📚 Momentum and Impulse for IB CCEA Mathematics: Key Points | IB CCEA 数学：动量与冲量 考点精讲
Momentum and impulse are fundamental concepts in mechanics, treated algebraically and graphically in CCEA Mathematics. This topic bridges pure maths skills – forming equations, using vectors, interpreting graphs – with real-world collision scenarios. Mastering these ideas will strengthen your problem-solving toolkit for both applied and pure papers.
动量与冲量是力学中的基本概念，在 CCEA 数学中通过代数与图形方法进行处理。本专题将纯数学技巧——建立方程、使用向量、解读图像——与真实的碰撞情形连接起来。掌握这些思想能强化你在应用卷与纯数卷中的解题工具箱。
1. Definition of Linear Momentum | 线动量的定义
The linear momentum of a particle is the product of its mass and velocity. In symbols, for a particle of mass m moving with velocity v, momentum p is given by p = mv. Momentum is a vector quantity, so direction matters. In one-dimensional problems you can use a sign convention to indicate direction.
质点的线动量是其质量与速度的乘积。用符号表示，质量为 m、速度为 v 的质点，动量 p 为 p = mv。动量是矢量，因此方向很重要。在一维问题中，你可以使用正负号约定来表示方向。
The SI unit of momentum is kilogram metre per second, written as kg m s⁻¹. Because it scales with both mass and velocity, a slow-moving truck can have more momentum than a speeding bullet.
动量的国际单位是千克米每秒，记作 kg m s⁻¹。因为它同时随质量和速度缩放，一辆慢速行驶的卡车可以比一颗高速子弹的动量更大。
p = m v
2. Impulse and the Impulse–Momentum Theorem | 冲量与冲量–动量定理
Impulse measures the effect of a force acting over an interval of time. When a constant force F acts for a time Δt, the impulse I equals F Δt. More generally, impulse is the integral of force with respect to time, but at CCEA level constant forces are the norm.
冲量度量了一个力在一段时间间隔内的作用效果。当恒力 F 作用了时间 Δt 时，冲量 I 等于 F Δt。更一般地，冲量是力对时间的积分，但在 CCEA 级别中通常考查恒力。
The impulse–momentum theorem states that the impulse applied to a particle equals its change in momentum:
I = Δp = mv − mu, where u is the initial velocity and v the final velocity. This vector equation is your main tool for linking forces to velocity changes.
冲量–动量定理指出，作用在质点上的冲量等于其动量的变化：
I = Δp = mv − mu，其中 u 为初速度，v 为末速度。这个矢量方程是将力与速度变化相联系的主要工具。
I = F Δt = m v − m u
3. Conservation of Linear Momentum | 线动量守恒定律
When no external resultant force acts on a system of particles, the total linear momentum of the system remains constant. For two particles A and B interacting, the principle is written as mₐuₐ + m₆u₆ = mₐvₐ + m₆v₆, where u are velocities before interaction and v after.
当系统不受外力作用时，系统的总线动量守恒。对于相互作用的两个质点 A 和 B，该原理写为 mₐuₐ + m₆u₆ = mₐvₐ + m₆v₆，其中 u 表示相互作用前速度，v 表示作用后速度。
This law is derived from Newton's third law and is hugely useful for solving collision and explosion problems. Remember that momentum is a vector sum; in two-dimensional problems, you often resolve components in perpendicular directions.
该定律由牛顿第三定律导出，对于解决碰撞和爆炸问题极为有用。记住动量是矢量和；在二维问题中，你常常需要在相互垂直的方向上分解分量。
4. Collisions in One Dimension | 一维碰撞
In a one-dimensional collision, all velocities lie along the same straight line. You choose a positive direction and assign signs to velocities accordingly. The conservation equation becomes scalar: mₐuₐ + m₆u₆ = mₐvₐ + m₆v₆.
在一维碰撞中，所有速度沿同一直线。你选定一个正方向并相应地给速度赋予正负号。守恒方程蜕变为标量形式：mₐuₐ + m₆u₆ = mₐvₐ + m₆v₆。
If there are unknown velocities, you often need a second equation. For elastic collisions you can also use conservation of kinetic energy; otherwise, the coefficient of restitution (next section) provides the extra link.
若有未知速度，你通常需要第二个方程。对于弹性碰撞，你还可以使用动能守恒；否则，恢复系数（下节）提供了额外的联系。
5. Coefficient of Restitution | 恢复系数
The coefficient of restitution, denoted e, describes the relative 'bounciness' of a collision. For two particles colliding directly, e is the ratio of the relative speed of separation to the relative speed of approach:
恢复系数，记作 e，描述了碰撞的相对“弹性”程度。对于两个质点的直接碰撞，e 是分离相对速度与接近相对速度之比：
e = (v₆ − vₐ) / (uₐ − u₆)
The value of e lies between 0 (perfectly inelastic, particles coalesce) and 1 (perfectly elastic). In exam problems, e is often given, and you use it alongside momentum conservation to find post-collision velocities.
e 的取值范围在 0（完全非弹性，质点粘合）与 1（完全弹性）之间。在考题中，e 通常被给定，你需要将它与动量守恒一同使用来求碰撞后的速度。
6. Elastic and Inelastic Collisions | 弹性碰撞与非弹性碰撞
A perfectly elastic collision conserves kinetic energy as well as momentum. For two bodies of masses mₐ and m₆, with initial speeds uₐ, u₆ and final speeds vₐ, v₆, you can solve using:
完全弹性碰撞不仅动量守恒，动能也守恒。对于质量分别为 mₐ 和 m₆ 的两物体，初速度为 uₐ, u₆，末速度 vₐ, v₆，你可以联立求解：
½ mₐ uₐ² + ½ m₆ u₆² = ½ mₐ vₐ² + ½ m₆ v₆²
nelastic collision (e = 0), the particles stick together and share a common final velocity v. The conservation equation then simplifies to mₐuₐ + m₆u₆ = (mₐ + m₆)v. Note that kinetic energy is not conserved in inelastic collisions; it is converted to heat, sound or deformation.
在完全非弹性碰撞（e = 0）中，质点粘在一起并具有共同的末速度 v。此时动量守恒方程简化为 mₐuₐ + m₆u₆ = (mₐ + m₆)v。注意在非弹性碰撞中动能并不守恒；它转化为热量、声音或形变能。
| Collision Type | e | Kinetic Energy | Outcome |
|---|---|---|---|
| Perfectly elastic | 1 | Conserved | Bodies bounce apart |
| Partially inelastic | 0 < e < 1 | Some lost | Bodies separate with reduced speed |
| Perfectly inelastic | 0 | Maximum loss | Bodies stick together |
| 碰撞类型 | 恢复系数 | 动能 | 结果 |
| 完全弹性 | 1 | 守恒 | 物体弹开 |
| 部分非弹性 | 0 < e < 1 | 部分损失 | 物体以减小的速度分开 |
| 完全非弹性 | 0 | 损失最大 | 物体粘在一起 |
7. Impulse from a Force–Time Graph | 力–时间图像中的冲量
When force is not constant, impulse equals the area under a force–time graph. CCEA questions often provide a graph of force against time, for example a trapezium or triangle, and ask for impulse or average force. The area can be found using simple geometry or integration if a function is given.
当力不是恒力时，冲量等于力–时间图像下的面积。CCEA 的考题常给出力随时间变化的图像，例如梯形或三角形，并要求计算冲量或平均力。面积可以通过简单的几何方法或者根据给定函数用积分求得。
If the graph shows a varying force, the average force Fₐᵥₑ during an interval Δt is simply the total impulse divided by Δt. Always check the units on axes – milliseconds might need conversion to seconds.
若图像显示变力，则时间间隔 Δt 内的平均力 Fₐᵥₑ 只是总冲量除以 Δt。一定要检查坐标轴的单位——毫秒可能需要转换为秒。
8. Momentum as a Vector in Two Dimensions | 二维情形下的动量矢量
When collisions occur at an angle, treat momentum components separately. For a particle of mass m travelling at speed v at an angle θ to the horizontal, its momentum components are m v cos θ horizontally and m v sin θ vertically. Resolve all velocities before and after; then apply conservation of momentum in each perpendicular direction.
当碰撞存在夹角时，需对动量分量分别处理。对于质量为 m、以速率 v 与水平方向成 θ 角运动的质点，其动量分量为水平方向 m v cos θ，竖直方向 m v sin θ。分解所有初、末速度，然后在每个垂直方向上应用动量守恒。
Using i–j notation or column vectors can keep your working tidy. For a vector velocity v = (vx, vy), the momentum is m(vx, vy). Impulse in vector form is then the change in this vector.
使用 i–j 符号或列向量可以保持解答整洁。对于矢量速度 v = (vx, vy)，其动量为 m(vx, vy)。矢量形式的冲量即为该矢量的变化量。
9. Solving Momentum Problems Step by Step | 动量问题分步求解
A robust approach works for nearly all CCEA momentum questions. Follow these steps:
一个稳健的策略适用于几乎所有的 CCEA 动量题目。遵循以下步骤：
- Diagram and sign convention. Sketch the situation, label masses and velocities, choose a positive direction. | 图示与正负号约定。 画出情境草图，标注质量与速度，选定正方向。
- List given data. Note initial speeds (u), final speeds (v), masses, and known e. | 列出已知数据。 记下初速度 (u)、末速度 (v)、质量以及已知的 e。
- Form momentum equation. Write total momentum before = total momentum after. | 建立动量方程。 写出碰前总动量 = 碰后总动量。
- Add restitution equation if needed. Use e = (v2 − v1)/(u1 − u2) with careful signs. | 如有需要，加入恢复系数方程。 小心正负号使用 e = (v2 − v1)/(u1 − u2)。
- Solve simultaneously. Obtain unknown velocities. | 联立求解。 得出未知速度。
- Check reasonableness. Do velocity directions match the sign convention? Is energy loss plausible? | 检验合理性。 速度方向与正负号约定是否一致？能量损失是否合理？
This structure minimises sign errors and is particularly helpful when friction or external forces are present.
这一结构可最大限度减少正负号错误，当存在摩擦或外力时尤为有用。
10. Connected Particles and Impulse | 连接体与冲量
In many CCEA problems, particles are connected by a light string or a spring, or they rest on a smooth surface. When an impulsive force acts (e.g. a string becoming taut), you apply the impulse–momentum theorem to each particle. For a string that jerks tight, the impulse acts on both particles simultaneously; they end up moving with the same velocity.
在许多 CCEA 问题中，物体由轻绳或弹簧相连，或者放置在光滑平面上。当脉冲力作用时（例如绳子猛然绷紧），你会对每个物体应用冲量–动量定理。对于猛然拉紧的绳子，冲量同时作用在两个物体上；它们最终以相同速度运动。
If impulsive tension appears, denote it by I and write separate impulse equations for each particle. Eliminate I to find the common final speed. Always treat the instant before and after the impulse separately – displacements are ignored during the infinitesimal time of the impulse.
若出现脉冲张力，用 I 表示，并对每个质点分别列出冲量方程。消去 I 以求得共同末速度。务必把冲量作用瞬间的前后状态分开处理——在冲量无穷小的时间间隔内，位移可忽略不计。
11. Typical Exam Questions and Pitfalls | 典型考题与常见陷阱
Q: ving at 3 m s⁻¹ in the same direction. Given e = 0.8, find the velocities after collision.
陷阱： 相对接近速度为 (5 − 3) = 2 m s⁻¹。分离速度 = e × 2 = 1.6 m s⁻¹。联立动量守恒：2×5 + 1×3 = 2v₁ + 1v₂，且 v₂ − v₁ = 1.6。解出 v₁ = 4.2 m s⁻¹, v₂ = 5.8 m s⁻¹。
Q: A 3 kg ball strikes a wall at 10 m s⁻¹ and rebounds at 6 m s⁻¹. The contact time is 0.05 s. Find the average force exerted by the wall.
Pitfall: mpulse = change in momentum = 3×(6) − 3×(−10) = 48 N s. Average force = 48 ÷ 0.05 = 960 N.
Q: 一个 2 kg 的质点以 5 m s⁻¹ 的速度追上前方同向运动的 1 kg 质点（3 m s⁻¹）。已知 e = 0.8，求碰后速度。
Pitfall: 相对接近速度为 (5 − 3) = 2 m s⁻¹。分离速度 = e × 2 = 1.6 m s⁻¹。动量守恒：2×5 + 1×3 = 2v₁ + 1v₂，且 v₂ − v₁ = 1.6。解得 v₁ = 4.2 m s⁻¹, v₂ = 5.8 m s⁻¹。
Common mistakes: forgetting that momentum is a vector, mixing up u and v in the restitution formula, or ignoring direction when calculating relative speeds. Always draw a clear vector diagram and label all velocities with both magnitude and direction.
常见错误： 忘记动量是矢量；在恢复系数公式中混淆 u 和 v；或在计算相对速度时忽略方向。务必画出清晰的矢量图，并标注所有速度的大小和方向。
12. Summary of Key Formulae | 关键公式总结
- Momentum: p = m v | 动量: p = m v
- Impulse: I = F Δt = m v − m u | 冲量: I = F Δt = m v − m u
- Conservation of momentum: Σ m u = Σ m v | 动量守恒: Σ m u = Σ m v
- Restitution: e = (v₂ − v₁) / (u₁ − u₂) | 恢复系数: e = (v₂ − v₁) / (u₁ − u₂)
- Elastic collision: e = 1, kinetic energy conserved | 弹性碰撞: e = 1，动能守恒
- Perfectly inelastic: e = 0, particles move together | 完全非弹性: e = 0，质点共同运动
Regular practice with these vector equations, particularly in two dimensions, will build fluency. Translate each scenario into clear diagrams and algebraic equations, and you will confidently handle momentum and impulse problems in CCEA Mathematics.
经常练习这些矢量方程，尤其是二维情形，将帮助你熟练。将每个情景转化为清晰的示意图和代数方程，你就能自信地解决 CCEA 数学中的动量与冲量问题。
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