📚 Case Study Practical Workout for Year 11 CIE Physics | 案例分析实战演练
In Year 11 CIE Physics, case study questions test your ability to apply scientific principles to real-world situations. This article presents a series of practical workout examples that mirror the style of IGCSE Paper 4 (Extended) and Paper 6 (Alternative to Practical). Each case study combines theory, calculation and experimental reasoning, helping you build confidence in handling unfamiliar contexts.
在Year 11 CIE物理中，案例分析题考查你将科学原理应用于实际情境的能力。本文提供一系列实战演练案例，模拟IGCSE试卷4（扩展）和试卷6（实验替代）的出题风格。每个案例都融合了理论、计算和实验推理，帮助你从容应对陌生情境。
1. Understanding Case Study Analysis | 理解案例分析
A case study in physics usually presents a scenario, data and a problem to solve. The key steps are: identify known quantities, select the correct formula, carry out the calculation with proper units, and evaluate the result in the context of the question. Common pitfalls include unit conversion errors and misapplication of equations. Always ask yourself: ‘What is the physics behind this situation?’
物理案例分析通常会给出一个场景、数据和待解决的问题。关键步骤是：识别已知量，选择正确的公式，用合适的单位进行计算，并在题目情境中评估结果。常见的错误包括单位换算出错和公式误用。要始终问自己：“这个情境背后的物理原理是什么？”
2. Case Study 1: Stopping Distances on a Wet Road | 案例1：潮湿路面的刹车距离
A driver sees a hazard ahead while travelling at 20 m/s on a wet road. The driver’s reaction time is 0.7 s. The car then decelerates at 4.5 m/s² due to reduced friction. Calculate (a) the thinking distance, and (b) the braking distance. Hence find the total stopping distance and discuss why it is greater than on a dry road.
司机以20 m/s的速度在潮湿路面上行驶时看到前方有危险。反应时间为0.7 s。由于摩擦力降低，车子随后以4.5 m/s²的减速度减速。计算(a)反应距离，(b)制动距离。进而求出总停车距离，并讨论为何该距离大于干燥路面。
Thinking distance is the distance travelled during the driver’s reaction time. Using uniform motion: s = vt = 20 m/s × 0.7 s = 14 m. Braking distance uses the equation v² = u² + 2as. The final speed v = 0. Rearranging: 0 = (20)² + 2 × (–4.5) × s, giving s = 400 / 9 ≈ 44.4 m. Total stopping distance = 14 m + 44.4 m = 58.4 m. On a dry road with higher deceleration (e.g. 7 m/s²), braking distance would be only about 28.6 m, so wet conditions nearly double the stopping distance.
反应距离是司机反应时间内行驶的距离。用匀速运动计算：s = vt = 20 m/s × 0.7 s = 14 m。制动距离用公式 v² = u² + 2as。末速度 v = 0。整理得：0 = (20)² + 2 × (–4.5) × s，得 s = 400 / 9 ≈ 44.4 m。总停车距离 = 14 m + 44.4 m = 58.4 m。干燥路面上减速度更大（如7 m/s²），制动距离仅约28.6 m，因此潮湿条件几乎使停车距离翻倍。
3. Case Study 2: Series Circuit Fault Diagnosis | 案例2：串联电路故障诊断
A circuit has a battery, a switch, an ammeter and two lamps L1 and L2 connected in series. When the switch is closed, the ammeter reads zero and neither lamp lights. A student suggests two possible faults: (i) L1 is blown (open circuit), or (ii) there is a short circuit across L2. Explain how to use a voltmeter to identify which fault is present.
一个电路包含电池、开关、电流表和两个灯泡L1和L2串联。闭合开关后，电流表读数为零，两灯都不亮。某学生提出两种可能的故障：(i) L1烧断（断路），或(ii) L2被短路。解释如何用电压表判断存在哪种故障。
For an open circuit, the broken filament creates an infinite resistance; no current flows. A voltmeter connected across the open lamp will read the full battery voltage (e.g. 6 V) because the voltmeter provides a high-resistance path. Across a short-circuited lamp, the voltmeter reads nearly zero, as there is a conducting bypass. So if the voltmeter across L1 reads 6 V and across L2 reads 0 V, L1 is open. If across L1 reads 0 V and across L2 also 0 V, L2 might be shorted but current would still flow – however, ammeter reads zero, so that can’t be the only fault; perhaps both faults exist. A realistic diagnosis: measure voltage across L1; if 6 V, L1 open. If 0 V, check across battery; if battery voltage present, a break elsewhere. This systematic approach helps isolate the problem.
断路时灯丝断裂造成无穷大电阻，没有电流。电压表并联在开路灯泡两端会读得全电池电压（如6 V），因为电压表提供高阻通路。而在短路的灯泡两端，电压表读数近乎为零，因为有导线旁路。所以若L1两端电压为6 V，L2两端为0 V，则L1断路。若L1两端为0 V，L2也为0 V，但电流表为零，则可能L2短路且另有断路。合理的诊断是：测量L1电压；若为6 V则L1断路；若为0 V，检查电池两端；如果电池有电压，则别处有断路。这种系统方法有助于定位故障。
4. Case Study 3: Energy Efficiency of an Electric Kettle | 案例3：电热水壶的能量效率
An electric kettle rated at 2200 W heats 1.5 kg of water from 22 °C to 100 °C. It takes 4 minutes and 30 seconds to boil. The specific heat capacity of water is 4200 J/(kg K). Calculate (a) the electrical energy supplied, (b) the useful thermal energy gained by the water, and (c) the efficiency of the kettle. Account for energy losses.
一个额定功率2200 W的电热水壶将1.5 kg水从22 °C加热至100 °C。沸腾耗时4分30秒。水的比热容为4200 J/(kg K)。计算(a) 提供的电能，(b) 水获得的有用热能，(c) 水壶的效率。并解释能量损失。
Electrical energy supplied: E = Pt = 2200 W × (4 × 60 + 30) s = 2200 × 270 = 594,000 J. Useful energy: Q = mcΔθ = 1.5 × 4200 × (100 – 22) = 1.5 × 4200 × 78 = 491,400 J. Efficiency = (useful output / energy input) × 100% = (491,400 / 594,000) × 100% ≈ 82.7%. About 17% of the energy is lost to heating the kettle body, surrounding air and by evaporation.
电能：E = Pt = 2200 W × (4×60+30) s = 2200 × 270 = 594,000 J。有用能量：Q = mcΔθ = 1.5 × 4200 × (100 – 22) = 1.5 × 4200 × 78 = 491,400 J。效率 =（有用输出/输入能量）×100% = (491,400/594,000)×100% ≈ 82.7%。约17%的能量损失在了加热壶体、周围空气以及蒸发上。
5. Case Study 4: Measuring Density of an Irregular Stone | 案例4：测量不规则石块的密度
A student uses a displacement method: a measuring cylinder contains 50 cm³ of water. After gently lowering in a stone on a thin thread, the water level rises to 68 cm³. The mass of the stone is 45 g. Calculate the density in g/cm³ and in kg/m³. Discuss why the string should be thin and the stone fully submerged without splashing.
某学生用排水法测量：量筒中原有50 cm³水。用细线轻轻放入石块后，水面升至68 cm³。石块质量为45 g。计算密度，以g/cm³和kg/m³表示。讨论为什么细线要细、石块要完全浸没且不能溅水。
Volume of stone = 68 cm³ – 50 cm³ = 18 cm³. Density ρ = mass/volume = 45 g / 18 cm³ = 2.5 g/cm³. In kg/m³: 2.5 g/cm³ = 2.5 × 1000 = 2500 kg/m³. A thick string would displace extra water, giving an overestimate of volume and lower density. If the stone is not fully submerged or splashes water out, volume reading is inaccurate. Hence careful technique is vital for reliable results.
石块的体积 = 68 cm³ – 50 cm³ = 18 cm³。密度 ρ = 质量/体积 = 45 g / 18 cm³ = 2.5 g/cm³。换算：2.5 g/cm³ = 2.5 × 1000 = 2500 kg/m³。粗线会额外排水，高估体积并降低密度值。若石块未完全浸没或溅出水，体积读数就不准确。因此细致的操作对可靠结果至关重要。
6. Case Study 5: Hooke’s Law Spring Extension | 案例5：胡克定律弹簧伸长
A spring is hung vertically with a pointer indicating extension. Weights added give forces of 1.0 N, 2.0 N, 3.0 N, 4.0 N and 5.0 N. The measured extensions are 2.1 cm, 4.0 cm, 6.0 cm, 8.1 cm and 10.0 cm respectively. Plot a force-extension graph, determine the spring constant k, and identify if the elastic limit was exceeded. Also find the extension for a 2.5 N load.
一根弹簧竖直悬挂，指针指示伸长量。依次加挂钩码产生力1.0 N, 2.0 N, 3.0 N, 4.0 N, 5.0 N。测量的伸长量分别为2.1 cm, 4.0 cm, 6.0 cm, 8.1 cm, 10.0 cm。绘制力-伸长量图，求出劲度系数k，判断是否超过弹性限度。并求2.5 N负载下的伸长量。
Hooke’s Law: F = kx, where x is extension. From the data, ratio F/x is roughly constant: 1.0/0.021 ≈ 47.6 N/m, 2.0/0.040 = 50 N/m, 3.0/0.060 = 50 N/m, 4.0/0.081 ≈ 49.4 N/m, 5.0/0.100 = 50 N/m. Average k ≈ 49.6 N/m. The linear relationship holds well, so elastic limit is not exceeded. For F = 2.5 N, x = F/k = 2.5 / 50 = 0.050 m = 5.0 cm. A graph would show a straight line through origin, confirming proportionality.
胡克定律：F = kx，x为伸长量。数据中F/x比值大致恒定：1.0/0.021 ≈ 47.6 N/m, 2.0/0.040 = 50 N/m, 3.0/0.060 = 50 N/m, 4.0/0.081 ≈ 49.4 N/m, 5.0/0.100 = 50 N/m。平均k ≈ 49.6 N/m。线性关系保持良好，因此未超过弹性限度。当F = 2.5 N，x = F/k = 2.5/50 = 0.050 m = 5.0 cm。过原点的直线图可确认比例关系。
7. Case Study 6: Refraction of Light Through a Glass Block | 案例6：光线通过玻璃砖的折射
A ray of light enters a rectangular glass block (refractive index n = 1.50) from air at an angle of incidence of 30°. Calculate the angle of refraction inside the glass and the angle at which it emerges back into air. Explain why the emergent ray is parallel to the incident ray.
一束光线从空气以30°入射角射入矩形玻璃砖（折射率n = 1.50）。计算玻璃内的折射角以及光线回到空气时的出射角。解释为何出射光线与入射光线平行。
Using Snell’s law: n₁ sin θ₁ = n₂ sin θ₂. At first surface: 1.00 × sin 30° = 1.50 × sin r. sin 30° = 0.5, so sin r = 0.5 / 1.50 = 0.3333, r ≈ 19.5°. The ray then travels to the opposite parallel face. The angle of incidence at the second surface equals r (alternate angles). Applying Snell’s law again: 1.50 × sin 19.5° = 1.00 × sin e. Since sin 19.5° = 0.3333, sin e = 1.50 × 0.3333 = 0.5, e = 30°. Therefore the emergent ray is parallel to the incident ray, though laterally displaced. This is due to the parallel faces of the block.
用斯涅尔定律：n₁ sin θ₁ = n₂ sin θ₂。第一界面：1.00 × sin 30° = 1.50 × sin r。sin 30°=0.5，所以 sin r = 0.5 / 1.50 = 0.3333，r ≈ 19.5°。光线射向平行的对边，在第二界面的入射角等于r（内错角）。再次应用斯涅尔定律：1.50 × sin 19.5° = 1.00 × sin e。因sin 19.5°=0.3333，sin e = 1.50 × 0.3333 = 0.5，e = 30°。因此出射光线与入射光线平行，但有侧向位移。这是由于玻璃砖两平行面所致。
8. Case Study 7: Terminal Velocity of a Parachutist | 案例7：跳伞者的终极速度
A 70 kg skydiver jumps from a plane. Initially the only force is weight, causing acceleration. As speed increases, air resistance grows until it equals weight. At this point forces balance and the diver reaches terminal velocity. Explain the energy changes and why opening a parachute greatly reduces terminal velocity.
一名70 kg的跳伞者从飞机跳下。最初仅有重力作用，产生加速。随着速度增加，空气阻力增大直至与重力相等。此时力平衡，跳伞者达到终极速度。解释能量变化以及为何打开降落伞会大幅降低终极速度。
Weight = mg = 70 × 10 = 700 N (taking g = 10 m/s²). At terminal velocity, air resistance = 700 N. The diver’s gravitational potential energy is converted to kinetic energy and thermal energy (due to air friction). Terminal velocity v_t depends on the balance: when parachute opens, cross-sectional area increases dramatically, air resistance rises for the same speed, so a new lower terminal velocity is quickly established. For a parachute, v_t might drop from about 50 m/s to 5 m/s, reducing impact force.
重量 = mg = 70 × 10 = 700 N（取g = 10 m/s²）。终极速度时，空气阻力 = 700 N。跳伞者的重力势能转化为动能和热能（由空气摩擦产生）。终极速度 v_t 取决于平衡：打开降落伞时，截面积显著增加，相同速度下空气阻力增大，从而迅速建立新的、较低的终极速度。降落伞可使 v_t 从约50 m/s降至5 m/s，减小冲击力。
9. Case Study 8: Pressure in a Hydraulic Jack | 案例8：液压千斤顶中的压力
A hydraulic jack has a small piston of area 5.0 × 10⁻⁴ m² and a large piston of area 2.0 × 10⁻² m². A force of 150 N is applied to the small piston. Calculate the pressure transmitted in the fluid and the maximum load the large piston can lift. Discuss the practical advantage and state one limitation.
一个液压千斤顶的小活塞面积为5.0×10⁻⁴ m²，大活塞面积为2.0×10⁻² m²。在小活塞上施加150 N的力。计算流体中传递的压强和大活塞能举起的最大负荷。讨论实际优点并说出一个局限。
Pressure P = Force / Area (small piston) = 150 N / (5.0×10⁻⁴ m²) = 300,000 Pa. The same pressure acts on the large piston, so Force on large piston F = P × A = 300,000 Pa × (2.0×10⁻² m²) = 6000 N. Thus a small effort can lift a heavy load. The trade-off is that the small piston must move a greater distance: work input equals work output (ignoring friction). A limitation is that the fluid must be incompressible and well-sealed to transmit pressure effectively.
压强P = 力/面积（小活塞）= 150 N / (5.0×10⁻⁴ m²) = 300,000 Pa。同样压强作用在大活塞上，大活塞受力 F = P × A = 300,000 Pa × (2.0×10⁻² m²) = 6000 N。因此较小的力可举起重物。代价是小活塞必须移动更长距离：输入功等于输出功（忽略摩擦）。一个局限是液体必须不可压缩且密封良好，以有效传递压强。
10. Key Skills for Practical Case Studies | 案例分析的关键技能
Across all these examples, certain skills repeatedly prove essential: careful reading of data, correct unit conversions, selecting the right equation (e.g. P = F/A, E = Pt, F = kx), and interpreting graphs. You should also practise giving explanations in terms of physics principles rather than just stating results. Always check that your answer is sensible in the real-world context given.
纵观所有案例，一些技能反复证明是核心：仔细阅读数据，正确转换单位，选用正确公式（如P=F/A，E=Pt，F=kx），以及解读图表。你还应练习从物理原理角度给出解释，而非仅陈述结果。最后务必检查你的答案在给定实际情境中是否合理。
A successful case study answer often includes a clear statement of the physics law involved, a step-by-step calculation, and a concluding remark connecting back to the problem. Mastering this approach will not only boost your exam performance but also develop your analytical thinking.
一个成功的案例分析答案通常包括：对涉及的物理定律的清晰陈述，逐步的计算过程，以及联系回问题的总结性评论。掌握这一方法不仅会提高你的考试成绩，还将培养你的分析思维。
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