Case Study Practice for Year 12 Cambridge Physics | Year 12 Cambridge 物理：案例分析实战演练

📚 Case Study Practice for Year 12 Cambridge Physics | Year 12 Cambridge 物理：案例分析实战演练

In the Cambridge AS Physics examination, case study questions challenge your ability to apply physical principles to unfamiliar, real-world situations. These problems often require estimation, modelling, and critical evaluation of assumptions. Mastering case study skills builds your confidence in tackling both structured and unstructured numerical problems. This guide provides practical strategies and worked examples to help you excel.

在剑桥AS物理考试中，案例分析题考察你将物理原理应用于陌生、真实场景的能力。这类问题往往需要估算、建模以及对假设的批判性评估。掌握案例分析技能能提升你处理结构化和非结构化计算问题的信心。本指南提供实用的策略和例题，助你取得优异成绩。

1. What is a Case Study? | 什么是案例分析？

A case study in physics presents a scenario drawn from everyday life, technology, or nature. You must identify relevant physical quantities, make reasonable estimates, and perform calculations to answer questions. The goal is not exact precision but sensible order-of-magnitude reasoning.

物理中的案例分析提出了一个来自日常生活、技术或自然的场景。你需要识别相关的物理量，做出合理的估算，并通过计算来回答问题。目标不是精确无误，而是合理的数量级推理。

Typical tasks include estimating the energy consumption of a household device, the force exerted by a mosquito, or the number of atoms in a grain of sand. Unlike traditional textbook problems, case studies may provide excess or insufficient data. You must select relevant information and justify your assumptions.

常见的任务包括估算家用电器的能耗、蚊子施加的力，或者一粒沙中的原子数。与传统的教科书问题不同，案例分析可能提供过多或不足的数据。你必须选择相关信息并证明你的假设是合理的。

2. Estimation Strategies | 估算策略

Effective estimation begins with identifying key variables and breaking down the problem into manageable steps. Use physical laws as a framework. For example, to estimate the power output of a wind turbine, consider the kinetic energy of air passing through the rotor area per second.

有效的估算从识别关键变量并将问题分解为可处理的步骤开始。以物理定律为框架。例如，要估算风力发电机的输出功率，可考虑每秒通过转子区域空气的动能。

A systematic approach: 1) Draw a simple diagram showing the geometry or energy flow. 2) List the relevant equations. 3) Estimate the order of magnitude of each quantity. 4) Compute and check whether the result is plausible.

系统方法：1) 绘制简单的示意图显示几何或能量流动。2) 列出相关方程。3) 估算每个量的数量级。4) 计算并检验结果是否合理。

When estimating, memorising a few benchmark values speeds up your work. Below is a set of commonly used quantities.

在估算时，记住一些基准值能提高效率。下面是一组常用量。

Quantity	Typical value
Mass of an adult	70 kg
Walking speed	1.5 m s⁻¹
Car speed (urban)	13 m s⁻¹ (≈ 30 mph)
Density of water	1000 kg m⁻³
Atmospheric pressure	1.0 × 10⁵ Pa
g, gravitational acceleration	9.8 m s⁻²
Boltzmann constant k	1.38 × 10⁻²³ J K⁻¹
Avogadro constant Nₐ	6.02 × 10²³ mol⁻¹

Use these numbers as anchors to derive other estimates quickly, but always state the values you have assumed.

使用这些数值作为锚点可快速推导其他估算值，但务必声明你所假设的数值。

3. Orders of Magnitude | 数量级推理

Often, the answer is expected to the nearest power of ten. This reduces calculation demands and helps check internal consistency. For instance, the number of heartbeats in a lifetime can be estimated as (70 beats min⁻¹) × (525,600 min year⁻¹) × (80 years) ≈ 3 × 10⁹.

答案常要求精确到最近的十次幂。这降低了计算量并有助于检验自洽性。例如，一生中心跳次数可估算为（70次/分）×（525,600分/年）×（80年）≈ 3 × 10⁹。

Practice converting any approximate quantity into standard form: 0.0005 m = 5 × 10⁻⁴ m, and combining exponents using index laws. If a calculated number is unexpectedly large or small, revisit your assumptions.

练习将任何近似量转换为标准形式：0.0005米 = 5 × 10⁻⁴米，并使用指数律合并指数。如果算出的数字异常大或小，就要重新审视你的假设。

A rough order-of-magnitude check is often sufficient to eliminate implausible multiple-choice options in an exam.

粗略的数量级检验通常足以排除考试中不合理的选择题选项。

4. Estimating Common Quantities | 日常物理量估计

Common AS-level case studies ask you to estimate the mass of a textbook, the volume of a room, or the pressure exerted by a stiletto heel. To estimate volume, you can model the object as a simple shape, e.g., a cylinder or cuboid.

常见的AS级案例分析会要求你估计一本教科书的质量、房间的体积，或细高跟施加的压强。估算体积时，你可以将物体建模为简单形状，例如圆柱体或长方体。

For example, estimate the mass of air in a classroom. Room dimensions: 10 m × 8 m × 3 m, volume = 240 m³. Density of air ≈ 1.2 kg m⁻³, so mass ≈ 288 kg. The calculation is straightforward once the appropriate density is recalled.

例如，估算教室中空气的质量。房间尺寸：10 m × 8 m × 3 m，体积 = 240 m³。空气密度 ≈ 1.2 kg m⁻³，因此质量 ≈ 288 kg。一旦记起合适的密度，计算就很简单。

mass = density × volume

Remember to check unit consistency; here all units are SI, so the result is in kilograms.

记住检查单位一致性；这里所有单位都是国际单位制，因此结果以千克为单位。

5. Geometric Modeling and Assumptions | 几何建模与假设

Real objects are rarely perfect geometric shapes, so the key is to make reasonable approximations. For example, a human body can be modelled as a cylinder of height 1.7 m and radius 0.15 m for thermal radiation estimates. Clearly state your assumptions.

真实物体很少是完美的几何形状，因此关键在于做出合理的近似。例如，在估算热辐射时，人体可建模为高1.7米、半径0.15米的圆柱体。要明确陈述你的假设。

When estimating the area of a leaf, you might treat it as a rectangle or ellipse. Include justifications: ‘I assume the leaf is elliptical because its shape is roughly symmetrical.’

估算树叶面积时，可将其视为矩形或椭圆形。要包括理由：“我假设叶子是椭圆形的，因为它大致对称。”

Drawing a clear diagram with labelled dimensions earns marks and reduces errors. Even a simple sketch helps you visualise the geometry and choose the right formula.

绘制带有标注尺寸的清晰简图能得分并减少错误。即使一个简单的草图也有助于你想象几何结构并选择合适的公式。

If the problem asks for the volume of a water droplet, assume a sphere of diameter 2 mm. V = 4/3 π r³ ≈ 4/3 × 3.14 × (10⁻³ m)³ ≈ 4.2 × 10⁻⁹ m³.

如果问题询问水滴的体积，假设直径为2 mm的球体。V = 4/3 π r³ ≈ 4/3 × 3.14 × (10⁻³ m)³ ≈ 4.2 × 10⁻⁹ m³。

6. Energy and Power Analysis | 能量与功率分析

Energy case studies often involve calculating the energy content of food, the power of a climbing athlete, or the efficiency of a solar panel. Use P = E/t and consider conversions between joules, kilowatt-hours, and calories.

能量案例分析通常涉及计算食物中的能量、攀岩运动员的功率，或太阳能电池板的效率。使用P = E/t，并考虑焦耳、千瓦时和卡路里之间的换算。

Estimate the chemical energy in a slice of pizza. Assume 30 g fat, 50 g carbohydrate, 15 g protein. Energy from fat: 30 × 37 kJ ≈ 1110 kJ; carbohydrate: 50 × 17 kJ ≈ 850 kJ; protein: 15 × 17 kJ ≈ 255 kJ. Total ≈ 2.2 × 10⁶ J.

估算一片比萨饼中的化学能。假设含30 g脂肪、50 g碳水化合物、15 g蛋白质。脂肪提供的能量：30 × 37 kJ ≈ 1110 kJ；碳水化合物：50 × 17 kJ ≈ 850 kJ；蛋白质：15 × 17 kJ ≈ 255 kJ。总计 ≈ 2.2 × 10⁶ J。

E = m_fat × 37 kJ g⁻¹ + m_carb × 17 kJ g⁻¹ + m_protein × 17 kJ g⁻¹

To estimate climbing power, use P = mgh/t. A 70 kg person climbing 5 m in 10 s delivers P = 70 × 9.8 × 5 / 10 ≈ 343 W.

估算攀爬功率，使用P = mgh/t。一个70 kg的人用10 s爬高5 m，其功率P = 70 × 9.8 × 5 / 10 ≈ 343 W。

Always comment on efficiency. Biological systems are rarely 100% efficient; the actual metabolic power is much higher.

永远要评论效率。生物系统很少达到100%效率；实际的代谢功率要高得多。

7. Estimating Microscopic Quantities | 微观现象估计

Case studies at the atomic scale require Avogadro’s number and molar masses. For instance, estimate the number of water molecules in a glass of water. Water volume = 200 cm³ = 2 × 10⁻⁴ m³, mass = ρV = 1000 kg m⁻³ × 2 × 10⁻⁴ m³ = 0.2 kg. Number of moles = 200 g / 18 g mol⁻¹ ≈ 11.1 mol. Molecules = 11.1 × 6.02 × 10²³ ≈ 6.7 × 10²⁴.

原子尺度的案例分析需要阿伏伽德罗常数和摩尔质量。例如，估算一杯水中的水分子数。水体积 = 200 cm³ = 2 × 10⁻⁴ m³，质量 = ρV = 1000 kg m⁻³ × 2 × 10⁻⁴ m³ = 0.2 kg。摩尔数 = 200 g / 18 g mol⁻¹ ≈ 11.1 mol。分子数 = 11.1 × 6.02 × 10²³ ≈ 6.7 × 10²⁴。

Similarly, you could estimate the thickness of an oil film using molecular size assumptions. A drop of oil of volume 0.05 cm³ spreads into a film of diameter 0.3 m. Area ≈ πr² ≈ 7.1 × 10⁻² m², so thickness = volume/area ≈ 7 × 10⁻¹⁰ m, close to a molecular monolayer.

类似地，你可利用分子大小假设估算油膜厚度。一滴体积为0.05 cm³的油扩散成直径0.3 m的薄膜。面积 ≈ πr² ≈ 7.1 × 10⁻² m²，因此厚度 = 体积/面积 ≈ 7 × 10⁻¹⁰ m，接近一个分子单层。

When dealing with extremely small or large numbers, use standard form throughout to avoid mistakes with zeros.

处理极小的或极大的数字时，全程使用标准形式以避免零的个数出错。

8. Case Study in Electromagnetism | 电磁学案例分析

Consider estimating the resistance of a toaster from its power rating and mains voltage. A typical toaster is 800 W at 230 V. Using P = V²/R, we get R = V²/P = (230)²/800 ≈ 66 Ω. Then, you might be asked to estimate the length of nichrome wire needed, given its resistivity.

考虑从功率额定值和市电电压估算烤面包机的电阻。典型烤面包机功率800 W，电压230 V。根据P=V²/R，得R=V²/P=(230)²/800≈66 Ω。接着，你可能需要估计给定电阻率时所需的镍铬丝长度。

Resistivity of nichrome ≈ 1.1 × 10⁻⁶ Ω m. For a wire of cross-sectional area 1.0 × 10⁻⁷ m², length L = R A / ρ = 66 × 1.0 × 10⁻⁷ / (1.1 × 10⁻⁶) ≈ 6.0 m.

镍铬电阻率 ≈ 1.1 × 10⁻⁶ Ω m。对于截面面积1.0 × 10⁻⁷ m²的导线，长度L = R A / ρ = 66 × 1.0 × 10⁻⁷ / (1.1 × 10⁻⁶) ≈ 6.0 m。

Another common case: estimate the magnetic force on a current-carrying wire in a school experiment. With B ≈ 0.02 T (a typical permanent magnet), I = 2.0 A, and wire length 0.10 m perpendicular to the field, F = B I L = 0.02 × 2.0 × 0.10 = 4.0 × 10⁻³ N.

另一个常见案例：估算学校实验中载流导线所受的磁力。B ≈ 0.02 T（典型永磁体），I = 2.0 A，导线长度0.10 m且垂直于磁场，则F = B I L = 0.02 × 2.0 × 0.10 = 4.0 × 10⁻³ N。

Such forces are small but measurable. Always note the direction using Fleming’s left-hand rule if required.

这样的力很小但可测。若有需要，务必使用弗莱明左手定则注明方向。

9. Interpreting Experimental Data | 实验数据解读

Case studies may present graphs, tables, or raw experimental data with uncertainties. You must calculate gradients, intercepts, or use the data to determine a physical constant. Always consider percentage uncertainty and error propagation.

案例分析可能给出图形、表格或带有不确定度的原始实验数据。你必须计算斜率、截距，或使用数据确定物理常数。始终考虑百分不确定度和误差传播。

Example: using a

Published by TutorHao | Year 12 Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Case Study Practice for Year 12 Cambridge Physics | Year 12 Cambridge 物理：案例分析实战演练