📚 Case Study Practice for Year 10 CAIE Physics | Year 10 CAIE 物理：案例分析实战演练
In Year 10 CAIE Physics, applying theoretical knowledge to real-world scenarios is essential for mastering the subject. This article presents a series of case studies that mirror the types of problems found in exams, guiding you through step-by-step analyses. By working through these examples, you will strengthen your problem-solving skills and deepen your understanding of mechanics, energy, waves, electricity, and thermal physics.
在 Year 10 CAIE 物理课程中，将理论知识应用于实际场景是掌握这门学科的关键。本文通过一系列案例分析，模拟考试中常见的问题类型，并逐步引导你进行分析。通过演练这些案例，你将提升解题技巧，加深对力学、能量、波、电学和热物理的理解。
1. Free Fall of a Tennis Ball | 网球的自由落体
A tennis ball is dropped from a balcony 20 m above the ground. Ignoring air resistance, we want to determine the time it takes to reach the ground and its impact speed. Assume g = 10 m/s².
一个网球从离地面 20 m 高的阳台自由落下。忽略空气阻力，我们需要求出它落地所需时间和撞击地面的速度。取 g = 10 m/s²。
For an object falling from rest, initial velocity u = 0. We apply the equation of motion: s = ut + ½at². Since u = 0, s = ½gt². Substituting s = 20 m and g = 10 m/s² gives 20 = ½ × 10 × t², so 5t² = 20, hence t² = 4, t = 2 s.
对于从静止开始下落的物体，初速度 u = 0。我们使用运动方程：s = ut + ½at²。由于 u = 0，则 s = ½gt²。代入 s = 20 m 和 g = 10 m/s² 得到 20 = ½ × 10 × t²，因此 5t² = 20，t² = 4，t = 2 s。
Next, we find the final velocity v using v = u + gt. Thus v = 0 + 10 × 2 = 20 m/s. Alternatively, use v² = u² + 2gs, so v² = 0 + 2 × 10 × 20 = 400, v = 20 m/s.
接下来，利用 v = u + gt 求末速度。v = 0 + 10 × 2 = 20 m/s。或者用 v² = u² + 2gs，则 v² = 0 + 2 × 10 × 20 = 400，v = 20 m/s。
v² = u² + 2as
This analysis assumes constant acceleration and no air resistance, a common simplification in introductory physics.
这个分析假设加速度恒定且没有空气阻力，这是基础物理中常见的简化假设。
2. Braking Distance on a Wet Road | 潮湿路面的刹车距离
A car travelling at 15 m/s on a wet road decelerates uniformly at 3.0 m/s² when the brakes are applied. The driver’s reaction time is 0.8 s. Calculate the total stopping distance.
一辆汽车以 15 m/s 的速度在潮湿路面上行驶，刹车时匀减速，减速度为 3.0 m/s²。司机的反应时间为 0.8 s。计算总停车距离。
Stopping distance consists of thinking distance (distance travelled during reaction time) and braking distance. Thinking distance = speed × reaction time = 15 × 0.8 = 12 m.
停车距离由反应距离（反应时间内行驶的距离）和制动距离组成。反应距离 = 速度 × 反应时间 = 15 × 0.8 = 12 m。
For braking distance, initial velocity u = 15 m/s, final velocity v = 0, acceleration a = -3.0 m/s². Using v² = u² + 2as, we rearrange: s = (v² – u²) / (2a). Substitute: s = (0 – 15²) / (2 × -3.0) = (-225) / (-6.0) = 37.5 m.
对于制动距离，初速度 u = 15 m/s，末速度 v = 0，加速度 a = -3.0 m/s²。利用 v² = u² + 2as，整理得 s = (v² – u²) / (2a)。代入得 s = (0 – 15²) / (2 × -3.0) = (-225) / (-6.0) = 37.5 m。
Total stopping distance = 12 m + 37.5 m = 49.5 m. This case highlights the importance of reaction time and road conditions in real-world safety.
总停车距离 = 12 m + 37.5 m = 49.5 m。这个案例凸显了反应时间和路面状况在实际安全中的重要性。
3. Energy Conservation in a Pendulum | 单摆的能量守恒
A pendulum bob of mass 0.2 kg is raised to a height of 0.1 m above its lowest point and released. Assuming no energy losses, find its maximum speed. g = 10 N/kg.
一个质量为 0.2 kg 的摆锤被拉至最低点上方 0.1 m 处释放。假设无能量损失，求它的最大速度。g = 10 N/kg。
At the highest point, the bob has gravitational potential energy (GPE) = mgh = 0.2 × 10 × 0.1 = 0.2 J. At the lowest point, all GPE is converted to kinetic energy (KE). KE = ½mv². So ½ × 0.2 × v² = 0.2, giving 0.1 v² = 0.2, v² = 2, v = √2 ≈ 1.41 m/s.
在最高点，摆锤具有重力势能 GPE = mgh = 0.2 × 10 × 0.1 = 0.2 J。在最低点，所有重力势能转化为动能 KE = ½mv²。因此 ½ × 0.2 × v² = 0.2，得 0.1 v² = 0.2，v² = 2，v = √2 ≈ 1.41 m/s。
½mv² = mgh
In reality, air resistance and pivot friction cause the amplitude to decay, but the energy analysis remains a powerful tool for idealised systems.
现实中，空气阻力和支点摩擦会导致振幅衰减，但在理想化系统中能量分析法仍然是一个强有力的工具。
4. Investigating a Faulty Circuit | 电路故障调查
A student builds a series circuit with a battery, a switch, a lamp, and an ammeter. When the switch is closed, the lamp does not light and the ammeter reads 0 A. The battery is known to work. Diagnose the fault.
一名学生搭建了一个由电池、开关、灯泡和电流表组成的串联电路。闭合开关后，灯泡不亮且电流表读数为 0 A。已知电池正常。请诊断故障。
A zero current reading indicates an open circuit somewhere. Possible causes: a broken filament in the lamp, loose connection, or a faulty switch. To test, use a voltmeter across each component. If the voltmeter reads the full battery voltage across the lamp, the lamp is open-circuit (blown). If zero voltage across the lamp but voltage across switch, the switch is open when closed, indicating a faulty switch.
电流为零表明电路中存在断路。可能原因：灯泡灯丝断裂、连接松动或开关故障。可用电压表跨接每个元件测试。如果电压表在灯泡两端测得满电池电压，则灯泡断路（烧毁）。如果灯泡两端电压为零而开关两端有电压，说明闭合开关时开关仍断开，表明开关故障。
| Component | Voltmeter reading | Conclusion |
|---|---|---|
| Lamp | Full voltage | Open-circuit lamp |
| Switch (closed) | Full voltage | Faulty switch |
This systematic approach is a cornerstone of circuit troubleshooting.
这种系统化的方法是电路故障排查的基础。
5. Image Formation by a Convex Lens | 凸透镜成像
An object of height 5 cm is placed 30 cm from a convex lens of focal length 10 cm. Determine the image distance, magnification, and image characteristics.
一个高 5 cm 的物体放在焦距为 10 cm 的凸透镜前 30 cm 处。求像距、放大率和像的性质。
Using the lens formula 1/f = 1/u + 1/v, where f = 10 cm, u = 30 cm (real object, positive). Rearranging: 1/v = 1/f – 1/u = 1/10 – 1/30 = (3 – 1)/30 = 2/30 = 1/15, so v = 15 cm.
利用透镜公式 1/f = 1/u + 1/v，其中 f = 10 cm，u = 30 cm（实物，正）。整理得 1/v = 1/f – 1/u = 1/10 – 1/30 = (3 – 1)/30 = 2/30 = 1/15，所以 v = 15 cm。
Magnification m = v/u = 15/30 = 0.5. Image height = m × object height = 0.5 × 5 = 2.5 cm. Since v is positive, the image is real and inverted. The image is smaller than the object, placed 15 cm on the opposite side of the lens.
放大率 m = v/u = 15/30 = 0.5。像高 = m × 物高 = 0.5 × 5 = 2.5 cm。由于 v 为正，像是倒立的实像。像比物体小，位于透镜另一侧 15 cm 处。
1/f = 1/u + 1/v
This case illustrates how lens equations predict image properties.
这个案例说明了透镜公式如何预测像的性质。
6. Measuring Specific Heat Capacity of Aluminium | 铝的比热容测量
A 0.5 kg aluminium block is heated by a 50 W electric heater for 5 minutes. Its temperature rises from 20 °C to 32 °C. Calculate the specific heat capacity of aluminium. Assume no heat loss.
一个 0.5 kg 的铝块用 50 W 的电加热器加热 5 分钟。温度从 20 °C 升至 32 °C。计算铝的比热容。假设无热量损失。
Electrical energy supplied E = P × t = 50 W × (5 × 60 s) = 50 × 300 = 15000 J. This energy is absorbed as thermal energy: Q = mcΔθ, where Δθ = 32 – 20 = 12 °C.
提供的电能 E = P × t = 50 W × (5 × 60 s) = 50 × 300 = 15000 J。这部分能量被吸收为热能：Q = mcΔθ，其中 Δθ = 32 – 20 = 12 °C。
Rearrange: c = Q / (mΔθ) = 15000 / (0.5 × 12) = 15000 / 6 = 2500 J/(kg °C). In practice, the experimental value may be higher due to heat loss to the surroundings.
整理得 c = Q / (mΔθ) = 15000 / (0.5 × 12) = 15000 / 6 = 2500 J/(kg °C)。实际上，由于向周围散热，实验值可能偏高。
This experiment also tests understanding of power, time conversion, and energy conservation.
这个实验也考察了对功率、时间换算和能量守恒的理解。
7. Ripple Tank Investigation of Waves | 水波槽波动研究
In a ripple tank, plane waves of frequency 15 Hz travel across a boundary from deep to shallow water. The wavelength in deep water is 3 cm, and in shallow water it is 2 cm. Calculate the wave speeds and explain the change.
在水波槽中，频率为 15 Hz 的平面波从深水区越过边界进入浅水区。深水中波长为 3 cm，浅水中波长为 2 cm。计算波速并解释变化。
Wave speed v = fλ. In deep water: v₁ = 15 × 3 = 45 cm/s. In shallow water: v₂ = 15 × 2 = 30 cm/s. Frequency remains constant as waves cross the boundary; thus the speed decreases because wavelength decreases.
波速 v = fλ。深水中：v₁ = 15 × 3 = 45 cm/s。浅水中：v₂ = 15 × 2 = 30 cm/s。当波越过边界时频率保持不变，因此由于波长减小，波速减小。
The wave equation v = fλ is fundamental. If the wave is refracted, the direction changes according to Snell’s law, but here we focus on the speed-wavelength relationship.
波动方程 v = fλ 是基础。如果波发生折射，方向会根据斯涅尔定律改变，但这里我们重点讨论速度与波长的关系。
v = fλ
Careful measurements in a ripple tank require a stroboscope or video to freeze the wave pattern.
在水波槽中精确测量需要使用频闪仪或视频来冻结波形。
8. Determining Density of an Irregular Stone | 不规则石块的密度测定
A stone of irregular shape has a mass of 180 g. When immersed in a measuring cylinder containing water, the water level rises from 50 cm³ to 110 cm³. Calculate the density of the stone in g/cm³ and kg/m³.
一块形状不规则的石块质量为 180 g。将其浸入盛水的量筒中，水面从 50 cm³ 上升至 110 cm³。计算石块的密度，单位用 g/cm³ 和 kg/m³。
Volume of stone = change in water volume = 110 – 50 = 60 cm³. Density ρ = mass/volume = 180 g / 60 cm³ = 3.0 g/cm³.
石块的体积 = 排水体积变化 = 110 – 50 = 60 cm³。密度 ρ = 质量/体积 = 180 g / 60 cm³ = 3.0 g/cm³。
To convert to kg/m³: 1 g/cm³ = 1000 kg/m³, so 3.0 g/cm³ = 3000 kg/m³. This method uses Archimedes’ principle indirectly, assuming the stone sinks completely.
转换为 kg/m³：1 g/cm³ = 1000 kg/m³，因此 3.0 g/cm³ = 3000 kg/m³。这个方法间接利用了阿基米德原理，假设石块完全沉没。
Irregular volumes are often measured by displacement, a skill required in practical assessments.
不规则物体的体积通常用排水法测量，这是实验考核中要求的技能。
Published by TutorHao | Physics Revision Series | aleveler.com
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