度。经过的半衰期个数 = 16/8 = 2。每经过一个半衰期，活度减半。因此活度 = 40 × (½)² = 10 MBq。
Iodine-131 is suitable because the body naturally absorbs iodine in the thyroid gland, so the radiation targets only the cancerous cells. Beta particles are moderately penetrating and can destroy the tumour while causing limited damage to surrounding tissue. The relatively short half-life ensures the radioactivity does not remain in the body for too long, reducing the risk of prolonged exposure.
碘-131 之所以合适，是因为身体会自然将碘吸收到甲状腺，因此放射线仅靶向癌细胞。β 粒子穿透力适中，能摧毁肿瘤且对周围组织损伤有限。较短的半衰期可确保放射性不会在体内存留过久，从而降低长时间暴露的风险。
When evaluating, consider that the actual biological clearance rate may differ, and the dose must be carefully calculated to balance therapeutic effect and safety.
评估时，应考虑实际生物清除率可能不同，且剂量需精确计算，以平衡疗效与安全。
6. Case Study: Energy Transfers on a Roller Coaster | 案例：过山车上的能量转换
A roller coaster carriage of total mass 800 kg is lifted to the top of the first hill, which is 40 m above the ground
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