📚 Year 11 CIE Physics: Case Study Practical Drills | Year 11 CIE 物理：案例分析实战演练
Case studies are a powerful way to apply your physics knowledge to real-world situations. In CIE IGCSE Physics, you are often required to analyse given data, interpret graphs, and use formulas to solve problems. This article presents 10 case study drills covering key topics such as kinematics, forces, energy, pressure, thermal physics, waves, electricity, electromagnetism, nuclear physics, and astronomy. Each drill includes a scenario, guiding questions, and a step-by-step solution to help you develop strong analytical skills.
案例分析是将物理知识应用于实际情境的有力方式。在 CIE IGCSE 物理考试中，你经常需要分析给定数据、解读图表，并使用公式解决问题。本文提供 10 个案例实战演练，涵盖运动学、力、能量、压强、热学、波动、电路、电磁学、核物理和天文学等重点主题。每个案例包括情景设定、引导性问题和分步解答，帮助你培养扎实的分析能力。
1. Kinematics – Safe Following Distance on Motorway | 运动学 – 高速公路安全距离分析
A driver is travelling at 30 m/s on a motorway when an obstacle appears ahead. After a reaction time of 0.7 s, the driver brakes hard, causing a uniform deceleration of 6.0 m/s². You need to calculate the total stopping distance, which is the sum of the thinking distance and the braking distance. This type of analysis is vital for road safety and understanding stopping distances under different conditions.
一名司机以 30 m/s 的速度在高速公路上行驶，前方出现障碍物。经过 0.7 秒的反应时间后，司机猛踩刹车，产生 6.0 m/s² 的匀减速。需要计算总停车距离，即反应距离与刹车距离之和。这类分析对道路安全以及理解不同条件下的停车距离至关重要。
, use v² = u² + 2as, where v = 0, u = 30 m/s, a = -6.0 m/s². Rearranging: 0 = 30² + 2(-6)s ⇒ 0 = 900 - 12s ⇒ s = 900/12 = 75 m. Total stopping distance = 21 + 75 = 96 m.
反应距离 = 速度 × 反应时间 = 30 × 0.7 = 21 m。刹车距离利用公式 v² = u² + 2as，其中 v = 0，u = 30 m/s，a = -6.0 m/s²。代入得：0 = 30² + 2(-6)s ⇒ 0 = 900 - 12s ⇒ s = 75 m。总停车距离 = 21 + 75 = 96 m。
Notice how the braking distance depends on the square of the initial speed, while the thinking distance is directly proportional to speed. This is why motorway speeds require much greater separation between vehicles.
注意，刹车距离与初速度的平方成正比，而反应距离与速度成正比。这就是高速公路上车辆之间需要保持更大间距的原因。
2. Forces and Motion – Forces at the Bottom of a Roller Coaster Loop | 力与运动 – 过山车最低点受力分析
A roller coaster car of mass 500 kg enters a vertical loop. At the bottom of the loop, its speed is 20 m/s and the radius of curvature is 10 m. Determine the normal contact force exerted by the track on the car at this point. This problem requires combining circular motion concepts with Newton's second law.
一辆质量 500 kg 的过山车通过一个竖直圆环。在圆环最低点，车速为 20 m/s，曲率半径为 10 m。求此时轨道对过山车的法向支持力。这道题需要将圆周运动概念与牛顿第二定律相结合。
ovides the centripetal force: N - mg = mv²/r. Therefore, N = mg + mv²/r = 500×9.8 + 500×(20²)/10 = 4900 + 500×400/10 = 4900 + 20000 = 24900 N. The passenger would experience an apparent weight of about 5 times their actual weight.
在最低点，作用力有向上的支持力 N 和向下的重力 mg。指向圆心的净力提供向心力：N - mg = mv²/r。因此，N = mg + mv²/r = 500×9.8 + 500×(20²)/10 = 4900 + 20000 = 24900 N。乘客会感受到大约实际体重 5 倍的视重。
This case illustrates why roller coaster loops are designed with a larger radius at the bottom to reduce the extreme forces on riders.
这个案例说明了为何过山车圆环底部设计成较大的半径，以降低对乘客的极端作用力。
3. Energy Conservation – Bungee Jump Elastic Potential Energy Conversion | 能量守恒 – 蹦极跳的弹性势能转化
A bungee jumper of mass 65 kg uses a cord with unstretched length 25 m and spring constant 150 N/m. The jumper falls from a bridge and the cord begins to stretch when the jumper is 25 m below the bridge. Calculate the maximum distance the jumper falls before momentarily coming to rest, assuming no non-conservative forces. This is a classic conservation of energy problem linking gravitational potential energy to elastic potential energy.
一名质量 65 kg 的蹦极者使用一条自然长度 25 m、弹性系数 150 N/m 的绳子。蹦极者从桥上跳下，绳在桥下 25 m 处开始伸长。假设没有非保守力，计算蹦极者在瞬时静止前下落的最大距离。这是一道经典的将重力势能转化为弹性势能的能量守恒问题。
ational PE = gain in elastic PE: mg(25 + x) = ½kx². Substitute: 65×9.8×(25 + x) = ½×150×x² ⇒ 637(25 + x) = 75x². Simplify: 15925 + 637x = 75x² ⇒ 75x² - 637x - 15925 = 0. Solve the quadratic: x = [637 ± √637² + 4×75×15925] / (2×75) = [637 ± √405769 + 4777500 king the positive root: x ≈ 2913.7/150 ≈ 19.4 m. Thus total distance = 25 + 19.4 = 44.4 m.
设总下落距离 H = 25 + x，其中 x 是超出原长的伸长量。减少的重力势能 = 增加的弹性势能：mg(25 + x) = ½kx²。代入数据：65×9.8×(25 + x) = ½×150×x² ⇒ 637(25 + x) = 75x² ⇒ 15925 + 637x = 75x² ⇒ 75x² - 637x - 15925 = 0。解二次方程得：x ≈ 19.4 m，因此总距离约为 44.4 m。
Energy methods simplify problems where acceleration is not constant. In reality, air resistance and internal cord friction would reduce this distance.
能量方法简化了加速度非均匀的问题。现实中，空气阻力和绳内摩擦会减小这个距离。
4. Density and Pressure – Hydraulic Jack Principle | 密度与压强 – 液压千斤顶的工作原理
A hydraulic jack has two connected cylinders: a small piston of area 2.0 cm² and a large piston of area 50 cm². A force of 40 N is applied to the small piston. The jack is used to lift a car part. Calculate the weight of the load that can be lifted, assuming the pistons are at the same height and Pascal's principle applies. This demonstrates the force multiplication in hydraulic systems.
一个液压千斤顶有两个相连的油缸：小活塞面积 2.0 cm²，大活塞面积 50 cm²。小活塞上施加 40 N 的力。千斤顶用于抬起汽车部件。假设两活塞在同一高度且帕斯卡原理适用，计算能被抬起的重物重量。这展示了液压系统的力放大作用。
Pascal' m². This same pressure acts on the large piston, so F₂ = P × A₂ = 20 N/cm² × 50 cm² = 1000 N. The load that can be lifted is 1000 N (mass ≈ 102 kg).
帕斯卡原理指出，施加于密闭液体的压强能大小不变地传递。小活塞上的压强：P = 40 N / 2.0 cm² = 20 N/cm²。大活塞上受到同样的压强，因此力 F₂ = 20 N/cm² × 50 cm² = 1000 N。能被抬起的载荷为 1000 N（质量约 102 kg）。
Note that the ratio of forces equals the ratio of areas. Hydraulic jacks enable a small effort to lift heavy loads by trading distance moved for force.
注意，力之比等于面积之比。液压千斤顶通过以移动距离换取力的方式，实现以小力抬起重物。
5. Thermal Physics – Mixture Method to Find Specific Heat Capacity | 热学 – 混合法测比热容
In an experiment, a 0.200 kg metal block is heated to 100 °C and then dropped into a calorimeter containing 0.500 kg of water at 20.0 °C. The final temperature of the mixture is 25.0 °C. The specific heat capacity of water is 4200 J/(kg °C) and the calorimeter has negligible heat capacity. Determine the specific heat capacity of the metal. This method relies on the principle of conservation of energy: heat lost by the metal equals heat gained by the water.
一个实验中，一个 0.200 kg 的金属块被加热到 100 °C，然后放入装有 0.500 kg、20.0 °C 水的量热器中。混合物的最终温度是 25.0 °C。水的比热容为 4200 J/(kg °C)，量热器的热容可忽略。求金属的比热容。该方法基于能量守恒原理：金属放出的热量等于水吸收的热量。
Heat lost by metal: Qmetal = mₘ cₘ (100 - 25) = 0.2 × cₘ × 75 = 15 cₘ. Heat gained by water: Qwater = mw cw (25 - 20) = 0.5 × 4200 × 5 = 10500 J. Equating: 15 cₘ = 10500 ⇒ heat capacity around 700 J/(kg °C) include aluminium; this could be aluminium or an alloy.
金属放热：Qmetal = 0.2 × cₘ × (100 - 25) = 15 cₘ。水吸热：Qwater = 0.5 × 4200 × (25 - 20) = 10500 J。相等得：15 cₘ = 10500 ⇒ cₘ = 700 J/(kg °C)。比热容约为 700 J/(kg °C) 的常见材料有铝，因此该金属可能是铝或铝合金。
Accurate measurements require good insulation and stirring to ensure uniform temperature. In practice, heat losses to the surroundings are minimised by a lagged calorimeter.
精确测量需要良好的隔热和搅拌以确保温度均匀。实践中，用绝热量热器尽可能地减少环境热损失。
6. Waves – Double-slit Interference to Measure Wavelength | 波动 – 双缝干涉测量波长
sured to be 5.2 mm. Calculate the wavelength of the laser light. This experiment is a classic demonstration of wave interference and allows precise determination of wavelength using the formula λ = ax/D.
一束波长未知的激光照射缝距为 0.25 mm 的双缝。屏幕放在 2.00 m 外，测得条纹间距为 5.2 mm。计算激光的波长。该实验是波动干涉的经典演示，可利用公式 λ = ax/D 精确测定波长。
⁻⁶) / 2.00 = 6.5 × 10⁻⁷ m = 650 nm. This wavelength corresponds to red light in the visible spectrum.
确定各量：a = 0.25 mm = 2.5 × 10⁻⁴ m；D = 2.00 m；x = 5.2 mm = 5.2 × 10⁻³ m。干涉公式：λ = ax / D = (2.5 × 10⁻⁴ × 5.2 × 10⁻³) / 2.00 = 6.5 × 10⁻⁷ m = 650 nm。这个波长对应可见光中的红光。
If the slit separation were decreased, the fringe spacing would increase, demonstrating the inverse relationship. This method is widely used in spectroscopy and optical metrology.
若减小缝距，条纹间距会增大，证明了反比关系。该方法广泛用于光谱学和光学计量。
7. Electric Circuits – Short Circuit Fault in Household Wiring | 电路 – 家庭电路短路故障分析
A household circuit is protected by a 13 A fuse and operates at 230 V mains voltage. An appliance with a resistance of 50 Ω is plugged into the socket and works normally. Suddenly, a fault causes the live and neutral wires to touch (short circuit) directly across the power supply, bypassing the appliance. The combined resistance of the connecting wires in the short circuit is 0.5 Ω. Determine the current drawn from the mains and whether the fuse blows.
一个家庭电路由 13 A 保险丝保护，工作电压为 230 V 市电。一台电阻为 50 Ω 的电器插入插座正常工作。突然，故障使火线和零线直接接触（短路），绕过电器直接连接到电源。短路路径的连接线总电阻为 0.5 Ω。计算从市电抽取的电流并判断保险丝是否熔断。
A rating. During the short circuit, the effective resistance drops to 0.5 Ω, so current Ishort = 230 V / 0.5 Ω = 460 A. This enormous current far exceeds the fuse' ng it to blow almost instantly, thereby breaking the circuit and preventing a fire hazard.
正常工作时，电流 I = 230 V / 50 Ω = 4.6 A，远低于 13 A 额定值。短路时，有效电阻降为 0.5 Ω，因此电流 Ishort = 230 V / 0.5 Ω = 460 A。这一巨大电流远超保险丝额定值，致使其瞬间熔断，从而切断电路并防止火灾危险。
This case highlights the vital role of fuses and circuit breakers in electrical safety. The low resistances in short circuits can produce currents capable of melting wires and starting fires.
该案例突显了保险丝和断路器在电气安全中的关键作用。短路中的低电阻会产生足以熔化导线和引发火灾的电流。
8. Electromagnetism – Working Principle of a Dynamic Loudspeaker | 电磁学 – 电动扬声器的工作原理
A dynamic loudspeaker uses the motor effect to convert electrical signals into sound waves. A coil is wound around a cylindrical former that sits within a radial magnetic field. When an alternating current flows through the coil, it experiences a force that pushes the cone backward and forward. Use Fleming's left-hand rule to determine the direction of the force and explain how this produces sound.
电动扬声器利用电动机效应将电信号转换为声波。一个绕在圆筒形骨架上的线圈置于径向磁场内。当交流电通过线圈时，线圈受到力的作用，推动音盆前后运动。使用弗莱明左手定则判断力的方向，并解释其如何产生声音。
In the radial magnetic field, the magnetic field lines are always approximately perpendicular to the current-carrying coil sides. By Fleming's left-hand rule, if the current enters upward, and the magnetic field is radially outward, the force on the coil is to one side, moving the cone. When the current reverses, the force reverses, pulling the cone back. The coil vibrates at the frequency of the audio signal, producing compressions and rarefactions in the air – sound waves.
在径向磁场中，磁感线大致总是与通电线圈的边垂直。根据左手定则，如果电流向上，磁场径向向外，则线圈受到指向一侧的力，推动音盆。当电流反向时，力也反向，拉回音盆。线圈以声频信号的频率振动，在空气中产生疏密相间的纵波——声波。
The frequency of the alternating current determines the pitch of the sound, while the amplitude determines the loudness. This efficient transducer is the basis of most modern speakers.
交流电的频率决定声音的音调，振幅决定响度。这种高效换能器是大多数现代扬声器的基础。
9. Nuclear Physics – Radiocarbon Dating | 核物理 – 放射性碳定年法
Radiocarbon dating uses the decay of carbon-14 (half-life 5730 years) to estimate the age of organic materials. A sample of ancient wood has a carbon-14 activity that is 12.5% of the activity of a living tree of the same mass. Calculate the age of the wood. This practical application of radioactive decay relies on the exponential decay law and the constant half-life.
放射性碳定年法利用碳-14（半衰期 5730 年）的衰变来估算有机物的年代。一块古老木头的碳-14 放射性活度为同等质量活树的 12.5%。计算木头的年龄。这一放射性衰变的实际应用依赖于指数衰变定律和恒定的半衰期。
12.5% efore, the wood is 3 half-lives old: age = 3 × 5730 = 17190 years. This indicates the wood dates to approximately 17000 years ago.
12.5% 意味着剩余比例 = 0.125 = 1/8。经过 n 个半衰期后，剩余比例为 (1/2)ⁿ。需要 (1/2)ⁿ = 1/8，因此 n = 3。所以木头已历经 3 个半衰期：年龄 = 3 × 5730 = 17190 年。这表明该木头大约可追溯至 17000 年前。
The method assumes the carbon-14 atmospheric concentration has remained constant over time, but small corrections are needed for precise dating beyond a few thousand years.
该方法假设大气中碳-14 浓度随时间保持恒定，但若要精确测定数千年以上的年代，需做微小校正。
10. Astronomy – Redshift of Stellar Spectra and Universe Expansion | 天文学 – 恒星光谱红移与宇宙膨胀
of hydrogen from a distant galaxy at a wavelength of 680 nm. The same line measured in the laboratory has a wavelength of 656 nm. Calculate the recession velocity of the galaxy and, using Hubble' mate its distance. This case links the Doppler effect for light with the cosmic expansion.
天文学家观测到遥远星系的一条氢谱线波长为 680 nm。同一谱线在实验室中测量的波长为 656 nm。计算该星系的退行速度，并利用哈勃常数 H₀ = 70 km s⁻¹ Mpc⁻¹ 估算其距离。本案例将光的多普勒效应与宇宙膨胀联系起来。
ion velocity v ≈ c z = 3.00 × 10⁸ m/s × 0.0366 ≈ 1.10 × 10⁷ m/s = 11000 km/s. Using Hubble' ce d = v / H₀ = 11000 km/s / 70 km s⁻¹ Mpc⁻¹ ≈ 157 Mpc (or about 510 million light-years).
红移 z = (λ_observed - λ_rest) / λ_rest = (680 - 656)/656 = 24/656 ≈ 0.0366。低速近似下，退行速度 v ≈ c z = 3.00 × 10⁸ m/s × 0.0366 ≈ 1.10 × 10⁷ m/s = 11000 km/s。利用哈勃定律 v = H₀ d，距离 d = v / H₀ = 11000 / 70 ≈ 157 Mpc（约 5.1 亿光年）。
Redshift measurements provide strong evidence that the universe is expanding. The further away a galaxy is, the faster it appears to be moving away from us.
红移的测量为宇宙膨胀提供了有力证据。星系距离我们越远，看起来远离我们的速度就越快。
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