Understanding Collisions | 理解碰撞

📚 Understanding Collisions | 理解碰撞

Collisions are central to A-Level Physics because they show how momentum and energy behave when objects interact over a very short time. In this article, we build a clear, exam-focused understanding of elastic and inelastic collisions, the coefficient of restitution, impulse, and one- and two-dimensional collision problems.

碰撞是 A-Level 物理的核心内容之一，因为它们展示了物体在极短时间相互作用时动量和能量如何变化。本文将从考试重点出发，系统讲解弹性碰撞与非弹性碰撞、恢复系数、冲量以及一维和二维碰撞问题。

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1. Linear Momentum | 线动量

Linear momentum p is defined as the product of an object’s mass and its velocity: p = m v. Since velocity is a vector, momentum is also a vector, and its direction is the same as the velocity direction.

线动量 p 定义为物体质量与速度的乘积：p = m v。由于速度是矢量，动量也是矢量，其方向与速度方向相同。

The SI unit of momentum is kg m s⁻¹, which is equivalent to N s. In a collision, each object carries momentum into the interaction, and the vector sum of all momenta describes the total momentum of the system.

动量的国际单位是 kg m s⁻¹，也等同于 N s。在碰撞中，每个物体都带着动量进入相互作用，所有动量的矢量和描述了系统的总动量。

p = m v

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2. Conservation of Linear Momentum | 线动量守恒

When no external resultant force acts on a system, the total linear momentum of that system remains constant. This is the principle of conservation of momentum, and it is a direct consequence of Newton’s third law: the forces between colliding bodies are equal and opposite, so their impulses cancel.

当系统不受合外力作用时，系统的总线动量保持不变。这就是动量守恒定律，它是牛顿第三定律的直接结果：碰撞物体之间的力大小相等、方向相反，因此它们的冲量相互抵消。

For a two-body collision, we can write the conservation equation as follows, where u represents initial velocities and v represents final velocities:

对于两个物体的碰撞，我们可以写出如下守恒方程，其中 u 表示初速度，v 表示末速度：

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

It is essential to choose a positive direction and use the correct signs for all velocities. A common mistake is to treat momentum as a scalar and ignore the direction of motion.

必须选择一个正方向，并对所有速度使用正确的符号。一个常见错误是把动量当作标量处理，忽略运动方向。

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3. Elastic vs Inelastic Collisions | 弹性碰撞与非弹性碰撞

An elastic collision is one in which both momentum and total kinetic energy are conserved. No kinetic energy is transformed into internal energy, sound, or permanent deformation. In macroscopic everyday life, perfectly elastic collisions are rare, but collisions between hard steel balls or gas molecules can be approximately elastic.

弹性碰撞是指动量和总动能都守恒的碰撞。没有动能转化为内能、声能或永久形变。在日常宏观世界中，完全弹性碰撞很少见，但硬钢球或气体分子之间的碰撞可以近似为弹性碰撞。

An inelastic collision conserves momentum but does not conserve total kinetic energy. Some kinetic energy is converted into other forms. If the two bodies stick together and move with a common velocity after the collision, it is called a perfectly inelastic collision.

非弹性碰撞动量守恒，但总动能不守恒。部分动能转化为其他形式的能量。如果两个物体碰撞后粘在一起并以共同速度运动，则称为完全非弹性碰撞。

The table below summarises the key differences:

下表总结了主要区别：

Type | 类型	Momentum | 动量	Kinetic energy | 动能	Final state | 末状态
Elastic | 弹性	Conserved | 守恒	Conserved | 守恒	Separate | 分离
Inelastic | 非弹性	Conserved | 守恒	Not conserved | 不守恒	Separate | 分离
Perfectly inelastic | 完全非弹性	Conserved | 守恒	Maximum KE lost | 动能损失最大	Stick together | 粘在一起

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4. Coefficient of Restitution | 恢复系数

The coefficient of restitution e measures how elastic a collision is. It is defined as the ratio of the relative speed of separation after the collision to the relative speed of approach before the collision.

恢复系数 e 用来衡量碰撞的弹性程度。它定义为碰撞后分离的相对速度与碰撞前接近的相对速度之比。

e = (v₂ − v₁) ÷ (u₁ − u₂)

The value of e ranges from 0 to 1 for ordinary collisions. If e = 1, the collision is perfectly elastic. If e = 0, the collision is perfectly inelastic. For most real collisions, 0 < e < 1, meaning the collision is partially elastic.

对于普通碰撞，e 的取值范围为 0 到 1。若 e = 1，碰撞为完全弹性碰撞。若 e = 0，碰撞为完全非弹性碰撞。对于大多数真实碰撞，0 < e < 1，表示碰撞是部分弹性的。

When using this formula in one dimension, you must choose one positive direction and substitute all velocities with their correct signs. Reversing the order of subtraction will reverse the sign and give an incorrect result.

在一维情况下使用该公式时，必须选择一个正方向，并将所有速度带正确符号代入。颠倒减法顺序会反转符号并导致错误结果。

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5. Impulse and Force in a Collision | 碰撞中的冲量与力

Impulse J is the product of the average force acting on an object and the time interval over which it acts. According to Newton’s second law in its impulse-momentum form, impulse equals the change in momentum of the object.

冲量 J 是作用在物体上的平均力与其作用时间间隔的乘积。根据牛顿第二定律的冲量-动量形式，冲量等于物体动量的变化量。

J = F Δt = Δp = m(v − u)

During a collision, the force is usually not constant: it rises rapidly to a peak and then falls to zero. The area under a force-time graph gives the impulse, and therefore the change in momentum. This is extremely useful when the force varies during the impact.

碰撞过程中，力通常不是恒定的：它迅速增大到峰值，然后降为零。力-时间图下的面积表示冲量，也就是动量变化量。当碰撞过程中力发生变化时，这一点非常有用。

In safety design, increasing the collision time for the same momentum change reduces the average force. This is why cars have crumple zones and why athletes bend their knees when landing.

在安全设计中，对于相同的动量变化，延长碰撞时间可以减小平均力。这就是为什么汽车有碰撞缓冲区，以及运动员落地时弯曲膝盖的原因。

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6. One-Dimensional Collision Calculations | 一维碰撞计算

To solve a one-dimensional collision problem, use a clear step-by-step method. First, choose a positive direction and label all velocities with signs. Then write the conservation of momentum equation. If the coefficient of restitution is known, write the restitution equation and solve the two equations simultaneously.

解决一维碰撞问题时，应使用清晰的步骤。首先，选择一个正方向并标注所有速度的符号。然后写出动量守恒方程。如果已知恢复系数，写出恢复系数方程并联立两个方程求解。

Worked example: A 2 kg mass moving at 4 m s⁻¹ collides head-on with a 3 kg mass moving at −2 m s⁻¹. If e = 0.5, find the final velocities.

例题：一个 2 kg 的物体以 4 m s⁻¹ 的速度与一个以 −2 m s⁻¹ 运动的 3 kg 物体正碰。若 e = 0.5，求两个物体的末速度。

Momentum conservation gives 2(4) + 3(−2) = 2v₁ + 3v₂, so 2v₁ + 3v₂ = 2. The restitution equation gives v₂ − v₁ = 0.5 × (4 − (−2)) = 3. Solving gives v₁ = −1.4 m s⁻¹ and v₂ = 1.6 m s⁻¹.

动量守恒给出 2(4) + 3(−2) = 2v₁ + 3v₂，因此 2v₁ + 3v₂ = 2。恢复系数方程给出 v₂ − v₁ = 0.5 × (4 − (−2)) = 3。解得 v₁ = −1.4 m s⁻¹，v₂ = 1.6 m s⁻¹。

Always check that your final velocities are physically reasonable. For example, in a head-on collision the two objects should not pass through each other unless the restitution coefficient is very high and the masses are unusual.

始终要检查末速度是否在物理上合理。例如，在正碰中，两个物体不应相互穿过，除非恢复系数非常高且质量情况特殊。

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7. Two-Dimensional Collisions | 二维碰撞

In a two-dimensional collision, momentum is a vector quantity, so conservation of momentum must be applied separately to two perpendicular directions, usually the x-direction and the y-direction. This is essential when objects collide at an angle, such as billiard balls colliding off-centre.

在二维碰撞中，动量是矢量，因此必须将动量守恒分别应用于两个垂直方向，通常为 x 方向和 y 方向。这对于物体斜向碰撞至关重要，例如台球偏心碰撞。

If one object is initially at rest and the moving object travels along the x-axis before impact, the x-component equation and y-component equation can be written as follows:

如果一个物体最初静止，运动物体碰撞前沿 x 轴运动，则 x 分量方程和 y 分量方程可写为：

x: m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂
y: 0 = m₁v₁ sin θ₁ + m₂v₂ sin θ₂

Angles θ₁ and θ₂ are measured relative to the original direction of motion. The signs of the sine and cosine terms automatically account for whether each object moves above or below the axis, but you must still define your positive directions clearly.

角度 θ₁ 和 θ₂ 相对于原始运动方向测量。正弦和余弦项的符号会自动说明每个物体是向轴上方还是下方运动，但你仍必须明确定义正方向。

In many exam problems, you are given one angle and asked to find another, or you are given enough information to determine an unknown speed. The key is to set up two independent equations and solve them simultaneously.

在许多考试题中，会给出一个角度并要求求另一个角度，或者给出足够信息来确定未知速度。关键是建立两个独立方程并联立求解。

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8. Energy Changes in Collisions | 碰撞中的能量变化

The change in total kinetic energy during a collision is calculated by subtracting the total kinetic energy before the collision from the total kinetic energy after the collision.

碰撞过程中总动能的变化量，等于碰撞后总动能减去碰撞前总动能。

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

For an elastic collision, ΔKE = 0. For an inelastic collision, ΔKE is negative, meaning the system loses kinetic energy. The lost energy is usually converted into heat, sound, or permanent deformation. In a perfectly inelastic collision, the loss of kinetic energy is the maximum possible for the given initial momentum.

对于弹性碰撞，ΔKE = 0。对于非弹性碰撞，ΔKE 为负值，说明系统损失了动能。损失的能量通常转化为热能、声能或永久形变。在完全非弹性碰撞中，对于给定的初始动量，动能损失为最大可能值。

There is also the reverse situation: an explosion or a recoil event can increase the total kinetic energy of the fragments because stored chemical or elastic potential energy is released. Momentum is still conserved, but the kinetic energy of the system increases.

还存在相反的情况：爆炸或反冲事件可以增加碎片的系统总动能，因为储存的化学势能或弹性势能被释放出来。动量仍然守恒，但系统的动能增加了。

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9. Experimental Methods and Common Pitfalls | 实验方法与常见误区

In the laboratory, collisions are often studied using an air track to reduce friction, or using motion sensors, light gates, or video analysis to measure velocities before and after impact. The coefficient of restitution can be determined by measuring the separation speed and approach speed of two gliders.

在实验室中，碰撞研究通常使用气垫导轨以减少摩擦，或使用运动传感器、光电门或视频分析来测量碰撞前后的速度。通过测量两个滑块分离速度和接近速度，可以确定恢复系数。

A common experiment for e uses a ball dropped from a known height h₁ onto a hard surface, measuring the rebound height h₂. Since the kinetic energy just before impact is proportional to h₁ and just after impact is proportional to h₂, the coefficient of restitution is e = √(h₂/h₁).

测量 e 的一个常见实验是将球从已知高度 h₁ 落到坚硬表面，测量反弹高度 h₂。由于碰撞前动能正比于 h₁，碰撞后动能正比于 h₂，恢复系数为 e = √(h₂/h₁)。

Common pitfalls in collision questions include forgetting that momentum is a vector, mixing up the direction signs in the restitution formula, assuming kinetic energy is always conserved, and failing to separate x and y components in two-dimensional problems. Always draw a diagram and define the positive direction before writing any equation.

碰撞题中的常见误区包括：忘记动量是矢量、在恢复系数公式中混淆方向符号、误以为动能总是守恒，以及在二维问题中没有分解 x 和 y 分量。在写任何方程之前，务必画出示意图并定义正方向。

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