📚 Case Study Practice for AQA GCSE Physics | AQA 物理案例分析实战演练
Case study questions challenge you to apply physics concepts to real-world situations. In the AQA GCSE exam, these questions often combine calculations, data analysis, and written explanations. This revision guide presents eight detailed case studies covering key topics from forces, electricity, energy, waves, and matter. For each scenario, you will work through the problem step by step and learn how to structure your answers to gain full marks. Use this practice to sharpen your skills and build confidence for the final exam.
案例分析题要求你将物理概念应用于实际情境。在 AQA GCSE 考试中，这类题目常结合计算、数据分析和文字解释。本复习指南呈现了八个详细案例，涵盖力、电、能量、波与物质等关键主题。每个场景都会引导你逐步分析问题，并学习如何组织答案以获得满分。通过练习，你将提升解题技巧，增强应考信心。
1. Stopping Distances – Urban Road Scenario | 停车距离 – 城市道路情景
A car is travelling at 13 m/s in a built-up area. The driver reacts to a child stepping into the road ahead. His reaction time is 0.7 seconds. The car can decelerate at 6.5 m/s². The distance to the child is 24 metres. Determine whether the driver can stop in time. Use the equations for thinking distance and braking distance. Take the final velocity as 0 m/s.
一辆汽车以 13 m/s 的速度在城区行驶。司机发现前方有儿童冲上马路并作出反应。他的反应时间为 0.7 秒。汽车可以以 6.5 m/s² 的减速度刹车。儿童距离车 24 米。请判断司机能否及时停车。使用思考距离和刹车距离的方程，取末速度为 0 m/s。
me = 13 m/s × 0.7 s = 9.1 m. This is the distance travelled before the brakes are applied.
首先计算思考距离：思考距离 = 速度 × 反应时间 = 13 m/s × 0.7 s = 9.1 m。这是踩下刹车前车辆行驶的距离。
Next, use the equation v² = u² + 2as to find the braking distance. Rearrange to s = (v² – u²) / (2a). Substituting u = 13 m/s, v = 0, a = –6.5 m/s² gives s = (0 – 169) / (–13) = 13 m.
然后使用方程 v² = u² + 2as 求刹车距离。变形为 s = (v² – u²) / (2a)。代入 u = 13 m/s，v = 0，a = –6.5 m/s² 得 s = (0 – 169) / (–13) = 13 m。
22.1 m is less than the 24 m available, the car stops just in time, avoiding a collision.
总停车距离 = 思考距离 + 刹车距离 = 9.1 m + 13 m = 22.1 m。由于 22.1 m 小于可用的 24 m，汽车恰好及时停下，避免碰撞。
Be careful to use a negative value for deceleration in the equation. A common mistake is forgetting to subtract the thinking distance before comparing with the obstacle distance.
注意在方程中要使用负值表示减速度。常见错误是在与障碍物距离比较时忘记了先减去思考距离。
2. Series and Parallel Circuits – Christmas Lights | 串联与并联电路 – 圣诞彩灯
A string of 40 decorative lamps is connected in series to a 230 V mains supply. Each lamp has a resistance of 2.5 Ω when lit. A second string of 40 lamps is connected in parallel with the first string. Analyse the circuit: calculate the total resistance of one series string, the current through it, and the total power consumed when both strings are operating. Explain why a single failed lamp in a series string causes all the lamps to go out, but a modern parallel-wired set does not have this problem.
一串由 40 个装饰灯泡串联而成的彩灯连接到 230 V 市电。每个灯泡点亮时的电阻为 2.5 Ω。另一串 40 个灯泡与第一串并联。分析电路：计算一串串联灯的总电阻、通过的电流，以及两串同时工作时的总功耗。解释为什么串联灯串中一个灯泡损坏会导致全部熄灭，而现代并联接线灯组不存在此问题。
For one series string, the total resistance is the sum of individual resistances: Rstring = 40 × 2.5 Ω = 100 Ω. The current through one string is I = V / R = 230 V / 100 Ω = 2.3 A.
对于一串串联灯组，总电阻等于各电阻之和：R串联 = 40 × 2.5 Ω = 100 Ω。通过一串的电流为 I = V / R = 230 V / 100 Ω = 2.3 A。
The total current from the mains is therefore 2.3 A + 2.3 A = 4.6 A. Total power P = V × Itotal = 230 V × 4.6 A = 1058 W.
当第二串同样的灯泡并联时，每个支路仍然承受 230 V 电压且各吸取 2.3 A 电流。因此市电总电流为 2.3 A + 2.3 A = 4.6 A。总功率 P = V × I总 = 230 V × 4.6 A = 1058 W。
In a series circuit, the same current flows through all components. If a single lamp burns out, the circuit is broken and current stops everywhere. In a parallel circuit, each lamp has its own direct path to the supply; if one path fails, current can still flow through the others.
在串联电路中，所有元件通过同一电流。若一个灯泡烧毁，电路断开，各处电流均停止。在并联电路中，每个灯泡有独立的供电回路；某一条回路断开，电流仍可通过其他回路流通。
When tackling circuit analysis, always redraw the circuit to identify series and parallel branches. Remember that for parallel branches, voltage is the same across each branch, while current divides.
进行电路分析时，务必重新画图以辨别串联与并联支路。记住，并联支路两端电压相等，而电流会分流。
3. Energy Efficiency of a Kettle | 电热水壶的能效
An electric kettle is rated at 2500 W. It takes 180 seconds to heat 1.0 kg of water from 20 °C to 100 °C. The specific heat capacity of water is 4200 J/kg°C. Calculate the electrical energy input, the useful thermal energy gained by the water, and the efficiency of the kettle. Give your efficiency as a percentage.
一只电热水壶额定功率为 2500 W。它将 1.0 kg 水从 20 °C 加热到 100 °C 耗时 180 秒。水的比热容为 4200 J/kg°C。计算输入的电能、水获得的有用热能，以及水壶的效率。效率用百分比表示。
Energy input from the mains is Einput = power × time = 2500 W × 180 s = 450 000 J (or 450 kJ).
市电输入的电能为 E输入 = 功率 × 时间 = 2500 W × 180 s = 450 000 J（即 450 kJ）。
Useful energy gained by the water is Euseful = m c Δθ = 1.0 kg × 4200 J/kg°C × (100 – 20)°C = 1.0 × 4200 × 80 = 336 000 J (336 kJ).
水获得的有用能量为 E有用 = m c Δθ = 1.0 kg × 4200 J/kg°C × (100 – 20)°C = 1.0 × 4200 × 80 = 336 000 J（336 kJ）。
Efficiency = (useful energy output / total energy input) × 100% = (336 000 / 450 000) × 100% = 74.7% ue to heat escaping to the kettle body and air, and some water evaporating before boiling.
效率 = (有用输出能量 / 总输入能量) × 100% = (336 000 / 450 000) × 100% = 74.7%。效率相对较低，主要因为热量散逸到壶体和空气中，且部分水在沸腾前已经蒸发。
In AQA exams, always show the working formula and substitute values with units. Express efficiency to 3 significant figures unless otherwise stated. Watch out for traps like power given in kW – remember to convert to watts when using seconds.
在 AQA 考试中，要写出所用公式并代入带单位的数值。除非另有说明，效率应取三位有效数字。当心陷阱，如功率以 kW 给出时，要记得在使用秒作单位时转换为瓦特。
4. Specific Heat Capacity – Engine Cooling | 比热容 – 发动机冷却
A car engine block has a mass of 120 kg and is made of cast iron with a specific heat capacity of 460 J/kg°C. During a cold start, its temperature rises from 10 °C to 90 °C. The coolant used is a water-glycol mixture with a specific heat capacity of 3800 J/kg°C. The coolant mass in the jacket is 8.0 kg. Calculate the thermal energy absorbed by the engine block and by the coolant. Explain why a substance with a high specific heat capacity makes an effective coolant.
一台汽车发动机缸体质量为 120 kg，材质为铸铁，比热容为 460 J/kg°C。在冷启动过程中，其温度从 10 °C 升至 90 °C。所用冷却液为水-乙二醇混合物，比热容为 3800 J/kg°C，水套内冷却液质量为 8.0 kg。计算发动机缸体和冷却液各自吸收的热能。解释为何高比热容的物质能成为有效的冷却剂。
For the engine block: ΔEblock = m c Δθ = 120 kg × 460 J/kg°C × (90 – 10)°C = 120 × 460 × 80 = 4 416 000 J (about 4.42 MJ).
对于发动机缸体：ΔE缸体 = m c Δθ = 120 kg × 460 J/kg°C × (90 – 10)°C = 120 × 460 × 80 = 4 416 000 J（约 4.42 MJ）。
For the coolant, assuming it absorbs the same energy until thermal equilibrium, but more realistically in many problems they ask for temperature rise of coolant given the energy lost by the engine. However, here we simply calculate the energy that can be absorbed by the coolant for a given temperature rise of, say, 20 °C difference: ΔEcoolant 00 J/kg°C × 20°C = 608 000 J. Even a small mass of coolant can take away significant heat.
对于冷却液，假设其吸收相同的能量至热平衡，但更切合实际的做法是，在许多问题中会给出发动机损失的能量并求冷却液的温升。这里我们简单计算冷却液在温度上升 20°C 时可吸收的能量：ΔE冷却液 = 8.0 kg × 3800 J/kg°C × 20°C = 608 000 J。即使质量不大，冷却液也能带走可观的热量。
A high specific heat capacity means the substance can absorb a large amount of heat for each kilogram per degree rise in temperature. This effectively stabilises the engine temperature, preventing overheating while the coolant remains at a lower temperature for longer.
高比热容意味着每千克该物质每升高一度能吸收大量热量。这能有效稳定发动机温度，防止过热，且冷却液自身温度能更长时间保持在较低水平。
Remember that in AQA questions, you may be given a table of materials and asked to select the material with the highest specific heat capacity for a cooling application. Be prepared to justify your choice with a calculation.
请记住，在 AQA 试题中，可能会给出材料表格并要求你选择比热容最高的材料用于冷却。准备好用计算说明你的选择。
5. Radioactive Decay and Half-Life – Medical Tracer | 放射性衰变与半衰期 – 医用示踪剂
A hospital uses iodine‑131 as a tracer in thyroid investigations. I‑131 has a half‑life of 8 days. A patient is given an injection with an initial activity of 1200 Bq. Calculate the activity remaining after 24 days. Determine how many days it takes for the activity to fall below 75 Bq. Explain why a short half‑life is crucial for a medical tracer.
一家医院在甲状腺检查中使用碘‑131 作为示踪剂。I‑131 的半衰期为 8 天。某患者接受初始活度为 1200 Bq 的注射。计算 24 天后的剩余活度。求活度降至 75 Bq 以下需要的天数。解释为何短半衰期对医用示踪剂至关重要。
24 days represents three half‑lives (24 ÷ 8 = 3). After each half‑e the activity halves: 1200 → 600 → 300 → 150 Bq. So after 24 days the activity is 150 Bq.
24 天为三个半衰期（24 ÷ 8 = 3）。每经过一个半衰期活度减半：1200 → 600 → 300 → 150 Bq。因此 24 天后活度为 150 Bq。
To fall below 75 Bq, we continue the halving: after 4 half‑ below 75 Bq, just over 32 days are needed. So at 33 days the activity is less than 75 Bq.
要降至 75 Bq 以下，我们继续减半：4 个半衰期（32 天）后活度为 1200 × (½)⁴ = 75 Bq。要低于 75 Bq，需要稍超过 32 天。因此在 33 天时活度低于 75 Bq。
A short half‑life ensures the radiation dose is limited and that the radioactive material does not remain in the body for too long, reducing damage to healthy tissue. At the same time, it must be long enough to complete the scan.
短半衰期可确保辐射剂量有限，且放射性物质不会在体内停留过长时间，从而减少对健康组织的损害。同时，半衰期又必须足够长以完成扫描。
Use the half‑life equation or a simple table for these calculations. Always underline the half‑life value you use and check whether the question asks for the activity after a given time or the time to reach a certain activity.
此类计算可使用半衰期公式或简单表格。务必在你所使用的半衰期数值下划线，并确认题目问的是给定时间后的活度还是达到某一活度所需的时间。
6. Waves and Echo Sounding – Sonar on a Ship | 波与回声测深 – 船载声纳
A survey vessel sends a sonar pulse downwards. The pulse travels through seawater at a speed of 1520 m/s and the reflected echo is detected 0.25 seconds later. Calculate the depth of the seabed. The sonar frequency is 50 kHz. Determine the wavelength of the sound in water. Explain why the pulse must be very short and why the speed of sound must be accurately known.
一艘测量船向下发射声纳脉冲。脉冲在海水中的传播速度为 1520 m/s，0.25 秒后接收到反射回声。计算海底深度。声纳频率为 50 kHz。求该声波在水中的波长。解释为何脉冲必须极短以及为何需要准确知晓声速。
The total travel time for the pulse to go down and back is 0.25 s, so the one‑way time is 0.125 s. Depth = speed × time = 1520 m/s × 0.125 s = 190 m.
脉冲往返总时间为 0.25 s，因此单程时间为 0.125 s。深度 = 速度 × 时间 = 1520 m/s × 0.125 s = 190 m。
e wave equation v = f λ, we find λ = v / f = 1520 m/s ÷ 50 000 Hz = 0.0304 m (or 3.04 cm).
利用波动方程 v = f λ，可得 λ = v / f = 1520 m/s ÷ 50 000 Hz = 0.0304 m（即 3.04 cm）。
A short pulse is needed to clearly separate the transmitted signal from the returning echo, allowing precise timing. If the pulse were long, the outgoing and returning waves could overlap, making it hard to measure the round‑trip time accurately. Accurate speed is essential because depth is directly proportional to it; a 1% error in speed leads to a 1% error in depth.
脉冲必须短，才能使发射信号与返回回声清晰分离，从而实现精确定时。若脉冲较长，出射波与反射波可能重叠，导致难以准确测量往返时间。声速准确至关重要，因为深度与声速成正比；声速 1% 的误差将导致深度 1% 的误差。
Watch out for the difference between the round‑trip time and the one‑way time. Many students forget to divide the echo time by two. Also convert all frequencies to hertz before substitution.
注意往返时间和单程时间的区别。许多学生忘记将回声时间除以二。同时，代入公式前应将所有频率换算为赫兹。
7. Hooke’s Law and Spring Energy – Bungee Jump Model | 胡克定律与弹性势能 – 蹦极模型
nstant k of the cord. The jumper of mass 80 kg jumps and stretches the cord by an additional 2.0 m beyond the equilibrium extension due to kinetic energy. Calculate the elastic potential energy stored at this maximum extension. Use g = 9.8 N/kg.
一根蹦极绳索性质类似弹簧。当悬挂 50 kg 的测试质量时，其原长 25 m 伸长了 8.0 m。求绳索的劲度系数 k。质量为 80 kg 的蹦极者跳跃后，由于动能，绳索在平衡伸长量的基础上又多伸长了 2.0 m。计算在此最大伸长量下储存的弹性势能。取 g = 9.8 N/kg。
mg = k e. k = mg / e = (50 kg × 9.8 N/kg) / 8.0 m = 490 N / 8.0 m = 61.25 N/m ≈ 61.3 N/m.
对于处于平衡的测试质量，重力等于绳索张力。因此 mg = k e。k = mg / e = (50 kg × 9.8 N/kg) / 8.0 m = 490 N / 8.0 m = 61.25 N/m ≈ 61.3 N/m。
The jumper of 80 kg would have an equilibrium extension ejumper ond equilibrium, so the total extension is 12.8 + 2.0 = 14.8 m. Elastic potential energy Eel = ½ k x² = 0.5 × 61.25 × (14.8)² = 0.5 × 61.25 × 219.04 ≈ 6706 J (≈ 6.71 kJ).
80 kg 的蹦极者平衡伸长量 e蹦极者 = m g / k = (80 × 9.8) / 61.25 = 784 / 61.25 ≈ 12.8 m。但题目说明绳索在平衡基础上又伸长了 2.0 m，因此总伸长量为 12.8 + 2.0 = 14.8 m。弹性势能 E弹 = ½ k x² = 0.5 × 61.25 × (14.8)² = 0.5 × 61.25 × 219.04 ≈ 6706 J（≈ 6.71 kJ）。
Always check that the extension x is measured from the unstretched length. In energy problems, the maximum extension includes the static extension due to weight plus any dynamic overshoot. Be consistent with units.
始终检查伸长量 x 是否从原长量起。在能量问题中，最大伸长量包括由于重量产生的静态伸长再加上任何动态超程。注意单位一致。
8. Pressure in Liquids – Deep-Sea Vessel | 液体压强 – 深海探测器
A deep‑ rvation window of diameter 30 cm. Calculate the pressure of the water at this depth. Then find the force exerted on the window. Explain why the window must be constructed with very thick, tapered glass and reinforced frames. Use g = 9.8 N/kg.
一艘深潜器下潜至太平洋 7000 米深处。海水密度取 1030 kg/m³。深潜器有一个直径 30 cm 的圆形观察窗。计算该深度的水压。然后求窗口所受的力。解释为何观察窗须用极厚的锥形玻璃和加强框架建造。取 g = 9.8 N/kg。
The pressure due to a liquid column is p = h ρ g = 7000 m × 1030 kg/m³ × 9.8 N/kg = 70 000 ? ospheric pressure at the surface is about 1.0 × 10⁵ Pa, which is negligible by comparison.
液柱产生的压强为 p = h ρ g = 7000 m × 1030 kg/m³ × 9.8 N/kg = 70 658 000 Pa，即约 7.07 × 10⁷ Pa（70.7 MPa）。海面大气压约为 1.0 × 10⁵ Pa，相比之下可忽略不计。
million newtons). This is an enormous force, equivalent to the weight of about 500 tonnes.
窗口面积 A = π r² = π × (0.15 m)² ≈ 0.0707 m²。力 F = p × A = 70 658 000 Pa × 0.0707 m² ≈ 5.0 × 10⁶ N（约 500 万牛顿）。如此巨大的力，相当于约 500 吨的重量。
Thick, tapered glass is needed so that the pressure compresses the glass against the frame, improving sealing. The conical shape prevents the window from popping inward. Reinforced frames distribute the stress safely. Without these features, the immense pressure difference would shatter a thin, flat pane.
需要厚锥形玻璃，以便压力将玻璃压向框架，改善密封性。锥形设计防止窗口向内爆裂。加强框架可安全分散应力。没有这些设计，巨大的压差会将薄平板玻璃压碎。
In calculations, always convert centimetres to metres before substituting. Remember that pressure in a liquid acts equally in all directions and depends only on depth, density and gravitational field strength.
计算时，要先将厘米换算为米再代入。记住，液体压强向各个方向同等作用，且仅取决于深度、密度和重力场强度。
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