📚 Year 8 CCEA Physics: Case Study Practice | Year 8 CCEA 物理：案例分析实战演练
This article presents a series of real-world physics case studies designed to help Year 8 CCEA students apply key concepts to problem-solving situations. Each case study is broken down into a scenario, relevant data, and step-by-step analysis. By working through these examples, you will strengthen your ability to interpret data, perform calculations, and explain physical phenomena using correct scientific terminology.
本文提供了一系列基于真实情景的物理案例分析，旨在帮助 Year 8 CCEA 学生将关键概念应用于问题解决。每个案例都分解为情境、相关数据和分步解析。通过练习这些例题，你将增强解读数据、完成计算以及使用正确科学术语解释物理现象的能力。
1. Case Study: Calculating Average Speed | 案例一：计算平均速度
A student runs once around a 400 m school track. She starts at 08:00:00 and crosses the finish line at 08:02:30. Calculate her average speed in metres per second.
一名学生绕学校 400 m 跑道跑了一圈。她在 08:00:00 出发，在 08:02:30 冲过终点线。请计算她的平均速度，单位为米/秒。
30 seconds. Convert this into seconds: 2 × 60 + 30 = 150 s. The distance covered is 400 m.
我们首先计算所用时间。结束时间为 08:02:30，因此经过时间为 2 分 30 秒。将其换算成秒：2 × 60 + 30 = 150 s。跑过的距离为 400 m。
The formula for average speed is:
平均速度的公式为：
average speed = distance ÷ time
50 s = 2.67 m/s (rounded to two decimal places). The student’s average speed was 2.67 m/s.
代入数值：平均速度 = 400 m ÷ 150 s = 2.67 m/s（保留两位小数）。该学生的平均速度为 2.67 m/s。
This case shows that even if the runner sped up or slowed down during the lap, average speed only depends on total distance and total time.
这个案例表明，即使跑者在跑圈过程中加速或减速，平均速度也仅取决于总距离和总时间。
2. Case Study: Density and Floating | 案例二：密度与浮沉
A solid plastic block has a mass of 36 g and a volume of 40 cm³. Will it float or sink in water? The density of water is 1.0 g/cm³.
一块实心塑料块的质量为 36 g，体积为 40 cm³。它在水中会浮起还是下沉？水的密度为 1.0 g/cm³。
Density is calculated using:
密度的计算公式为：
density (ρ) = mass ÷ volume
For the plastic: ρ = 36 g ÷ 40 cm³ = 0.9 g/cm³.
对于塑料块：ρ = 36 g ÷ 40 cm³ = 0.9 g/cm³。
Since 0.9 g/cm³ is less than the density of water (1.0 g/cm³), the plastic block will float. Objects with density lower than the fluid’s density rise to the surface.
由于 0.9 g/cm³ 小于水的密度（1.0 g/cm³），该塑料块会浮起。密度小于流体密度的物体会浮上表面。
If we placed the same block in a liquid with density 0.8 g/cm³, the block would sink because its density is greater.
若将同一块塑料放入密度为 0.8 g/cm³ 的液体中，它就会下沉，因为其密度更大。
3. Case Study: Series Circuit Fault | 案例三：串联电路故障
Three identical bulbs are connected in series with a switch and a battery. When the switch is closed, none of the bulbs light up. A student measures the voltage across the battery and finds 6 V. What could be the fault?
三只相同的灯泡与一个开关和电池串联。闭合开关后，没有一只灯泡发光。一位学生测量了电池两端电压，读数为 6 V。可能是什么故障？
In a series circuit, the current is the same everywhere. If one bulb is blown (filament broken), it creates an open circuit, stopping current through all components. A voltmeter placed across the faulty bulb would read the full battery voltage (6 V), while across the others it would read 0 V.
在串联电路中，各处电流相同。若一只灯泡烧坏了（灯丝断开），则形成开路，电流无法通过所有元件。将电压表接在故障灯泡两端，会显示整个电池电压（6 V），而接在其他灯泡两端则显示 0 V。
Another possibility is a loose connection. Checking each connection and the switch can help identify the break. The case reminds us that a single fault in a series circuit breaks the entire loop.
另一种可能是接线松动。检查各个接线点及开关有助于找出断路处。该案例提醒我们，串联电路中一处故障便会使整个回路断开。
4. Case Study: Sound Propagation | 案例四：声音传播
During a thunderstorm, a student sees a lightning flash and hears the thunder 3 seconds later. Given that the speed of sound in air is approximately 340 m/s, estimate how far away the lightning struck.
在一次雷暴中，一名学生看到闪电，3 秒后听到雷声。已知声音在空气中的速度约为 340 m/s，估算闪电发生处的距离。
Light travels almost instantly to the observer, so the time delay is entirely due to the sound travel time. The formula linking distance, speed and time is:
光几乎瞬间到达观察者，因此延迟完全来自声音的传播时间。连接距离、速度和时间的公式为：
distance = speed × time
Substitute: distance = 340 m/s × 3 s = 1020 m, approximately 1 km.
代入：距离 = 340 m/s × 3 s = 1020 m，大约 1 km。
This case demonstrates how sound speed can be used for distance estimation. In reality, the speed of sound varies slightly with air temperature.
这个案例展示了如何利用声速估算距离。实际上，声音的速度会随气温略有变化。
5. Case Study: Reflection of Light | 案例五：光的反射
A ray of light strikes a plane mirror at an angle of 30° to the mirror surface. Draw a labelled diagram and state the angle of incidence and angle of reflection.
一束光线以与镜面成 30° 角射向平面镜。请画出带标注的光路图，并说明入射角和反射角。
icular to the mirror. Therefore, the angle between the incident ray and the normal is: 90° – 30° = 60°. This is the angle of incidence (i).
首先，题目给出的是光线与镜面的夹角，而非与法线的夹角。法线是垂直于镜面的假想直线。因此，入射光线与法线的夹角为：90° – 30° = 60°。这就是入射角（i）。
r). Thus, the angle of reflection is also 60°. In the diagram, both angles should be labelled next to the normal.
反射定律表明：入射角等于反射角（i = r）。因此，反射角同样为 60°。在图中，两个角都应标注在法线旁。
This case highlights the common mistake of measuring from the mirror surface rather than the normal. Always remember the normal is the reference line.
这个案例强调了从镜面而非从法线测量角度的常见错误。始终记住法线是参考线。
6. Case Study: Energy Transfers | 案例六：能量转换
A child drops a ball from a height. Describe the main energy transfers that occur from the moment the ball is released until it hits the ground (assume negligible air resistance).
一个孩子从高处放开一个球。描述从球被释放至撞击地面的过程中发生的主要能量转换（假设空气阻力可忽略）。
When the ball is held at height, it has stored gravitational potential energy (GPE) relative to the ground. As it falls, GPE decreases and kinetic energy (KE) increases because the ball speeds up.
当球被举在高处时，它具有相对于地面的重力势能（GPE）。下落过程中，重力势能减小，动能（KE）增加，因为球的速度加快。
At any point during the fall, the decrease in GPE equals the increase in KE, so total mechanical energy is conserved (ignoring air resistance). Just before impact, nearly all the initial GPE has been converted to KE.在下落过程中的任意时刻，重力势能的减少量等于动能的增加量，因此总机械能守恒（忽略空气阻力）。在即将撞击地面的瞬间，几乎所有的初始重力势能都转化成了动能。
The energy transfer can be summarised as:
能量转换可概括为：
gravitational potential energy → kinetic energy
This case reinforces the principle of conservation of energy.
这个案例强化了能量守恒原理。
7. Case Study: Friction and Motion | 案例七：摩擦力与运动
A wooden block is pulled along a horizontal bench using a forcemeter. When the block moves at a steady speed, the forcemeter reads 2.5 N. Explain the forces acting and why the reading is constant during steady motion.
使用测力计沿水平台面拉动一个木块。当木块匀速运动时，测力计读数为 2.5 N。解释作用力的情况，并说明为何在稳定运动时读数保持恒定。
When the block moves at steady speed, the forces are balanced. The pulling force from the forcemeter equals the frictional force opposing motion. Since the speed is constant, there is no net force, and the reading stays steady at 2.5 N, which equals the kinetic friction force.
当木块匀速运动时，作用力保持平衡。测力计的拉力等于阻碍运动的摩擦力。由于速度恒定，合力为零，因此读数稳定在 2.5 N，等于动摩擦力的大小。
If the pulling force increased, the block would accelerate because the forces would become unbalanced. This case helps distinguish between static friction (before motion starts) and kinetic friction (during motion).
如果拉力增大，木块将会加速，因为力不再平衡。这个案例有助于区分静摩擦力（开始运动前）和动摩擦力（运动过程中）。
8. Case Study: Stretching a Spring | 案例八：弹簧拉伸
A student hangs different masses on a spring and measures the total length each time. The unloaded length of the spring is 10.0 cm. The results are shown in the table below.
一名学生在弹簧上悬挂不同质量并测量每次的总长度。弹簧空载时的长度为 10.0 cm。结果如下表所示。
| Mass (g) | Total length (cm) |
|---|---|
| 0 | 10.0 |
| 100 | 12.0 |
| 200 | 14.0 |
| 300 | 16.0 |
| 400 | 18.5 |
Calculate the extension for each load and explain whether the spring obeys Hooke’s law within this range.
计算每次负载下的伸长量，并解释该弹簧在此范围内是否遵循胡克定律。
Extension = total length – unloaded length. For 100 g: 12.0 – on is directly proportional to force (weight), provided the elastic limit is not exceeded.
伸长量 = 总长度 – 空载长度。100 g 时：12.0 – 10.0 = 2.0 cm；200 g 时：4.0 cm；300 g 时：6.0 cm。可以看到，当质量翻倍时，伸长量也翻倍。胡克定律指出，只要不超过弹性极限，伸长量与力（重量）成正比。
However, at 400 g the extension is 18.5 – han 8.0 cm predicted by doubling. This suggests the spring has exceeded its elastic limit and is stretching permanently. Hooke’s law was obeyed up to 300 g.
然而，400 g 时的伸长量为 18.5 – 10.0 = 8.5 cm，略高于按比例预测的 8.0 cm。这表明弹簧已经超出了弹性极限，正在发生永久伸长。在 300 g 以内弹簧遵循胡克定律。
Published by TutorHao | Physics Revision Series | aleveler.com
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