📚 Case Study Practice for KS3 CIE Physics | KS3 CIE 物理：案例分析实战演练
Case studies are a powerful way to link physics principles with real-world situations. In KS3 CIE Physics, you are often asked to interpret data, perform calculations, and explain phenomena using scientific reasoning. This article provides a series of practical case study exercises, covering key topics such as energy, electricity, forces, density, waves, and thermal physics. Each case is presented with a scenario, followed by guided analysis and step-by-step solutions. By working through these examples, you will strengthen your problem-solving skills and deepen your understanding of how physics applies to everyday life.
案例分析是将物理原理与现实情境联系起来的有效方法。在 KS3 CIE 物理中，你经常需要解释数据、进行计算并运用科学推理解释现象。本文提供了一系列实用的案例分析练习，涵盖能量、电学、力、密度、波和热物理等关键主题。每个案例都提供了场景，然后是引导性分析和逐步解答。通过这些练习，你将增强解决问题的能力，加深对物理学如何应用于日常生活的理解。
1. How Case Study Questions Work in KS3 CIE Physics | KS3 CIE 物理案例分析题是如何设计的
A case study question is not simply a calculation question. It usually gives you a short realistic context - a kettle, a lamp, a cyclist, a cup of tea - and then asks you to do three different things with it: read information carefully, use an equation correctly, and explain the result using a physics idea. The context is there to test whether you can transfer a law you learned in a textbook to a situation you have never seen before.
案例分析题并不只是计算题。它通常给出一个简短的真实情境——水壶、灯泡、骑车的人、一杯茶——然后要求你围绕它做三件事：仔细读取信息、正确使用公式、并用物理概念解释结果。情境的作用是检验你能否把课本上学到的规律迁移到一个从未见过的新场景中。
Most KS3 CIE case studies follow a predictable pattern of command words. Recognising these words tells you exactly how much detail is required and how many marks are available, so you never write too little or waste time writing too much.
大多数 KS3 CIE 案例题都遵循可预测的指令词模式。识别这些指令词可以告诉你需要写多少细节、能拿到多少分，从而既不会写得太少，也不会浪费时间写太多。
| Command word | What the examiner wants | 中文说明 |
| State / Give | One short fact, no explanation | 写一个简短事实，无需解释 |
| Calculate | Equation, substitution, answer with unit | 写公式、代入、带单位作答 |
| Explain | Cause and effect, linked physics reason | 写因果链，用物理原因连接 |
| Suggest | Apply physics to a new context | 把物理应用到新情境 |
| Evaluate / Compare | Weigh two options with data as evidence | 用数据作证据比较两种方案 |
Every worked case below is written in the same four-step routine: identify the physics, choose the equation, substitute the values in base units, then write a final sentence that answers the question in words. Use this routine on every case study you meet.
下面每一个案例都采用同样的四步法：识别物理量、选择公式、用基本单位代入数值、最后用一句话回答题目。遇到任何案例分析题，都套用这套流程。
2. The Electric Kettle – Efficiency in Action | 电水壶案例 – 效率的实际计算
A student uses an electric kettle rated at 2000 W to heat 0.5 kg of water. The water temperature rises from 20°C to 100°C. The kettle is switched on for exactly 2 minutes (120 seconds). The specific heat capacity of water is 4200 J/(kg °C). Let's calculate the energy efficiency of the kettle.
一位学生使用额定功率为 2000 W 的电水壶加热 0.5 kg 的水。水温从 20°C 上升到 100°C。水壶恰好接通 2 分钟（120 秒）。水的比热容为 4200 J/(kg °C)。我们来计算水壶的能量效率。
The electrical energy supplied to the kettle is calculated using E = P × t: E₍input₎ = 2000 W × 120 s = 240,000 J.
供给水壶的电能可以用 E = P × t 计算：E_输入 = 2000 W × 120 s = 240,000 J。
The useful thermal energy gained by the water is Q = m × c × Δθ: Δθ = 100°C - 20°C = 80°C. So Q = 0.5 kg × 4200 J/(kg °C) × 80°C = 168,000 J.
水获得的有用热能为 Q = m × c × Δθ：Δθ = 100°C - 20°C = 80°C。因此 Q = 0.5 kg × 4200 J/(kg °C) × 80°C = 168,000 J。
Efficiency = (useful energy output ÷ total energy input) × 100%
Efficiency = (168,000 J ÷ 240,000 J) × 100% = 70%
Therefore the efficiency of the kettle is 70%. The missing 30% - 72,000 J - is not lost. It has been transferred by conduction and convection to the kettle body, the air around it and the worktop, warming them slightly. Energy is always conserved; it is only the useful fraction that changes.
因此，水壶的效率为 70%。缺少的 30%——72,000 J——并没有消失，而是通过传导和对流转移到水壶外壳、周围空气和台面上，使它们稍微变热。能量总是守恒的，变化的只是"有用的"那一部分的比例。
A good case study answer adds one evaluation sentence. For example: the kettle would be more efficient if it were insulated or if it had a lid that trapped steam, because less energy would be transferred to the surroundings before the water boils.
优秀的案例分析答案还会再加一句评价。例如：如果水壶有保温层，或者盖子能锁住水蒸气，它的效率会更高，因为在沸腾之前向周围环境转移的能量会更少。
3. The Torch Battery – Charge, Current and Energy | 手电筒电池 – 电荷、电流与能量
A torch uses two 1.5 V cells in series, giving a total potential difference of 3.0 V. The bulb draws a steady current of 0.25 A and the torch is switched on for 600 seconds. Work out the charge that flows, the energy transferred, and the chemical energy used from each cell.
一只手电筒串联使用两节 1.5 V 电池，总电压为 3.0 V。灯泡的电流稳定在 0.25 A，手电筒持续点亮 600 秒。请计算流过的电荷量、转移的能量，以及每节电池消耗的化学能。
Charge is the product of current and time, Q = I × t. Substituting: Q e ampere of current flowing for one second, so 150 C means 150 A s of charge passed through the bulb.
电荷量是电流与时间的乘积，Q = I × t。代入得 Q = 0.25 A × 600 s = 150 C。1 库仑就是 1 安培的电流持续 1 秒，因此 150 C 表示有 150 A·s 的电荷通过灯泡。
Energy transferred is found with E = V × Q, or equivalently E = V × I × t. Both give E = 3.0 V × 150 C = 3.0 V × 0.25 A × 600 s = 450 J.
转移的能量用 E = V × Q 计算，也可以写成 E = V × I × t。两种写法结果相同：E = 3.0 V × 150 C = 3.0 V × 0.25 A × 600 s = 450 J。
Each cell provides 1.5 V of the total, so each cell transfers E into electrical energy. The two cells together therefore supply 450 J, which matches the total - a useful check that your answer is consistent.
每节电池提供总电压中的 1.5 V，因此每节电池把 E = 1.5 V × 150 C = 225 J 的化学能转化为电能。两节电池合计提供 450 J，与总量一致，这是检验答案是否自洽的好方法。
In the torch, the useful output is light and a little heat from the filament. In an incandescent bulb only about 5% of the 450 J becomes visible light; about 427 J heats the filament and the surrounding air. This is why LED torches are far more efficient: they convert a much larger fraction of the electrical energy into light instead of thermal energy.
在手电筒中，有用的输出是光以及灯丝产生的少量热。白炽灯泡大约只有 5% 的 450 J 变成可见光，约 427 J 用于加热灯丝和周围空气。这就是 LED 手电筒效率高得多的原因：它们把更大比例的的电能转化为光而不是热能。
4. The Skateboarder on the Ramp – Forces and Energy | 坡道上的滑板少年 – 力与能量
A skateboarder of mass 60 kg starts from rest at the top of a smooth ramp, 5 m above the ground. Use g ravitational potential energy at the top, and the speed at the bottom assuming no energy losses.
一位质量为 60 kg 的滑板少年从离地 5 m 高的光滑坡道顶端由静止出发。取 g = 10 N/kg。请计算他的重力、顶端时的重力势能，以及假设没有能量损失时到达底端的速度。
Weight is a force and is calculated with W = m × g: W arefully: mass is measured in kilograms, weight in newtons, and the two are not the same quantity.
重力是一个力，用 W = m × g 计算：W = 60 kg × 10 N/kg = 600 N。注意单位：质量以千克计，重力以牛顿计，两者不是同一个物理量。
The gravitational potential energy stored at the top is Eₚ = m × g × h = 60 kg × 10 N/kg × 5 m = 3000 J.
顶端储存的重力势能为 Eₚ = m × g × h = 60 kg × 10 N/kg × 5 m = 3000 J。
With no friction or air resistance, all of this becomes kinetic energy at the bottom, so Eₖ = ½ × m × v² = 3000 J. Rearranging: v² = 2 × 3000 J ÷ 60 kg = 100 m²/s², so v = 10 m/s.
在没有摩擦和空气阻力的情况下，这些势能在底端全部转化为动能，所以 Eₖ = ½ × m × v² = 3000 J。整理得 v² = 2 × 3000 J ÷ 60 kg = 100 m²/s²，因此 v = 10 m/s。
In reality the skateboarder arrives more slowly, because friction in the bearings and air resistance transfer some of the 3000 J to the thermal energy store of the wheels, the ramp and the air. The energy has not disappeared - it has been dissipated to the surroundings, exactly as in the kettle case.
实际上滑板少年到达底端的速度会慢一些，因为轴承中的摩擦和空气阻力把 3000 J 中的一部分转移到了轮子、坡道和空气的热能储存中。能量并没有消失——它被耗散到周围环境中，与水壶案例完全一致。
A common exam extension is to ask how the answer changes if the mass of the skateboarder doubles. The potential energy doubles to 6000 J, but because kinetic energy also contains the mass, v² = 2gh does not contain mass at all, so the final speed stays 10 m/s. Heavier objects are not faster on a frictionless ramp.
常见的考试拓展问题是：如果滑板少年的质量加倍，答案会如何变化？重力势能加倍到 6000 J，但由于动能中也含有质量，v² = 2gh 根本不含质量，所以最终速度仍是 10 m/s。在无摩擦坡道上，更重的物体并不会更快。
5. The Unknown Metal – Density by Displacement | 未知金属 – 排水法测密度
A student is given an irregular piece of metal. She finds its mass on a balance: 540 g. She then lowers it into a measuring cylinder containing 60 cm³ of water. The water level rises to 260 cm³. Identify the metal by calculating its density.
一位学生拿到一块形状不规则的金属。她用天平测得其质量为 540 g。然后她把金属放入装有 60 cm³ 水的量筒中，水面上升到 260 cm³。请通过计算密度来判断这是什么金属。
The volume of the metal equals the volume of water displaced: V nly if the metal sinks completely and no water splashes out; a student should always read the bottom of the meniscus at eye level to avoid a parallax error.
金属的体积等于被排开水的体积：V = 260 cm³ - 60 cm³ = 200 cm³。这一方法成立的前提是金属完全沉入且水没有溅出；读数时应让视线与凹液面最低处齐平，以避免视差误差。
Density = mass ÷ volume
ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³
1 g/cm³ = 1000 kg/m³. Comparing with a data table, 2700 kg/m³ is the density of aluminium.
换算成国际单位：2.7 g/cm³ = 2700 kg/m³，因为 1 g/cm³ = 1000 kg/m³。与数据表对照，2700 kg/m³ 正是铝的密度。
- Practical checks: dry the metal before weighing it, use enough water to cover the sample, and repeat the reading three times to reduce random error. / 实验要点：称量前把金属擦干，水量要没过样品，读数重复三次以减小随机误差。
- Why displacement works: two pieces of matter cannot occupy the same space, so the metal pushes exactly its own volume of water out of the way. / 为什么排水法有效：两物体不能同时占据同一空间，因此金属会把与自己体积相等的水排开。
- Fair-test extension: if you compare different metals, keep the measuring cylinder and the starting water volume the same. / 控制变量拓展：比较不同金属时应使用同一个量筒和相同的起始水量。
6. The Ripple Tank – Wave Speed and Frequency | 水波槽 – 波速与频率
A ripple tank produces straight water waves with a frequency of 5 Hz. The distance between successive wave crests is measured as 4 cm. Calculate the wave speed, the period, and how long a wave takes to travel 60 cm across the tank.
水波槽产生频率为 5 Hz 的直线水波。测得相邻两个波峰之间的距离为 4 cm。请计算波速、周期，以及波穿过 60 cm 长的水槽所需的时间。
First convert the wavelength into metres: λ = 4 cm = 0.04 m. The wave speed is given by v = f × λ = 5 Hz × 0.04 m = 0.20 m/s.
先把波长换算成米：λ = 4 cm = 0.04 m。波速由 v = f × λ 给出：v = 5 Hz × 0.04 m = 0.20 m/s。
T = 1 ÷ f = 1 ÷ 5 Hz = 0.2 s
The period is the time for one complete wave, 0.2 s. The time to cross 60 cm (0.60 m) is t .0 s. As a check, 3.0 s is exactly 15 periods, so 15 wavefronts would arrive at the far edge.
周期是完成一个完整波的时间，为 0.2 s。穿过 60 cm（0.60 m）所需时间为 t = 距离 ÷ 速度 = 0.60 m ÷ 0.20 m/s = 3.0 s。检验一下：3.0 s 恰好是 15 个
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