Year 10 CCEA Physics: Case Study Practice | 案例分析实战训练

📚 Year 10 CCEA Physics: Case Study Practice | 案例分析实战训练

Case study questions are a vital part of the CCEA Physics course. They test your ability to apply scientific knowledge to real-world situations, analyse data and carry out calculations in context. This practice set walks you through ten common case studies, building skills in motion, energy, electricity, waves, radioactivity and more. Each section presents a scenario, then breaks down the solution step by step, with clear English and Chinese explanations paired throughout.

案例分析题是 CCEA 物理课程的重要组成部分，考查你将科学知识应用于真实情境、分析数据并进行计算的能力。这套实战训练精选十个常见案例，带你演练运动、能量、电学、波、放射性等内容。每个小节先给出情景，再逐步拆解分析过程，全程中英双语紧密配对，帮助你扎实掌握解题思路。

1. Analysing Motion of a Cyclist | 分析骑行者的运动

A cyclist starts from rest and accelerates uniformly to 10 m/s in 5 seconds. He then maintains this constant speed for 20 seconds before braking uniformly to a stop in 4 seconds. We will analyse the motion by calculating acceleration in each phase, the distance covered in each segment and the total distance, then sketch a velocity–time graph.

一位骑行者从静止开始匀加速，5 秒内达到 10 m/s。然后保持该速度匀速行驶 20 秒，再均匀刹车，4 秒后停下。我们将计算各阶段的加速度、每段行驶距离和总距离，并画出速度–时间图。

Phase 1 (0–5 s): acceleration a = (v – u)/t = (10 – 0)/5 = 2 m/s². The cyclist accelerates at 2 m/s².

第 1 阶段 (0–5 s)：加速度 a = (v – u)/t = (10 – 0)/5 = 2 m/s²，骑行者以 2 m/s² 加速。

a = 2 m/s²

Distance in phase 1: using s = ut + ½at² = 0×5 + ½×2×5² = 25 m, or average velocity method: (0+10)/2 × 5 = 25 m.

第 1 阶段距离：利用 s = ut + ½at² = 0×5 + ½×2×5² = 25 m，或平均速度法：(0+10)/2 × 5 = 25 m。

Phase 2 (5–25 s): constant speed of 10 m/s. Distance s = v × t = 10 × 20 = 200 m.

第 2 阶段 (5–25 s)：匀速 10 m/s，距离 s = v × t = 10 × 20 = 200 m。

Phase 3 (25–29 s): braking. Deceleration a = (0 – 10)/4 = –2.5 m/s² (negative acceleration). Distance s = ut + ½at² = 10×4 + ½×(–2.5)×4² = 40 – 20 = 20 m.

第 3 阶段 (25–29 s)：刹车。减速度 a = (0 – 10)/4 = –2.5 m/s² (负加速度)。距离 s = ut + ½at² = 10×4 + ½×(–2.5)×4² = 40 – 20 = 20 m。

Total distance = 25 + 200 + 20 = 245 m. The velocity–time graph is a straight line rising from (0,0) to (5,10), horizontal to (25,10), then sloping down to (29,0).

总距离 = 25 + 200 + 20 = 245 m。速度–时间图是一条从 (0,0) 上升到 (5,10) 的直线，然后水平延伸至 (25,10)，再下降到 (29,0)。

2. Energy Transfers in a Roller Coaster | 过山车中的能量转换

A roller coaster car of mass 500 kg is lifted to the top of a hill 30 m above the ground. It then rolls down the track, converting gravitational potential energy to kinetic energy. Assume no energy is lost to friction or air resistance. Calculate the maximum speed of the car at the bottom of the hill and discuss energy changes.

一辆过山车质量为 500 kg，被提升到距地面 30 m 高的山顶，然后沿轨道滑下，重力势能转化为动能。假设没有摩擦和空气阻力，计算车在底部能达到的最大速度并讨论能量转换。

Gravitational potential energy at top: GPE = mgh = 500 × 10 × 30 = 150 000 J (using g = 10 N/kg).

顶部的重力势能：GPE = mgh = 500 × 10 × 30 = 150 000 J (取 g = 10 N/kg)。

GPE = mgh

At the bottom, all GPE is converted to kinetic energy (KE = ½mv²). So ½ × 500 × v² = 150 000. Solving: 250 v² = 150 000 → v² = 600 → v ≈ 24.5 m/s.

到达底部时，所有重力势能转化为动能 (KE = ½mv²)。因此 ½ × 500 × v² = 150 000。求解：250 v² = 150 000 → v² = 600 → v ≈ 24.5 m/s。

v = √(2gh) = √(2×10×30) ≈ 24.5 m/s

In practice, some energy is always transferred as thermal energy due to friction, so the real speed would be lower. The total energy of the system remains constant, changing only between stores.

实际上，摩擦总会将一部分能量转化为热能，因此真实速度会低一些。系统的总能量保持不变，只是在能量储存方式之间转换。

3. Investigating Ohm’s Law in a Lamp Circuit | 灯泡电路中探究欧姆定律

A student builds a circuit with a filament lamp, a variable resistor, an ammeter and a voltmeter. As the current increases, the resistance of the lamp changes. The recorded data are: at 0.20 A, 1.2 V; at 0.30 A, 2.7 V; at 0.40 A, 4.8 V; at 0.50 A, 7.5 V. Determine if the lamp obeys Ohm’s law and explain the trend.

学生用灯泡、可变电阻、电流表和电压表搭建电路。随着电流增大，灯泡电阻发生变化。记录数据为：0.20 A 时 1.2 V；0.30 A 时 2.7 V；0.40 A 时 4.8 V；0.50 A 时 7.5 V。判断灯泡是否遵从欧姆定律并解释变化趋势。

Ohm’s law states that the current through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant. This means resistance R = V/I should be constant.

欧姆定律指出，在温度保持不变的条件下，通过导体的电流与导体两端的电势差成正比，即电阻 R = V/I 应为定值。

Calculate resistance at each point: R₁ = 1.2/0.20 = 6.0 Ω; R₂ = 2.7/0.30 = 9.0 Ω; R₃ = 4.8/0.40 = 12.0 Ω; R₄ = 7.5/0.50 = 15.0 Ω. The resistance clearly increases as current rises.

计算各点电阻：R₁ = 1.2/0.20 = 6.0 Ω；R₂ = 2.7/0.30 = 9.0 Ω；R₃ = 4.8/0.40 = 12.0 Ω；R₄ = 7.5/0.50 = 15.0 Ω。可见电阻随电流增大而增大。

The filament lamp does not obey Ohm’s law because its resistance is not constant. The filament gets hotter as current increases; in metals, resistance rises with temperature because the metal ions vibrate more, increasing collisions with electrons.

该灯泡不遵从欧姆定律，因为它的电阻不是常数。随着电流增大，灯丝温度升高；对于金属，温度升高使金属离子振动加剧，电子与离子的碰撞增多，导致电阻增大。

4. Radioactive Decay of a Sample | 样品的放射性衰变

A laboratory sample of radon-222 has an initial activity of 800 Bq. Radon-222 has a half-life of 3.8 days. Calculate the activity after 11.4 days and the fraction of the original sample remaining. Describe the random nature of decay and the meaning of half-life.

某实验室的氡-222 样品初始活度为 800 Bq，半衰期为 3.8 天。计算经过 11.4 天后的活度以及剩余样品占初始的比例，并说明衰变的随机性及半衰期的意义。

First, find the number of half-lives elapsed: total time 11.4 days ÷ 3.8 days = 3 half-lives.

首先，计算经过的半衰期数目：总时间 11.4 天 ÷ 3.8 天 = 3 个半衰期。

After each half-life, the activity halves. Activity after 3 half-lives: 800 → 400 → 200 → 100 Bq. So the activity is 100 Bq.

每经过一个半衰期，活度减半。3 个半衰期后的活度：800 → 400 → 200 → 100 Bq，因此活度为 100 Bq。

A = A₀ × (½)ⁿ = 800 × (½)³ = 100 Bq

Fraction remaining = 100 / 800 = 1/8. Radioactive decay is a random process; we cannot predict which nucleus will decay next, but for a large number of nuclei, the overall rate follows a predictable pattern described by half-life.

剩余比例 = 100 / 800 = 1/8。放射性衰变是随机过程，无法预言哪个原子核会下一个发生衰变，但对于大量原子核，整体衰变速率遵循可预测的规律，即半衰期。

5. Wave Behaviour at a Water Surface | 水面波的特性

A ripple tank is used to investigate water waves. Straight waves of frequency 5 Hz travel from deep water into shallow water at an angle. In deep water, the wavelength is 4.0 cm. In shallow water, the wave speed drops to 12 cm/s. Calculate the wavelength in shallow water and describe the changes in direction and frequency.

利用水波槽研究水面波。频率为 5 Hz 的直波从深水区倾斜进入浅水区。深水区波长为 4.0 cm，浅水区波速降至 12 cm/s。计算浅水区波长，并描述传播方向和频率的变化。

Wave speed in deep water: v_deep = f × λ = 5 × 4.0 = 20 cm/s. Frequency remains unchanged when waves cross the boundary, so f = 5 Hz in shallow water as well.

深水区的波速：v_深 = f × λ = 5 × 4.0 = 20 cm/s。波越过界面时频率保持不变，因此浅水区频率也是 5 Hz。

Wavelength in shallow water: λ = v_shallow / f = 12 / 5 = 2.4 cm. The wavelength decreases because wave speed reduces.

浅水区波长：λ = v_浅 / f = 12 / 5 = 2.4 cm。由于波速下降，波长变短。

λ_shallow = 2.4 cm, frequency = 5 Hz

As the waves enter shallow water at an angle, they refract: the direction changes towards the normal because the wave slows down. The wavefronts become closer together but keep the same frequency.

当波以一定角度进入浅水区时，会发生折射：由于波速减慢，传播方向会偏向法线。波阵面变得更密，但频率不变。

6. Newton’s Laws in a Car Crash | 牛顿定律在汽车碰撞中的应用

A car of mass 1200 kg moving at 20 m/s crashes into a wall and comes to rest in 0.5 seconds. The driver, of mass 70 kg, is wearing a seatbelt that stretches slightly, bringing the driver to rest in 0.8 seconds. Use Newton’s second law to calculate the force on the car and the force on the driver. Discuss how the seatbelt reduces injury.

一辆质量 1200 kg 的汽车以 20 m/s 的速度撞上墙壁，在 0.5 秒内停下。质量为 70 kg 的司机系有安全带，安全带略有伸展，使司机在 0.8 秒内停下。运用牛顿第二定律计算汽车和司机所受的力，并讨论安全带如何减轻伤害。

For the car: initial velocity u = 20 m/s, final v = 0, time t = 0.5 s. Deceleration a = (v – u)/t = (0 – 20)/0.5 = –40 m/s². Force on car: F = ma = 1200 × 40 = 48 000 N (magnitude).

对汽车：初速度 u = 20 m/s，末速度 v = 0，时间 t = 0.5 s。减速度 a = (v – u)/t = (0 – 20)/0.5 = –40 m/s²。汽车受到的力大小：F = ma = 1200 × 40 = 48 000 N。

F_car = 48 kN

For the driver: same change in speed, but t = 0.8 s. Deceleration a = –20/0.8 = –25 m/s². Force on driver from seatbelt: F = 70 × 25 = 1750 N.

对司机：速度变化相同，但 t = 0.8 s，减速度 a = –20/0.8 = –25 m/s²。安全带施加给司机的力：F = 70 × 25 = 1750 N。

Without a seatbelt, the driver would hit the dashboard in approximately the same 0.5 s as the car, experiencing a force of 70 × 40 = 2800 N. The seatbelt increases the stopping time, reducing the force felt by the driver (impulse = change in momentum = F × t, so larger t gives smaller F). This prevents serious injuries.

若不系安全带，司机可能在约 0.5 s 内撞上仪表盘，受力为 70 × 40 = 2800 N。安全带延长了停止时间，从而减小了司机承受的力（冲量 = 动量变化 = F × t，t 增大则 F 减小），有助于防止重伤。

7. Calculating Household Energy Costs | 计算家庭能源费用

A household uses a 3 kW electric heater for 4 hours, ten 15 W LED bulbs for 6 hours each, and a 200 W television for 5 hours. Electricity costs 18 pence per kilowatt-hour (kWh). Calculate the total energy consumed in kWh and the total cost.

某家庭使用一台 3 kW 的电暖器 4 小时，10 个 15 W 的 LED 灯泡各开 6 小时，以及一台 200 W 的电视机 5 小时。电价为每千瓦时 (kWh) 18 便士。计算总消耗能量（kWh）和总费用。

Energy used by heater: power 3 kW × time 4 h = 12 kWh. Energy for bulbs: total power = 10 × 15 W = 150 W = 0.15 kW; energy = 0.15 kW × 6 h = 0.9 kWh. Television: 200 W = 0.2 kW × 5 h = 1.0 kWh.

电暖器耗能：功率 3 kW × 时间 4 h = 12 kWh。灯泡总功率 = 10 × 15 W = 150 W = 0.15 kW；能量 = 0.15 kW × 6 h = 0.9 kWh。电视机：200 W = 0.2 kW × 5 h = 1.0 kWh。

Total energy consumed = 12 + 0.9 + 1.0 = 13.9 kWh. Cost = 13.9 kWh × 18 p/kWh = 250.2 pence, which is about £2.50.

总消耗能量 = 12 + 0.9 + 1.0 = 13.9 kWh。费用 = 13.9 kWh × 18 p/kWh = 250.2 便士，约 £2.50。

Total cost = energy (kWh) × price per kWh

This calculation shows why devices with high power ratings or long usage times dominate the electricity bill. Using energy-efficient LEDs helps keep lighting costs low.

这一计算显示了高功率或长时间使用的电器为何在电费中占大头。使用节能的 LED 灯泡有助于控制照明开支。

8. Thermal Insulation and the Vacuum Flask | 保温隔热与真空瓶

A vacuum flask keeps hot soup hot for hours. Explain how the flask’s design reduces thermal energy transfer by conduction, convection and radiation. Use the idea of the flask’s double-walled glass, vacuum, silvered surfaces and stopper.

真空瓶能让热汤保温数小时。请解释瓶的设计如何通过减少传导、对流和辐射来降低热能传递，需用到双层玻璃壁、真空层、镀银表面和瓶塞等知识。

Conduction is reduced because the two glass walls are separated by a vacuum. There are almost no particles in a vacuum to pass on kinetic energy, so conduction through the walls is minimal. The thin glass walls also conduct poorly.

传导被减少是因为双层玻璃壁之间为真空。真空中几乎没有粒子可以传递动能，因此通过壁面的传导极少。薄玻璃本身也是不良热导体。

Convection requires movement of particles. Since there is a vacuum between the walls, no convection currents can form there. The stopper at the top also prevents air from circulating into or out of the flask, stopping convection from the liquid surface.

对流需要粒子的运动。由于壁间是真空，无法形成对流。顶部的瓶塞还阻止空气进出瓶内，从而防止液体表面上方形成对流。

Radiation is reduced because the inner and outer glass surfaces are silvered. Shiny surfaces are poor emitters and poor absorbers of infrared radiation. The silvered inner wall reflects thermal radiation back into the liquid, while the outer wall reflects external radiation away.

辐射通过镀银表面来减少。光亮表面是红外辐射的不良发射体和吸收体。内壁镀银将热辐射反射回液体，外壁则反射外界辐射。

Together, these features keep the drink hot for many hours, with only a small amount of heat escaping slowly through the stopper and the glass neck.

这些设计共同使饮品长时间保温，只有极少量热量通过瓶塞和玻璃颈部缓慢散失。

9. Density and Buoyancy of a Toy Boat | 玩具船的密度与浮力

A child’s toy boat has a mass of 0.25 kg and floats on a pond. The boat displaces 0.00025 m³ of water. The density of water is 1000 kg/m³. Calculate the weight of the boat, the upthrust (buoyancy force) acting on it, and determine if the boat is floating high or low in the water.

一艘儿童玩具船质量为 0.25 kg，浮在池塘中，排开水的体积为 0.00025 m³。水的密度为 1000 kg/m³。计算船的重力、所受的上推力（浮力），并判断船是浮得高还是低。

Weight of the boat: W = m × g = 0.25 × 10 = 2.5 N (taking g = 10 N/kg).

船的重力：W = m × g = 0.25 × 10 = 2.5 N (取 g = 10 N/kg)。

According to Archimedes’ principle, the upthrust equals the weight of the water displaced. Mass of water displaced = density × volume = 1000 × 0.00025 = 0.25 kg. Its weight = 0.25 × 10 = 2.5 N. So upthrust = 2.5 N.

根据阿基米德原理，浮力等于排开水的重力。排开水的质量 = 密度 × 体积 = 1000 × 0.00025 = 0.25 kg，其重力 = 0.25 × 10 = 2.5 N。因此上推力 = 2.5 N。

Upthrust = ρ × V_displaced × g

Because the upthrust exactly equals the weight, the boat floats in equilibrium. The volume of water displaced equals the volume of the boat that is submerged. For this boat, the submerged volume yields an upthrust equal to its weight, so it floats stably. If it carried extra load, it would displace more water and sit lower in the pond.

由于浮力恰好等于重力，船处于浮力平衡状态。排水体积等于船体浸没部分的体积。对该船而言，浸没体积产生的浮力等于自身重力，因此它稳定漂浮。如果船上增加负载，排

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Year 10 CCEA Physics: Case Study Practice | 案例分析实战训练