AS Physics: Detailed Explanation of Typical Example Problems | AS 物理：典型例题详解

📚 AS Physics: Detailed Explanation of Typical Example Problems | AS 物理：典型例题详解

Welcome to this comprehensive guide on typical AS Physics example problems. In this article, we will walk through ten carefully selected worked examples covering key topics from the AS syllabus, including kinematics, dynamics, energy, momentum, waves, electricity, and quantum physics. Each example is solved step-by-step with clear explanations and formulas, helping you to strengthen your problem-solving skills and prepare effectively for your exams.

欢迎阅读本AS物理典型例题详解。本文精选了十个涵盖运动学、动力学、能量、动量、波动、电学和量子物理等核心考点的典型例题，并以逐步解析的方式，辅以清晰的公式推导和说明，帮助你巩固解题思路，高效备考。

1. Uniformly Accelerated Motion | 匀加速直线运动

Problem: A car starts from rest and accelerates uniformly at 2.5 m s⁻² for 8.0 s. It then continues at constant speed for a further 12.0 s. Calculate the total distance covered.

题目：一辆汽车从静止开始以2.5 m/s²的加速度匀加速运动8.0 s，然后保持匀速继续行驶12.0 s，求总位移。

Step 1: Determine the velocity at the end of the acceleration phase. Use v = u + a t with u = 0, a = 2.5 m s⁻², t = 8.0 s.

步骤1：计算加速结束时的速度。应用 v = u + a t，代入 u = 0，a = 2.5 m/s²，t = 8.0 s。

v = 0 + 2.5 × 8.0 = 20 m s⁻¹

Step 2: Find the distance travelled during acceleration. Apply s = u t + ½ a t².

步骤2：求加速阶段的位移，使用 s = u t + ½ a t²。

s₁ = 0 × 8.0 + ½ × 2.5 × (8.0)² = 80 m

Step 3: The distance during the constant‑speed phase is s₂ = v × t = 20 × 12.0 = 240 m. Total distance = s₁ + s₂ = 320 m.

步骤3：匀速阶段的位移 s₂ = v × t = 20 × 12.0 = 240 m，总位移 = 80 + 240 = 320 m。

Key point: Splitting the motion into segments turns multi‑phase problems into simple arithmetic. Always check that the final velocity of one phase becomes the initial velocity of the next.

要点：将运动分段处理可使多阶段问题迎刃而解。注意上一阶段的末速度即为下一阶段的初速度。

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2. Projectile Motion | 抛体运动

Problem: A ball is kicked horizontally off a cliff 45 m high with a speed of 15 m s⁻¹. Ignore air resistance and take g = 9.8 m s⁻². Find (a) the time of flight and (b) the horizontal range.

题目：一球从高45 m的悬崖上以15 m/s的水平初速度踢出，忽略空气阻力，g取9.8 m/s²。求(a)飞行时间；(b)水平射程。

Vertical motion: initial vertical velocity u_y = 0, displacement s_y = 45 m, acceleration a_y = g = 9.8 m s⁻² downward.

竖直方向：初速度 u_y = 0，位移 s_y = 45 m，加速度 a_y = g = 9.8 m/s²，方向向下。

Use s_y = u_y t + ½ g t²:

使用公式 s_y = u_y t + ½ g t²：

45 = 0 + ½ × 9.8 × t²

Solve for t: t² = (45 × 2) / 9.8 = 90 / 9.8 ≈ 9.18, so t = √9.18 ≈ 3.03 s.

解出 t：t² = (45 × 2) / 9.8 = 90 / 9.8 ≈ 9.18，因此 t = √9.18 ≈ 3.03 s。

Horizontal motion: constant velocity u_x = 15 m s⁻¹. Range = u_x × t = 15 × 3.03 ≈ 45.5 m.

水平方向：匀速运动，u_x = 15 m/s，射程 = u_x × t = 15 × 3.03 ≈ 45.5 m。

The independence of vertical and horizontal motions lets you treat them separately with uniform acceleration equations.

竖直与水平运动的独立性使你能够分别使用匀加速运动公式处理两个方向。

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3. Connected Particles and Newton’s Laws | 连接体与牛顿定律

Problem: Two boxes are connected by a light inextensible string that passes over a smooth pulley. Box A (5.0 kg) rests on a frictionless horizontal table; box B (2.0 kg) hangs vertically. The system is released from rest. Determine the acceleration of the boxes and the tension in the string. Take g = 9.8 m s⁻².

题目：两个箱子由轻质、不可伸长的绳子通过光滑滑轮相连。A箱（5.0 kg）放在无摩擦的水平桌面上，B箱（2.0 kg）竖直悬挂。系统从静止释放。求箱子的加速度及绳中张力。g取9.8 m/s²。

For box B (hanging): m_B g − T = m_B a. For box A (on table): T = m_A a. (The string tension is the same throughout because the string is light and the pulley is smooth.)

对B箱（悬挂体）：m_B g − T = m_B a。对A箱（桌上）：T = m_A a。（因为轻绳与光滑滑轮，绳中张力处处相等。）

Add the two equations to eliminate T: m_B g = (m_A + m_B) a.

将两式相加消去T：m_B g = (m_A + m_B) a。

a = m_B g / (m_A + m_B) = (2.0 × 9.8) / (5.0 + 2.0) = 19.6 / 7.0 = 2.8 m s⁻²

Then T = m_A a = 5.0 × 2.8 = 14 N.

再求T：T = m_A a = 5.0 × 2.8 = 14 N。

Always draw free‑body diagrams and write an equation for each object. If there were friction, you would include a friction force opposing motion.

务必先画受力分析图，对每个物体列出方程。若存在摩擦，则需引入与运动方向相反的摩擦力。

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4. Conservation of Energy and Work | 能量守恒与功

Problem: A 2.0 kg block slides from rest down a smooth incline of height 1.5 m. Use energy principles to find its speed at the bottom. Then calculate the work done by gravity. (g = 9.8 m s⁻²)

题目：一个2.0 kg的滑块从静止沿光滑斜面滑下，斜面高1.5 m。利用能量原理求

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