📚 Pre-U CIE Physics: Case Study Practical Workouts | Pre-U CIE 物理：案例分析实战演练
In the Pre-U CIE Physics examination, case study questions demand much more than recalling formulas – they test your ability to interpret real-world scenarios, extract relevant data, apply physical principles across topics, and critically evaluate results. This article provides two in-depth case studies, one on mechanics and one on electricity, followed by general strategies, to sharpen your analytical skills and build the confidence needed for top-tier performance.
在 Pre-U CIE 物理考试中，案例分析题远不止是回忆公式——它们考查你解读真实情境、提取相关信息、跨主题应用物理原理以及批判性评估结果的能力。本文提供两个深入的案例，一个关于力学，一个关于电学，并随后给出通用策略，以磨炼你的分析技巧并建立取得优异成绩所需的信心。
1. Case Study 1: Braking Distance and Reaction Time | 案例一：制动距离与反应时间
A car is travelling at a constant speed of 20.0 m s⁻¹ (72 km h⁻¹) on a dry, level road. The driver suddenly notices a stationary obstacle ahead. After a reaction time of 0.70 s, the driver applies the brakes, locking all four wheels. The coefficient of sliding friction between the tyres and the road surface is 0.80. Assume the acceleration due to gravity g = 9.81 m s⁻². Your task is to determine the total distance the car travels from the instant the obstacle becomes visible until it comes to rest, and to discuss the sensitivity of the result to changing conditions.
一辆汽车在干燥的水平路面上以 20.0 m s⁻¹（72 km h⁻¹）的恒定速度行驶。驾驶员突然注意到前方有一个静止的障碍物。经过 0.70 s 的反应时间后，驾驶员踩下刹车，使四个车轮全部抱死。轮胎与路面之间的滑动摩擦系数为 0.80。假设重力加速度 g = 9.81 m s⁻²。你的任务是确定从障碍物出现到汽车停止所行驶的总距离，并讨论结果对条件变化的敏感程度。
2. Extracting Given Quantities and Assumptions | 数据提取与假设
Start by listing all numerical data with their symbols: initial speed u = 20.0 m s⁻¹, final speed v = 0 m s⁻¹, reaction time tᵣ = 0.70 s, coefficient of friction μ = 0.80, g = 9.81 m s⁻². The motion splits into two phases: a constant-velocity phase during the driver’s reaction time, and a uniformly decelerated phase under sliding friction. We assume the road is horizontal, air resistance is negligible, and the coefficient of friction remains constant during braking.
首先列出所有数值数据及其符号：初速度 u = 20.0 m s⁻¹，末速度 v = 0 m s⁻¹，反应时间 tᵣ = 0.70 s，摩擦系数 μ = 0.80，g = 9.81 m s⁻²。运动分为两个阶段：驾驶员反应时间内的匀速阶段，以及滑动摩擦下的匀减速阶段。我们假设道路水平，空气阻力忽略不计，且制动过程中摩擦系数保持不变。
3. Kinematics Analysis | 运动学分析
During the reaction time, the car travels at constant speed. The reaction distance s₁ is given by s₁ = u × tᵣ = 20.0 × 0.70 = 14.0 m. For the braking phase, we use the kinematic equation v² = u² + 2 a s. Here a is the deceleration. The only horizontal force is the kinetic friction f = μ N = μ m g, so the deceleration magnitude is a = f / m = μ g = 0.80 × 9.81 = 7.848 m s⁻². Since it decelerates, a = -7.848 m s⁻². Then 0 = (20.0)² + 2 × (-7.848) × s₂. Solving gives braking distance s₂ = 400 / (2 × 7.848) ≈ 25.5 m. Thus total stopping distance s_total = s₁ + s₂ = 14.0 + 25.5 = 39.5 m.
在反应时间内，汽车以恒定速度行驶。反应距离 s₁ 由 s₁ = u × tᵣ = 20.0 × 0.70 = 14.0 m 给出。制动阶段，我们使用运动学方程 v² = u² + 2 a s。此处 a 为减速度。唯一的水平力是动摩擦力 f = μ N = μ m g，因此减速度大小为 a = f / m = μ g = 0.80 × 9.81 = 7.848 m s⁻²。由于是减速，a = -7.848 m s⁻²。于是 0 = (20.0)² + 2 × (-7.848) × s₂，解得制动距离 s₂ = 400 / (2 × 7.848) ≈ 25.5 m。因此总制动距离 s_total = s₁ + s₂ = 14.0 + 25.5 = 39.5 m。
4. Newton’s Second Law and Friction | 牛顿第二定律与摩擦力
The deceleration derived from a = μ g can be verified by directly applying Newton’s second law. The net horizontal force is the kinetic friction f in the opposite direction to motion. Since f = μ m g, Newton’s second law gives m a = μ m g, so a = μ g. It is important to note that the braking distance s₂ is independent of the mass of the car, because both the inertia and the frictional force are proportional to m. This is a classic result that students should be able to explain clearly.
由 a = μ g 导出的减速度可以通过直接应用牛顿第二定律来验证。净水平力是方向与运动相反的动摩擦力 f。由于 f = μ m g，牛顿第二定律给出 m a = μ m g，因此 a = μ g。重要的是要注意，制动距离 s₂ 与汽车质量无关，因为惯性和摩擦力都与 m 成正比。这是一个经典结论，学生应该能够清晰地解释。
5. Energy Method Verification | 能量方法验证
The same braking distance can be obtained by equating the kinetic energy lost to the work done by friction. The initial kinetic energy is ½ m u². The work done by the constant friction force f over distance s₂ is f × s₂ = μ m g s₂. Setting ½ m u² = μ m g s₂ gives s₂ = u² / (2 μ g), which matches the kinematic result. Substituting values: s₂ = (20.0)² / (2 × 0.80 × 9.81) = 400 / 15.696 ≈ 25.5 m. This energy approach is often quicker and avoids explicitly calculating the deceleration.
相同的制动距离可以通过将损失的动能与摩擦力所做的功相等来获得。初始动能为 ½ m u²。恒定摩擦力 f 在距离 s₂ 上做的功为 f × s₂ = μ m g s₂。令 ½ m u² = μ m g s₂ 得出 s₂ = u² / (2 μ g)，这与运动学结果一致。代入数值：s₂ = (20.0)² / (2 × 0.80 × 9.81) = 400 / 15.696 ≈ 25.5 m。这种能量方法通常更快捷，且无需显式计算减速度。
6. Sources of Error and Experimental Design | 误差来源与实验设计
In reality, reaction time varies among drivers (typical range 0.5 s – 1.5 s). The coefficient of friction changes with road conditions: on a wet road μ may drop to 0.40, which would more than double the braking distance. Tyre wear, brake temperature, and slope are other factors. To measure reaction time, students could use a ruler-drop test. To determine μ, they could pull a block of known mass across the surface with a force sensor. In a Pre-U case study, you should always comment on the validity of assumptions and suggest how uncertainties propagate.
在现实中，反应时间因人而异（典型范围 0.5 s – 1.5 s）。摩擦系数随路面状况改变：在湿滑路面上 μ 可能降至 0.40，这会使制动距离增加一倍以上。轮胎磨损、刹车温度以及坡度是其他因素。要测量反应时间，学生可以使用直尺下落测试。要测定 μ，他们可以用力传感器拉着已知质量的物块在表面上滑动。在 Pre-U 案例分析中，你应该始终评论假设的有效性，并说明不确定度如何传播。
7. Case Study 2: Maximum Power Transfer from a Battery | 案例二：电池的最大功率传输
A student has a battery with an emf E = 12.0 V and an internal resistance r = 2.0 Ω. She connects a variable load resistor R across the terminals and aims to investigate the power delivered to the load. She varies R from 0.5 Ω to 10 Ω and records the current. The core question: At what value of R is the power dissipated in the load a maximum, and what is that maximum power? Additionally, discuss the efficiency of power delivery and the practical implications.
一位学生有一个电动势 E = 12.0 V、内阻 r = 2.0 Ω 的电池。她在电池两端连接一个可变负载电阻 R，旨在研究输送到负载的功率。她将 R 从 0.5 Ω 改变到 10 Ω，并记录电流。核心问题是：R 取何值时负载耗散的功率最大？最大功率是多少？此外，讨论功率传输的效率及其实际意义。
8. Circuit Analysis and Model | 电路分析与模型
The circuit is a simple series loop: the total resistance is R + r. The current I is given by I = E / (R + r). The power P dissipated in the load resistor is P = I² R = [E² R] / (R + r)². Write this clearly. Using the given values: E = 12.0 V, r = 2.0 Ω, so P = (144 R) / (R + 2)², with power in watts when R is in ohms. Before calculus, we can explore trends: when R is very small (0.5 Ω), denominator ≈ (2.5)² = 6.25, so P ≈ 144 × 0.5 / 6.25 ≈ 11.5 W. When R = 2 Ω, P = 144 × 2 / (4)² = 288 / 16 = 18.0 W. When R = 10 Ω, P = 1440 / 144 = 10.0 W. This suggests a maximum near R = r.
该电路是一个简单的串联回路：总电阻为 R + r。电流 I 由 I = E / (R + r) 给出。负载电阻上耗散的功率 P 为 P = I² R = [E² R] / (R + r)²。将式子写清楚。使用给定数值：E = 12.0 V，r = 2.0 Ω，因此 P = (144 R) / (R + 2)²，当 R 以欧姆为单位时功率以瓦特为单位。在进行微积分之前，我们可以探索其趋势：当 R 非常小（0.5 Ω）时，分母 ≈ (2.5)² = 6.25，因此 P ≈ 144 × 0.5 / 6.25 ≈ 11.5 W。当 R = 2 Ω 时，P = 144 × 2 / (4)² = 288 / 16 = 18.0 W。当 R = 10 Ω 时，P = 1440 / 144 = 10.0 W。这表明最大值出现在 R = r 附近。
9. Deriving Maximum Power Condition | 推导最大功率条件
To find the exact maximum, differentiate P with respect to R and set the derivative to zero. It is easier to differentiate P = E² R (R + r)⁻². Let dP/dR = E²[ (R + r)⁻² + R × (-2)(R + r)⁻³ ] = 0. Since E² ≠ 0 and (R + r)⁻³ ≠ 0, we have (R + r) – 2R = 0, so r – R = 0, hence R = r. The maximum power is therefore P_max = E² r / (2r)² = E² / (4r). Substituting numbers: P_max = (12.0)² / (4 × 2.0) = 144 / 8 = 18.0 W. This elegant result shows that maximum power is transferred when the load resistance matches the internal resistance of the source.
为了找到精确的最大值，将 P 对 R 求导并令导数为零。对 P = E² R (R + r)⁻² 求导更容易。令 dP/dR = E²[ (R + r)⁻² + R × (-2)(R + r)⁻³ ] = 0。由于 E² ≠ 0 且 (R + r)⁻³ ≠ 0，我们有 (R + r) – 2R = 0，因此 r – R = 0，即 R = r。因此最大功率为 P_max = E² r / (2r)² = E² / (4r)。代入数值：P_max = (12.0)² / (4 × 2.0) = 144 / 8 = 18.0 W。这一简洁的结果表明，当负载电阻与电源内阻匹配时，传输的功率最大。
10. Graphical Approach: P vs Load Resistance | 图解法：P – R 负载曲线
A graph of P against R is very revealing. Use a table of calculated values to sketch it:
P 随 R 变化的曲线图非常能说明问题。使用计算值的表格进行绘制：
| R / Ω | (R+2)² | P / W |
|---|---|---|
| 0.5 | 6.25 | 11.5 |
| 1.0 | 9.0 | 16.0 |
| 2.0 | 16.0 | 18.0 |
| 4.0 | 36.0 | 16.0 |
| 8.0 | 100 | 11.5 |
The curve rises steeply from R = 0, peaks at R = 2 Ω, and then decays gradually. The shape is asymmetric. In a case study, you should be prepared to sketch or interpret such a graph, identifying the maximum and showing that P → 0 as R → 0 and as R → ∞.
曲线从 R = 0 处急剧上升，在 R = 2 Ω 处达到峰值，然后逐渐下降。形状是不对称的。在案例分析中，你应该准备好绘制或解读此类曲线图，识别最大值，并显示当 R → 0 和 R → ∞ 时 P → 0。
11. Practical Application and Battery Efficiency | 实际应用与电池效率
Although R = r yields the maximum power, the efficiency of power transfer is only 50% at this point. Efficiency η is defined as power in load divided by total power supplied by the battery: η = I² R / (I² (R + r)) = R / (R + r). When R = r, η = 1/2. In high-power applications like mobile phone chargers and electric vehicles, we sacrifice maximum power for high efficiency by making R >> r. In contrast, in communication systems where signal power matters more than energy loss, impedance matching is desirable. In a case study, you must discuss such trade-offs.
虽然 R = r 能获得最大功率，但此时功率传输的效率仅为 50%。效率 η 定义为负载功率除以电池提供的总功率：η = I² R / (I² (R + r)) = R / (R + r)。当 R = r 时，η = 1/2。在手机充电器和电动汽车等高功率应用中，我们通过使 R >> r 来牺牲最大功率以获得高效率。相反，在信号功率比能量损耗更重要的通信系统中，阻抗匹配是可取的。在案例分析中，你必须讨论此类权衡。
12. General Strategies for Tackling Case Studies | 应对案例分析的通用策略
When facing an unfamiliar scenario, read the problem twice and underline key quantities and constraints. Identify the relevant physical principles – is it mechanics, thermodynamics, electricity, waves? Break the problem into steps and draw clear diagrams. Always state your assumptions and check whether they are justified. Use algebra before inserting numbers to spot cancellations. Reflect on the physical reasonableness of your answer and suggest how experimental errors could affect the result. Finally, link the case to real-world applications, which demonstrates deeper understanding.
当面对陌生的情境时，读两遍题目并在关键量和限制条件下划线。确定相关的物理原理——是力学、热力学、电学还是波动？将问题分解为步骤并绘制清晰的示意图。始终说明你的假设并检查它们是否合理。在代入数字之前先使用代数，以便发现简化。反思你的答案在物理上的合理性，并提出实验误差如何影响结果。最后，将该案例与现实世界的应用联系起来，这能展示更深层次的理解。
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