📚 Year 12 AQA Physics: Case Study Practical Workout | Year 12 AQA 物理：案例分析实战演练
Mastering AQA Year 12 Physics goes beyond memorising formulas; it demands the ability to apply concepts to real-world situations. Case study questions test your skill in linking theory to practical scenarios, from car crashes and bridge cables to noise-cancelling technology and radiocarbon dating. This workout walks you through five core case studies, demonstrating how to extract data, choose the right model, set up equations, and critically evaluate results – exactly as the exam expects.
掌握 AQA Year 12 物理不仅需要记住公式，更需要把概念应用到真实情境中。案例分析题考查你将理论与实际场景联系起来的能力，场景涵盖车祸碰撞、桥梁缆索、降噪技术乃至放射性碳定年。本实战演练将带你走过五个核心案例，展示如何提取数据、选择正确模型、建立方程并批判性地评估结果——正是考试所要求的技能。
1. Understanding the AQA Case Study Approach | 理解AQA案例分析题型
Case studies in AQA AS Physics are extended problem-solving tasks rooted in practical applications. You are given a passage with data, diagrams and context, followed by questions that require you to identify relevant principles, perform calculations and comment on assumptions. Success depends on a structured reading of the scenario, linking each piece of information to a topic from the specification.
AQA AS 物理中的案例研究是植根于实际应用的拓展性问题解决任务。试题会提供一段包含数据、示意图和背景信息的材料，随后的问题要求你识别相关原理、进行计算并对假设条件作出评论。成功的关键在于有条理地阅读情境，将每一条信息与考纲中的某个主题联系起来。
Start by highlighting quantitative data and qualitative constraints (e.g. ‘neglect air resistance’, ‘the wire obeys Hooke’s law’). Then map the problem to the appropriate part of the course – mechanics, materials, waves, electricity or particle physics. Finally, set out your solution stepwise, clearly stating the physics law used, the substitution and the answer with the correct unit and significant figures.
首先要标出定量数据和定性约束条件（如“忽略空气阻力”“导线遵循胡克定律”）。然后将问题映射到课程的相关部分——力学、材料、波、电学或粒子物理。最后逐步写出解答，明确说明所用的物理定律、代入过程和结果，并给出正确的单位与有效数字。
2. Case 1: Car Crash Safety – Momentum and Impulse | 案例一：汽车碰撞安全——动量与冲量
A highway safety report describes a head-on collision between a 1200 kg car travelling at 15 m s⁻¹ and a stationary 800 kg car. The cars crumple and move together after impact. The report asks whether the impact force exceeded the 90 kN limit that the human body can briefly tolerate for a 0.12 s collision time.
一份高速公路安全报告描述了一辆质量为 1200 kg、以 15 m s⁻¹ 行驶的轿车与一辆静止的 800 kg 轿车发生的正面碰撞。碰撞后两车变形并一起运动。报告询问在 0.12 s 的碰撞时间内，撞击力是否超过了人体能短暂承受的 90 kN 极限。
By conservation of momentum, total momentum before = (1200 × 15) + (800 × 0) = 18 000 kg m s⁻¹. After the collision, the combined mass is 2000 kg, so the common velocity v = 18 000 / 2000 = 9.0 m s⁻¹. The change in velocity of the 800 kg car is Δv = 9.0 m s⁻¹, giving an impulse FΔt = mΔv = 800 × 9.0 = 7200 N s.
根据动量守恒，碰撞前的总动量 = (1200 × 15) + (800 × 0) = 18 000 kg m s⁻¹。碰撞后总质量为 2000 kg，因此共同速度 v = 18 000 / 2000 = 9.0 m s⁻¹。800 kg 车辆的速度变化为 Δv = 9.0 m s⁻¹，由此得到冲量 FΔt = mΔv = 800 × 9.0 = 7200 N s。
With Δt = 0.12 s, the average force on the smaller car is F = 7200 / 0.12 = 60 000 N (60 kN). This is well below the 90 kN threshold. However, real forces are not uniform; peak force may be higher. Exam markers expect you to note the assumption of constant force and the neglect of external friction.
当 Δt = 0.12 s 时，小车的平均受力为 F = 7200 / 0.12 = 60 000 N（60 kN），远低于 90 kN 的阈值。然而实际受力并不均匀，峰值力可能更高。阅卷人期待你指出恒力假设以及忽略了外部摩擦。
3. Analyzing Forces and Energy in Collisions | 分析碰撞中的力和能量
Energy considerations reinforce the analysis. The initial kinetic energy is ½ × 1200 × 15² = 135 000 J. The final kinetic energy of the coupled wreckage is ½ × 2000 × 9.0² = 81 000 J. The missing 54 000 J is dissipated as heat, sound and deformation work. The crumple zone deliberately increases the collision time, reducing the average force.
能量分析能巩固上述推理。初始动能为 ½ × 1200 × 15² = 135 000 J；撞击后结合体的动能为 ½ × 2000 × 9.0² = 81 000 J。消失的 54 000 J 转化为热能、声能和形变功。溃缩区特意延长了碰撞时间，从而降低平均受力。
You can estimate the work done in deforming the cars: W = F_avg × d, where d is the compression distance. If the combined compression is 0.8 m, then F_avg = 54 000 / 0.8 = 67.5 kN. This checks reasonably with the impulse-based figure and highlights why modern vehicles are designed to crumple in a controlled manner.
可以估算使车辆变形的功：W = F_avg × d，其中 d 为压缩距离。若总压缩量为 0.8 m，则 F_avg = 54 000 / 0.8 = 67.5 kN，与冲量方法算出的数值基本吻合，也突显了现代汽车采用可控溃缩设计的原因。
4. Case 2: Stress and Strain in Engineering Materials | 案例二：工程材料中的应力与应变
A suspension bridge uses high-tensile steel cables of diameter 4.0 cm. Each cable supports a tensile load of 2.4 × 10⁵ N. The original cable length is 80 m, and the steel’s Young modulus is 2.0 × 10¹¹ Pa. The inspection team must verify that the strain does not exceed 0.1%, the safe limit for fatigue life.
某悬索桥采用直径为 4.0 cm 的高强度钢缆，每根钢缆承受 2.4 × 10⁵ N 的拉伸载荷。原长为 80 m，钢材的杨氏模量为 2.0 × 10¹¹ Pa。检测团队需验证应变是否超过 0.1% 的安全疲劳极限。
Cross-sectional area A = π × (d/2)² = π × (0.020)² = 1.257 × 10⁻³ m². Tensile stress σ = F / A = 2.4×10⁵ / 1.257×10⁻³ ≈ 1.91 × 10⁸ Pa. Strain ε = σ / E = 1.91×10⁸ / 2.0×10¹¹ = 9.55 × 10⁻⁴ = 0.0955%. This is just below 0.1%, so the cable is at the boundary of the safe limit. A safety factor of at least 2 would be advisable for prolonged use.
横截面积 A = π × (d/2)² = π × (0.020)² = 1.257 × 10⁻³ m²。拉伸应力 σ = F / A = 2.4×10⁵ / 1.257×10⁻³ ≈ 1.91 × 10⁸ Pa。应变 ε = σ / E = 1.91×10⁸ / 2.0×10¹¹ = 9.55 × 10⁻⁴ = 0.0955%，略低于 0.1%，因此钢缆正处于安全极限边缘。为长期使用考虑，建议至少采用 2 倍的安全系数。
5. Young Modulus and Safety Factors | 杨氏模量与安全系数
The Young modulus is derived from the linear portion of a stress-strain graph. For the bridge cable, extension ΔL = ε × L₀ = 9.55×10⁻⁴ × 80 = 0.0764 m (7.64 cm). Such an elongation is acceptable if the joints allow for thermal expansion. Engineers often check that the working stress is less than half of the yield stress.
杨氏模量来自应力-应变图的线性段。对于桥缆，伸长量 ΔL = ε × L₀ = 9.55×10⁻⁴ × 80 = 0.0764 m（7.64 cm）。如果接缝允许热膨胀，这样的伸长是可以接受的。工程师通常会检验工作应力是否低于屈服应力的一半。
| Quantity | Value | Unit |
|---|---|---|
| Tensile stress | 1.91 × 10⁸ | Pa |
| Strain | 9.55 × 10⁻⁴ | dimensionless |
| Extension | 7.64 × 10⁻² | m |
Note the importance of using consistent SI units and converting diameters to radii. A common exam pitfall is to forget to halve the diameter or to use cm instead of m when calculating area, which dramatically alters the stress value.
注意使用一致的 SI 单位并将直径转换为半径。考试中常见的一个陷阱是忘记将直径减半，或在计算面积时以 cm 代替 m，这会大幅改变应力结果。
6. Case 3: Complex DC Circuit Analysis | 案例三：复杂直流电路分析
A sensor network in a weather station contains two loops with a shared branch. Loop 1 has a 12 V battery and resistors 4.0 Ω and 6.0 Ω; Loop 2 has a 9.0 V battery and resistors 3.0 Ω and the same 6.0 Ω shared resistor. The task is to find the current through each branch using Kirchhoff’s laws.
某气象站的传感器网络包含两个共有一条支路的回路。回路 1 由 12 V 电池和 4.0 Ω、6.0 Ω 电阻组成；回路 2 由 9.0 V 电池、3.0 Ω 和同一个 6.0 Ω 公共电阻组成。题目要求用基尔霍夫定律求出各支路电流。
Draw the circuit and label currents I₁ (left branch of 4 Ω), I₂ (right branch of 3 Ω) and I₃ (middle branch of 6 Ω), with directions assumed. Apply Kirchhoff’s first law at the top junction: I₁ + I₂ = I₃. For Loop 1 (left): 12 – 4I₁ – 6I₃ = 0. For Loop 2 (right): –9 + 6I₃ + 3I₂ = 0, or equivalently 6I₃ + 3I₂ = 9.
画出电路并标记电流：I₁（左侧 4 Ω 支路）、I₂（右侧 3 Ω 支路）和 I₃（中间 6 Ω 支路），并假设方向。在上方节点应用基尔霍夫第一定律：I₁ + I₂ = I₃。对回路 1（左侧）：12 – 4I₁ – 6I₃ = 0。对回路 2（右侧）：–9 + 6I₃ + 3I₂ = 0，即 6I₃ + 3I₂ = 9。
7. Applying Kirchhoff’s Laws Step-by-Step | 逐步应用基尔霍夫定律
Substitute I₃ = I₁ + I₂ into the loop equations. From Loop 1: 12 = 4I₁ + 6(I₁ + I₂) = 10I₁ + 6I₂. From Loop 2: 9 = 6(I₁ + I₂) + 3I₂ = 6I₁ + 9I₂. Now solve simultaneously. Multiply the first equation by 3: 36 = 30I₁ + 18I₂. Multiply the second by 2: 18 = 12I₁ + 18I₂. Subtract: 18 = 18I₁ → I₁ = 1.0 A.
将 I₃ = I₁ + I₂ 代入回路方程。由回路 1 得：12 = 4I₁ + 6(I₁ + I₂) = 10I₁ + 6I₂。由回路 2 得：9 = 6(I₁ + I₂) + 3I₂ = 6I₁ + 9I₂。联立求解。将第一个方程乘以 3：36 = 30I₁ + 18I₂；第二个方程乘以 2：18 = 12I₁ + 18I₂。两式相减得 18 = 18I₁ → I₁ = 1.0 A。
Then from 36 = 30×1 + 18I₂, we get 18I₂ = 6, so I₂ = 0.333 A. Hence I₃ = 1.333 A. A negative value would indicate the assumed direction is opposite; here all currents are positive, confirming the initial choices. Always check that powers balance: power supplied by batteries = 12×1 + 9×0.333 = 15.0 W; power dissipated = 4×1² + 3×0.333² + 6×1.333² = 4 + 0.333 + 10.67 = 15.0 W. The energy audit gives confidence in the result.
然后由 36 = 30×1 + 18I₂ 得 18I₂ = 6，I₂ = 0.333 A，从而 I₃ = 1.333 A。若出现负值则表示实际方向与假设相反；此处所有电流均为正值，印证了初始方向的选择。务必验算功率：电池提供的功率 = 12×1 + 9×0.333 = 15.0 W；消耗的功率 = 4×1² + 3×0.333² + 6×1.333² = 4 + 0.333 + 10.67 = 15.0 W。能量审计增强了对结果的信心。
8. Case 4: Interference in Noise-Cancelling Headphones | 案例四：降噪耳机中的干涉现象
Noise-cancelling headphones use a microphone to capture ambient sound and a speaker to produce a sound wave that destructively interferes with the incoming noise. Suppose a 300 Hz persistent hum has a wavelength of 1.13 m in air. The internal electronics advance the cancelling wave by exactly half a period. Explain the principle and calculate the required time delay.
降噪耳机利用麦克风捕捉环境声音，并通过扬声器产生一个与传入噪声发生相消干涉的声波。假设持续的 300 Hz 嗡嗡声在空气中的波长为 1.13 m，内置电子线路将抵消波精确提前半个周期。请解释其原理并计算所需的时间延迟。
The microphone signal is quickly processed to generate a wave that is in antiphase (180° phase difference) with the noise. Destructive interference occurs when the crest of the noise meets the trough of the generated wave, minimising the amplitude at the ear. A half-period shift corresponds to a path difference of λ/2, but realised electronically through a time delay T/2.
麦克风信号经过快速处理，产生一个与噪声反相（相位差 180°）的波。当噪声的波峰遇到生成波的波谷时即发生相消干涉，使耳边的振幅降至最低。半周期偏移对应 λ/2 的光程差，但在电子学上通过 T/2 的时间延迟来实现。
Period T = 1/f = 1/300 = 3.33 × 10⁻³ s (3.33 ms). Required time delay Δt = T/2 ≈ 1.67 ms. The system must sample and output the inverted signal within this window, which is feasible with modern digital signal processors. Real headphones additionally employ adaptive algorithms to handle changing noise spectra.
周期 T = 1/f = 1/300 = 3.33 × 10⁻³ s（3.33 ms）。所需时间延迟 Δt = T/2 ≈ 1.67 ms。系统必须在此时间窗口内采样并输出反相信号，这对现代数字信号处理器而言是可行的。实际耳机还采用自适应算法来应对变化的噪声频谱。
9. Path Difference and Phase Relationship | 路程差与相位关系
The principle extends to all wave phenomena: for destructive interference, the path difference must be an odd multiple of λ/2. In the headphone, the effective path difference is introduced electronically, but the same condition governs two-source interference, diffraction gratings and thin-film colours.
这一原理适用于所有波动现象：发生相消干涉时，光程差必须是 λ/2 的奇数倍。在耳机中，有效光程差是由电子学方式引入的，但同样的条件也支配着双源干涉、衍射光栅和薄膜色彩。
Constructive interference demands a path difference nλ. When analysing a case study, always identify whether the waves are in phase at the source and consider any phase changes upon reflection (e.g. a π shift at a fixed end or at an optically denser medium). Such details often decide the answer.
相长干涉要求光程差为 nλ。分析案例时，务必识别波源是否同相，并考虑反射时可能引入的相位变化（如固定端反射或光密介质反射时的 π 相位跃变）。这些细节往往决定了最终答案。
10. Case 5: Radioactive Dating with Carbon-14 | 案例五：碳-14放射性定年法
A wooden artefact excavated from an archaeological site has a carbon‑14 activity of 0.15 Bq per gram of carbon, whereas a living sample shows 0.25 Bq g⁻¹. The half‑life of ¹⁴C is 5730 years. Determine the age of the artefact and discuss the reliability of the dating method for very old samples.
一件从考古遗址出土的木器，其碳-14 活度为 0.15 Bq 每克碳，而现存生物样品的活度为 0.25 Bq g⁻¹。¹⁴C 的半衰期为 5730 年。测定该木器的年代，并讨论此定年法对极古老样品的可靠性。
Radioactive decay follows the exponential law A = A₀ e^{–λt}, where λ = ln2 / T½. First compute λ = 0.693 / 5730 = 1.21 × 10⁻⁴ year⁻¹. Then set up the ratio: A/A₀ = 0.15/0.25 = 0.60. Take natural logarithms: ln(0.60) = –λt, so t = –ln(0.60) / λ = –(–0.511) / 1.21×10⁻⁴ ≈ 4220 years.
放射性衰变遵循指数规律 A = A₀ e^{–λt}，其中 λ = ln2 / T½。先计算 λ = 0.693 / 5730 = 1.21 × 10⁻⁴ 年⁻¹。建立比值 A/A₀ = 0.15/0.25 = 0.60。取自然对数：ln(0.60) = –λt，故 t = –ln(0.60) / λ = –(–0.511) / 1.21×10⁻⁴ ≈ 4220 年。
Thus the artefact is approximately 4200 years old. The uncertainty increases for ages beyond about 10 half‑lives because the activity becomes too small to measure accurately. Also, the assumption that the atmospheric ¹⁴C concentration has remained constant must be scrutinised; calibration curves from dendrochronology are used to adjust the raw radiocarbon age.
因此，该木器大约有 4200 年历史。当年代超过约 10 个半衰期时，由于活度太低而难以精确测量，不确定性随之增大。此外，还需审视线关于大气中 ¹⁴C 浓度恒定不变的假设；通常使用树木年代学的校准曲线来校正原始的放射性碳年龄。
11. Decay Equations and Half-Life Calculations | 衰变方程与半衰期计算
Practice with decay equations strengthens your ability to handle the mathematical demands of AQA papers. Key relationships you must be comfortable with:
多练习衰变方程能增强应对 AQA 试卷中数学要求的能力。你必须熟练掌握以下关键关系：
A = A₀ e^{-λt} λ = ln2 / T½ t = (ln(A₀/A)) / λ
Sometimes a question will give the number of nuclei N rather than activity; recall A = λN. For instance, if a sample initially has 8.0 × 10¹⁰ ¹⁴C nuclei, after two half‑lives (11 460 years) the remaining number is 2.0 × 10¹⁰. Always match units: if T½ is in years, λ must be in year⁻¹, giving t in years.
有时题目给出的不是活度而是原子核数 N；请记住 A = λN。例如，若某样品初始含有 8.0 × 10¹⁰ 个 ¹⁴C 核，经过两个半衰期（11 460 年）后，剩余核数为 2.0 × 10¹⁰。务必统一单位：若 T½ 以年为单位，则 λ 的单位为 年⁻¹，由此求出的 t 也是年。
12. Bridging Theory to Practical Scenarios | 从理论到实际场景的衔接
The five cases illustrate that every exam problem can be unpacked by systematically identifying the physics, extracting data, stating assumptions and then solving with careful algebra. Always ask yourself: What is the fundamental law here? What quantities are conserved? What simplifications are reasonable, and how do they limit the answer’s applicability?
这五个案例表明，每一道考题都可以通过系统识别物理原理、提取数据、陈述假设，然后用细致的代数运算来解答。永远要问自己：这里的基本定律是什么？哪些物理量守恒？哪些简化是合理的，它们又如何限制了答案的适用范围？
Build a personal toolkit: for mechanics, always check momentum and energy; for materials, sketch a free‑body force diagram and use stress‑strain relationships; for circuits, apply Kirchhoff’s rules and verify with a power check; for waves, recall path difference and phase conditions; for nuclear processes, stick to the exponential decay law and watch the time units. With disciplined practice, case studies become the most rewarding questions on the paper.
建立你自己的工具箱：力学问题，务必核查动量和能量；材料问题，画出受力图并运用应力-应变关系；电路问题，应用基尔霍夫定律并通过功率核算加以验证；波动问题，回顾光程差和相位条件；核过程问题，紧扣指数衰变律并注意时间单位。经过有素的练习，案例分析题将成为试卷中最有成就感的题目。
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