Case Study Practice and Drills for Year 12 Edexcel Physics | 案例分析实战演练

📚 Case Study Practice and Drills for Year 12 Edexcel Physics | 案例分析实战演练

This article presents a comprehensive case study on vehicle collision and safety, integrating key concepts from mechanics, materials, and practical skills as required in the Year 12 Edexcel Physics specification. By working through this real-world scenario, you will sharpen your problem-solving abilities and exam technique.

本文通过一个关于车辆碰撞与安全的综合案例，整合了 Year 12 Edexcel 物理大纲中力学、材料学及实验技能的核心概念。通过演练这个真实世界的情景，你将提升解题能力与考试技巧。

1. Setting the Scene: A Car Braking Incident | 场景设定：汽车刹车事故

Imagine a car of mass 1200 kg travelling at 20 m s⁻¹ on a dry road. The driver sees an obstacle and reacts after 0.70 s. The car then brakes, and the coefficient of friction between the tyres and the road is 0.80. We will analyse the motion from the moment the obstacle is seen to a possible collision.

假设一辆质量为 1200 kg 的汽车以 20 m s⁻¹ 的速度在干燥路面上行驶。驾驶员看到障碍物，0.70 秒后做出反应。随后汽车制动，轮胎与路面的摩擦系数为 0.80。我们将分析从看到障碍物到可能发生碰撞的运动过程。

Coefficient of friction μ = 0.80, g = 9.81 m s⁻²

2. Kinematic Analysis of Braking Distance | 刹车距离的运动学分析

First, calculate the thinking distance: s_think = u × t_reaction = 20 × 0.70 = 14 m. The braking deceleration is found from a = μg = 0.80 × 9.81 = 7.85 m s⁻². Using the equation v² = u² + 2as with v = 0, u = 20 m s⁻¹, a = -7.85 m s⁻² gives braking distance s_brake = u² / (2a) = 400 / (2 × 7.85) ≈ 25.5 m. Thus the total stopping distance is 39.5 m. If the obstacle is only 30 m away, the car will collide. The speed at impact is found from v² = 20² − 2 × 7.85 × 16, giving v ≈ 12.2 m s⁻¹.

首先，计算反应距离：s_反应 = u × t_反应 = 20 × 0.70 = 14 m。制动减速度可由 a = μg = 0.80 × 9.81 = 7.85 m s⁻² 求出。应用 v² = u² + 2as，取 v = 0, u = 20 m s⁻¹, a = -7.85 m s⁻² 得到制动距离 s_制动 = u² / (2a) = 400 / (2 × 7.85) ≈ 25.5 m。因此总停车距离为 39.5 m。若障碍物仅 30 m 远，汽车将发生碰撞。碰撞速度由 v² = 20² − 2 × 7.85 × 16 算出，v ≈ 12.2 m s⁻¹。

v² = u² + 2as

3. Newton’s Second Law and Friction | 牛顿第二定律与摩擦力

From Newton’s second law, the braking force provided by friction is F = μmg. For μ = 0.80, F = 0.80 × 1200 × 9.81 ≈ 9420 N. The resulting deceleration is a = F/m = μg = 7.85 m s⁻², as used above. If the road is wet and μ drops to 0.40, the deceleration halves to 3.92 m s⁻², and the braking distance increases to about 51 m – clearly demonstrating the danger of reduced friction.

根据牛顿第二定律，摩擦力提供的制动力为 F = μmg。当 μ = 0.80 时，F ≈ 9420 N。所产生的减速度 a = F/m = μg = 7.85 m s⁻²，即前面所用数值。若路面湿滑且 μ 降至 0.40，减速度减半为 3.92 m s⁻²，制动距离将增至约 51 m——这清楚地表明摩擦力降低的危险性。

The table below compares stopping distances for different road conditions:

下表比较了不同路面条件下的停车距离：