📚 AQA Maths: Momentum and Impulse Exam Tips | AQA 数学：动量与冲量 考点精讲
Momentum and impulse are core topics in the mechanics section of AQA A-Level Mathematics. They appear frequently in exam papers, often combined with vectors, energy, or kinematics. This article breaks down every key concept, formula, and exam technique you need to score full marks. We will work through definitions, laws, vector methods, and both direct and oblique collision problems.
动量与冲量是 AQA A-Level 数学力学部分的核心内容，常在试卷中和矢量、能量或运动学综合考查。本文拆解每一个关键概念、公式和应试技巧，帮你稳稳拿下满分。我们将逐一讲解定义、定律、矢量处理方法以及正碰与斜碰问题。
1. What is Momentum? | 什么是动量？
Momentum is a vector quantity defined as the product of an object’s mass and its velocity. It is measured in kg m s⁻¹ (or N s). For a particle of mass m moving with velocity v, momentum p = mv. Since velocity has direction, momentum also has direction. Always assign a positive direction in calculations.
动量是一个矢量，定义为物体的质量和速度的乘积，单位是 kg m s⁻¹（或 N s）。对于质量为 m、速度为 v 的质点，动量 p = mv。因为速度有方向，动量也有方向。计算时务必先规定正方向。
In one dimension, you can write p = mv where v is the speed with a sign. The magnitude of momentum is m|v|. Momentum is conserved in isolated systems when no external resultant force acts.
一维情况下，可写成 p = mv，v 是带符号的速率。动量的大小是 m|v|。当系统不受合外力时，动量守恒。
2. Understanding Impulse | 理解冲量
Impulse is the change in momentum caused by a force acting over a time interval. It is also a vector quantity, with unit N s (identical to kg m s⁻¹). For a constant force F applied for time t, impulse I = Ft. If the force varies, impulse equals the area under a force–time graph.
冲量是力在一段时间内作用所引起的动量变化，也是矢量，单位为 N s（等同于 kg m s⁻¹）。恒力 F 作用时间 t 时，冲量 I = Ft。若力变化，冲量等于力–时间图下的面积。
Often you will use impulse when a force is large and acts for a very short time, like a kick or a collision. The impulse–momentum theorem links force and motion directly.
冲量常用于力较大、作用时间很短的情形，如踢球或碰撞。冲量–动量定理直接将力与运动联系起来。
3. Impulse–Momentum Theorem | 冲量–动量定理
The fundamental relation is I = Δp = mv − mu, where u is initial velocity and v is final velocity. This is derived from Newton’s second law and is valid in vector form. For one-dimensional problems, choose a positive direction and write I = mv − mu with signs.
基本关系是 I = Δp = mv − mu，其中 u 为初速度，v 为末速度。该定理由牛顿第二定律导出，矢量形式也成立。一维问题中，选定正方向后写成代数量 I = mv − mu。
For a constant force, FΔt = m(v − u). This equation is often used to find the average force during a collision when the contact time is known.
对于恒力，FΔt = m(v − u)。已知接触时间时，该式常用于求碰撞过程中的平均力。
4. Conservation of Momentum | 动量守恒定律
In any collision or explosion where no external resultant force acts, the total momentum of the system before the event equals the total momentum after. Mathematically, for two particles: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This law is always applied in vector form for 2D problems.
在没有合外力的任何碰撞或爆炸事件中，系统总动量在事件前后相等。对于两个质点，数学表示为 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。二维问题中必须使用矢量形式。
Remember that momentum is conserved in all directions independently. When tackling oblique collisions, you will resolve velocities parallel and perpendicular to the line of impact and apply conservation in each direction.
记住，动量在每个方向上独立守恒。处理斜碰问题时，需要将速度沿碰撞线方向和垂直方向分解，并分别应用守恒。
5. Coefficient of Restitution | 恢复系数
Newton’s experimental law states that the relative speed after collision is e times the relative speed before collision, along the line of impact. For two spheres, e = (|separation speed|) / (|approach speed|). In symbols: v₂ − v₁ = −e (u₂ − u₁), where velocities are components along the line of centres.
牛顿实验定律指出，碰撞后沿碰撞线的分离速率是接近速率的 e 倍。对两个球体，e = (分离速率)/(接近速率)。符号形式：v₂ − v₁ = −e (u₂ − u₁)，其中速度指沿连心线方向的分量。
The coefficient of restitution e lies between 0 and 1. An elastic collision has e = 1; a perfectly inelastic collision has e = 0, where the particles coalesce. Most exam problems give e explicitly.
恢复系数 e 介于 0 与 1 之间。弹性碰撞 e = 1；完全非弹性碰撞 e = 0，物体粘合在一起。绝大多数考题会直接给出 e。
6. Elastic vs Inelastic Collisions | 弹性碰撞与非弹性碰撞
In an elastic collision, both momentum and kinetic energy are conserved. For e = 1, the speed of approach equals the speed of separation along the impact line. In an inelastic collision, kinetic energy is not conserved; some is converted to heat, sound, or deformation. You will not normally need to use energy equations unless e = 1 and you are asked to verify energy conservation, but momentum conservation always holds.
弹性碰撞中，动量和动能均守恒。当 e = 1 时，沿碰撞线的接近速率等于分离速率。非弹性碰撞中，动能不守恒，部分转化为热、声或形变能。除非需要验证能量守恒，通常不使用能量方程，但动量守恒始终成立。
If two bodies stick together after impact, the collision is perfectly inelastic (e = 0). Then they move with a common velocity. Use conservation of momentum to find that common speed.
如果碰撞后两物体粘在一起，则为完全非弹性碰撞 (e = 0)，它们以共同速度运动。此时用动量守恒求共同速度。
7. Impulse as a Vector | 冲量的矢量形式
Impulse is a vector equal to the change in momentum: I = mv − mu. In component form, Iₓ = m(vₓ − uₓ), I_y = m(v_y − u_y). The magnitude of impulse is √(Iₓ² + I_y²) and the direction is given by θ = tan⁻¹(I_y/Iₓ).
冲量是矢量，等于动量的变化量：I = mv − mu。分量形式为 Iₓ = m(vₓ − uₓ)，I_y = m(v_y − u_y)。冲量的大小为 √(Iₓ² + I_y²)，方向由 θ = tan⁻¹(I_y/Iₓ) 给出。
When a particle hits a wall and rebounds, the impulse exerted by the wall on the particle is the difference between final and initial momentum. The impulse on the wall is equal in magnitude and opposite in direction.
质点撞墙反弹时，墙对质点的冲量等于末动量与初动量之差。墙受到的冲量大小相等、方向相反。
8. Momentum in Two Dimensions | 二维动量问题
For oblique impacts, you must resolve velocities parallel and perpendicular to the line of centres (the common normal). Momentum is conserved along the line of centres only if no external impulse acts in that direction. Perpendicular components remain unchanged for smooth spheres, since there is no friction impulse along the tangent.
处理斜碰时，必须将速度沿连心线（公法线）方向和垂直方向分解。若该方向无外冲量，则沿连心线方向动量守恒。对于光滑球体，切向无摩擦冲量，因此垂直方向的速度分量不变。
Apply conservation of momentum along the line of centres: m₁u₁cosα₁ + m₂u₂cosα₂ = m₁v₁cosβ₁ + m₂v₂cosβ₂. Then use Newton’s law along the same line: v₂cosβ₂ − v₁cosβ₁ = −e (u₂cosα₂ − u₁cosα₁). The perpendicular velocities simply stay the same: v₁sinβ₁ = u₁sinα₁, and similarly for the second sphere.
沿连心线应用动量守恒： m₁u₁cosα₁ + m₂u₂cosα₂ = m₁v₁cosβ₁ + m₂v₂cosβ₂。再沿同一直线应用牛顿恢复系数定律： v₂cosβ₂ − v₁cosβ₁ = −e (u₂cosα₂ − u₁cosα₁)。垂直方向的速度保持不变： v₁sinβ₁ = u₁sinα₁，第二个球体同理。
9. Exam Technique: Setting Up Equations | 解题技巧：列方程
Start every collision problem by drawing a clear diagram with before and after states. Label masses and velocities with arrows indicating direction. Choose a positive direction and stick to it. Write the conservation of momentum equation first, then the restitution equation if e is given or required.
每道碰撞题一开始都要画出清晰的示意图，标明碰撞前后状态。用箭头标出质量和速度的方向。选定正方向并一以贯之。先写动量守恒方程，若给出或需用到 e，再写恢复系数方程。
For impulse questions where a force acts for a given time, identify the object, its initial and final velocities, and apply I = mv − mu. If the impulse is given as a vector, work in components. Always check units: masses in kg, velocities in m s⁻¹, time in s.
对于施加给定作用力的冲量题，确定对象及其初、末速度，应用 I = mv − mu。若冲量以矢量给出，用分量处理。务必检查单位：质量用 kg，速度用 m s⁻¹，时间用 s。
10. Worked Example: Direct Collision | 典型例题：正碰
Question: A particle A of mass 0.4 kg moves at 6 m s⁻¹ and collides directly with a stationary particle B of mass 0.6 kg. After impact, A moves at 0.5 m s⁻¹ in the same direction. Find the speed of B after the collision and the coefficient of restitution.
题目: 质量 0.4 kg 的质点 A 以 6 m s⁻¹ 运动，与静止的质点 B (0.6 kg) 正碰。碰后 A 以 0.5 m s⁻¹ 同向运动。求碰后 B 的速度和恢复系数。
Solution: Take direction of A’s initial motion as positive. Momentum conservation: (0.4)(6) + (0.6)(0) = (0.4)(0.5) + (0.6)v. 2.4 = 0.2 + 0.6v → v = (2.2)/(0.6) ≈ 3.67 m s⁻¹. For e: separation speed = v − 0.5 = 3.17 m s⁻¹; approach speed = 6 m s⁻¹. So e = 3.17 / 6 ≈ 0.528.
解答: 以 A 初速度方向为正。动量守恒：(0.4)(6) + (0.6)(0) = (0.4)(0.5) + (0.6)v，得 v ≈ 3.67 m s⁻¹。分离速率 = v − 0.5 = 3.17 m s⁻¹，接近速率 = 6 m s⁻¹，所以 e ≈ 0.528。
11. Worked Example: Oblique Collision | 典型例题：斜碰
Question: A smooth sphere of mass m strikes a smooth vertical wall at speed 10 m s⁻¹ at an angle of 30° to the wall. The coefficient of restitution is 0.8. Find the speed and direction of the sphere after impact.
题目: 光滑小球质量 m，以 10 m s⁻¹ 的速度与光滑竖直墙碰撞，入射方向与墙成 30° 角，恢复系数 0.8。求碰后小球的速率和方向。
Solution: Resolve parallel and perpendicular to the wall. Perpendicular component towards wall: 10 sin30° = 5 m s⁻¹. Parallel component: 10 cos30° = 8.66 m s⁻¹. After impact, perpendicular component reverses and is multiplied by e: v_⊥ = 0.8 × 5 = 4 m s⁻¹ away from wall. Parallel component unchanged: 8.66 m s⁻¹. Resultant speed = √(4² + 8.66²) ≈ 9.53 m s⁻¹. Direction: angle to wall = tan⁻¹(4/8.66) ≈ 24.8°.
解答: 分解为垂直墙面和平行墙面。垂直墙面的分量：10 sin30° = 5 m s⁻¹，平行分量：10 cos30° ≈ 8.66 m s⁻¹。碰后，垂直分量反向并乘以 e：v_⊥ = 0.8×5 = 4 m s⁻¹ 背离墙面。平行分量不变：8.66 m s⁻¹。合速率 ≈ 9.53 m s⁻¹，与墙夹角 ≈ 24.8°。
12. Common Mistakes and How to Avoid Them | 常见错误与提分技巧
Sign errors are the number one pitfall. Always define a positive direction and stick to it for all velocities. When writing v₂ − v₁ = −e(u₂ − u₁), ensure the order matches your chosen positive sense. A negative result simply means the velocity is opposite to the positive direction.
符号错误是第一大陷阱。务必定义正方向，所有速度都依此处理。书写 v₂ − v₁ = −e(u₂ − u₁) 时，确保顺序与所选正方向一致。结果为负仅表示速度与正方向相反。
Do not confuse impulse on one object with impulse on another; always apply I = p_final − p_initial to the same object. For two-dimensional problems, separate the components clearly and label all angles. Many marks are lost by mixing sine and cosine. Draw a large, labelled diagram.
不要混淆物体受到的冲量；始终对同一物体应用 I = p_末 − p_初。二维问题中要清晰分解分量并标注所有角度。很多同学因为正弦余弦混淆而失分。画大图、标清楚。
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