Edexcel A-Level Physics: Momentum and Impulse | Edexcel A-Level 物理：动量与冲量

📚 Edexcel A-Level Physics: Momentum and Impulse | Edexcel A-Level 物理：动量与冲量

Momentum and impulse are central to the Edexcel A-Level Physics mechanics section. They link Newton’s laws, force-time graphs, collisions and explosions. A clear understanding of vector direction and conservation conditions is essential for both calculation questions and explanation questions.

动量与冲量是 Edexcel A-Level 物理力学部分的核心内容。它们将牛顿定律、力-时间图像、碰撞和爆炸联系在一起。清晰理解矢量方向与守恒条件，对计算题和解释题都至关重要。

1. The Definition of Momentum | 动量的定义

Momentum is defined as the product of an object’s mass and its velocity. Since velocity is a vector, momentum is also a vector. Its direction is the same as the direction of the velocity.

动量定义为物体质量与速度的乘积。由于速度是矢量，动量也是矢量。动量的方向与速度方向相同。

p = mv

In SI units, mass is measured in kg and velocity in m s⁻¹, so momentum is measured in kg m s⁻¹. This unit is equivalent to N s, which is useful when dealing with impulse.

在国际单位制中，质量以 kg 为单位，速度以 m s⁻¹ 为单位，因此动量的单位是 kg m s⁻¹。这个单位等价于 N s，在处理冲量时非常有用。

For example, a 1200 kg car travelling at 15 m s⁻¹ due east has a momentum of 18 000 kg m s⁻¹ due east. If the same car turns and travels due west at 15 m s⁻¹, its momentum becomes -18 000 kg m s⁻¹ when east is taken as positive.

例如，一辆 1200 kg 的汽车以 15 m s⁻¹ 向正东行驶，其动量为正东方向 18 000 kg m s⁻¹。如果该车转向后以 15 m s⁻¹ 向正西行驶，且规定正东为正，则其动量变为 -18 000 kg m s⁻¹。

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2. Impulse as the Change in Momentum | 冲量作为动量变化

Impulse is the product of the resultant force acting on an object and the time for which the force acts. Newton’s second law can be written in terms of momentum, showing that impulse equals change in momentum.

冲量是作用在物体上的合力与力作用时间的乘积。牛顿第二定律可以用动量来表述，表明冲量等于动量的变化量。

F Δt = Δp

This relationship is extremely useful when forces vary with time, such as during collisions or when a ball is struck by a bat. The area under a force-time graph gives the impulse delivered to the object.

当力随时间变化时，例如在碰撞过程中或球被球拍击打时，这个关系非常有用。力-时间图像下方的面积表示传递给物体的冲量。

If a force-time graph has a curved shape, you can still find the impulse by estimating the area under the curve. In exam questions, the graph is often a triangle or rectangle for easier calculation.

如果力-时间图像是曲线形状，你仍然可以通过估算曲线下的面积来求冲量。在考试题中，图像通常为三角形或矩形，以便于计算。

For a constant force of 40 N acting for 0.20 s on a stationary 0.10 kg ball, the impulse is 8.0 N s. The ball’s speed after the impact is then v = Δp / m = 8.0 / 0.10 = 80 m s⁻¹.

对于一个 40 N 的恒力作用于静止的 0.10 kg 小球 0.20 s，冲量为 8.0 N s。碰后小球速度为 v = Δp / m = 8.0 / 0.10 = 80 m s⁻¹。

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3. The Law of Conservation of Momentum | 动量守恒定律

The total momentum of a closed system remains constant provided no external resultant force acts on the system. This is known as the principle of conservation of linear momentum.

若系统不受外力或所受外力的合力为零，则系统的总动量保持不变。这就是线性动量守恒定律。

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In this equation, u₁ and u₂ are the velocities of two objects before a collision, and v₁ and v₂ are their velocities after the collision. All velocities must be measured along the same straight line or resolved into components.

在这个方程中，u₁ 和 u₂ 是两物体碰撞前的速度，v₁ 和 v₂ 是碰撞后的速度。所有速度必须沿同一直线测量，或分解为分速度。

It is important to choose a positive direction before writing the equation. Objects moving in the opposite direction are assigned negative velocities. This sign convention prevents many common mistakes.

在列方程之前，必须先规定正方向。沿相反方向运动的物体取负速度。这一正负号规定可以避免许多常见错误。

If a 2.0 kg cart moving right at 3.0 m s⁻¹ collides with a 1.0 kg cart moving right at 1.0 m s⁻¹, the total momentum before is (2.0 × 3.0) + (1.0 × 1.0) = 7.0 kg m s⁻¹. This total must be the same after the collision, unless an external force acts.

如果一个 2.0 kg 的小车以 3.0 m s⁻¹ 向右运动，与一个 1.0 kg 以 1.0 m s⁻¹ 向右运动的小车碰撞，碰撞前总动量为 (2.0 × 3.0) + (1.0 × 1.0) = 7.0 kg m s⁻¹。除非有外力作用，碰撞后总动量必须保持相同。

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4. Elastic Collisions | 弹性碰撞

An elastic collision is one in which both momentum and kinetic energy are conserved. In macroscopic situations, collisions are rarely perfectly elastic, but collisions between atomic particles or idealised trolleys can be treated as elastic.

弹性碰撞是指动量和动能都守恒的碰撞。在宏观情况下，碰撞很少是完全弹性的，但原子粒子之间或理想化小车之间的碰撞可以视为弹性碰撞。

For an elastic collision, you can write two equations: the momentum conservation equation and the kinetic energy conservation equation. Solving them together gives the final velocities.

对于弹性碰撞，你可以列出两个方程：动量守恒方程和动能守恒方程。联立求解可得到末速度。

A useful special case occurs when two objects of equal mass collide head-on and elastically. They exchange velocities. If one object is initially at rest, the moving object stops and the stationary object moves off with the original velocity.

一个有用的特例是：两个质量相等的物体发生正碰且为弹性碰撞时，它们交换速度。若其中一个物体最初静止，则运动的物体停止，静止的物体以原来的速度运动离开。

In exam questions, elastic collisions often appear in particle physics or ideal gas calculations, where energy loss is assumed to be negligible.

在考试题中，弹性碰撞常出现在粒子物理或理想气体计算中，此时能量损失可以忽略不计。

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5. Inelastic Collisions | 非弹性碰撞

In an inelastic collision, momentum is conserved but kinetic energy is not conserved. Some kinetic energy is transformed into internal energy, sound or heat, and the objects may be deformed.

在非弹性碰撞中，动量守恒，但动能不守恒。一部分动能转化为内能、声能或热能，物体可能发生形变。

Although kinetic energy is lost, total energy is always conserved. The missing kinetic energy has simply been transferred to the surroundings or stored as potential energy in the deformed material.

尽管动能减少了，总能量始终守恒。减少的动能只是转移到周围环境中，或作为形变材料的势能储存起来。

When solving inelastic collision problems, you should still apply momentum conservation. You cannot apply kinetic energy conservation, so you will need another piece of information, such as one final velocity, to find the other final velocity.

求解非弹性碰撞问题时，你仍然可以应用动量守恒。由于不能应用动能守恒，你需要另一个已知条件，例如其中一个末速度，才能求出另一个末速度。

For example, a 3.0 kg object moving at 4.0 m s⁻¹ collides with a 2.0 kg object at rest. After the collision, the 3.0 kg object moves at 1.0 m s⁻¹ in the same direction. The velocity of the 2.0 kg object is found from momentum conservation: (3.0 × 4.0) = (3.0 × 1.0) + (2.0 × v), giving v = 4.5 m s⁻¹.

例如，一个 3.0 kg 的物体以 4.0 m s⁻¹ 运动，与静止的 2.0 kg 物体碰撞。碰撞后，3.0 kg 物体沿原方向以 1.0 m s⁻¹ 运动。根据动量守恒： (3.0 × 4.0) = (3.0 × 1.0) + (2.0 × v)，得到 v = 4.5 m s⁻¹。

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6. Perfectly Inelastic Collisions | 完全非弹性碰撞

A perfectly inelastic collision is one in which the two objects stick together and move with a common velocity after the collision. Momentum is conserved, but the maximum possible kinetic energy is lost for the given initial conditions.

完全非弹性碰撞是指两物体碰撞后粘在一起，并以共同速度运动。动量守恒，但对于给定的初始条件，动能损失最大。

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

This formula gives the common velocity after a perfectly inelastic collision. It follows directly from momentum conservation because the final combined mass is m₁ + m₂.

该公式给出完全非弹性碰撞后的共同速度。它直接由动量守恒导出，因为碰撞后的总质量为 m₁ + m₂。

Consider a 5.0 kg block moving at 6.0 m s⁻¹ striking a 10.0 kg block at rest. If they stick together, their common velocity is v = (5.0 × 6.0) / (5.0 + 10.0) = 30.0 / 15.0 = 2.0 m s⁻¹.

考虑一个 5.0 kg 的物体以 6.0 m s⁻¹ 撞击静止的 10.0 kg 物体。如果它们粘在一起，共同速度为 v = (5.0 × 6.0) / (5.0 + 10.0) = 30.0 / 15.0 = 2.0 m s⁻¹。

To find the kinetic energy lost, calculate the kinetic energy before and after. In this case, before gives ½ × 5.0 × 6.0² = 90 J, while after gives ½ × 15.0 × 2.0² = 30 J, so 60 J is lost to heat, sound and deformation.

要求动能损失，需计算碰撞前后的动能。本例中，碰撞前为 ½ × 5.0 × 6.0² = 90 J，碰撞后为 ½ × 15.0 × 2.0² = 30 J，因此有 60 J 转化为热、声和形变能。

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7. Coefficient of Restitution | 恢复系数

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

恢复系数通常写作 e，用来衡量碰撞的弹性程度。其定义为沿碰撞方向上分离的相对速度与接近的相对速度之比。

e = (v₂ – v₁) / (u₁ – u₂)

For a perfectly elastic collision, e = 1. For a perfectly inelastic collision where the objects stick together, e = 0. Most real collisions have a coefficient of restitution between 0 and 1.

对于完全弹性碰撞，e = 1。对于物体粘在一起的完全非弹性碰撞，e = 0。大多数实际碰撞的恢复系数介于 0 和 1 之间。

The coefficient of restitution depends on the materials of the colliding bodies. Glass and steel tend to have high values, while clay and putty have very low values.

恢复系数取决于碰撞物体的材料。玻璃和钢通常具有较高的恢复系数，而黏土和橡皮泥的恢复系数非常低。

Collision type	Momentum	Kinetic energy	Coefficient e
Elastic	Conserved	Conserved	e = 1
Inelastic	Conserved	Not conserved	0 < e < 1
Perfectly inelastic	Conserved	Maximum loss	e = 0

The table summarises the three main collision types. You should be able to identify which type applies from the wording of an exam question, especially whether the objects stick together or whether kinetic energy is stated to be conserved.

该表总结了三种主要的碰撞类型。你应能根据考试题的措辞判断属于哪种类型，尤其是物体是否粘在一起，或题目是否说明动能守恒。

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8. Explosions and Recoil | 爆炸与反冲

An explosion is the reverse of a perfectly inelastic collision. A single object breaks into two or more pieces. If the object is initially at rest, the total momentum before the explosion is zero, so the total momentum after must also be zero.

爆炸与完全非弹性碰撞相反。一个物体分裂成两个或多个碎片。如果物体最初静止，爆炸前总动量为零，因此爆炸后总动量也必须为零。

0 = m₁v₁ + m₂v₂

From this equation, the pieces move in opposite directions with momenta of equal magnitude. The heavier piece moves more slowly, while the lighter piece moves faster.

由这个方程可知，碎片沿相反方向运动，且动量大小相等。较重的碎片运动较慢，较轻的碎片运动较快。

Unlike collisions, kinetic energy is not conserved in an explosion. The total kinetic energy after the explosion is greater than before because chemical potential energy or stored elastic energy is converted into kinetic energy.

与碰撞不同，爆炸中动能不守恒。爆炸后的总动能大于爆炸前，因为化学势能或储存的弹性势能转化为动能。

A 0.50 kg firework initially at rest explodes into two pieces. One piece of mass 0.20 kg moves east at 30 m s⁻¹. The other piece has mass 0.30 kg, so its velocity is v = – (0.20 × 30) / 0.30 = -20 m s⁻¹, meaning 20 m s⁻¹ west.

一个最初静止的 0.50 kg 烟花爆炸成两片。一片质量为 0.20 kg，以 30 m s⁻¹ 向东运动。另一片质量为 0.30 kg，其速度为 v = – (0.20 × 30) / 0.30 = -20 m s⁻¹，即 20 m s⁻¹ 向西。

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9. Momentum in Two Dimensions | 二维动量

When collisions occur at an angle, momentum is conserved independently in two perpendicular directions, usually horizontal and vertical. You must resolve

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