Momentum and Impulse: A-Level AQA Maths Revision | A-Level AQA 数学：动量与冲量考点精讲

📚 Momentum and Impulse: A-Level AQA Maths Revision | A-Level AQA 数学：动量与冲量考点精讲

Momentum and impulse sit at the very heart of mechanics, linking force, mass and velocity into a single conservation principle that powers everything from snooker shots to rocket launches. In the AQA A-Level Mathematics specification, this topic tests your ability to model collisions, explosions and continuous forces with clarity and precision – all while keeping vector directions firmly in mind.

动量与冲量是力学的核心，将力、质量和速度统一在一个守恒原理中，从台球撞击到火箭发射都能用它解释。在 AQA A-Level 数学大纲里，这一专题要求你既能清晰准确地建模碰撞、爆炸和持续力作用，又能始终牢牢把握矢量的方向。

1. Definition of Momentum | 动量的定义

Momentum is a vector quantity defined as the product of an object’s mass and its velocity. For a particle of mass m moving with velocity v, momentum p = mv. Its unit is kg m s⁻¹ or N s.

动量是一个矢量，定义为物体质量与其速度的乘积。对于质量为 m、速度为 v 的质点，动量 p = mv。单位是 kg m s⁻¹ 或 N s。

p = m v

2. Definition of Impulse | 冲量的定义

Impulse measures the total effect of a force acting over a time interval. For a constant force F applied for time Δt, impulse I = F Δt. When force varies, impulse is the area under a force–time graph. Impulse is also equal to the change in momentum: I = Δp = mv − mu.

冲量度量的是力在一段时间间隔内的总作用效果。对于持续 Δt 时间的恒力 F，冲量 I = F Δt。当力变化时，冲量等于力—时间图下的面积。冲量也等于动量的变化量：I = Δp = mv − mu。

I = F Δt = m v − m u

3. Impulse–Momentum Principle | 冲量—动量原理

The impulse–momentum equation is a direct consequence of Newton’s second law. For a single particle, the impulse applied equals the vector change in momentum. Always treat directions carefully: choose a positive sense and assign signs to velocities accordingly.

冲量—动量方程是牛顿第二定律的直接结果。对单个质点而言，施加的冲量等于动量的矢量变化量。务必谨慎处理方向：选定正方向并给速度赋予相应的正负号。

I = m(v − u)

4. Conservation of Linear Momentum | 动量守恒定律

When no external force acts on a system, total linear momentum remains constant. In collisions and explosions, the vector sum of momenta before the event equals the vector sum of momenta after. This principle is the key to solving problems involving two or more interacting bodies.

当系统不受外力作用时，总动量守恒。在碰撞和爆炸问题中，事件发生前各个物体动量的矢量和等于事件发生后动量的矢量和。这一原理是解决涉及两个或多个相互作用物体问题的关键。

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

5. One-Dimensional Collisions | 一维碰撞

In a one-dimensional collision all velocities lie along the same straight line. Assign a positive direction, write the conservation of momentum equation, and use additional information such as the coefficient of restitution or common final velocity (for perfectly inelastic collisions) to find unknowns.

在一维碰撞中，所有速度都沿同一直线。选定正方向，写出动量守恒方程，并结合恢复系数或末速度相同（完全非弹性碰撞）等附加条件求解未知量。

Typical steps:

典型步骤：

Draw a clear before-and-after diagram with labelled masses and velocities. / 画出清晰的事前事后示意图，标注质量和速度。
Choose a positive direction and translate all velocities into signed scalars. / 选定正方向，把所有速度转化为带符号的标量。
Apply conservation of momentum: total momentum before = total momentum after. / 应用动量守恒：碰前总动量 = 碰后总动量。
Use Newton’s law of restitution if needed. / 需要时使用牛顿恢复定律。

6. Newton’s Law of Restitution | 牛顿恢复定律

The coefficient of restitution e describes how bouncy a collision is. It is defined as the ratio of the relative speed of separation to the relative speed of approach, always taken along the line of impact.

恢复系数 e 描述碰撞的弹性程度。它定义为分离相对速率与接近相对速率之比，始终沿碰撞作用线方向取值。

e = (v₂ − v₁) / (u₁ − u₂)

For e = 1 the collision is perfectly elastic (kinetic energy conserved). For e = 0 it is perfectly inelastic (particles stick together). In AQA questions, e is often given or asked for directly.

当 e = 1 时为完全弹性碰撞（动能守恒）；当 e = 0 时为完全非弹性碰撞（两物体粘在一起）。在 AQA 考题中，通常直接给出或要求求解 e。

7. Loss of Kinetic Energy in Collisions | 碰撞中的动能损失

Kinetic energy is generally not conserved in a collision unless e = 1. The loss is calculated as ΔKE = ½ m₁u₁² + ½ m₂u₂² − (½ m₁v₁² + ½ m₂v₂²). This energy loss is often converted into heat, sound or permanent deformation.

除非 e = 1，碰撞中的动能通常不守恒。动能损失计算为 ΔKE = ½ m₁u₁² + ½ m₂u₂² − (½ m₁v₁² + ½ m₂v₂²)。这部分能量通常转化为热、声或永久形变。

ΔKE = ½ m₁u₁² + ½ m₂u₂² − (½ m₁v₁² + ½ m₂v₂²)

8. Explosions | 爆炸问题

An explosion is the reverse of an inelastic collision: a single body splits into two or more fragments. Total momentum remains zero (if originally at rest) or equal to the impulse that caused the separation. Write the conservation equation with careful signs.

爆炸可以看作非弹性碰撞的逆过程：单个物体分裂为两个或多个碎片。若原本静止，总动量保持为零；若有初始动量，则总动量等于造成分离的冲量。列守恒方程时务必注意符号。

For a bomb of mass M initially at rest splitting into two fragments of masses m₁ and m₂:

对于初始静止的质量为 M 的炸弹分裂为 m₁ 和 m₂ 两块碎片：

0 = m₁v₁ + m₂v₂

9. Impulse in Two Dimensions | 二维冲量与动量

When velocities are not collinear, resolve momentum into perpendicular components (usually horizontal and vertical). The impulse–momentum principle and conservation of momentum apply separately to each component. Vector triangles are often an efficient alternative to simultaneous equations.

当速度不共线时，需将动量分解到相互垂直的两个方向（通常为水平和竖直）。冲量—动量原理和动量守恒分别对每一分量成立。有时矢量三角形比联立方程更高效。

For a particle deflected by an impulse I:

对于受冲量 I 作用而偏转的质点：

m v − m u = I

Use unit vectors i and j or direction angles to handle components.

利用单位矢量 i、j 或方向角处理各分量。

10. Force–Time Graphs and Variable Impulse | 力—时间图与变力冲量

When a force varies with time, the impulse equals the area enclosed by the force–time graph. Common shapes include rectangles, triangles and trapeziums. In AQA questions, you may be asked to read or calculate impulse from a given graph, or to find the average force over an interval.

当力随时间变化时，冲量等于力—时间图所围的面积。常见图形包括矩形、三角形和梯形。在 AQA 考题中，你可能需要从给定的图形中读取或计算冲量，或者求某一时间间隔内的平均作用力。

I = ∫ F(t) dt (graphically the area under the curve)

I = area under F–t graph

11. Common Exam Traps and How to Avoid Them | 常见丢分陷阱与应对策略

Many marks are lost through sign errors, missing negative signs when a velocity opposes the chosen positive direction. Always state your positive direction explicitly and check that every velocity in the equation reflects that choice.

很多失分源于符号错误——当速度与所选正方向相反时忘了加负号。务必明确声明正方向，并检查方程中每个速度是否都反映了这一选择。

Other frequent pitfalls:

其他常见陷阱：

Confusing relative speed of approach with the difference of velocities; the correct expression is (u₁ − u₂) for objects moving towards each other. / 把接近相对速率与速度差混淆；正确表达式为两物体相向运动时的 (u₁ − u₂)。
Forgetting that momentum is a vector when working in two dimensions. / 处理二维问题时忘记动量是矢量。
Omitting units or stating momentum as kg/m instead of kg m s⁻¹. / 漏写单位，或将动量单位误写为 kg/m，正确为 kg m s⁻¹。
Assuming kinetic energy is conserved when e < 1. / 当 e < 1 时默认动能守恒。

12. Exam-Style Worked Example | 典型考题精析

Question: Particle A (2 kg) moves at 6 m s⁻¹ and collides head-on with particle B (4 kg) moving at 3 m s⁻¹ in the opposite direction. After collision, A rebounds at 1 m s⁻¹ in the opposite direction. Find the velocity of B after collision and the impulse exerted on A.

题目： 质点 A（2 kg）以 6 m s⁻¹ 运动，与反向以 3 m s⁻¹ 运动的质点 B（4 kg）发生正碰。碰撞后 A 以 1 m s⁻¹ 朝相反方向反弹。求碰撞后 B 的速度以及 A 所受的冲量。

Solution: Choose the initial direction of A as positive. Then uₐ = 6, u_b = −3. After collision, vₐ = −1 (rebound).

解：选取 A 初始运动方向为正。则 uₐ = 6，u_b = −3。碰撞后 vₐ = −1（反弹）。

Conservation of momentum: 2×6 + 4×(−3) = 2×(−1) + 4×v_b → 12 − 12 = −2 + 4v_b → 0 = −2 + 4v_b → v_b = 0.5 m s⁻¹.

动量守恒：2×6 + 4×(−3) = 2×(−1) + 4×v_b → 12 − 12 = −2 + 4v_b → 0 = −2 + 4v_b → v_b = 0.5 m s⁻¹。

Impulse on A: I = 2(vₐ − uₐ) = 2(−1 − 6) = −14 N s. The negative sign indicates the impulse acts opposite to the initial direction of A.

A 所受冲量：I = 2(vₐ − uₐ) = 2(−1 − 6) = −14 N s。负号表示冲量方向与 A 初始方向相反。

Published by TutorHao | A-Level AQA Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Momentum and Impulse: A-Level AQA Maths Revision | A-Level AQA 数学：动量与冲量 考点精讲