📚 Case Study Practice for Year 13 CCEA Physics | CCEA物理案例分析实战演练
In the Year 13 CCEA Physics examination, case-based questions are crafted to push your understanding beyond simple recall. They immerse you in realistic applications – from sports engineering to medical imaging – demanding that you extract relevant data, link concepts across the AS specification, and construct logical, well-justified solutions. This revision guide walks you through authentic case study scenarios, unpacking the techniques that turn a high-mark question into a manageable challenge.
在CCEA物理Year 13考试中，案例题的设计目标是将你的理解从简单的记忆推向应用。它们将你置于真实的物理情景中——从运动工程到医学成像——要求你提取相关数据，串联AS阶段不同知识模块，并构建逻辑严密、论据充分的解答。本文通过实战演练，为你拆解那些让高分题变得易于掌握的技巧。
1. The Anatomy of a CCEA Case Study Question | CCEA案例题的结构剖析
A typical case study opens with a descriptive paragraph and often a diagram or data table. Embedded within are numerical values, implicit assumptions, and sometimes distractors. The subsequent parts – usually (a) to (d) – progress from straightforward calculations to explanation and evaluation. Your first job is to identify which physics principles are being tested before you touch a calculator.
典型的案例题以一段描述性文字开场，常伴有图表或数据表格。其中嵌入了数值、隐含假设，有时还有干扰信息。后续的小题（通常是(a)至(d)）从简单计算逐步过渡到解释与评价。你的首要任务是在动笔计算前，先判断出这道题在考查哪些物理原理。
2. Strategy: Read – Model – Solve – Check | 解题策略：阅读 – 建模 – 求解 – 检查
Approach every case study with a four-step method. First, Read the stem actively, underlining quantities and their symbols. Second, Model the scenario: decide whether energy conservation, Newton's laws, or wave optics applies, and draw a simplified sketch. Third, Solve algebraically before substituting numbers; this keeps units consistent and reduces rounding errors. Finally, Check your answer against physical plausibility – does a speed exceed 100 m s⁻¹? Does a time constant make sense?
用四步法应对每个案例。第一步，阅读题干，积极划出物理量及其符号。第二步，建模：判断该用能量守恒、牛顿定律还是波动光学，并画出简化草图。第三步，求解时先进行代数推导再代入数值，这能保持单位一致并减少舍入误差。最后，检查答案的物理合理性——速度是否超过100 m s⁻¹？时间常数合理吗？
3. Case Study 1: Bungee Jumping – Energy and Elasticity | 案例分析1：蹦极 – 能量与弹性
ls vertically and can be modelled as a point mass. The question asks for (a) the maximum speed reached, (b) the maximum extension of the rope, and (c) the net force on the jumper at the lowest point.
一位质量为75 kg的蹦极者从桥上跳下。弹性绳未拉伸长度L = 20 m，劲度系数k = 180 N m⁻¹。可将蹦极者视为质点。题目要求 (a) 最大速度，(b) 绳子的最大伸长量，以及 (c) 在最低点时蹦极者所受的净力。
Modelling the motion: : Once the rope becomes taut, the upward elastic force increases. Maximum speed occurs when the net acceleration is zero – (75 × 9.81) / 180 ≈ 4.09 m. The total fallen distance at this point is L + ΔL₁ = 24.09 m.
建立运动模型：当绳子拉直后，向上的弹力逐渐增大。当合加速度为零时，即重力等于弹力时，速度达到最大：mg = k ΔL₁。因此ΔL₁ = mg/k = (75 × 9.81) / 180 ≈ 4.09 m。此时下落的距离为 L + ΔL₁ = 24.09 m。
t of maximum speed: mg(L + ΔL₁) = ½ m v² + ½ k (ΔL₁)². Rearranging gives v = √[2g(L + ΔL₁) – (k/m)(ΔL₁)²]. Substituting values: v = √[2×9.81×24.09 – (180/75)×(4.09)²] ≈ √[472.6 – 40.1] ≈ 20.8 m s⁻¹.
从起跳点到最大速度点应用能量守恒：mg(L + ΔL₁) = ½ m v² + ½ k (ΔL₁)²。整理得 v = √[2g(L + ΔL₁) – (k/m)(ΔL₁)²]。代入数据：v ≈ √[472.6 – 40.1] ≈ 20.8 m s⁻¹。
on. Energy conservation gives: mg(L + ΔL₂) = ½ k (ΔL₂)². This yields a quadratic: ½ k ΔL₂² – mg ΔL₂ – mg)² + 2k mgL)] / k. This comes to approximately 15.2 m. The net upward force is F = k ΔL₂ – mg ≈ (180×15.2) – 736 ≈ 2000 N.
在最低点动能变为零。设最大伸长量为ΔL₂，能量守恒：mg(L + ΔL₂) = ½ k (ΔL₂)²。得到二次方程：½ k ΔL₂² – mg ΔL₂ – mgL = 0。取正根解得约15.2 m。此时净上力为 F = k ΔL₂ – mg ≈ (180×15.2) – 736 ≈ 2000 N。
4. Case Study 2: Camera Flash – Capacitor Discharge | 案例分析2：相机闪光灯 – 电容放电
A camera flash unit consists of a 470 μF capacitor charged to 300 V via a high-resistance path and then discharged through a xenon tube of effective resistance 0.25 Ω. The manufacturer specifies that a useful flash lasts for a time tflash until the voltage drops to 40 V. Analyse the circuit by calculating (a) the initial stored energy, (b) the time constant, and (c) the flash duration tflash.
某相机闪光灯使用一个470 μF的电容器，通过高阻路径充电至300 V，然后通过有效电阻为0.25 Ω的氙灯管放电。制造商指出有效闪光持续到电压降至40 V时结束。请通过计算下列参数分析该电路：(a) 初始储能，(b) 时间常数，以及 (c) 闪光持续时间 tflash。
is τ = RC = 0.25 × 470 × 10⁻⁶ = 1.175 × 10⁻⁴ s = 117.5 μs. The voltage decays as V = V₀ e-t/RC. Setting V = 40 V gives 40 = 300 e-t/τ, so e-t/τ = 40/300 ≈ 0.1333. Taking the natural logarithm: -t/τ = ln(0.1333) ≈ -2.0149, hence tflash ≈ 2.015 × 117.5 μs ≈ 237 μs.
初始储存的能量为 E = ½ C V₀² = 0.5 × 470 × 10⁻⁶ × 300² = 21.15 J。电阻放电的时间常数 τ = RC = 0.25 × 470 × 10⁻⁶ = 1.175 × 10⁻⁴ s = 117.5 μs。电压按 V = V₀ e-t/RC 衰减。令 V = 40 V，得 40 = 300 e-t/τ，因此 e-t/τ = 0.1333。取自然对数：-t/τ = ln(0.1333) ≈ -2.0149，故 tflash ≈ 2.015 × 117.5 μs ≈ 237 μs。
The short discharge time demands a very low resistance path; even small contact resistances can alter τ. In a follow-up 'evaluate' question, you would discuss how stray inductance might cause oscillations and how a larger capacitor could extend the flash but increase recharge time.
极短的放电时间要求极低电阻路径，即使微小的接触电阻都会改变τ。在后续的“评价”题中，你可能需要讨论杂散电感如何引起振荡，以及更大的电容虽可延长闪光却会加长充电时间。
5. Case Study 3: Young's Double-Slit in Forensics | 案例分析3：杨氏双缝在法医学中的应用
pattern is compared with that from a known slit. In a simplified setup, the hair is replaced by a double-slit arrangement with slit separation d = 0.25 mm. The distance to the screen is D = 1.80 m. Monochromatic light of wavelength λ = 632.8 nm is used. The separation between the central and fifth bright fringe is measured to be 22.7 mm. Determine the slit separation suggested by these data and discuss any discrepancy.
法医学家有时利用光学干涉测量人类头发的直径。将一根头发放置在激光束中，将其衍射图样与已知狭缝的图样进行比较。在简化装置中，用双缝代替头发，缝间距 d = 0.25 mm，屏幕距离 D = 1.80 m，使用波长 λ = 632.8 nm 的单色光。测得中央亮纹与第五亮纹的间距为 22.7 mm。请根据数据计算双缝间距并讨论可能的偏差。
he formula for the n-th bright fringe is d sinθ = nλ, and for small angles sinθ ≈ tanθ = yn / D. Therefore yn 95 × 10⁻⁶) / 0.0227 ≈ 2.51 × 10⁻⁴ m = 0.251 mm. This close agreement validates the method.
第n级亮纹的公式为 d sinθ = nλ，在小角度下 sinθ ≈ tanθ = yn / D，因此 yn ≈ nλD / d。对于 n = 5，y₅ = 5λD / d。根据测量值可得 d = 5λD / y₅。代入数值得 d ≈ (5 × 632.8 × 10⁻⁹ × 1.80) / (22.7 × 10⁻³) ≈ 0.251 mm，与标称值高度吻合。
Discrepancy could arise from the small-angle approximation error, uncertainty in the fringe position, or the finite width of the slits. You would then be expected to calculate the percentage uncertainty and suggest improvements, such as using a larger D or averaging over more fringes.
偏差可能来源于小角度近似误差、条纹位置的不确定度或狭缝的有限宽度。接着你可能需要计算百分比不确定度并提出改进措施，如增大D或对更多级条纹取平均值。
6. Integrating AS Practical Skills into Written Cases | 将AS实验技能融入书面案例
CCEA frequently weaves practical techniques into the written paper. You might be asked to design a circuit to measure the internal resistance of a solar cell, or to plot a graph of ln(V) against time for a capacitor discharge and find the gradient. Thorough knowledge of Unit AS 3 procedures – percentage differences, logarithmic plots, and systematic error identification – is essential.
CCEA经常将实验技能编织进笔试试卷。你可能被要求设计一个测量太阳能电池内阻的电路，或绘制电容放电的ln(V)–t图并求其斜率。透彻掌握AS第三单元的实验流程——百分差、对数作图及系统误差识别——至关重要。
| Practical Skill | 常見應用 |
|---|---|
| Logarithmic plotting | Capacitor discharge, radioactive decay |
| Error propagation | Combined uncertainties in multi-step calculations |
| Gradient analysis | Determining g from pendulum data, λ from fringe spacing |
| Percentage difference | Comparing measured vs accepted values |
实践技能：对数作图对应电容放电、放射性衰变；误差传递对应多步计算中的合成不确定度；斜率分析对应从单摆数据求g、从条纹间距求λ；百分差对应对测量值与公认值进行比较。
7. Graphical Analysis in Case Study Responses | 案例解题中的图表分析
When a case study provides a graph – such as a Force-Extension curve for a climbing rope or a I-V characteristic for a solar cell – do not merely describe it. Extract gradients, areas under the curve, and intercepts. For a force-extension graph, area up to the elastic limit equals stored elastic potential energy. For an I-V curve, the inverse gradient at any point gives the dynamic resistance.
当案例题提供图表时——如登山绳的力–伸长曲线或太阳能电池的I-V特性曲线——不能仅仅描述它。要提取斜率、曲线下面积和截距。对于力–伸长图，弹性限以内的面积等于储存的弹性势能；对于I-V曲线，任一点斜率的倒数为动态电阻。
You may be asked to linearise a relationship. For instance, if given T = 2π √m/k, plotting T² against m yields a slope of 4π²/k, from which k is derived. Always label axes fully and use symmetrical scales to minimise plotting errors.
你可能被要求将关系线性化。例如，若给出 T = 2π √m/k，绘制 T²–m 图可得到斜率为 4π²/k，由此可求出k。坐标轴务必标注完整并使用合理刻度以减少绘图误差。
8. Common Pitfalls and How to Avoid Them | 常见陷阱与应对方法
- Unit neglect: Forgetting to convert cm to m or μF to F can wreck an otherwise perfect answer. Write units beside every substituted value.
- Unit neglect：忘记将cm转换为m或μF转换为F可能毁掉原本完美的解答。在每个代数值旁写下单位。
- Sign errors in energy conservation k to it. In bungee problems, define h=0 at the lowest point or at the bridge consistently.
- 能量守恒中的符号错误：选定一个势能零点并保持一致。在蹦极问题中，统一规定最低点或桥面为h=0。
- Blindly memorising derived forms: Derivations for terminal velocity in fluids or capacitor discharge half-life should be practised, as the exam often varies the initial conditions.
- 盲目记忆推导结果：流体的终极速度或电容放电半衰期等推导过程需要练习，因为考试常会改变初始条件。
- Ignoring air resistance when asked to discuss: In modelling questions, even if you neglect air resistance for part (a), part (d) may ask how it would affect the graph – always link back to energy dissipation.
- 在要求讨论时忽略空气阻力：在建模题中，即使(a)部分忽略了空气阻力，(d)部分可能询问它对图表的影响——此时务必联系能量耗散作答。
9. Tackling the 'Evaluate and Suggest' Questions | 攻克“评价与建议”类问题
High-band questions ask you to evaluate a model or experimental procedure. Structure your response: state the limitation, explain its impact on the result, and propose a specific improvement. For instance, in a summer-sault diver’s rotation analysis: 'The model assumes a rigid body, but the diver can change moment of inertia by tucking. This reduces the predicted angular speed. Using video analysis with markers on joints would quantify shape changes.'
高分题目要求你评价模型或实验步骤。组织答案的思路是：说明局限性，解释它对结果的影响，并提出具体的改进方案。例如，分析跳水运动员翻转时：“该模型假设刚体，但运动员通过抱膝可改变转动惯量，这会使预測角速度偏低。使用带有关节标记点的视频分析可以量化身体姿态变化。”
10. Time Management and Mark Allocation | 时间管理与分值分配
A typical case study carries 10–15 marks spread over several parts. Spend no more than 2 minutes per mark. If you are stuck on a calculation, record the relevant equation – this often gains a method mark – and move on. Reserve the final 5 minutes to re-read the stem; candidates frequently miss a key phrase like 'starts from rest' or 'initially uncharged capacitor'.
一个典型的案例题总分值在10–15分，分列于多个小题。每分值不超过2分钟。如果计算卡住，就写下相关公式——这通常能拿到方法分——然后继续。最后留出5分钟重读题干；考生常会遗漏诸如“从静止开始”或“电容初始不带电”等关键短语。
11. Bringing It All Together | 综合演练
Use past-paper case studies to practise the full cycle: annotate the stem, sketch a model, derive the algebra, compute the numbers, and critically evaluate. With each attempt, you will become quicker at recognising patterns – whether it is damped harmonic motion in car suspension or pulse-echo techniques in ultrasound imaging.
使用历年真题的案例题进行全流程练习：注释题干、绘制模型、推导公式、计算数值，最后批判性评价。通过每一次练习，你会更快识别出规律——无论涉及汽车悬挂的阻尼简谐运动还是超声成像中的脉冲回波技术。
12. Summary and Final Tips | 总结与终极技巧
Case studies reward structured thinking and deep conceptual understanding. Always link the physics to the context, show your working step by step, and check your answer’s realism. Remember that the examiner wants to see your thought process, not just the final number.
案例题奖励结构化的思维和深入的概念理解。永远将物理原理与情境关联起来，逐步展示解题步骤，并检验答案的现实性。记住，考官希望看到你的思维过程，而不仅仅是最终数字。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图