📚 Year 12 CIE Physics: Case Study Practice | Year 12 CIE 物理：案例分析实战演练
In CIE AS Physics, case study questions ask you to apply your knowledge to unfamiliar situations, combining theory, experimental data and logical reasoning. This article walks you through five carefully selected cases, showing how to break down problems, use models and present your solutions clearly.
在CIE AS物理中，案例分析题要求你把知识应用到陌生的情境中，融合理论、实验数据和逻辑推理。本文将通过五个精心挑选的案例，带你一步步拆解问题、使用模型并清晰地呈现解答。
1. Understanding CIE Case Study Questions | 理解CIE案例分析题型
Case study questions often present a short narrative followed by data, graphs or diagrams. They test your ability to identify relevant physics principles, extract information and carry out calculations or qualitative explanations. Marks are awarded for structured working, correct units and valid conclusions.
案例分析题通常会先给出一段简短叙述，再配上数据、图像或示意图。它们考查你识别相关物理原理、提取信息并进行计算或定性解释的能力。解题过程的结构化书写、正确单位以及合理结论都能得分。
Typical topics for Year 12 include mechanics, waves, electricity and thermal physics. The key is to treat the scenario as a real investigation: state assumptions, write down equations, substitute values and comment on the physical meaning of your results.
Year 12 的典型考点涵盖力学、波动、电学和热物理。关键是把场景当成真实研究：写明假设，列出方程，代入数值，并说明结果所对应的物理意义。
2. Case 1: Free Fall with Drag – Problem Statement | 案例1: 空气阻力下的自由落体–问题描述
A 2.0 kg object is dropped from rest. It falls vertically through air, experiencing a drag force that is directly proportional to its instantaneous speed: Fdrag = 0.5v (in N, where v is in m s⁻¹). The following velocity–time data was recorded.
一个 2.0 kg 的物体由静止释放，竖直下落。它在空气中运动时受到的空气阻力与瞬时速率成正比：Fdrag = 0.5v（单位为 N，v 的单位为 m s⁻¹）。记录的速率–时间数据如下。
| Time / s | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
| Velocity / m s⁻¹ | 0 | 8.2 | 14.1 | 18.4 | 21.5 | 23.8 |
Use the data and Newton’s laws to: (i) calculate the terminal velocity; (ii) find the initial acceleration; (iii) determine the acceleration when v = 10 m s⁻¹; (iv) sketch a labelled velocity–time graph and explain the shape using forces.
利用数据和牛顿定律：(i) 计算终端速率；(ii) 求初始加速度；(iii) 求 v = 10 m s⁻¹ 时的加速度；(iv) 绘制带有标注的速率–时间图并用力解释图线形状。
3. Case 1: Analysis and Solution | 案例1: 分析与解答
At terminal velocity the net force is zero, so weight = drag. Using mg = kvt gives vt = (2.0 × 9.81) / 0.5 = 39.24 m s⁻¹. The table suggests the object is still accelerating at 5.0 s (23.8 m s⁻¹), but it will eventually approach 39.2 m s⁻¹.
终端速率时合外力为零，因此重力等于阻力。由 mg = kvt 得 vt = (2.0 × 9.81) / 0.5 = 39.24 m s⁻¹。表中数据显示物体在 5.0 s 时仍在加速（23.8 m s⁻¹），但最终将趋近 39.2 m s⁻¹。
At t = 0, drag is zero, so the only force is weight. The initial acceleration is a₀ = mg / m = 9.81 m s⁻². This matches the steepest tangent on a v–t graph.
t = 0 时没有阻力，合力仅为重力，因此初始加速度 a₀ = mg / m = 9.81 m s⁻²。这与速率–时间图上最陡的切线斜率吻合。
When v = 10 m s⁻¹, net force F = mg − 0.5 × 10 = 19.62 − 5.0 = 14.62 N. Hence a = F / m = 14.62 / 2.0 = 7.31 m s⁻². As the object speeds up, drag increases, reducing the resultant force and therefore the acceleration, causing the gradient of the v–t graph to decrease until it becomes zero at terminal velocity.
当 v = 10 m s⁻¹ 时，合外力 F = mg − 0.5 × 10 = 19.62 − 5.0 = 14.62 N，因此加速度 a = 14.62 / 2.0 = 7.31 m s⁻²。随着物体速度增加，阻力变大，合力减小，从而加速度减小，导致速率–时间图的斜率逐渐减小，直到终端速率处斜率变为零。
The graph starts with a steep slope of 9.81 m s⁻², curves towards a horizontal asymptote at 39.24 m s⁻¹, and the area under the curve represents the distance fallen.
图线以 9.81 m s⁻² 的陡峭斜率起始，逐渐弯曲向 39.24 m s⁻¹ 的水平渐近线，图线下的面积表示下落距离。
4. Case 2: Projectile Motion – Problem | 案例2: 抛体运动–问题描述
A ball is kicked from ground level with an initial speed of 20.0 m s⁻¹ at an angle of 30.0° above the horizontal. Air resistance is negligible. Determine: (a) the time of flight; (b) the maximum height reached; (c) the horizontal range.
一个足球从地面以 20.0 m s⁻¹ 的初速率、与水平方向成 30.0° 的仰角踢出，忽略空气阻力。求：(a) 飞行时间；(b) 达到的最大高度；(c) 水平射程。
5. Case 2: Calculations and Reasoning | 案例2: 计算与推理
Resolve the initial velocity: ux = 20.0 cos30.0° = 17.3 m s⁻¹; uy = 20.0 sin30.0° = 10.0 m s⁻¹. The vertical motion is symmetric and governed by constant acceleration g = 9.81 m s⁻² downward.
分解初速度：ux = 20.0 cos30.0° = 17.3 m s⁻¹；uy = 20.0 sin30.0° = 10.0 m s⁻¹。竖直方向运动具有对称性，受恒定的向下加速度 g = 9.81 m s⁻² 支配。
Time to reach the highest point is tup = uy / g = 10.0 / 9.81 = 1.02 s. Total flight time T = 2 × 1.02 = 2.04 s.
到达最高点的时间 tup = uy / g = 10.0 / 9.81 = 1.02 s。总飞行时间 T = 2 × 1.02 = 2.04 s。
Maximum height H = (uy)² / (2g) = (10.0)² / (2 × 9.81) = 100 / 19.62 = 5.10 m.
最大高度 H = (uy)² / (2g) = (10.0)² / (2 × 9.81) = 100 / 19.62 = 5.10 m。
Horizontal range R = ux × T = 17.3 × 2.04 = 35.3 m. Because air resistance is ignored, horizontal velocity remains constant, and the trajectory is a parabola.
水平射程 R = ux × T = 17.3 × 2.04 = 35.3 m。由于忽略空气阻力，水平方向的速度保持不变，轨迹为抛物线。
6. Case 3: Internal Resistance – Experimental Setup | 案例3: 内阻–实验设置
A student investigates the internal resistance of a dry cell using a variable resistor, an ammeter and a voltmeter. The circuit is drawn with the voltmeter connected directly across the cell terminals. As the external resistance is changed, pairs of terminal voltage V and current I are recorded.
一位学生用可调电阻、电流表和电压表研究干电池的内阻。电路连接为电压表直接跨接在电池两端。改变外电阻时，记录下端电压 V 和电流 I 的成对数据。
| I / A | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
| V / V | 1.38 | 1.26 | 1.14 | 1.02 | 0.90 |
Using the data, plot a graph of V against I and determine the e.m.f. and internal resistance of the cell. Explain why the terminal voltage falls as current increases.
利用数据，绘制端电压 V 随电流 I 变化的图线，求出电池的电动势和内阻，并解释为什么端电压会随电流增大而下降。
7. Case 3: Data Interpretation and Graph | 案例3: 数据解读与图像
The relationship for a real cell is V = E − I r, where E is the e.m.f. and r is the internal resistance. Plotting V on the y‑axis and I on the x‑axis should give a straight line with gradient = −r and y‑intercept = E.
实际电池满足关系式 V = E − I r，其中 E 为电动势，r 为内阻。将 V 放在 y 轴、I 放在 x 轴绘图，应得到一条直线，其斜率 = −r，y 轴截距 = E。
Using the data, the points are almost collinear. Calculating the gradient: ΔV / ΔI = (0.90 − 1.38) / (0.50 − 0.10) = (−0.48) / 0.40 = −1.2 V A⁻¹. Hence internal resistance r = 1.2 Ω. Extrapolating the line to I = 0 gives intercept V = 1.50 V, so E ≈ 1.50 V.
从数据看，各点几乎共线。计算斜率：ΔV / ΔI = (0.90 − 1.38) / (0.50 − 0.10) = (−0.48) / 0.40 = −1.2 V A⁻¹，因此内阻 r = 1.2 Ω。将直线外推至 I = 0，得到截距 V = 1.50 V，故 E ≈ 1.50 V。
As current increases, more energy is dissipated inside the cell (
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