A-Level Physics: Typical Example Questions Explained in Detail | A-Level 物理：典型例题详解

📚 A-Level Physics: Typical Example Questions Explained in Detail | A-Level 物理：典型例题详解

A-level Physics papers reward method as much as memory. This article walks through twelve typical exam-style questions covering mechanics, electricity, waves, quantum physics and simple harmonic motion, showing the exact working an examiner expects: resolve the vectors, choose the equation that contains one unknown, substitute with units, and finish with a sensible check. Every worked example is given first in English and then in Chinese, so you revise the physics and the language together.

A-Level 物理试卷不仅考记忆，更考方法。本文详细讲解 12 道典型考题，覆盖力学、电学、波、量子物理与简谐运动，演示考官期待的完整步骤：分解矢量、选择只含一个未知量的方程、带单位代入、最后做合理性检验。每道例题都先给英文再给中文，让你一边复习物理，一边熟悉英文表达。

1. How to Attack a Typical Question | 典型题目的通用解题框架

Before touching any numbers, set up a routine. List the data in SI units and label the unknown. Choose a positive direction and keep it for the whole question, especially with vectors and with current. Select an equation that contains exactly one unknown. Substitute with units, then check the size, the sign and the significant figures. Most lost marks in A-level physics come from the second and fourth steps, not from the algebra.

动笔计算之前，先建立固定流程。把已知量全部换算成国际单位，并标出未知量。选定正方向，并在整道题中保持一致，矢量与电流尤其如此。选择只含一个未知量的方程。带单位代入，然后检查数量级、正负号与有效数字。A-Level 物理的丢分大多发生在第二步和第四步，而不是代数运算本身。

• Write "given / required / equation" at the top of the answer. | 在答题开头写清“已知 / 求 / 所用公式”。
• Convert cm, g, mA and minutes into m, kg, A and s first. | 先把 cm、g、mA、分钟换算成 m、kg、A、s。
• State the direction you call positive before using signs. | 在使用正负号之前，先声明你把哪个方向定为正。
• Finish with one sentence of physical sense-checking. | 最后用一句话做物理意义上的合理性检验。

2. Projectile Motion: Resolving the Velocity | 抛体运动：速度分解

maximum height, (b) the total time of flight, and (c) the horizontal range. Projectiles become simple once you accept that the velocity into horizontal and vertical motions are completely independent.

一个小球以 20 m/s 的初速度从地面斜向上抛出，与水平方向成 30° 角，取 g = 9.8 m/s²。求：(a) 最大高度，(b) 总飞行时间，(c) 水平射程。只要接受“水平运动与竖直运动彼此独立”，抛体问题就变得非常简单。

Resolve the initial velocity into components. Horizontal: u_h = u cos θ = 20 cos 30° = 20 × (√3/2) ≈ 17.32 m/s. Vertical: u_v = u sin θ = 20 sin 30° = 10 m/s.

先把初速度分解。水平分量：u_h = u cos θ = 20 cos 30° = 20 × (√3/2) ≈ 17.32 m/s。竖直分量：u_v = u sin θ = 20 sin 30° = 10 m/s。

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u_h = u cos θ　　u_v = u sin θ

Horizontally there is no acceleration if air resistance is ignored, so u_h stays constant for the whole flight. Vertically the acceleration is always g downwards. Writing these two components down as a separate line is often worth a mark on its own.

忽略空气阻力时，水平方向没有加速度，因此 u_h 在整个飞行过程中保持不变。竖直方向的加速度始终为 g，方向向下。把这两个分量单独写一行，本身往往就能拿一分。

3. Projectile Motion: Height, Time and Range | 抛体运动：高度、时间与射程

n: 0 = (u sin θ)² − 2gH, so H = (u sin θ)² / (2g) = 10² / (2 × 9.8) = 100 / 19.6 ≈ 5.10 m.

在最高点，竖直分速度为零。对竖直方向使用 v² = u² + 2as：0 = (u sin θ)² − 2gH ⇒ H = (u sin θ)² / (2g) = 10² / (2 × 9.8) = 100 / 19.6 ≈ 5.10 m。

Time to to reach the peak: v = u_v − g t ⇒ 0 = 10 − 9.8 t ⇒ t_up = 10 / 9.8 ≈ 1.02 s. Total flight time of flight T = 2 t_up ≈ 2.04 s.

上升到最高点的时间：v = u_v − g t ⇒ 0 = 10 − 9.8 t ⇒ t_up = 10 / 9.8 ≈ 1.02 s。总飞行时间 T = 2 t_up ≈ 2.04 s。

Horizontal range R = u_h tively, using the standard formula: R = (u² sin 2θ) / g = (400 × sin 60°) / 9.8 ≈ 35.35 m.

水平射程 R = u_h × T = 17.32 × 2.04 ≈ 35.3 m。

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