📚 A-Level OCR Physics: Momentum Exam Points | A-Level OCR 物理：动量 考点精讲
Linear momentum lies at the heart of OCR A-Level Physics, bridging forces, motion and the fundamental conservation laws. Mastering momentum means understanding not only how to calculate p = mv but also how to apply the principle of conservation to collisions, explosions and impulse scenarios. This article walks you through every essential topic, with worked examples and typical exam pitfalls, so you can score full marks on the momentum questions.
线动量是 OCR A-Level 物理的核心内容，它连接了力、运动和基本的守恒定律。掌握动量不仅仅意味着会计算 p = mv，更重要的是能把守恒原理应用到碰撞、爆炸和冲量情景中。本文将带你梳理所有关键知识点，配合例题和常见考试陷阱，助力你在动量题目上拿到满分。
1. Definition and Calculation of Momentum | 动量的定义与计算
Linear momentum p is the product of an object’s mass and its velocity. It is a vector quantity, so direction must always be considered. The formula is p = m v, where m is measured in kg, v in m s⁻¹, and p in kg m s⁻¹ (or N s).
线动量 p 是物体质量与速度的乘积。它是一个矢量，因此必须始终考虑方向。公式为 p = m v，质量 m 的单位是 kg，速度 v 的单位是 m s⁻¹，动量 p 的单位是 kg m s⁻¹（或 N s）。
For example, a 1500 kg car travelling north at 20 m s⁻¹ has momentum p = 1500 × 20 = 30 000 kg m s⁻¹ north. Because p depends on velocity, any change in speed or direction changes the momentum.
例如，一辆 1500 kg 的汽车以 20 m s⁻¹ 向北行驶，其动量 p = 1500 × 20 = 30 000 kg m s⁻¹ 向北。由于 p 依赖于速度，速度大小或方向的任何变化都会改变动量。
- Always assign a positive direction in calculations (e.g. right is +, left is −).
- 计算时始终规定正方向（如向右为正，向左为负）。
2. Conservation of Linear Momentum | 动量守恒定律
In a closed system with no external resultant forces, the total momentum before an interaction equals the total momentum after the interaction. This is a direct consequence of Newton’s third law: internal forces between objects are equal and opposite, so momentum changes cancel out.
在没有合外力的封闭系统中，相互作用前的总动量等于相互作用后的总动量。这是牛顿第三定律的直接结果：物体间的内力大小相等、方向相反，因此动量变化相互抵消。
Mathematically: Σ p_before = Σ p_after. For two bodies colliding, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂, where u represents initial velocities and v final velocities. The conservation law applies to all collisions and explosions.
数学表达式为：Σ p_前 = Σ p_后。对于两个物体碰撞，m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂，其中 u 表示初速度，v 表示末速度。此守恒定律适用于所有碰撞和爆炸。
Common exam tip: if a problem says ‘a railway truck shunts another stationary truck’, always write the momentum equation, substituting with signs. Many marks are lost by ignoring the direction of velocities after impact.
常见考试提示：如果题目说“一节铁路货车撞上另一节静止的货车”，一定要写出动量方程，代入带符号的速度。很多失分都是因为忽略了碰撞后速度的方向。
3. Elastic Collisions | 弹性碰撞
An elastic collision is one in which both momentum and kinetic energy are conserved. This means Σp = constant and ΣEₖ = constant throughout the impact. No energy is dissipated as heat, sound or permanent deformation.
弹性碰撞是指动量和动能都守恒的碰撞。即 Σp 不变且 ΣEₖ 在碰撞过程中不变。没有能量以热、声或永久形变的形式耗散。
The conditions are: m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ and ½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂². Solving these simultaneous equations yields neat relationships, such as the relative speed of approach equalling the relative speed of separation: u₁ − u₂ = v₂ − v₁ (when masses are equal and head-on).
其条件为：m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ 且 ½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²。解这两个联立方程会得到简洁的关系，例如接近时的相对速度等于分离时的相对速度：u₁ − u₂ = v₂ − v₁（当两物体质量相等且正面碰撞时）。
In reality, perfectly elastic collisions only occur on the atomic scale or in idealised laboratory conditions. In OCR exams, they are often used to check whether a collision is elastic: calculate total kinetic energy before and after; if equal, the collision is elastic.
实际上，完全弹性碰撞只在原子尺度或理想化实验条件下发生。在 OCR 考试中，常要求学生判断碰撞是否为弹性碰撞：计算碰撞前后的总动能；若相等，则为弹性碰撞。
4. Inelastic Collisions | 非弹性碰撞
An inelastic collision is any collision where some kinetic energy is converted into other forms, such as internal energy (heat) or sound. Momentum is still conserved, but kinetic energy is not.
非弹性碰撞是指有部分动能转化为其他形式（如内能或声能）的碰撞。动量依然守恒，但动能不守恒。
The typical exam task is to calculate the loss of kinetic energy: calculate KE_before and KE_after, then find ΔEₖ = KE_before − KE_after. A positive ΔEₖ indicates an inelastic collision. For example, a 3 kg ball moving at 4 m s⁻¹ hits a 2 kg stationary ball; after collision the 3 kg ball continues at 1 m s⁻¹ in the same direction. Momentum gives v₂ = 4.5 m s⁻¹. KE before = 24 J, after = 1.5 + 20.25 = 21.75 J, so 2.25 J is lost.
典型考题是计算动能的损失：先计算碰撞前动能和碰撞后动能，再求 ΔEₖ = 动能前 − 动能后。正的 ΔEₖ 表明碰撞是非弹性的。例如，一个 3 kg 的球以 4 m s⁻¹ 运动，撞上一个 2 kg 静止的球；碰撞后 3 kg 的球以 1 m s⁻¹ 沿原方向运动。由动量守恒得 v₂ = 4.5 m s⁻¹。碰撞前动能为 24 J，碰撞后为 1.5 + 20.25 = 21.75 J，因此损失了 2.25 J。
5. Perfectly Inelastic Collisions | 完全非弹性碰撞
A perfectly inelastic collision is a special case where the colliding bodies stick together and move with a common velocity after impact. This yields the maximum possible kinetic energy loss.
完全非弹性碰撞是一种特殊情况，碰撞物体粘在一起并以共同速度运动。此时动能损失最大。
The momentum equation simplifies to m₁ u₁ + m₂ u₂ = (m₁ + m₂) v, where v is the final common velocity. Kinetic energy is not conserved; the ‘lost’ kinetic energy is usually the largest for a given set of initial conditions.
动量方程简化为 m₁ u₁ + m₂ u₂ = (m₁ + m₂) v，其中 v 是最终的共同速度。动能不守恒；“损失”的动能通常是在给定初始条件下最大的。
A classic example is a bullet embedding itself in a block. Momentum gives the speed of the block+bullet immediately after impact; energy calculations can then find how high a ballistic pendulum rises.
经典例子是子弹射入木块并留在其中。动量给出碰撞后木块+子弹的共同速度；随后用能量计算可以求出弹道摆上升的高度。
6. Impulse and Change in Momentum | 冲量与动量变化
Impulse J is defined as the product of the average force F and the time Δt for which it acts. It equals the change in momentum: J = F Δt = Δp = m v_final − m v_initial. Impulse is also a vector.
冲量 J 定义为平均力 F 与其作用时间 Δt 的乘积。它等于动量的变化：J = F Δt = Δp = m v_末 − m v_初。冲量也是矢量。
When a tennis racket strikes a ball, the large force over a very short time changes the ball’s momentum drastically. The same impulse can be achieved with a small force over a long time (like crumple zones in cars) or a large force over a short time (like a golf club strike). This is why airbags increase the stopping time to reduce the force on the driver.
当网球拍击球时，巨大的力在极短时间内显著改变了球的动量。同样的冲量可以用较小的力作用较长时间（如汽车的溃缩区）来实现，也可以用较大的力作用很短时间（如高尔夫球杆的击球）。这就是为什么安全气囊通过延长停止时间来减小对驾驶员的作用力。
Units: N s, which is equivalent to kg m s⁻¹. Exam questions often ask for the average force given the contact time and change in velocity.
单位：N s，与 kg m s⁻¹ 等效。考题常给出接触时间和速度变化，要求计算平均力。
7. Area Under Force-Time Graphs | 力-时间图像下的面积
The impulse delivered by a varying force is found from the area under a force-time graph. For a constant force, the area is simply F × Δt. When the force varies, the area must be calculated using squares, triangles or rectangles.
变力产生的冲量可由力-时间图像下的面积求出。对于恒力，面积即为 F × Δt。当力变化时，面积需要用数格子、三角形或矩形的方法来计算。
An OCR question might show a graph of force against time for a football kick. The peak force and contact time give an area that equals the impulse, from which the change in momentum and final velocity can be deduced. Remember to check the scales carefully.
OCR 考题可能给出足球被踢时的力-时间图像。峰值力和接触时间所对应的面积等于冲量，由此可推出动量变化和末速度。记得仔细看清坐标轴的比例。
| Graph feature | Impulse interpretation |
|---|---|
| Rectangle | F Δt (constant force) |
| Triangle | ½ F_max Δt |
| Trapezium / curve | Count squares or use integration idea |
| 图像特征 | 冲量解读 |
|---|---|
| 矩形 | F Δt（恒力） |
| 三角形 | ½ F_max Δt |
| 梯形 / 曲线 | 数格子或使用积分思想 |
8. Newton’s Second Law in Terms of Momentum | 牛顿第二定律的动量形式
Newton originally stated his second law in terms of momentum: the resultant force equals the rate of change of momentum. F = Δp / Δt. This formulation is more general than F = m a because it holds even when mass changes (e.g., rockets losing fuel mass).
牛顿最初用动量来表述第二定律：合外力等于动量的变化率。F = Δp / Δt。这种形式比 F = m a 更普遍，因为即使质量改变（如火箭损失燃料质量）也成立。
For constant mass, F = m × (Δv / Δt) = m a, which is the familiar form. In exam questions about forces acting over a short interval, apply F = Δp / Δt directly, using the impulse concept.
对于质量不变的情况，F = m × (Δv / Δt) = m a，这就是我们熟悉的形式。在涉及短时间内力的作用的考题中，可直接应用 F = Δp / Δt，利用冲量的概念。
A typical application: water hitting a wall. If water of mass m arrives horizontally with speed v and falls vertically after impact, the horizontal momentum change is m v (since final horizontal momentum is zero). The force on the wall is F = Δp / Δt = (m v) / Δt.
一个典型应用：水柱冲击墙壁。如果质量为 m 的水以水平速度 v 撞击墙壁后垂直落下，水平动量变化就是 m v（因为最终水平动量为零）。墙所受的力为 F = Δp / Δt = (m v) / Δt。
9. Solving One-Dimensional Collision Problems | 一维碰撞问题解析
Most OCR momentum questions involve motion along a straight line. The method is systematic: (1) Draw a diagram and label masses and velocities. (2) Choose a positive direction and assign +/− signs. (3) Write the momentum conservation equation. (4) If the collision is elastic, also write the kinetic energy equation or use relative speed of approach = relative speed of separation. (5) Solve the equations.
大多数 OCR 动量题目涉及直线运动。解题步骤系统化：(1) 画出示意图，标出质量和速度。(2) 选定正方向，为速度赋予正负号。(3) 写出动量守恒方程。(4) 如果碰撞是弹性的，还要写出动能方程，或利用接近相对速度 = 分离相对速度。(5) 解方程。
Example: A 2 kg trolley moving at +3 m s⁻¹ collides elastically with a stationary 1 kg trolley. Find the final velocities. Momentum: 2×3 + 0 = 2v₁ + 1 v₂ → 6 = 2v₁ + v₂. Elastic condition: 3 − 0 = v₂ − v₁ → v₂ = 3 + v₁. Substitute gives 6 = 2v₁ + 3 + v₁ → 3v₁ = 3 → v₁ = 1 m s⁻¹, v₂ = 4 m s⁻¹. The 2 kg trolley slows down and the 1 kg moves off faster.
例题：一辆 2 kg 小车以 +3 m s⁻¹ 运动，与一辆静止的 1 kg 小车发生弹性碰撞。求末速度。动量：2×3 + 0 = 2v₁ + 1v₂ → 6 = 2v₁ + v₂。弹性条件：3 − 0 = v₂ − v₁ → v₂ = 3 + v₁。代入得 6 = 2v₁ + 3 + v₁ → 3v₁ = 3 → v₁ = 1 m s⁻¹，v₂ = 4 m s⁻¹。2 kg 小车减速，1 kg 小车以更快速度离去。
10. Introduction to Two-Dimensional Collision Problems | 二维碰撞问题简介
When particles collide and move off at angles, momentum must be conserved independently in perpendicular directions (usually horizontal x and vertical y). Resolve velocities into components, then apply conservation of momentum in x and y separately.
当粒子碰撞后以一定角度运动时，动量必须在相互垂直的方向（通常为水平 x 和垂直 y 方向）上分别守恒。将速度分解为分量，然后分别对 x 和 y 方向应用动量守恒。
If a snooker ball strikes a stationary identical ball and they go off at angles, the equations are: m u = m v₁ cosθ₁ + m v₂ cosθ₂ (x-direction) and 0 = m v₁ sinθ₁ − m v₂ sinθ₂ (y-direction, assuming symmetry). OCR exams occasionally feature simple two-dimensional conservation, sometimes involving a glancing collision.
如果一颗台球撞击另一颗相同的静止球，而后两者以一定角度散开，方程为：m u = m v₁ cosθ₁ + m v₂ cosθ₂（x 方向）和 0 = m v₁ sinθ₁ − m v₂ sinθ₂（y 方向，假设对称）。OCR 考试偶尔会出现简单的二维守恒问题，有时涉及偏转碰撞。
A useful check: in an elastic collision between equal masses, if one is initially at rest, the angle between the two outgoing paths is 90°. This is a classic result worth remembering.
一个有用的检验：两相等质量的物体发生弹性碰撞，若一物体最初静止，则两物体运动方向的夹角为 90°。这是一个值得记忆的经典结果。
11. Momentum Conservation in Explosions | 爆炸现象中的动量守恒
Explosions are reversed collisions: a single object breaks into fragments. If the object is initially at rest, the total initial momentum is zero. By conservation, the vector sum of the fragment momenta must also be zero.
爆炸是碰撞的逆过程：单个物体分裂成碎片。如果物体最初静止，总初动量为零。根据守恒定律，碎片动量的矢量和也必为零。
For two fragments: 0 = m₁ v₁ + m₂ v₂, so m₁ v₁ = − m₂ v₂. The fragments move in opposite directions, with speeds inversely proportional to their masses. Heavier fragment gets smaller speed.
对于两个碎片：0 = m₁ v₁ + m₂ v₂，因此 m₁ v₁ = − m₂ v₂。碎片向相反方向运动，速度大小与质量成反比。较重的碎片获得较小的速度。
A common exam scenario: a stationary cannon fires a cannonball. The cannon recoils backwards. Momentum conservation gives m_cannon v_cannon = − m_ball v_ball. The resulting kinetic energies are not equal; energy comes from the chemical explosion.
常见考试情景：一门静止的大炮发射炮弹。大炮向后反冲。动量守恒给出 m_炮 v_炮 = − m_炮弹 v_炮弹。由此产生的动能并不相等；能量来自化学爆炸。
12. Typical Exam Questions and Tips | 典型考题与解题技巧
OCR tends to blend momentum with energy, projectiles and materials. Here are key tips: (1) Always state the direction of momentum with a sign or compass bearing. (2) Distinguish between speed and velocity—momentum needs velocity. (3) When checking if a collision is elastic, calculate total KE, not just compare speeds. (4) In impulse problems, use the area method or F_avg Δt as appropriate. (5) For graph problems, carefully read units on axes and check time intervals. (6) If asked to explain safety features, link increased stopping time to reduced force via F = Δp / Δt.
OCR 考试常常将动量与能量、抛体运动和材料结合起来。以下是重要技巧：(1) 始终用正负号或方位表示动量的方向。(2) 区分速率和速度——动量需要速度。(3) 判断碰撞是否为弹性时，要计算总动能，而不是仅仅比较速率。(4) 在冲量问题中，酌情使用面积法或平均力 F_avg Δt。(5) 对于图像问题，仔细读取坐标轴的单位并检查时间间隔。(6) 如果要求解释安全特性，通过 F = Δp / Δt 将延长的停止时间与减小的力联系起来。
Finally, keep your working clear, showing the conservation equation in symbolic form before plugging in numbers. This minimises sign errors and ensures method marks even if arithmetic goes wrong.
最后，保持解题过程清晰，先写出符号形式的守恒方程，再代入数字。这样能最大限度地减少正负号错误，并且即使计算有误，也能确保得到方法分。
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