A-Level Physics: Momentum Decomposition in Two-Dimensional Collisions | A-Level 物理：二维碰撞的动量分解

📚 A-Level Physics: Momentum Decomposition in Two-Dimensional Collisions | A-Level 物理：二维碰撞的动量分解

When two objects collide in a plane rather than along a single straight line, the collision is described as two-dimensional. Unlike one-dimensional cases, the velocities before and after impact are not collinear, so momentum conservation must be applied separately along two perpendicular axes. This article explains how to decompose momentum in 2D collisions, a core topic in the CIE A-Level Physics syllabus.

当两个物体在平面内而非沿同一直线发生碰撞时，这种碰撞被称为二维碰撞。与一维情形不同，碰撞前后的速度不在同一条直线上，因此动量守恒必须分别在两个相互垂直的轴方向上单独应用。本文围绕 CIE A-Level 物理考纲，系统讲解如何在二维碰撞中进行动量分解。

1. Momentum as a Vector and Component Decomposition | 动量作为矢量及其分量分解

Momentum is defined as the product of mass and velocity, p = m v. Since velocity is a vector, momentum is also a vector. In a two-dimensional collision, each object’s momentum can be resolved into two independent components, typically along the x-axis and y-axis.

动量的定义为质量与速度的乘积，即 p = m v。由于速度是矢量，动量也是矢量。在二维碰撞中，每个物体的动量都可以分解为两个相互独立的分量，通常取 x 轴和 y 轴方向。

pₓ = m vₓ = m v cos θ, pᵧ = m vᵧ = m v sin θ

Here θ is the angle between the velocity vector and the x-axis. The magnitude of the total momentum is found using the Pythagorean theorem:

这里 θ 是速度矢量与 x 轴之间的夹角。总动量的大小由勾股定理求得：

|p| = √(pₓ² + pᵧ²)

This decomposition is essential because momentum conservation holds independently in each direction when no external force acts along that direction.

这种分解至关重要，因为在没有外力作用的方向上，动量守恒在每一个方向上分别成立。

2. The Geometry of the Collision | 碰撞的几何关系

In a typical two-dimensional collision problem, a particle of mass m₁ moving with velocity u₁ strikes a stationary particle of mass m₂. After the collision, the two particles move off at angles θ and φ relative to the initial direction of m₁.

在一个典型的二维碰撞问题中，质量为 m₁ 的粒子以速度 u₁ 撞击一个静止的质量为 m₂ 的粒子。碰撞后，两个粒子分别相对于 m₁ 的初始运动方向偏离角度 θ 和 φ 运动。

It is convenient to set the initial direction of m₁ as the positive x-axis. The y-component of the total initial momentum is then zero. This choice simplifies the algebra considerably.

为了方便计算，通常取 m₁ 的初始运动方向为正 x 轴。此时系统总的初始动量的 y 分量为零。这样的坐标选择可以大大简化代数运算。

It is important to note that the angles are measured from the original direction of motion, not from the line joining the centres at the moment of impact. In CIE exam questions, the angles given are usually the angles between the final velocities and the incident direction.

需要特别注意，角度是相对于入射方向测量的，而不是相对于碰撞瞬间两球心连线方向。在 CIE 考试题目中，给出的角度通常是末速度与入射方向之间的夹角。

3. Conservation of Momentum Along Each Axis | 沿各轴的动量守恒

For a collision between two objects in the absence of external forces, the total momentum of the system is conserved. In two dimensions, this yields two independent scalar equations:

在没有外力作用的碰撞系统中，总动量守恒。在二维情形下，这给出两个独立的标量方程：

m₁ u₁ = m₁ v₁ cos θ + m₂ v₂ cos φ (x-direction)

0 = m₁ v₁ sin θ − m₂ v₂ sin φ (y-direction)

The negative sign in the y-equation arises because the two particles typically move to opposite sides of the x-axis. If both were on the same side, the signs would adjust accordingly.

y 方向方程中的负号是因为两个粒子通常运动到 x 轴的两侧。如果两个粒子在同侧，则符号需要相应调整。

If an external impulse acts during the collision, such as a wall exerting a force, momentum is conserved only along the direction parallel to the wall. Perpendicular to the wall, the momentum may change.

如果碰撞过程中存在外冲量，例如墙壁施加的力，则只有平行于墙壁方向的动量守恒。垂直于墙壁的方向上动量可能改变。

4. The Coefficient of Restitution in Two Dimensions | 二维碰撞中的恢复系数

The coefficient of restitution, denoted e, relates the relative speeds before and after impact along the line of centres (the normal direction). In two-dimensional collisions, e is defined only for the component of velocity along the common normal at the point of contact:

恢复系数用 e 表示，它描述碰撞前后沿两球心连线方向（法线方向）的相对速度之间的关系。在二维碰撞中，e 仅定义在接触点公法线方向的速度分量上：

e = (relative speed of separation along normal) / (relative speed of approach along normal)

e = (v₂ₙ − v₁ₙ) / (u₁ₙ − u₂ₙ)

For a perfectly elastic collision, e = 1; for a perfectly inelastic collision, e = 0. The component of velocity perpendicular to the normal (tangential direction) is unchanged during the collision if the surfaces are smooth.

对于完全弹性碰撞，e = 1；对于完全非弹性碰撞，e = 0。如果表面光滑，垂直于法线的切向速度分量在碰撞过程中保持不变。

In exam problems, you will often need to resolve the initial and final velocities into normal and tangential components before applying the restitution equation.

在考试题目中，通常需要先将碰撞前后的速度分解为法向和切向分量，然后再应用恢复系数方程。

5. Perfectly Inelastic Two-Dimensional Collision | 完全非弹性二维碰撞

In a perfectly inelastic collision, the two objects stick together and move with a common final velocity v. Momentum conservation gives:

在完全非弹性碰撞中，两个物体粘在一起，以共同的末速度 v 运动。动量守恒给出：

m₁ u₁ₓ = (m₁ + m₂) vₓ, m₁ u₁ᵧ = (m₁ + m₂) vᵧ

Therefore the common final velocity has components vₓ = m₁ u₁ₓ / (m₁ + m₂) and vᵧ = m₁ u₁ᵧ / (m₁ + m₂). Its direction is given by tan α = vᵧ / vₓ, where α is measured from the x-axis.

因此，共同末速度的分量为 vₓ = m₁ u₁ₓ / (m₁ + m₂) 和 vᵧ = m₁ u₁ᵧ / (m₁ + m₂)。其方向由 tan α = vᵧ / vₓ 确定，其中 α 是相对于 x 轴的角度。

This is a common question type in CIE Paper 4. You are not asked to find v₁ and v₂ separately; instead, you solve for the combined velocity vector directly.

这是 CIE Paper 4 中常见的题型。这类题目不要求分别求出 v₁ 和 v₂，而是直接求解结合后的速度矢量。

6. Equal Masses and Right-Angle Deflection | 等质量与直角偏转

An interesting special case arises when two equal masses collide elastically, one of which is initially at rest. In such a collision, the two final velocity vectors are perpendicular to each other. That is, if m₁ = m₂ and e = 1, then θ + φ = 90°.

一个有趣的特殊情况是：两个质量相等的物体发生弹性碰撞，其中一个初始静止。在这种碰撞中，两个末速度矢量相互垂直，即若 m₁ = m₂ 且 e = 1，则 θ + φ = 90°。

This result can be derived from combining momentum conservation with energy conservation. It is a useful check for numerical answers in exam questions. If you obtain θ + φ ≠ 90° for equal masses with e = 1, your solution likely contains an error.

这一结论可以通过联立动量守恒和能量守恒推导得出。它是检查数值答案是否有误的有效方法。如果在等质量且 e = 1 的情况下算出 θ + φ ≠ 90°，那你的解答很可能存在错误。

However, note that this property applies only when the target particle is initially at rest. If both particles are initially moving, the angle relation does not hold.

但需注意，这个性质仅当靶粒子初始静止时才成立。如果两个粒子初始都在运动，角度关系不成立。

7. Step-by-Step Problem-Solving Framework | 解题步骤框架

To approach a two-dimensional collision problem systematically, follow these steps:

为了系统地解决二维碰撞问题，请遵循以下步骤：

Step 1: Draw a clearly labelled diagram showing the initial and final velocity vectors and all given angles.

第 1 步：画一张清晰标注的示意图，标出碰撞前后的速度矢量和所有已知角度。
Step 2: Choose a coordinate system, usually with the x-axis along the initial direction of the moving object.

第 2 步：选择坐标系，通常取 x 轴沿运动物体的初始方向。
Step 3: Resolve all velocities into x and y components using sine and cosine.

第 3 步：利用正弦和余弦将所有速度分解为 x 和 y 分量。
Step 4: Write the momentum conservation equation for each axis separately.

第 4 步：分别写出每个轴的动量守恒方程。
Step 5: If needed, apply the restitution equation along the line of centres.

第 5 步：如果需要，沿球心连线方向应用恢复系数方程。
Step 6: Solve the simultaneous equations for the unknown speeds and angles.

第 6 步：联立求解未知速度和角度。

Always check whether the number of unknowns matches the number of independent equations. In most CIE problems, you are given the masses and one of the final angles, and asked to find the final speeds.

始终检查未知数的个数是否与独立方程的个数一致。在多数 CIE 题目中，已知质量和其中一个末角度，要求求出末速度。

8. Worked Numerical Example | 数值例题

A sphere of mass 2.0 kg moving at 4.0 m s⁻¹ collides with a stationary sphere of mass 3.0 kg. After the collision, the 2.0 kg sphere moves at an angle of 30° above its original direction, and the 3.0 kg sphere moves at an angle of 45° below the original direction. Calculate the final speeds of both spheres.

一个质量为 2.0 kg 的小球以 4.0 m s⁻¹ 的速度撞击一个静止的 3.0 kg 小球。碰撞后，2.0 kg 的小球沿原方向上方 30° 运动，3.0 kg 的小球沿原方向下方 45° 运动。求两个小球的末速度。

Solution: Let v₁ be the final speed of the 2.0 kg sphere and v₂ be that of the 3.0 kg sphere.

解：设 v₁ 为 2.0 kg 小球的末速度，v₂ 为 3.0 kg 小球的末速度。

Along the x-axis:

沿 x 轴方向：

2.0 × 4.0 = 2.0 v₁ cos 30° + 3.0 v₂ cos 45°

8.0 = 1.732 v₁ + 2.121 v₂ (1)

Along the y-axis:

沿 y 轴方向：

0 = 2.0 v₁ sin 30° − 3.0 v₂ sin 45°

0 = 1.0 v₁ − 2.121 v₂ (2)

From (2): v₁ = 2.121 v₂. Substituting into (1):

由 (2) 得：v₁ = 2.121 v₂。代入 (1)：

8.0 = 1.732 × 2.121 v₂ + 2.121 v₂ = 5.795 v₂

Therefore v₂ = 1.38 m s⁻¹ and v₁ = 2.93 m s⁻¹.

因此 v₂ = 1.38 m s⁻¹，v₁ = 2.93 m s⁻¹。

9. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Several errors frequently appear in student solutions to two-dimensional collision problems. Being aware of these can save valuable marks in the exam.

在二维碰撞问题的解答中，学生常犯若干典型错误。了解这些错误可以帮助你在考试中避免失分。

Forgetting the sign of the y-component: When particles move to opposite sides of the x-axis, their y-components have opposite signs. Always define a positive y-direction and use it consistently.

忘记 y 分量的符号：当粒子运动到 x 轴两侧时，它们的 y 分量符号相反。务必设定正 y 方向并始终一致使用。
Using the restitution equation in the wrong direction: The coefficient of restitution applies only to the component along the line of centres, not to the total velocities.

恢复系数用错方向：恢复系数只适用于球心连线方向上的分量，而不是总速度。
Confusing angles: The angles in CIE problems are usually measured from the incident direction, not from the normal. Read the question carefully.

角度混淆：CIE 题目中的角度通常是相对于入射方向测量的，而不是相对于法线。请仔细审题。
Applying momentum conservation to the whole system when an external impulse exists: If a collision occurs against a wall, momentum perpendicular to the wall is not conserved.

存在外冲量时仍对整个系统应用动量守恒：如果碰撞涉及墙壁，垂直于墙壁方向的动量并不守恒。

Always check your final answers for physical plausibility: speeds must be positive, and for elastic collisions the total kinetic energy must remain the same.

始终检查最终答案的物理合理性：速度必须为正，对于弹性碰撞，总动能必须保持不变。

10. Kinetic Energy in Two-Dimensional Collisions | 二维碰撞中的动能

In an elastic two-dimensional collision, total kinetic energy is conserved. The energy equation is:

在弹性二维碰撞中，总动能守恒。能量方程为：

½ m₁ u₁² = ½ m₁ v₁² + ½ m₂ v₂²

This equation is scalar, not vector, so it does not contain angle terms directly. It can be combined with the two momentum conservation equations to solve for unknown speeds and angles.

该方程是标量方程，不直接涉及角度项。它可以与两个动量守恒方程联立，求解未知速度和角度。

If the collision is inelastic, kinetic energy is lost. The loss is given by:

如果碰撞是非弹性的，动能会有损失。损失量由下式给出：

ΔEₖ = ½ m₁ u₁² − (½ m₁ v₁² + ½ m₂ v₂²)

You may be asked to calculate this loss in CIE questions. Note that momentum is always conserved, but kinetic energy is conserved only for perfectly elastic collisions.

CIE 题目中可能会要求计算这个损失量。请注意，动量总是守恒的，但动能仅在完全弹性碰撞中守恒。

11. Collisions with a Fixed Surface | 与固定表面的碰撞

When an object collides with a fixed wall or surface, the wall’s mass is effectively infinite, so the wall’s momentum change is not part of the analysis. Instead, we consider the impulse exerted on the object.

当物体与固定墙壁或表面碰撞时，墙壁的质量可视为无穷大，因此墙壁的动量变化不参与分析。我们转而考虑作用在物体上的冲量。

For a smooth wall, the component of velocity parallel to the wall is unchanged, while the component perpendicular to the wall reverses. If the collision is elastic, the angle of reflection equals the angle of incidence relative to the normal:

对于光滑墙壁，平行于墙壁的速度分量不变，垂直于墙壁的速度分量反向。若为弹性碰撞，反射角相对于法线等于入射角：

θᵣ = θᵢ

If the collision is inelastic, the normal component of velocity is reduced by the coefficient of restitution: vₙ = e uₙ, with the direction reversed.

如果碰撞是非弹性的，法向速度分量按恢复系数衰减：vₙ = e uₙ，方向反向。

This type of problem tests your ability to resolve velocities into normal and tangential components, a skill equally important in two-body collision problems.

这类问题考查你将速度分解为法向和切向分量的能力，这一技能在双体碰撞问题中同样重要。

12. Exam Tips from a Cambridge Perspective | 剑桥视角的考试技巧

To score well on two-dimensional collision questions in CIE A-Level Physics Paper 4, keep the following strategic points in mind:

要在 CIE A-Level 物理 Paper 4 的二维碰撞问题上获得高分，请牢记以下策略性要点：

Always define axes explicitly: State which direction is positive. The examiner awards method marks for clear notation.

始终明确设定坐标轴：说明哪个方向为正。清晰的符号表示可以帮助你获得方法分。
Write equations in symbols first: Substitute numbers only at the final stage. This minimises calculation errors and makes your reasoning visible.

先用符号写方程：最后一步才代入数值。这样可以减少计算错误，也让推理过程清晰可见。
Use the exact angle orientation: If you accidentally swap sin and cos for the wrong component, you will lose both method and answer marks.

注意角度的正确取向：如果不小心把 sin 和 cos 用错了分量，你会同时失去方法分和答案分。
Practise eliminating variables: Most problems require solving two linear equations in two unknowns. Quick substitution is usually more efficient than matrix methods.

练习变量消元：大多数问题需要解两个未知数的两个线性方程。快速代入通常比矩阵方法更高效。

Remember that the total number of marks for this topic across the paper is modest but predictable: one structured question of 5–8 marks is typical. Mastery of the fundamental technique therefore yields high marks per hour of revision.

请记住，这个主题在整份试卷中的分值虽不多但可预测：通常是一道 5–8 分的结构性题目。因此，掌握基本技巧后，每小时的复习产出是很高的。

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A-Level Physics: Momentum Decomposition in Two-Dimensional Collisions | A-Level 物理：二维碰撞的动量分解