📚 Year 11 AQA Physics: Case Study Practice | Year 11 AQA 物理：案例分析实战演练
In AQA GCSE Physics, applying your knowledge to real-world scenarios is a key skill tested in exams. Case study questions require you to identify the relevant physics, carry out calculations, analyse data, and evaluate outcomes. This article walks you through four practical case studies covering stopping distances, domestic electricity, thermal insulation, and radioactivity. Each section builds your confidence in tackling structured questions, using equations correctly, and justifying your answers with clear scientific reasoning.
在 AQA 物理考试中，将知识应用于真实场景是一项关键技能。案例分析题要求你识别相关物理原理、进行计算、分析数据并评估结果。本文将通过四个实用案例，涵盖停止距离、家庭电路、保温隔热和放射性，带你逐步分析。每个部分都能帮助你提升解题信心，熟练运用公式，并用清晰的科学推理证明你的答案。
1. Understanding Stopping Distances | 理解停止距离
t is made up of two parts: thinking distance and braking distance. Thinking distance = speed × reaction time. Braking distance depends on the car's speed, mass, and the braking force applied.
汽车的停止距离是指从驾驶员发现危险到车辆完全停下的总行驶距离。它由两部分组成：思考距离和制动距离。思考距离 = 速度 × 反应时间。制动距离则取决于车速、质量和施加的制动力。
When the brakes are applied, the friction between the brake pads and discs does work to reduce the car's kinetic energy to zero. The work done by the braking force equals the initial kinetic energy of the car, so we can write: braking force × braking distance = ½mv². This relationship allows us to calculate the braking distance if we know the force and the car's initial speed.
踩下刹车时，刹车片与刹车盘之间的摩擦力做功，将汽车的动能降为零。制动力做的功等于汽车的初始动能，因此可表示为：制动力 × 制动距离 = ½mv²。通过这个关系，只要知道制动力和初始速度，就能计算出制动距离。
Factors that increase thinking distance include tiredness, alcohol, drugs, and distractions – they all lengthen reaction time. Braking distance is affected by worn tyres, wet or icy roads, and faulty brakes, all of which reduce the maximum braking force. Higher speed increases both thinking and braking distance dramatically.
增加思考距离的因素包括疲劳、饮酒、药物和分心等，这些都会延长反应时间。影响制动距离的因素有磨损的轮胎、湿滑或结冰路面以及刹车故障，这些都会减小最大制动力。速度的增加会显著增大思考距离和制动距离。
2. Calculating Braking Distance | 计算制动距离
Let's work through a typical exam-style case. A car of mass 1200 kg is travelling at 30 m/s. The brakes can apply a constant braking force of 6000 N. Calculate the braking distance.
我们来看一道典型的考试题。一辆质量为 1200 kg 的汽车正以 30 m/s 的速度行驶。刹车能施加 6000 N 的恒定制动力。计算制动距离。
ranging gives d = (½mv²) / F. Substitute the numbers: m = 1200 kg, v = 30 m/s, F = 6000 N.
首先，使功与动能相等：F × d = ½mv²。整理得 d = (½mv²) / F。代入数值：m = 1200 kg, v = 30 m/s, F = 6000 N。
d = (0.5 × 1200 × 30²) ÷ 6000 = (600 × 900) ÷ 6000 = 540 000 ÷ 6000 = 90 m
ed × time = 30 × 0.7 = 21 m. So the total stopping distance = 90 m + 21 m = 111 m. Always show your working step by step.
制动距离为 90 米。在实际情况中，驾驶员还会行驶一段思考距离。若反应时间为 0.7 s，则思考距离 = 30 × 0.7 = 21 m。因此总停止距离 = 90 m + 21 m = 111 m。务必按步骤展示计算过程。
3. Evaluating Factors that Affect Stopping | 评估影响停止距离的因素
If the car' d. Using the same braking force of 6000 N, the new braking distance would be (0.5 × 1200 × 60²) ÷ 6000 = 360 m. This illustrates why high-speed crashes are so severe.
若车速加倍至 60 m/s，制动距离将变为原来的四倍，因为动能（½mv²）与速度的平方成正比。使用同样的制动力 6000 N，新的制动距离为 (0.5 × 1200 × 60²) ÷ 6000 = 360 m。这表明为什么高速碰撞后果如此严重。
Road conditions also play a critical role. A wet surface can halve the friction between tyres and road, reducing the maximum braking force. If the braking force dropped to 3000 N at 30 m/s, the braking distance would double to 180 m. The table below compares typical stopping distances under different conditions.
路况也起着关键作用。潮湿路面可能使轮胎与路面的摩擦力减半，从而降低最大制动力。如果以 30 m/s 行驶时制动力降至 3000 N，制动距离将翻倍至 180 m。下表比较了不同条件下的典型停止距离。
| Speed (m/s) | Braking Force (N) | Braking Distance (m) | Thinking + Braking (m)* |
| 30 | 6000 | 90 | 111 |
| 30 | 3000 | 180 | 201 |
| 60 | 6000 | 360 | 402 |
* assuming reaction time of 0.7 s
A responsible driver should therefore leave a larger gap when driving fast or in poor weather. In exams, you may be asked to suggest reasons for a longer stopping distance and link them back to the equations.
因此，负责任的驾驶员在高速行驶或恶劣天气下应保持更大的车距。在考试中，可能会要求你解释停止距离变长的原因，并联系公式作答。
4. Diagnosing Electrical Faults | 诊断电路故障
Domestic appliances can stop working for several reasons. Consider a 2000 W hairdryer connected to the UK mains supply (230 V). If it fails to turn on, possible faults include a blown fuse in the plug, a tripped circuit breaker, or a broken heating element.
家用电器可能因多种原因停止工作。假设一个 2000 W 的电吹风连接到英国市电（230 V），如果无法启动，可能的故障包括插头保险丝熔断、断路器跳闸或加热元件断裂。
30 V ≈ 8.7 A. A 3 A fuse would blow under this current, so the correct fuse rating should be 13 A. If a 3 A fuse were installed by mistake, replacing it with the appropriate 13 A fuse would solve the problem.
为诊断故障，电工首先会检查保险丝。电吹风正常工作电流用 I = P ÷ V = 2000 W ÷ 230 V ≈ 8.7 A。若使用 3 A 保险丝，会在此电流下熔断，因此正确保险丝额定值应为 13 A。如果误装了 3 A 保险丝，更换为合适的 13 A 保险丝即可解决问题。
If the fuse is intact, the heating element might have broken. A continuity test using a multimeter can check for an open circuit. In a series circuit, if the element breaks, no current flows and the hairdryer remains cold.
如果保险丝完好，加热元件可能断裂。用万用表进行通断测试可以检查是否断路。在串联电路中，若元件断裂，则无电流通过，电吹风不会发热。
5. Selecting Fuses and Power Calculations | 选择保险丝与功率计算
The fuse in a plug protects the appliance and the cable from overheating by melting and breaking the circuit if the current exceeds a safe limit. Standard fuse ratings in the UK are 3 A, 5 A, and 13 A. The chosen fuse should be just above the normal operating current of the device.
插头中的保险丝通过熔断来保护电器和电线，在电流超过安全极限时断开电路。英国的标准保险丝额定值有 3 A、5 A 和 13 A。所选保险丝的额定值应略高于电器的正常工作电流。
26 A, so a 3 A fuse is appropriate. Using a fuse rating that is too high (e.g. 13 A for a lamp) would not protect against small overloads and could cause a fire if a fault develops.
对于一个 1000 W 的微波炉（230 V），I = 1000 ÷ 230 ≈ 4.35 A。略高于此的标准保险丝是 5 A，因此应选用 5 A 保险丝。对于一个 60 W 台灯，I = 60 ÷ 230 ≈ 0.26 A，使用 3 A 保险丝合适。如果使用额定值过高的保险丝（如台灯用 13 A），则无法防范较小的过载，一旦发生故障可能引发火灾。
R shows that high currents cause significant heating in the cables (I²R losses). Thick wires and correct fuses prevent overheating and maintain safety.
此外，功率公式 P = I²R 表明大电流会在电线中引起显著发热（I²R 损耗）。使用粗导线和正确保险丝能防止过热并保证安全。
6. Electrical Safety in the Home | 家庭用电安全
Many appliances have metal cases. If a live wire touches the metal case due to a fault, the case becomes live and can deliver a fatal electric shock. The earth wire provides a low-resistance path to the ground. This causes a large current to flow, melting the fuse or tripping the circuit breaker instantly, and disconnecting the appliance.
许多电器有金属外壳。如果因故障使火线接触金属外壳，外壳就会带电，可能导致致命触电。接地线提供了一条通往大地的低电阻通路，这会引发大电流，迅速熔断保险丝或使断路器跳闸，从而断开电器。
Double-insulated appliances, marked with a square-within-a-square symbol, have a plastic casing that cannot become live. They do not require an earth wire. A hairdryer with a plastic body is often double-insulated. In case studies, you may be asked to identify whether an appliance needs earthing and to explain the role of the earth wire.
带有“回”字标志的双层绝缘电器具有不会带电的塑料外壳，不需要接地线。常见的塑料壳电吹风通常是双层绝缘的。在案例题中，可能会要求你判断电器是否需要接地，并解释接地线的作用。
7. Designing a Thermal Insulation Experiment | 设计保温实验
To compare the insulation properties of different materials, a classic experiment involves wrapping identical containers of hot water with the materials and measuring the temperature drop over time. Key variables to control include the initial water temperature, the volume of water, the thickness of the insulating layer, and the room temperature.
为了比较不同材料的保温性能，经典实验是用这些材料包裹相同装有热水的容器，并测量温度随时间下降的情况。需控制的关键变量包括初始水温、水量、保温层厚度和室温。
The independent variable is the type of insulating material (e.g. foam, cotton wool, aluminium foil, vacuum flask). The dependent variable is the temperature of the water, recorded every minute for 20 minutes. A control setup with no insulation provides a baseline. Using a thermometer with 0.5 °C precision and ensuring the lids are sealed helps improve accuracy.
自变量是保温材料的种类（如泡沫、棉絮、铝箔、真空瓶）。因变量是水的温度，每 1 分钟记录一次，持续 20 分钟。一个不加保温层的对照组可提供参照。使用精度为 0.5 °C 的温度计并确保盖子密封，可提高准确性。
A table should be drawn up with columns for time, temperature for material A, material B, etc. This allows a direct comparison of cooling rates.
应绘制一个表格，包含时间、材料 A 的温度、材料 B 的温度等列，以便直接比较冷却速率。
8. Analysing Cooling Data | 分析冷却数据
eat capacity of the water are known. For 100 g of water (c = 4200 J/(kg °C)), the heat energy lost over a time interval is Q = m c Δθ.
记录温度后，即可绘制每种材料的冷却曲线。若已知水的质量和比热容，还可计算热量散失速率。对于 100 g 水（c = 4200 J/(kg°C)），在一段时间内散失的热量 Q = m c Δθ。
J/(kg°C) × 15°C = 6300 J. The average power loss = 6300 J ÷ 600 s = 10.5 W. Comparing this with foam insulation, which might lose only 4200 J (7 W) over the same period, shows that foam is a better insulator.
例如，若棉絮包裹的水在 10 分钟（600 s）内从 80 °C 降至 65 °C，温度变化 Δθ = 15 °C。散失热量 Q = 0.1 kg × 4200 J/(kg°C) × 15°C = 6300 J。平均散热功率 = 6300 J ÷ 600 s = 10.5 W。与之相比，泡沫保温层在相同时间内可能仅散失 4200 J（7 W），这表明泡沫是更好的保温材料。
Plotting the temperature against time reveals the shapes of the curves: a steeper line indicates a faster rate of cooling. The data also allows you to predict the temperature after a longer period by extrapolating the graph.
绘制温度-时间图可显示曲线形状：斜率越陡，冷却速率越快。通过延长曲线还能
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