📚 Year 10 Cambridge Physics: Case Study Practice | 剑桥10年级物理案例分析实战演练
Welcome to a hands-on, case-study-based revision session tailored for Year 10 Cambridge Physics. In this article, you will work through ten carefully chosen scenarios that mirror the style of IGCSE structured questions. Each case targets a core topic, from energy conservation and circuit diagnostics to waves and radioactivity. By analysing real-world problems step by step, you will strengthen your ability to interpret data, apply equations, and justify your reasoning — exactly what examiners look for. Let's dive in and turn theory into confident problem-solving skill.
欢迎来到专为剑桥10年级物理设计的案例分析实战演练。本文精选了十个贴近IGCSE考题风格的场景，逐一拆解核心知识点，涵盖能量守恒、电路故障、波动和放射性等内容。通过逐步分析真实问题，你将提升解读数据、运用公式以及合理解释的能力——这正是考试评分的关键。让我们开始，把理论知识转化为从容的解题自信。
1. Roller Coaster Energy Conversion | 过山车能量转换
A roller coaster car of mass 520 kg starts from rest at the top of a track, 25 m above the lowest point. Assuming friction and air resistance are negligible, we can analyse the energy conversion between gravitational potential energy (GPE) and kinetic energy (KE).
一辆质量为 520 kg 的过山车从轨道最高点静止释放，最高点距离最低点 25 m。假设摩擦和空气阻力忽略不计，我们可以分析重力势能和动能之间的转换。
Step 1 – GPE: Eₚ = mgh = 520 × 9.8 × 25 = 127 400 J. This is the total mechanical energy available.
步骤1——计算初始重力势能：Eₚ = mgh = 520 × 9.8 × 25 = 127 400 J。这是可利用的总机械能。
Step 2 – int, all GPE has become KE. Using Eₖ = ½mv², set 127 400 = ½ × 520 × v². Solving gives v = √2 × 127 400 / 520 ≈ 22.1 m/s. The speed does not depend on mass.
步骤2——在最低点，所有重力势能转化为动能。利用 Eₖ = ½mv²，列出 127 400 = ½ × 520 × v²。解出 v = √2 × 127 400 / 520 ≈ 22.1 m/s。速度与质量无关。
Now consider a more realistic case where 15% lost to friction. The useful KE becomes 0.85 × 127 400 = 108 290 J. The new speed would be √2 × 108 290 / 520 . Always show that you understand energy cannot be created or destroyed, only transferred.
现在考虑更贴近实际的情况：15% 的能量因摩擦损失。有用的动能变为 0.85 × 127 400 = 108 290 J。此时速度约为 √2 × 108 290 / 520 ≈ 20.4 m/s。牢记能量不能凭空产生或消失，只能转化或转移。
2. Spring Compression and Stopping Distance | 弹簧压缩与停止距离
A dynamics trolley of mass 0.80 kg moves at 1.5 m/s on a frictionless track before hitting a spring of stiffness 300 N/m. We need to find the maximum compression x.
一辆质量为 0.80 kg 的动力学小车在无摩擦轨道上以 1.5 m/s 运动，撞上一个劲度系数为 300 N/m 的弹簧。求弹簧的最大压缩量 x。
ₑ = ½kx². So 0.90 = ½ × 300 × x², which gives x² = 0.0060 and x = 0.0775 m (about 7.8 cm).
初始动能为 Eₖ = ½ × 0.80 × (1.5)² = 0.90 J。在最大压缩处，动能全部转化为弹性势能：Eₑ = ½kx²。于是 0.90 = ½ × 300 × x²，解得 x² = 0.0060，x = 0.0775 m（约 7.8 cm）。
ts: N/m × m² = N·m = J, consistent. In an exam, always state the energy conversion clearly — kinetic → elastic potential — and show the substitution step before rounding the final answer.
检查单位：N/m × m² = N·m = J，保持一致。在考试中，务必清晰地写明能量转换方向——动能 → 弹性势能——并展示代入步骤，最后再取近似值。
3. Series Circuit Troubleshooting | 串联电路故障诊断
A simple series circuit consists of a 6.0 V battery, a switch, an ammeter, and two identical lamps rated at 6.0 V. When the switch is closed, the ammeter reads 0 A and neither lamp lights. Using a voltmeter, you measure 6.0 V across the battery, 6.0 V across one lamp, and 0 V across the other. Diagnose the fault.
一个简单的串联电路包含一节 6.0 V 电池、一个开关、一个电流表和两盏额定电压 6.0 V 的相同灯泡。闭合开关后，电流表读数为 0 A，两灯都不亮。用电压表测量，电池两端为 6.0 V，其中一盏灯两端为 6.0 V，另一盏灯为 0 V。请诊断故障。
The 0 A reading confirms a break (open circuit) somewhere. The fact that the full battery voltage appears across one lamp indicates that the break is in that lamp — its filament is blown. There is a complete potential drop across the break, so the rest of the circuit has 0 V. The other lamp is intact but has no current.
电流读数为 0 A 确认电路中存在断路。整节电池的电压都出现在其中一盏灯两端，说明断路就在这盏灯——灯丝烧断。断路处承担了全部电势降落，因此电路其余部分电压为 0 V。另一盏灯完好，只是没有电流通过。
ou read the supply voltage. Always reference current = 0 to justify your fault conclusion.
这个案例突显了电压表诊断技巧：跨接在连续导体上电压接近 0 V；跨接在断路上会读到电源电压。务必结合电流为零的事实来佐证你的故障判断。
4. Motor Efficiency when Lifting a Load | 电动机吊起重物效率
An electric motor lifts a 2.5 kg mass through a vertical height of 1.8 m in 3.0 s. The motor operates at 12 V and draws a current of 2.2 A. Calculate the input power, output power, and efficiency.
一台电动机将 2.5 kg 的重物在 3.0 s 内竖直提升 1.8 m。电动机工作电压为 12 V，电流为 2.2 A。计算输入功率、输出功率和效率。
Input power Pin 12 × 2.2 = 26.4 W. The useful work done is mgh = 2.5 × 9.8 × 1.8 = 44.1 J. Output power Pout = work / time = 44.1 / 3.0 = 14.7 W. Efficiency η = (Pout / Pin) × 100% = (14.7 / 26.4) × 100% ≈ 55.7%. The remaining 44% is dissipated as heat and sound.
输入功率 Pin = V × I = 12 × 2.2 = 26.4 W。有用功为 mgh = 2.5 × 9.8 × 1.8 = 44.1 J。输出功率 Pout = 有用功 / 时间 = 44.1 / 3.0 = 14.7 W。效率 η = (Pout / Pin) × 100% = (14.7 / 26.4) × 100% ≈ 55.7%。剩下约 44% 以热和声的形式耗散。
In an efficiency problem, always make it clear which energy is useful and which is wasted. Even though the lifting is vertical, you do not need to consider kinetic energy because the mass moves at a slow, steady speed — the exam assumes a constant velocity or negligible KE change.
处理效率问题时，务必明确区分哪部分能量是有用的、哪部分是浪费的。尽管是竖直提升，由于重物缓慢匀速运动，考试中通常假定动能变化可忽略。
5. Convex Lens Image Formation | 凸透镜成像分析
A convex lens has a focal length of 12 cm. An object 4.0 cm tall is placed 18 cm from the lens. Determine the image distance, magnification, and describe the image.
一凸透镜焦距为 12 cm，一个 4.0 cm 高的物体置于距透镜 18 cm 处。求像距、放大率并描述像的性质。
age is real (v positive) and on the opposite side. Magnification m = v/u = 36/18 = 2. Image height = m × object height = 8.0 cm, inverted.
运用透镜公式：1/f = 1/u + 1/v，采用实正符号约定。f = 12 cm，u = 18 cm。计算 1/v = 1/12 − 1/18 = (3 − 2)/36 = 1/36，得 v = +36 cm。像为实像（v 为正），位于透镜另一侧。放大率 m = v/u = 36/18 = 2。像高 = 2 × 4.0 cm = 8.0 cm，倒立。
Check the ray diagram: the object is between f and 2f, so the image must be beyond 2f, magnified, and inverted — consistent with our calculation. Memorising this pattern helps catch errors quickly.
用光路图检验：物体位于 f 和 2f 之间，像必然在 2f 之外、放大且倒立——与计算结果一致。记住这个规律有助于快速发现错误。
6. Pressure in a Liquid | 液体压强案例分析
iver at the bottom, given water density 1000 kg/m³ and atmospheric pressure 1.01 × 10⁵ Pa.
一淡水水库水深 15 m，水的密度为 1000 kg/m³，大气压为 1.01 × 10⁵ Pa。求水底潜水员所受到的压强。
Pressure due to the water column: pliquid = ρgh = 1000 × 9.8 × 15 = 1.47 × 10⁵ Pa. Total pressure ptotal = patm + pliquid pprox. 2.5 atm). The pressure difference between the bottom and the surface is simply ρgh.
水柱产生的压强：pliquid = ρgh = 1000 × 9.8 × 15 = 1.47 × 10⁵ Pa。总压强 ptotal = patm + pliquid = 1.01 × 10⁵ + 1.47 × 10⁵ = 2.48 × 10⁵ Pa（约 2.5 个大气压）。底部与水面间的压强差即 ρgh。
Remember that liquid pressure depends only on depth, density, and g, not on the shape or total volume of the container. A common mistake is to multiply by mass or weight directly without converting to force per unit area.
注意液体压强仅取决于深度、密度和重力加速度 g，与容器形状或总水量无关。常见错误是直接用质量或重量相乘，而没有转换为力除以面积。
7. Specific Heat Capacity Investigation | 比热容实验分析
A student heats 0.50 kg of water using a 50 W immersion heater. The temperature rises from 22°C to 34°C in 8 minutes. Calculate the specific heat capacity of water from this data and comment on the accuracy.
一名学生用 50 W 浸没式加热器给 0.50 kg 水加热，温度在 8 分钟内从 22°C 升至 34°C。根据这些数据计算水的比热容，并评价准确性。
(kg·°C). The accepted value is 4 200 J/(kg·°C). The experiment underestimates c because heat is lost to the surroundings and the cup.
提供的能量 E = P × t = 50 W × (8 × 60 s) = 24 000 J。温升 Δθ = 12°C。假设无热损失，由 Q = mcΔθ 得 c = Q / (m Δθ) = 24 000 / (0.50 × 12) = 4 000 J/(kg·°C)。公认值为 4 200 J/(kg·°C)。实验值偏低，因为热量散失到环境和容器中。
To improve, the experiment should include insulation, stirring, and shielding from draughts. Understanding systematic errors is just as important as performing the calculation — exam questions often ask why the result is lower than the true value.
改进措施包括加强隔热、搅拌和挡风。理解系统误差与实际计算同等重要——考试中常会问为什么测得的比热容低于真实值。
8. Echo Sounding and Wave Speed | 回声测距与波速
A sonar system on a ship emits a sound pulse vertically downwards. The echo returns after 0.60 s. The speed of sound in seawater is 1500 m/s. Calculate the depth of the seabed.
船上的声呐垂直向下发射声脉冲，0.60 s 后收到回声。海水中声速为 1500 m/s。求海底深度。
m. Clarity is crucial: always state that the time is for the round trip and divide by two.
脉冲往返海底，故总路程为 2 倍深度。总路程 = 速度 × 时间 = 1500 × 0.60 = 900 m。所以深度 = 450 m。表述务必清晰：说明时间是往返时间，然后除以 2。
If the seabed is sloped or the boat is moving, the reading may be distorted, but at Year 10 level you can assume vertical transmission and a stationary vessel. This principle is also used in ultrasound scanning and radar.
若海底倾斜或船在移动，测量值会失真，但10年级水平可假定垂直发射且船静止。此原理同样应用于超声扫描和雷达。
9. Radioactive Half-Life Calculation | 放射性半衰期计算
A sample of iodine-131 has an initial mass of 80 mg. Its half-life is 8 days. How much remains after 24 days, and what mass has decayed?
一份碘-131 样品的初始质量为 80 mg，其半衰期为 8 天。24 天后剩余多少？衰变了多少？
0 − 10 = 70 mg. The exponential decay pattern can be shown directly: 80 → 40 → 20 → 10 mg.
经过的半衰期个数 n = 总时间 / T½ = 24 / 8 = 3。剩余比例 = (½)³ = ⅛。剩余质量 = 80 mg × ⅛ = 10 mg。已衰变质量 = 80 − 10 = 70 mg。可以通过逐代递减直观展示：80 → 40 → 20 → 10 mg。
Avoid the common slip-up of thinking the mass disappears after a few half-lives — it never reaches exactly zero; it becomes negligible for practical purposes. Examiners expect you to use the halving method or the formula and to express the answer to an appropriate number of significant figures.
避免常见误区：认为几个半衰期后质量就完全消失——实际上永远不会精确归零，只是实用上可以忽略。考官期望你用折半法或公式计算，并保留适当的有效数字。
10. Moments and Equilibrium | 力矩与平衡
A uniform metre rule is pivoted at its 50 cm mark. A 2.0 N weight hangs at the 20 cm mark. At what distance from the pivot on the other side must a 3.0 N weight be suspended to balance the rule?
一根均匀的米尺在 50 cm 刻度处支起。一个 2.0 N 的重物挂在 20 cm 刻度处。另一侧应在距支点多远处悬挂一个 3.0 N 的重物，使米尺水平平衡？
t = anticlockwise moment, so 3.0 × d₂ = 0.60, giving d₂ = 0.20 m = 20 cm from the pivot, on the 70 cm mark. Since the rule is uniform, its own weight acts through the centre and produces zero moment about the 50 cm pivot.
2.0 N 重物产生逆时针力矩：距离 d₁ = 50 − 20 = 30 cm = 0.30 m。力矩 = 2.0 × 0.30 = 0.60 N·m。平衡时，顺时针力矩 = 逆时针力矩，故 3.0 × d₂ = 0.60，得 d₂ = 0.20 m = 20 cm，即挂在 70 cm 处。由于米尺均匀，其自身重力通过中心，对 50 cm 支点不产生力矩。
Exam tip: always convert distances to metres when using SI units unless the question specifically uses cm in final answers. Draw a clear diagram of forces and label the pivot — it makes the direction of moments far easier to visualise.
考试提示：在 SI 单位制中，除非题目明确要求用 cm，否则应将距离转换为米。画出一张清晰的受力示意图并标注支点——这能让你更容易判定力矩方向。
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