📚 Case Study Practical Exercises for Year 10 CCEA Physics | Year 10 CCEA 物理：案例分析实战演练
In Year 10 CCEA Physics, applying concepts to real-world situations is just as important as learning the theory. This article presents a series of case study practical exercises that cover mechanics, energy, electricity and waves. Each case study asks you to analyse a scenario, identify the relevant physics and perform simple calculations using standard equations. Working through these examples will strengthen your problem-solving skills and prepare you for exam-style questions.
在 Year 10 CCEA 物理中，将概念应用到现实情境与学习理论同样重要。本文提供了一系列案例分析实战演练，涵盖力学、能量、电学和波。每个案例都要求你分析一个场景，找出相关的物理原理，并用标准方程进行简单计算。通过这些练习，可以增强解题能力，为应对考试题型做好准备。
1. Case Study 1: Car Braking System | 案例一：汽车制动系统
A driver travelling at 20 m/s sees a hazard ahead. The total stopping distance is made up of thinking distance (distance travelled during the driver’s reaction time) and braking distance (distance travelled under braking). The car’s mass is 1200 kg.
一名司机以 20 m/s 的速度行驶时，发现前方有危险。总停车距离由反应距离（司机反应时间内行驶的距离）和刹车距离（制动过程中行驶的距离）组成。汽车质量为 1200 kg。
2. 1.1 Thinking Distance and Reaction Time | 反应距离与反应时间
The driver’s reaction time is 0.6 s. Thinking distance = speed × reaction time = 20 m/s × 0.6 s = 12 m. The law requires drivers to be fit to respond; alcohol or tiredness increases reaction time, making the thinking distance longer.
司机的反应时间为 0.6 s。反应距离 = 速度 × 反应时间 = 20 m/s × 0.6 s = 12 m。法律要求驾驶员具备良好的反应能力；酒精或疲劳会延长反应时间，增加反应距离。
Even when the brakes are applied instantly, the car does not stop immediately because of inertia. The thinking distance is only the first part of stopping.
即便立即踩下刹车，汽车也因惯性不会立刻停下。反应距离只是停车过程的第一部分。
3. 1.2 Braking Distance and Kinetic Energy | 刹车距离与动能
Once the brakes are applied, a constant friction force of 6000 N acts against the motion. The initial kinetic energy is transferred into thermal energy in the brakes and tyres. Initial kinetic energy Eₖ = ½mv² = ½ × 1200 kg × (20 m/s)² = 240 000 J.
刹车后，恒定的 6000 N 摩擦力阻碍汽车运动。初始动能转化为刹车片和轮胎中的热能。初始动能 Eₖ = ½mv² = ½ × 1200 kg × (20 m/s)² = 240 000 J。
Using work done = force × distance, braking distance d = work done / force = 240 000 J / 6000 N = 40 m. So total stopping distance = 12 m + 40 m = 52 m. On a wet road friction reduces, increasing braking distance.
根据做功 = 力 × 距离，刹车距离 d = 做功 / 力 = 240 000 J / 6000 N = 40 m。因此总停车距离 = 12 m + 40 m = 52 m。在湿滑路面上，摩擦力减小，刹车距离会增加。
4. Case Study 2: Roller Coaster Loop | 案例二：过山车回环
A roller coaster car of mass 500 kg is released from rest at the top of a hill 30 m high. The track includes a circular loop of radius 10 m. Friction and air resistance can be ignored.
一辆质量为 500 kg 的过山车从高 30 m 的山顶由静止释放。轨道包含一个半径 10 m 的圆形回环。忽略摩擦和空气阻力。
5. 2.1 Energy Transformations on the Track | 轨道上的能量转换
At the top, the car has gravitational potential energy Eₚ = mgh = 500 kg × 10 m/s² × 30 m = 150 000 J. As it descends, this energy changes into kinetic energy. At the bottom of the hill, all the initial potential energy has become kinetic energy, so speed is maximum there.
在顶部，过山车具有重力势能 Eₚ = mgh = 500 kg × 10 m/s² × 30 m = 150 000 J。下滑时，这部分能量转化为动能。到达山脚时，所有初始势能都转变为动能，因此速度最大。
Going up into the loop, kinetic energy is converted back into potential energy. The car must have enough speed at the base to reach the highest point of the loop without falling.
进入回环时，动能又转化为势能。过山车在底部必须具有足够的速度，才能到达环的最高点而不坠落。
6. 2.2 Calculating Maximum Height and Speed | 计算最大高度与速度
Using conservation of energy, we can find the speed at the bottom: v = √(2gh) = √(2 × 10 × 30) = √600 ≈ 24.5 m/s. At the top of the 10 m radius loop (height = 20 m from ground), the speed will be lower because some energy has become potential energy.
利用能量守恒，可得底部速度：v = √(2gh) = √(2 × 10 × 30) = √600 ≈ 24.5 m/s。在半径 10 m 的回环顶部（距地面高度 20 m），速度会减小，因为部分能量已转化为势能。
Eₖ at top = Eₖ at bottom – mgh_loop = 150 000 J – 500×10×20 J = 50 000 J → v_top = √(2×50 000/500) ≈ 14.1 m/s
Eₖ 顶部 = Eₖ 底部 – mgh_环 = 150 000 J – 500×10×20 J = 50 000 J → v_顶部 = √(2×50 000/500) ≈ 14.1 m/s
The coaster stays on track because at the very top the centripetal force needed is provided by weight and the normal reaction. A safe design ensures the speed is high enough to keep passengers secure.
过山车能留在轨道上，是因为在最高点所需的向心力由重力和法向反作用力提供。安全的设计要确保速度足够快，以保障乘客安全。
7. Case Study 3: Household Electrical Safety | 案例三：家庭电路安全
A student is wiring a 230 V mains plug for an electric heater rated at 2.3 kW. The plug contains a fuse that protects the flex from overheating if a fault occurs. The flex is rated at 13 A.
一名学生正为一个额定功率 2.3 kW 的电暖气连接 230 V 电源插头。插头内装有保险丝，一旦发生故障可保护导线不过热。该导线的额定电流为 13 A。
8. 3.1 Choosing the Correct Fuse | 选择合适的保险丝
The operating current is calculated from P = IV: I = P / V = 2300 W / 230 V = 10 A. The standard fuse sizes available are 3 A, 5 A and 13 A. A fuse slightly above the normal current is chosen, so a 13 A fuse is appropriate.
工作电流由 P = IV 计算：I = P / V = 2300 W / 230 V = 10 A。常用的标准保险丝规格有 3 A、5 A 和 13 A。应选用略高于正常工作电流的保险丝，因此 13 A 保险丝是合适的。
If a 5 A fuse were used, it would blow during normal operation. The fuse must be in the live wire so that when it melts, the appliance is disconnected from the high voltage.
如果使用 5 A 保险丝，正常工作时就会熔断。保险丝必须接在火线上，这样一旦熔断，用电器便能从高电压中断开。
9. 3.2 Power and Current Calculation | 功率与电流计算
A second appliance, a 0.5 kW microwave oven, is used on the same ring circuit. Its operating current is I = 500 W / 230 V ≈ 2.2 A. This appliance should be protected by a 3 A or 5 A fuse. Using a 13 A fuse in a low-power device is dangerous because a fault current might not be high enough to blow the fuse quickly.
第二个用电器是 0.5 kW 的微波炉，接在同一环形电路上。其工作电流 I = 500 W / 230 V ≈ 2.2 A。该电器应使用 3 A 或 5 A 保险丝保护。在低功率设备中使用 13 A 保险丝是危险的，因为故障电流可能不足以快速熔断保险丝。
Remember that double insulation means an appliance does not need an earth wire. The symbol for double insulation is a square inside a square.
请注意，双重绝缘意味着用电器不需要接地线。双重绝缘的标志是一个方框内套一个方框。
10. Case Study 4: Investigating Waves with a Ripple Tank | 案例四：水波槽研究波
A ripple tank is used to generate straight water waves in shallow water. The waves pass from a deep region into a shallow region at an angle. The frequency of the wave source is 12 Hz, and the wavelength in deep water is 2.5 cm.
用水波槽在浅水中产生平直的水波。这些波以一定角度从深水区进入浅水区。波源的频率为 12 Hz，深水中的波长为 2.5 cm。
11. 4.1 Reflection and Refraction of Water Waves | 水波的反射与折射
When plane waves strike a straight barrier at an angle, the reflected waves obey the law of reflection: angle of incidence equals angle of reflection. This is shown by measuring the angle between the incident ray and the normal.
平面波以一定角度射向平直障碍物时，反射波遵守反射定律：入射角等于反射角。这可以通过测量入射光线与法线之间的夹角来展示。
Refraction occurs when waves pass from deep to shallow water. The speed and wavelength decrease, but the frequency remains constant. Because speed reduces, the wavefronts bend towards the normal.
波从深水进入浅水时会发生折射。波速和波长减小，但频率保持不变。由于速度降低，波前向法线方向弯折。
12. 4.2 Measuring Wavelength and Frequency | 测量波长与频率
Using the wave equation c = fλ, the wave speed in deep water is c = 12 Hz × 0.025 m = 0.30 m/s. In the shallow region, the wavelength is measured as 1.5 cm (0.015 m). Since frequency does not change, the new speed is c = 12 Hz × 0.015 m = 0.18 m/s.
利用波动方程 c = fλ，深水中波速 c = 12 Hz × 0.025 m = 0.30 m/s。在浅水区，测得波长为 1.5 cm (0.015 m)。由于频率不变，新的波速为 c = 12 Hz × 0.015 m = 0.18 m/s。
A stroboscope can be used to ‘freeze’ the wave pattern, allowing an accurate measurement of wavelength. If the stroboscope frequency is adjusted to exactly match the wave frequency, the waves appear stationary.
可使用频闪仪’冻结’波形图案，从而精确测量波长。如果频闪仪的频率调节到与波频完全一致，波看起来就是静止的。
Published by TutorHao | Physics Revision Series | aleveler.com
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