📚 Case Study Mastery: Cambridge Year 12 Physics Problem-Solving | Year 12 剑桥物理案例分析实战演练
Case study questions in Cambridge Year 12 Physics challenge you to apply concepts to real-world scenarios. This article provides practical drills, breaking down how to analyse data, select the right equations, and avoid common mistakes. By working through carefully chosen examples—from projectile motion to photoelectric effect—you will sharpen your problem-solving skills and boost your confidence for the AS examination.
剑桥 Year 12 物理中的案例分析题要求你将概念应用于真实场景。本文通过实战演练，拆解如何分析数据、选择正确方程并避免常见错误。通过精心挑选的例题——从抛体运动到光电效应——你将提升解题技巧，为 AS 考试增强信心。
1. Understanding the Case Study Approach | 理解案例分析的方法
A case study typically presents a scenario with multiple pieces of information: diagrams, graphs, tables, and descriptions. Your first step is to identify the relevant physics principles. Underline key quantities and units, convert them to SI, and note any assumptions the problem expects you to make.
案例分析通常会呈现一个包含多种信息的场景：图表、数据表和描述。第一步是识别相关的物理原理。标注关键量和单位，转换成国际单位制，并注意题目希望你做出的假设。
- Always draw a labelled diagram if one is not provided. | 若无示意图，务必绘制带标注的草图。
- List known and unknown variables in symbolic form. | 用符号列出已知量和未知量。
- Select the principle—Newton’s laws, energy conservation, or wave theory—that links the data. | 选择能关联数据的原理——牛顿定律、能量守恒或波动理论。
2. Case 1: Projectile Motion with Air Resistance | 案例一：考虑空气阻力的抛体运动
A cricket ball is struck with an initial speed of 28 m s⁻¹ at 35° to the horizontal. The case study provides a drag force proportional to velocity, F_drag = 0.15v. You are asked to estimate the reduction in range compared to the ideal case. Begin by solving the drag‑free trajectory: horizontal range R₀ = (u² sin 2θ)/g = (28² sin 70°)/9.81 ≈ 75.2 m.
一个板球被以 28 m s⁻¹ 的初速、与水平面成 35° 击出。案例中给出与速度成正比的阻力 F_drag = 0.15v。要求估算相较于理想情况射程的减小量。先求解无阻力轨迹：水平射程 R₀ = (u² sin 2θ)/g = (28² sin 70°)/9.81 ≈ 75.2 m。
Because the drag force reduces both horizontal and vertical velocity components, the ball follows a shorter, asymmetric path. A first approximation uses an energy argument: the work done by drag is about 0.15 × average speed × path length. Taking average speed ≈ 0.9u, estimated work ≈ 0.15 × 25.2 × 75.2 ≈ 284 J. The initial kinetic energy is ½mu², so for m = 0.16 kg, KE = ½ × 0.16 × 28² ≈ 62.7 J. Wait—this implies work exceeds initial KE, which is impossible. This signals that a simple average‑speed approach is flawed; you must use differential equations or numerical methods, often provided in the data. The case study may give a graph of velocity against time. Read the area under the horizontal‑velocity curve to obtain the actual range, typically about 60 m, a 20% reduction.
阻力同时减小水平和竖直速度分量，球沿更短且不对称的路径飞行。用能量法近似：阻力做功约 0.15 × 平均速率 × 路程。取平均速率 ≈ 0.9u，估计做功 ≈ 0.15 × 25.2 × 75.2 ≈ 284 J。初动能 ½mu²，球质量 m = 0.16 kg，动能 ≈ 62.7 J。咦，做功竟然超过了初动能，说明简单的平均速率法行不通；必须借助微分方程或数值方法，通常题目会提供。案例可能给出速度–时间图。读取水平速度曲线下的面积得到实际射程，典型值约 60 m，减小约 20%。
Key skill: Recognise when an analytical shortcut fails and how to extract information from a supplied graph or dataset.
关键技能：识别解析捷径何时失效，以及如何从提供的图表或数据集中提取信息。
3. Case 2: Energy Conservation in a Roller Coaster | 案例二：过山车的能量守恒
A roller coaster carriage (mass 500 kg) starts from rest at a height of 40 m. The track includes a circular loop of radius 8 m. You must determine whether the carriage stays on the track at the top of the loop. Friction is negligible. At the top, speed v_top is found from energy conservation: mgH = mg(2R) + ½mv_top², giving v_top = √(2g(H − 2R)) = √(2 × 9.81 × (40 − 16)) ≈ √(470.9) ≈ 21.7 m s⁻¹.
过山车车厢（质量 500 kg）从 40 m 高处由静止出发，轨道包含半径 8 m 的圆形回环。需判断车厢在环顶是否不脱离轨道。忽略摩擦。环顶速度 v_top 由能量守恒求得：mgH = mg(2R) + ½mv_top²，得 v_top = √(2g(H − 2R)) = √(2 × 9.81 × (40 − 16)) ≈ √(470.9) ≈ 21.7 m s⁻¹。
The centripetal force required is mv_top²/R = 500 × (21.7²)/8 ≈ 29400 N. Weight alone is mg = 4905 N. Since the required force exceeds weight, the track must push down on the carriage; thus the carriage is safe. This case study might ask you to discuss how a real roller coaster includes friction wheels or magnetic braking. You can comment that energy loss to friction would reduce v_top, so the designer must increase the initial height accordingly.
所需向心力 mv_top²/R = 500 × (21.7²)/8 ≈ 29400 N。车厢重力仅为 mg = 4905 N。因所需向心力大于重力，轨道必须向下压车厢，故车厢安全。此案例可能要求讨论真实过山车如何加装摩擦轮或磁力刹车。可指出摩擦能量损失会降低 v_top，因此设计者需相应提高起始高度。
| Quantity | Value |
|---|---|
| Initial height H | 40 m |
| Loop radius R | 8 m |
| Speed at top | 21.7 m s⁻¹ |
| Centripetal force | 2.94 × 10⁴ N |
| Weight | 4.91 × 10³ N |
4. Case 3: Moments and Equilibrium in a Crane | 案例三：起重机中的力矩与平衡
A crane boom of length 12 m and mass 200 kg is hinged at its lower end and held at 50° to the horizontal by a cable attached 3 m from the upper end. The cable makes an angle of 30° with the boom. A load of 800 kg hangs from the top. Taking moments about the hinge eliminates the hinge force. The boom’s weight acts at its centre (6 m from hinge). Calculate the tension T in the cable.
起重机吊臂长 12 m，质量 200 kg，其下端铰接，距顶端 3 m 处由缆绳拉住，与水平成 50°；缆绳与吊臂成 30°。800 kg 负载挂在顶端。对铰点取矩可消去铰接力。吊臂自重作用于中点（距铰 6 m）。计算缆绳张力 T。
Perpendicular distance for the load: 12 m × cos 50°, for the boom weight: 6 m × cos 50°, and for the tension: the distance from the hinge to the cable attachment is 12 − 3 = 9 m. The perpendicular component of T is T sin 30°, so its moment arm = 9 m × sin 30°. Clockwise moments = (800g × 12 cos 50°) + (200g × 6 cos 50°). Anticlockwise moment = T × 9 sin 30°. Equating: T × 9 × 0.5 = g cos 50° (800 × 12 + 200 × 6). Solve: T = [9.81 × 0.6428 × (9600 + 1200)] / 4.5 ≈ (9.81 × 0.6428 × 10800)/4.5 ≈ (68100)/4.5 ≈ 15100 N.
负载的垂直距离：12 m × cos 50°，吊臂自重垂直距离：6 m × cos 50°，张力垂直分量的力臂：吊臂上缆绳附着点距离铰点 12 − 3 = 9 m，T 的分量 T sin 30°，力臂 = 9 m × sin 30°。顺时针力矩 = (800g × 12 cos 50°) + (200g × 6 cos 50°)。逆时针力矩 = T × 9 sin 30°。相等：T × 9 × 0.5 = g cos 50° (800 × 12 + 200 × 6)。解得 T = [9.81 × 0.6428 × (9600 + 1200)] / 4.5 ≈ (9.81 × 0.6428 × 10800)/4.5 ≈ 68100/4.5 ≈ 15100 N。
Examiner’s trick: They may give the tension in the cable and ask for the maximum load before the crane topples. In that case, take moments about the pivot edge of the base.
考官套路：可能给出缆绳张力，要求求倾倒前的最大负载。此时需对底座边缘取矩。
5. Case 4: Wave Interference in Noise-Cancelling Headphones | 案例四：降噪耳机中的波干涉
Active noise-cancelling headphones use destructive interference. The case study provides a waveform of ambient noise with frequency 400 Hz and amplitude 0.05 mm at the microphone. The headphone speaker must generate a wave exactly out of phase. You calculate the time delay required: for a phase difference of π rad (180°), time shift Δt = ½T = 1/(2f) = 1/(800) = 1.25 × 10⁻³ s.
主动降噪耳机利用相消干涉。案例提供环境噪声波形，频率 400 Hz，麦克风处振幅 0.05 mm。耳机扬声器需生成相位完全相反的声波。计算所需时间延迟：相位差 π rad (180°)，时移 Δt = ½T = 1/(2f) = 1/800 = 1.25 × 10⁻³ s。
However, the microphone and speaker are separated by 2.5 cm. Sound speed is 340 m s⁻¹, so the travel time from mic to speaker is 0.025/340 ≈ 7.35 × 10⁻⁵ s. The processor must add an extra delay so that the total delay equals 1.25 ms plus any integer multiple of the period. The required electronic delay ≈ 1.25 × 10⁻³ − 7.35 × 10⁻⁵ ≈ 1.18 × 10⁻³ s. You may also discuss limitations: the cancellation is perfect only for a single frequency and direction; real headphones use adaptive filters.
但麦克风与扬声器相距 2.5 cm，声速 340 m s⁻¹，声音由麦克风传至扬声器耗时 0.025/340 ≈ 7.35 × 10⁻⁵ s。处理器须增加额外延时，使总延时等于 1.25 ms 加上周期的整数倍。所需电子延时 ≈ 1.25 × 10⁻³ − 7.35 × 10⁻⁵ ≈ 1.18 × 10⁻³ s。还可讨论局限性：仅对单一频率和方向完全抵消；真实耳机采用自适应滤波器。
The principle is expressed as path difference = (n + ½)λ for destructive interference. Here the electronic delay creates the equivalent path difference.
原理表述为相消干涉的波程差 = (n + ½)λ。此处电子延迟产生等效波程差。
6. Case 5: Resistive Networks and Kirchhoff’s Laws | 案例五：电阻网络与基尔霍夫定律
A sensor circuit contains three resistors (10 Ω, 20 Ω, 30 Ω) arranged in a mixed network with a 9.0 V battery. The 20 Ω and 30 Ω resistors are in parallel, and that combination is in series with the 10 Ω resistor. You must find the current through the 20 Ω resistor. First, parallel equivalent: 1/R_para = 1/20 + 1/30 = 5/60 ⇒ R_para = 12 Ω. Total resistance = 10 + 12 = 22 Ω. Main current I = 9.0/22 ≈ 0.409 A. Voltage across parallel pair = I × 12 = 4.91 V. Current through 20 Ω = 4.91/20 ≈ 0.246 A.
传感器电路含三只电阻（10 Ω, 20 Ω, 30 Ω）组成混合网络，接 9.0 V 电池。20 Ω 与 30 Ω 并联，再与 10 Ω 串联。求通过 20 Ω 电阻的电流。并联等效电阻：1/R_para = 1/20 + 1/30 = 5/60，R_para = 12 Ω。总电阻 = 10 + 12 = 22 Ω。主电流 I = 9.0/22 ≈ 0.409 A。并联组端电压 = I × 12 = 4.91 V。通过 20 Ω 的电流 = 4.91/20 ≈ 0.246 A。
The case study may give ammeter readings at different points and ask you to identify a faulty component. Apply Kirchhoff’s current law at junctions: ΣI_in = ΣI_out. If a measured current violates this, the fault may be a short circuit or open circuit. For example, if the current entering the parallel pair equals the current leaving but the branch currents do not add up correctly, suspect a meter error or an internal short.
案例可能给出各点电流表读数，要求判断故障元件。应用基尔霍夫电流定律于节点：ΣI_in = ΣI_out。若实测电流违反此律，故障可能是短路或断路。例如，若流入并联组的电流等于流出，但支路电流和不正确，则怀疑电流表误差或内部短路。
V = IR, ΣEmf = ΣIR, ΣI_in = ΣI_out
7. Case 6: Photoelectric Effect Application | 案例六：光电效应应用
A photoelectric sensor uses potassium (work function φ = 2.3 eV). Light of wavelength 420 nm is incident. Determine whether electrons are emitted and find their maximum kinetic energy. Photon energy E = hc/λ. Using hc = 1240 eV·nm, E = 1240/420 ≈ 2.95 eV. Since E > φ, emission occurs. K_max = E − φ = 2.95 − 2.3 = 0.65 eV. In joules: 0.65 × 1.6 × 10⁻¹⁹ = 1.04 × 10⁻¹⁹ J.
光电传感器使用钾（功函数 φ = 2.3 eV）。波长 420 nm 的光入射。判断能否发射电子并求最大动能。光子能量 E = hc/λ。利用 hc = 1240 eV·nm，E = 1240/420 ≈ 2.95 eV。因 E > φ，发射发生。K_max = E − φ = 2.95 − 2.3 = 0.65 eV。换算为焦耳: 0.65 × 1.6 × 10⁻¹⁹ = 1.04 × 10⁻¹⁹ J。
The case study might provide a graph of kinetic energy against frequency. The gradient is Planck’s constant h, and the x‑intercept is the threshold frequency f₀ = φ/h. A common question: ‘How would the graph change if a metal with a larger work function were used?’ The line would shift to the right, giving a larger threshold frequency, but the gradient remains unchanged.
案例可能提供动能–频率图。斜率为普朗克常数 h，x 轴截距为截止频率 f₀ = φ/h。常见问题：‘若改用功函数更大的金属，图像如何变化？’ 直线将右移，截止频率变大，但斜率不变。
Stopping potential equation: eV_s = hf − φ. In an experiment, V_s is measured and plotted; the slope gives h/e.
遏止电压方程：eV_s = hf − φ。实验中测定 V_s 并作图，斜率得 h/e。
8. Error Analysis and Experimental Design | 误差分析与实验设计
Case studies often include experimental data and ask you to evaluate uncertainty. For a set of repeated measurements, calculate the mean, range, and percentage uncertainty. If the accepted value lies outside the range of experimental values, systematic error is present. For instance, in a free‑fall experiment using a stopwatch, reaction time introduces a systematic error; using a light gate and data‑logger reduces this.
案例分析常包含实验数据，要求评估不确定度。对一组重复测量，计算平均值、极差和百分不确定度。若公认值落在实验值范围之外，则存在系统误差。例如，用秒表的自由落体实验，反应时间引入系统误差；改用光门和数据采集器可减少之。
| Measurement | Value (s) |
|---|---|
| 1 | 0.56 |
| 2 | 0.59 |
| 3 | 0.54 |
| Mean | 0.563 |
| Range | 0.05 |
Percentage uncertainty = (range/2)/mean × 100% = (0.025/0.563) × 100% ≈ 4.4%.
百分不确定度 = (极差/2)/平均值 × 100% = (0.025/0.563) × 100% ≈ 4.4%。
9. Common Pitfalls in Case Studies | 案例分析常见误区
Students frequently lose marks by ignoring units, misapplying sign conventions, or using the wrong g value. In projectile problems, always resolve into components and treat vertical and horizontal motions independently. In circuit analysis, forgetting that an ideal voltmeter has infinite resistance can lead to incorrect conclusions about loading effects.
学生常因忽视单位、错用符号约定或 g 值错误而失分。抛体问题中，务必分解为分量并独立处理竖直和水平运动。电路分析中，忘记理想电压表内阻无穷大会导致对负载效应的错误结论。
- Using 10 m s⁻² for g without checking if the question specifies 9.81 m s⁻². | 未检查题目是否指定 g = 9.81 m s⁻² 而直接使用 10 m s⁻²。
- Assuming tension in a cable is equal to the weight of the object when it acts at an angle. | 当缆绳有角度时，误认为张力等于物体重力。
- Misinterpreting the area under a force–extension graph as work done (correct) but forgetting that the graph must be force vs. extension, not length. | 误将力–伸长图下面积理解为做功（正确），但忘记必须是力对伸长作图，而非对长度。
10. Practice Drills and Self-Assessment | 练习与自我评估
To master case studies, regularly attempt past‑paper scenarios under timed conditions. Start by reading the question twice: first for an overview, second to extract data. Always show your working stepwise; even if the final answer is wrong, method marks are awarded. After solving, compare your approach with the mark scheme and note where you lost efficiency.
要精通案例分析，应定时练习往年真题场景。先阅读两遍：第一遍概览，第二遍提取数据。始终分步书写过程；即便最终答案错误，仍能得到过程分。解完后将自己的方法与评分方案对比，标注效率不足之处。
Try this drill: A ball is kicked from a cliff 15 m high with speed 22 m s⁻¹ at 40° above the horizontal. Air resistance is negligible. Find the time of flight, horizontal range, and the velocity vector just before impact. (Answer: time = 3.8 s, range = 64 m, v_y = −15.7 m s⁻¹, v_x = 16.8 m s⁻¹, speed ≈ 23.0 m s⁻¹ at 43° below horizontal.)
尝试此练习：一球从 15 m 高悬崖以 22 m s⁻¹、与水平成 40° 踢出，忽略空气阻力。求飞行时间、水平射程和撞击前的速度矢量。（答案：时间 = 3.8 s，射程 = 64 m，v_y = −15.7 m s⁻¹，v_x = 16.8 m s⁻¹，速率 ≈ 23.0 m s⁻¹，与水平夹角 43° 向下。）
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