Typical Example Problems in IB WJEC Physics | IB WJEC 物理：典型例题详解

📚 Typical Example Problems in IB WJEC Physics | IB WJEC 物理：典型例题详解

Mastering physics requires not only understanding concepts but also solving a variety of problems. This article presents typical example problems commonly encountered in IB and WJEC physics syllabi, with detailed step-by-step solutions. Whether you are tackling mechanics, electricity, waves, or modern physics, these worked examples will help you sharpen your problem-solving skills and prepare for exams effectively.

掌握物理学不仅需要理解概念，还需要通过大量解题来巩固。本文精选了 IB 和 WJEC 物理课程中常见的典型例题，并提供详细的分步解答。无论你面对的是力学、电学、波动还是近代物理，这些范例将帮助你提升解题能力，有效备考。

1. Projectile Motion | 抛体运动

A ball is kicked from the ground with an initial speed of 20 m/s at an angle of 30° to the horizontal. Calculate (a) the maximum height reached, (b) the time of flight, and (c) the horizontal range. Take g = 9.8 m/s².

一球以20 m/s的初速度、与水平面成30°角从地面踢出。计算：(a) 达到的最大高度，(b) 飞行时间，(c) 水平射程。取 g = 9.8 m/s²。

Step 1: Resolve the initial velocity. u_x = u cosθ = 20 cos30° ≈ 17.32 m/s, u_y = u sinθ = 20 sin30° = 10 m/s.

步骤1：分解初速度。u_x = u cosθ = 20 cos30° ≈ 17.32 m/s，u_y = u sinθ = 20 sin30° = 10 m/s。

(a) At maximum height, v_y = 0. Using v_y² = u_y² – 2gH, we get H = u_y²/(2g) = 10²/(2×9.8) ≈ 5.10 m.

(a) 在最高点 v_y = 0。利用 v_y² = u_y² – 2gH，得 H = u_y²/(2g) = 10²/(2×9.8) ≈ 5.10 m。

(b) Time to reach the peak: t_up = u_y/g = 10/9.8 ≈ 1.02 s. Total time of flight T = 2 t_up ≈ 2.04 s.

(b) 到达最高点时间 t_up = u_y/g = 10/9.8 ≈ 1.02 s。总飞行时间 T = 2 t_up ≈ 2.04 s。

(c) Horizontal range R = u_x × T = 17.32 × 2.04 ≈ 35.3 m.

(c) 水平射程 R = u_x × T = 17.32 × 2.04 ≈ 35.3 m。

2. Connected Bodies and Newton’s Laws | 连接体与牛顿定律

Two blocks of masses m₁ = 5 kg and m₂ = 3 kg are connected by a light inextensible string over a frictionless pulley. m₁ rests on a smooth horizontal table, while m₂ hangs vertically. The system is released from rest. Find the acceleration of the system and the tension in the string. (g = 9.8 m/s²)

两个质量分别为 m₁ = 5 kg 和 m₂ = 3 kg 的物体通过轻质不可伸长的绳子相连，跨过一个光滑滑轮。m₁ 放在光滑水平桌面上，m₂ 竖直悬挂。系统由静止释放。求系统的加速度和绳中张力。(g = 9.8 m/s²)

For m₂: weight – T = m₂ a → 3g – T = 3a. For m₁: T = m₁ a = 5a. Substitute: 3×9.8 – 5a = 3a → 29.4 = 8a → a = 3.675 m/s². Then tension T = 5 × 3.675 = 18.375 N.

对 m₂：重力 – T = m₂ a → 3g – T = 3a。对 m₁：T = m₁ a =

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