📚 Collisions in Further Mechanics | 碰撞问题（进阶力学）
Collisions form a cornerstone of Further Mechanics at A Level, combining conservation laws with the concept of restitution. This article provides a thorough treatment of direct and oblique collisions, impulse, energy loss, and multi-collision problems — everything you need for the Edexcel AS and A Level Further Mathematics Further Mechanics specification.
碰撞问题是A Level进阶力学（Further Mechanics）的核心内容之一，它将动量守恒定律与恢复系数（restitution）概念紧密结合。本文将系统地讲解正向碰撞、斜碰撞、冲量、能量损失以及多次碰撞问题——全面覆盖Edexcel AS和A Level进阶数学中进阶力学部分的所有考点。
1. Conservation of Linear Momentum | 线性动量守恒
For a system of particles with no external forces acting upon it, the total linear momentum remains constant. For two colliding particles of masses m₁ and m₂ with initial velocities u₁ and u₂ and final velocities v₁ and v₂, the conservation law is expressed as:
对于一个不受外力作用的粒子系统，其总线性动量保持恒定。对于质量分别为 m₁ 和 m₂ 的两个碰撞粒子，若初速度为 u₁ 和 u₂，末速度为 v₁ 和 v₂，动量守恒定律可表示为：
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Velocities are treated as signed quantities: a velocity to the right is taken as positive, and to the left as negative. This sign convention is essential for solving collision problems correctly.
速度必须作为带符号的量处理：向右的速度取正值，向左的速度取负值。这一符号约定对于正确求解碰撞问题至关重要。
- External forces must be absent or negligible during the brief impact duration.
- Momentum is a vector quantity — always consider direction.
- The equation holds regardless of whether the collision is elastic or inelastic.
- 碰撞瞬间必须没有外力作用，或外力可忽略不计。
- 动量是矢量——必须始终考虑方向。
- 无论碰撞是弹性的还是非弹性的，动量守恒方程均成立。
2. Newton's Experimental Law of Restitution | 牛顿碰撞恢复定律（恢复系数）
When two bodies collide, their relative velocity after impact is related to their relative velocity before impact. Newton's law of restitution states that the ratio of the relative speed of separation to the relative speed of approach is a constant, denoted by e — the coefficient of restitution.
当两个物体碰撞时，碰撞后的相对速度与碰撞前的相对速度之间存在确定关系。牛顿碰撞恢复定律指出：分离相对速率与接近相对速率之比为常数，记作 e——称为恢复系数。
e = (v₂ − v₁) / (u₁ − u₂)
Here, u₁ > u₂ indicates that particle 1 approaches particle 2, and v₂ > v₁ indicates separation after impact. The coefficient e satisfies 0 ≤ e ≤ 1 for real materials.
在此，u₁ > u₂ 表示粒子1向粒子2接近，v₂ > v₁ 表示碰撞后两者分离。对于实际材料，恢复系数满足 0 ≤ e ≤ 1。
- e = 1: perfectly elastic collision — kinetic energy is conserved.
- e = 0: perfectly inelastic collision — particles coalesce and move together.
- 0 < e < 1: partially elastic — some kinetic energy is lost.
- e = 1：完全弹性碰撞——动能守恒。
- e = 0：完全非弹性碰撞——粒子粘合在一起以相同速度运动。
- 0 < e < 1：部分弹性碰撞——部分动能损失。
3. Impulse and Impact | 冲量与碰撞冲击
The impulse of a force acting over a time interval is equal to the change in momentum it produces. For a particle of mass m whose velocity changes from u to v, the impulse I is:
力在时间间隔内产生的冲量等于它所引起的动量变化。对于质量为 m、速度从 u 变为 v 的粒子，冲量 I 为：
I = m(v − u)
Impulse is a vector quantity measured in newton-seconds (N·s). During a collision, each body experiences an equal and opposite impulse, consistent with Newton's third law.
冲量是矢量，单位为牛顿秒（N·s）。在碰撞过程中，每个物体受到大小相等、方向相反的冲量，这与牛顿第三定律一致。
- Impulse on body A = −(Impulse on body B) during any collision.
- Impulse can also be expressed as I = F·Δt for a constant force.
- In oblique collisions, treat x- and y-components of impulse separately.
- 碰撞过程中，物体A所受冲量 = −(物体B所受冲量)。
- 对于恒力，冲量也可表示为 I = F·Δt。
- 在斜碰撞中，需分别处理冲量的 x 分量和 y 分量。
4. Direct Collisions in One Dimension | 一维正向碰撞
A direct collision occurs when two bodies move along the same straight line before and after impact. Solving a direct collision problem requires two equations: the conservation of linear momentum and Newton's law of restitution.
正向碰撞指两个物体在碰撞前后均沿同一直线运动。求解正向碰撞问题需要两个方程：线性动量守恒方程和牛顿恢复定律方程。
Worked example | 典例解析: A sphere A of mass 2 kg moves at 5 m·s⁻¹ and collides with a stationary sphere B of mass 3 kg. Given e = 0.6, find the velocities after impact.
典例解析：质量为 2 kg 的球A以 5 m·s⁻¹ 的速度运动，与质量为 3 kg 的静止球B发生碰撞。已知 e = 0.6，求碰撞后各球的速度。
By conservation of momentum: 2(5) + 3(0) = 2v₁ + 3v₂, so 10 = 2v₁ + 3v₂. By restitution: e = (v₂ − v₁)/(u₁ − u₂) = 0.6, giving v₂ − v₁ = 0.6 × 5 = 3. Solving simultaneously: v₁ = 0.2 m·s⁻¹ and v₂ = 3.2 m·s⁻¹.
由动量守恒：2(5) + 3(0) = 2v₁ + 3v₂，即 10 = 2v₁ + 3v₂。由恢复定律：e = (v₂ − v₁)/(u₁ − u₂) = 0.6，得 v₂ − v₁ = 0.6 × 5 = 3。联立求解：v₁ = 0.2 m·s⁻¹，v₂ = 3.2 m·s⁻¹。
- Always define a positive direction before writing equations.
- Check that v₂ > v₁ after impact; otherwise, the sign of e is used incorrectly.
- For e = 0, set v₁ = v₂ and solve momentum alone.
- 列方程前务必规定正方向。
- 验证碰撞后 v₂ > v₁；否则说明 e 的符号使用有误。
- 当 e = 0 时，令 v₁ = v₂，仅用动量守恒求解即可。
5. Collision with a Fixed Barrier | 与固定障碍物的碰撞
When a particle collides with a fixed wall or floor, the wall's mass is effectively infinite, and its velocity remains zero. The coefficient of restitution relates the rebound speed to the approach speed: v = e·u, where u is the speed of approach and v is the speed of separation.
当粒子与固定墙壁或地面碰撞时，墙的有效质量视为无穷大，其速度保持为零。恢复系数将回弹速率与接近速率联系起来：v = e·u，其中 u 为接近速率，v 为分离速率。
Key points | 要点: For a particle dropped from height h and rebounding to height h':
要点：对于从高度 h 下落并回弹至高度 h' 的粒子：
u = √2gh, v = e·u, h' = v²/(2g) = e²·h
Successive rebound heights follow a geometric progression: after n rebounds, the height is hₙ = e²ⁿh. Similarly, the times between successive impacts also form a geometric sequence with common ratio e.
连续回弹高度构成等比数列：第 n 次回弹的高度为 hₙ = e²ⁿh。类似地，连续碰撞之间的时间间隔也构成公比为 e 的等比数列。
- The velocity of the wall is always taken as zero in both before and after equations.
- For oblique impact with a wall, decompose velocity into normal and tangential components.
- The tangential component of velocity is unchanged by the collision.
- 在碰撞前后的方程中，墙的速度始终取为零。
- 对于与墙壁的斜碰撞，需将速度分解为法向分量和切向分量。
- 速度的切向分量在碰撞前后保持不变。
6. Oblique Collisions | 斜碰撞
An oblique collision occurs when the velocity vectors are not collinear with the line of centres at the instant of impact. The standard approach is to resolve velocities into two perpendicular components: along the line of centres (normal direction, n) and perpendicular to it (tangential direction, t).
斜碰撞是指碰撞瞬间速度矢量与两球心连线（碰撞线）不在同一直线上。标准解法是将速度分解为两个垂直分量：沿碰撞线方向（法向 n）和垂直于碰撞线方向（切向 t）。
Fundamental results | 基本结论:
- Momentum is conserved along the line of centres: m₁u₁ₙ + m₂u₂ₙ = m₁v₁ₙ + m₂v₂ₙ.
- Newton's law applies only to normal components: e = (v₂ₙ − v₁ₙ)/(u₁ₙ − u₂ₙ).
- Tangential components are unchanged: v₁ₜ = u₁ₜ and v₂ₜ = u₂ₜ.
- 沿碰撞线方向动量守恒：m₁u₁ₙ + m₂u₂ₙ = m₁v₁ₙ + m₂v₂ₙ。
- 牛顿恢复定律仅适用于法向分量：e = (v₂ₙ − v₁ₙ)/(u₁ₙ − u₂ₙ)。
- 切向分量不变：v₁ₜ = u₁ₜ，v₂ₜ = u₂ₜ。
To solve an oblique collision problem, first resolve all velocities into components along n and t using trigonometry, then apply the three sets of equations above, and finally recombine components to find the final speed and direction of each particle.
求解斜碰撞问题时，首先利用三角函数将所有速度分解为沿 n 和 t 方向的分量，然后应用上述三组方程，最后将分量重新合成以求出每个粒子的最终速度大小和方向。
7. Oblique Collision with a Smooth Wall | 与光滑斜壁的斜碰撞
Consider a particle moving with speed u at an angle α to a smooth fixed wall. After impact, the particle rebounds at speed v at an angle β to the wall. The tangential velocity component is unchanged, while the normal component is reversed and scaled by e.
考虑一个以速率 u、与光滑固定壁面成角 α 运动的粒子。碰撞后，粒子以速率 v、与壁面成角 β 反弹。切向速度分量保持不变，而法向分量反向并按比例 e 缩放。
u·cos α = v·cos β and v·sin β = e·u·sin α
Dividing these equations yields the angle relationship tan β = e·tan α. Note that angles are measured to the wall, not to the normal. This is a common source of error in examinations.
将两式相除可得角度关系 tan β = e·tan α。注意：这里的角度是相对壁面而非法线度量的。这是考试中常见的错误来源。
- If angles are given relative to the normal, convert them first.
- When e = 1, the angle of reflection equals the angle of incidence (β = α).
- The rebound speed is v = u√cos²α + e²·sin²α.
- 若给出的角度是相对法线的，请先进行转换。
- 当 e = 1 时，反射角等于入射角（β = α）。
- 反弹速率 v = u√cos²α + e²·sin²α。
8. Kinetic Energy Loss in Collisions | 碰撞中的动能损失
For any collision with e < 1, some kinetic energy is converted into heat, sound, and internal deformation energy. The kinetic energy before and after impact are:
对于任何 e < 1 的碰撞，一部分动能会转化为热量、声音和内变形能。碰撞前后的动能分别为：
KE_initial = ½m₁u₁² + ½m₂u₂² , KE_final = ½m₁v₁² + ½m₂v₂²
The energy loss ΔKE = KE_initial − KE_final can be expressed in terms of the relative velocity before collision:
能量损失 ΔKE = KE_initial − KE_final 可以用碰撞前的相对速度表示：
ΔKE = ½·(m₁m₂)/(m₁+m₂)·(u₁ − u₂)²·(1 − e²)
This compact formula is extremely useful. For a perfectly elastic collision (e = 1), the energy loss is zero; for a perfectly inelastic collision (e = 0), the maximum possible energy is lost.
这个简洁的公式非常实用。对于完全弹性碰撞（e = 1），能量损失为零；对于完全非弹性碰撞（e = 0），动能损失达到最大值。
- For e = 0 (coalescing particles), ΔKE = ½·(m₁m₂)/(m₁+m₂)·(u₁ − u₂)².
- Energy is always lost in real collisions — never gained.
- Use energy loss to determine whether particles can reach a given height in projectile-after-collision problems.
- 对于 e = 0（粒子粘合），ΔKE = ½·(m₁m₂)/(m₁+m₂)·(u₁ − u₂)²。
- 实际碰撞中能量总是损失的——永远不会增加。
- 在碰撞后抛体问题中，利用能量损失可判断粒子能否达到给定高度。
9. Successive Collisions | 多次连续碰撞
Many exam problems involve a particle colliding successively with two other particles or with the same barrier multiple times. The key strategy is to treat each collision as a separate event, carefully updating velocities after each step.
许多考试题目涉及一个粒子依次与另外两个粒子碰撞，或与同一障碍物多次碰撞。关键策略是将每次碰撞视为独立事件，在每一步后仔细更新速度。
Standard procedure | 标准解题步骤:
- Step 1: Solve the first collision completely to find the velocity of each body.
- Step 2: Determine which bodies collide next by comparing their velocities and positions.
- Step 3: Apply momentum and restitution equations to the next collision, then repeat.
- Step 4: For alternating collisions, look for recurrence relations or geometric patterns.
- 步骤1：完整求解第一次碰撞，得出每个物体的速度。
- 步骤2：比较各物体的速度和位置，判断下一次碰撞发生在哪两个物体之间。
- 步骤3：对下一次碰撞应用动量和恢复方程，然后重复。
- 步骤4：对于交替碰撞，寻找递推关系或等比规律。
A common pattern is a particle bouncing alternately between two walls or between a wall and another particle. In such cases, the speed after each bounce is multiplied by e, producing geometric progressions in speed, height, and time intervals.
常见模式是粒子在两墙之间或在一墙与另一粒子之间交替反弹。此时，每次反弹后速率乘以 e，从而在速率、高度和时间间隔上产生等比数列规律。
10. Collision of Particles Connected by a String | 用轻绳连接的粒子碰撞
When one particle is attached to a string (e.g., connected to a second particle hanging over a pulley or fixed at a point), a collision can cause an impulsive tension in the string. The analysis requires combining the collision equations with the impulsive tension equation.
当一个粒子与轻绳相连（例如，通过定滑轮与另一个悬挂粒子相连，或固定在某点），碰撞会在绳中产生冲量张力。分析时需将碰撞方程与冲量张力方程结合。
For a particle of mass m attached to a light inextensible string that becomes taut, the impulsive tension T produces a change in velocity. If the particle is momentarily brought to rest or its velocity is redirected, the impulse equation is:
对于连接在不可伸长轻绳上的质量为 m 的粒子，当绳子突然绷紧时，冲量张力 T 引起速度变化。如果粒子瞬时停止或其速度被重新定向，冲量方程为：
T·ΔT = m(v − u)
In such problems, remember that momentum is conserved for the collision itself, but the impulsive tension may rapidly alter velocities immediately after. Treat the collision and the string becoming taut as separate stages.
在此类问题中，记住碰撞本身动量守恒，但碰撞后绳子的突然绷紧可能在瞬间改变速度。将碰撞和绳子绷紧视为两个独立阶段来处理。
11. Worked Examination-Style Problem | 考试风格综合例题
Problem: A particle P of mass 0.5 kg is projected with speed 10 m·s⁻¹ at an angle of 30° to a smooth horizontal floor. The coefficient of restitution between P and the floor is 0.5. Find: (a) the speed of P immediately after the first impact; (b) the angle at which it rebounds; (c) the total horizontal distance travelled before the second impact.
题目：质量为 0.5 kg 的粒子 P 以速率 10 m·s⁻¹、与光滑水平地面成 30° 角被抛出。P 与地面间的恢复系数为 0.5。求：(a) 第一次碰撞后 P 的速率；(b) 反弹角度；(c) 第二次碰撞前水平方向的总位移。
Solution | 解答: (a) At impact, the velocity components are: horizontal uₓ = 10·cos 30° = 8.66 m·s⁻¹; vertical (downward) uy = 10·sin 30° = 5 m·s⁻¹. After impact: vₓ = uₓ = 8.66 m·s⁻¹; vy = e·uy = 0.5 × 5 = 2.5 m·s⁻¹ upward. Speed v = √8.66² + 2.5² = √75 + 6.25 = √81.25 ≈ 9.01 m·s⁻¹.
解答：(a) 碰撞时刻的速度分量为：水平 uₓ = 10·cos 30° = 8.66 m·s⁻¹；竖直（向下）uy = 10·sin 30° = 5 m·s⁻¹。碰撞后：vₓ = uₓ = 8.66 m·s⁻¹；vy = e·uy = 0.5 × 5 = 2.5 m·s⁻¹（向上）。速率 v = √8.66² + 2.5² = √75 + 6.25 = √81.25 ≈ 9.01 m·s⁻¹。
(b) The angle of rebound to the horizontal is β where tan β = vy/vₓ = 2.5/8.66 = 0.2887, giving β ≈ 16.1°. (c) Time to reach maximum height after impact: t = vy/g = 2.5/9.8 ≈ 0.255 s. Flight time to next impact: T = 2t ≈ 0.510 s. Horizontal distance: d = vₓ × T = 8.66 × 0.510 ≈ 4.42 m.
(b) 反弹角相对水平面为 β，tan β = vy/vₓ = 2.5/8.66 = 0.2887，得 β ≈ 16.1°。(c) 碰撞后到达最高点的时间：t = vy/g = 2.5/9.8 ≈ 0.255 s。到达下一次碰撞的飞行时间：T = 2t ≈ 0.510 s。水平位移：d = vₓ × T = 8.66 × 0.510 ≈ 4.42 m。
12. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Students frequently lose marks on collision questions due to a small number of recurring errors. Avoid them with these reminders:
学生在碰撞问题上失分往往源于少数几个反复出现的错误。请注意以下提醒以避免失分：
- Always assign a consistent positive direction and stick to it throughout the computation.
- In oblique collisions, apply Newton's law only to the component along the line of centres — never to the full velocity.
- When angles are given, confirm whether they are measured to the wall or to the normal.
- For energy calculations, use speeds (not velocities) — kinetic energy is always positive.
- Check your final answers: v₂ > v₁ after direct impact, rebound speeds less than approach speeds when e < 1.
- Draw a clear labelled diagram before writing any equations; define every variable you use.
- 始终规定一致的正方向，并在整个计算中坚持使用。
- 在斜碰撞中，牛顿定律仅适用于沿碰撞线方向的分量——切勿对完整速度使用。
- 题目给出角度时，确认是相对壁面还是相对法线度量的。
- 在能量计算中使用速率（而非速度矢量）——动能始终为正。
- 检查最终答案：正向碰撞后 v₂ > v₁；当 e < 1 时反弹速率小于接近速率。
- 在写任何方程之前画一张清晰的标注图；定义使用的每个变量。
Mastering collisions requires practice with both direct and oblique cases, including rebounds from walls and successive impacts. Work through problems systematically: resolve, apply the three core equations, recombine, and verify.
掌握碰撞问题需要同时练习正向和斜碰撞情况，包括墙壁反弹和连续碰撞。系统化地解题：分解速度、应用三个核心方程、重新合成、最后验证结果。
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