Case Study Practical Drills for CAIE AS Physics | CAIE AS物理案例分析与实战演练

📚 Case Study Practical Drills for CAIE AS Physics | CAIE AS物理案例分析与实战演练

In the CAIE AS Physics examination, case study questions test the ability to apply physical principles to real-world scenarios. This article presents a series of practical drills covering mechanics, materials, waves, electricity, and nuclear physics. Each case includes a description, key concepts, calculations, and error analysis to sharpen your problem-solving skills.

在 CAIE AS 物理考试中，案例分析题考查将物理原理应用于真实情景的能力。本文提供一套涵盖力学、材料、波、电学和原子核物理的实战演练。每个案例包括情景描述、核心概念、计算步骤和误差分析，以提升你的解题技巧。

1. Car Braking Distance Analysis | 汽车制动距离分析

A car travels at 20 m s⁻¹ on a dry road. The driver reacts in 0.50 s and then brakes with a constant deceleration of 6.0 m s⁻². Calculate the total distance covered before the car stops.

一辆汽车在干燥路面上以20 m s⁻¹行驶。驾驶员反应时间为0.50 s，随后以6.0 m s⁻²的恒定减速度刹车。计算汽车停下所经过的总距离。

on). We use s = v t for constant speed and v² = u² + 2as for uniformly accelerated motion.

总停车距离是思考距离（反应时间内）与制动距离（减速过程）之和。匀速段用 s = v t，匀加速段用 v² = u² + 2as。

Thinking distance: s₁ = 20 × 0.50 = 10 m

思考距离：s₁ = 20 × 0.50 = 10 m

Braking: 0² = 20² + 2(−6.0)s₂ ⇒ s₂ = 400 / 12 ≈ 33.3 m

制动阶段：0² = 20² + 2(−6.0)s₂ ⇒ s₂ = 400 / 12 ≈ 33.3 m

on may drop to 3.0 m s⁻², nearly doubling the braking distance. Always consider tyre and road conditions in real-world safety assessments.

总停车距离 = 10 m + 33.3 m = 43.3 m。在湿滑路面，减速度可能降至3.0 m s⁻²，制动距离几乎翻倍。实际安全评估中必须考虑轮胎和路面条件。

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2. Block on an Inclined Plane | 斜面上的滑块

A 5.0 kg block slides from rest down a plane inclined at 30° to the horizontal. The coefficient of dynamic friction between the block and the surface is 0.20. Determine the acceleration of the block.

一个5.0 kg的滑块从静止沿与水平面成30°的斜面下滑。滑块与表面之间的动摩擦因数为0.20。求滑块的加速度。

ion R balances mg cosθ, and the net force down the slope is mg sinθ minus friction f = μR.

将重力 mg 分解为沿斜面和垂直斜面的分量。法向反作用力 R 平衡 mg cosθ，沿斜面的合力为 mg sinθ 减去摩擦力 f = μR。

Parallel force = mg sin30° = 5 × 9.81 × 0.5 = 24.525 N

平行分量 = mg sin30° = 5 × 9.81 × 0.5 = 24.525 N

Normal: R = mg cos30° = 5 × 9.81 × 0.866 = 42.48 N

法向力：R = mg cos30° = 5 × 9.81 × 0.866 = 42.48 N

Friction f = μR = 0.20 × 42.48 = 8.50 N

摩擦力 f = 0.20 × 42.48 = 8.50 N

Net force F = 24.525 − 8.50 = 16.025 N ⇒ a = F/m = 3.21 m s⁻²

净合力 F = 24.525 − 8.50 = 16.025 N ⇒ 加速度 a = F/m = 3.21 m s⁻²

acceleration would be g sin30° = 4.9 m s⁻², showing friction significantly reduces motion.

滑块加速度约为3.2 m s⁻²。若斜面光滑，加速度为 g sin30° = 4.9 m s⁻²，可见摩擦力显著减缓运动。

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3. Basketball Projectile Motion | 篮球投篮的抛体运动

s 3.0 m above the ground and is 6.0 m away horizontally. Will the ball go through the hoop? (Ignore air resistance, g = 9.81 m s⁻²)

一名篮球运动员从2.0 m高度以8.0 m s⁻¹的初速度、与水平方向成45°角投篮。篮框离地3.0 m，水平距离6.0 m。球能否入框？（忽略空气阻力，g = 9.81 m s⁻²）

Resolve initial velocity: vₓ = v₀ cosθ, v_y to reach the horizontal distance and then find the vertical displacement at that instant.

分解初速度：vₓ = v₀ cosθ，v_y = v₀ sinθ。计算水平飞行所需时间，再求该时刻的竖直位移。

vₓ = 8.0 cos45° = 5.657 m s⁻¹, v_y = 8.0 sin45° = 5.657 m s⁻¹

vₓ = 8.0 cos45° = 5.657 m s⁻¹，v_y = 8.0 sin45° = 5.657 m s⁻¹

Time to hoop: t = 6.0 / 5.657 = 1.061 s

到达篮框时间：t = 6.0 / 5.657 = 1.061 s

Vertical displacement: y = v_y t − ½gt² = 5.657×1.061 − 4.905×(1.061)²

竖直位移：y = v_y t − ½gt² = 5.657×1.061 − 4.905×(1.061)²

y ≈ 6.001 − 5.522 = 0.479 m above release point

y ≈ 6.001 − 5.522 = 0.479 m（高于出手点）

hich is below the 3.0 m hoop. The shot is too short. To score, the player must increase speed or release angle slightly.

最终高度 = 2.0 m + 0.479 m = 2.48 m，低于3.0 m的篮框。此次投篮过短。要进球，球员需略微增加速度或出手角度。

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4. Stress and Strain in a Suspension Bridge Cable | 吊桥缆索的应力与应变

kN. The Young modulus of the steel is 2.0 × 10¹¹ Pa. Calculate the extension of the cable.

悬索桥的一根竖直钢缆长50 m，直径20 mm，承受80 kN荷载。钢的杨氏模量为2.0 × 10¹¹ Pa。计算缆索的伸长量。

e cross-sectional area, then stress, strain, and finally extension using ΔL = (F L)/(A E).

首先求横截面积，再求应力、应变，最后用 ΔL = (F L)/(A E) 计算伸长。

Area A = πd²/4 = π(0.020)²/4 = 3.14 × 10⁻⁴ m²

横截面积 A = πd²/4 = π(0.020)²/4 = 3.14 × 10⁻⁴ m²

Stress σ = F/A = 80 000 / (3.14 × 10⁻⁴) = 2.55 × 10⁸ Pa

应力 σ = F/A = 80 000 / (3.14 × 10⁻⁴) = 2.55 × 10⁸ Pa

Strain ε = σ/E = (2.55 × 10⁸) / (2.0 × 10¹¹) = 1.275 × 10⁻³

应变 ε = σ/E = (2.55 × 10⁸) / (2.0 × 10¹¹) = 1.275 × 10⁻³

Extension ΔL = ε × L = 1.275 × 10⁻³ × 50 = 0.0638 m (≈6.4 cm)

伸长量 ΔL = ε × L = 1.275 × 10⁻³ × 50 = 0.0638 m（约6.4 cm）

This small extension is within the elastic limit. In design, a safety factor is applied by using cables with larger diameter or multiple strands to accommodate dynamic loads like wind and traffic.

该微小伸长在弹性限度内。设计中通过使用更大直径的缆索或多股钢索引入安全系数，以应对风荷载和交通等动态荷载。

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5. LED Circuit with Potential Divider | LED分压器电路

An LED with forward voltage 2.0 V and maximum current 20 mA is to be powered by a 9.0 V battery. Design a suitable potential divider circuit to provide exactly 2.0 V across the LED, even if the battery voltage drops slightly. Use two fixed resistors R₁ and R₂.

一个正向电压为2.0 V、最大电流20 mA的LED需由9.0 V电池供电。设计一个合适的分压器电路，为LED提供恰好2.0 V电压，即使电池电压略有下降。使用两个固定电阻R₁和R₂。

The potential divider formula V_out = V_in Choose R₂ = 220 Ω, then solve for R₁. Also ensure the current through the LED is limited.

分压公式 V_out = V_in × R₂/(R₁ + R₂) 得 2.0 = 9.0 × R₂/(R₁ + R₂)。选择 R₂ = 220 Ω，再求解 R₁。同时需确保LED电流受限制。

2.0/9.0 = R₂/(R₁ + R₂) ⇒ R₁ = 3.5 R₂

2.0/9.0 = R₂/(R₁ + R₂) ⇒ R₁ = 3.5 R₂

With R₂ = 220 Ω, R₁ = 3.5 × 220 = 770 Ω (use 750 Ω or 820 Ω standard value)

若 R₂ = 220 Ω，则 R₁ = 3.5 × 220 = 770 Ω（使用标准值750 Ω或820 Ω）

A current-limiting resistor may be added in series with the LED to

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