📚 Collisions in Two Dimensions | 二维碰撞
When two objects collide, they don't always move along the same straight line before or after the impact. On pool tables, in particle physics experiments, and in everyday traffic accidents, collisions frequently occur in two dimensions. In your CIE A-Level Physics course, extending momentum conservation to two dimensions is a key skill that builds on your understanding of vectors. You will learn to resolve velocities into perpendicular components, apply conservation of momentum separately along each axis, and analyse both elastic and inelastic interactions using the coefficient of restitution.
当两个物体发生碰撞时，它们在碰撞前后并不总是沿同一直线运动。在台球桌上、粒子物理实验中以及日常交通事故里，碰撞经常发生在二维空间。在 CIE A-Level 物理课程中，将动量守恒扩展到二维是建立在矢量理解基础上的一项关键技能。你将学习如何将速度分解为互相垂直的分量，沿每个坐标轴分别应用动量守恒定律，并利用恢复系数分析弹性和非弹性碰撞。
1. Why Two-Dimensional Collisions Matter | 二维碰撞为何重要
One-dimensional collisions simplify the mathematics because all velocities lie along a single axis. In reality, glancing blows produce motion in a plane. Understanding two-dimensional collisions is essential for analysing car crashes where vehicles rebound at angles, for predicting the paths of snooker balls after a cue strike, and for interpreting data from cloud chambers or particle detectors. The same vector principles apply, but you must be methodical about choosing coordinate axes and resolving quantities correctly.
一维碰撞之所以简化了数学处理，是因为所有速度都沿单一轴线。但在实际情况中，斜碰会在平面内产生运动。理解二维碰撞对于分析车辆以一定角度反弹的交通事故、预测台球受击后的行进路径，以及解读云室或粒子探测器的数据都至关重要。同样的矢量原理依然适用，但你必须系统地选择坐标轴并正确分解各个物理量。
CIE exam questions often present a known incident velocity and ask you to find unknown final speeds and directions. They may provide some angles, or you may need to deduce an angle from kinetic energy conditions. This topic reinforces your ability to handle vectors, trigonometry, and simultaneous equations under pressure.
CIE 考试题目通常会给出已知的入射速度，要求你求出未知的末速度和方向。题目可能提供部分角度，也可能需要你根据动能条件推导出某个角度。这一主题能够强化你在压力下处理矢量、三角学和联立方程的能力。
2. Conservation of Momentum in Vector Form | 动量守恒的矢量形式
The total momentum of a closed system remains constant in both magnitude and direction. For two colliding bodies A and B, the vector sum of their momenta before the collision equals the vector sum after:
封闭系统的总动量在大小和方向上均保持恒定。对于两个碰撞物体 A 和 B，它们碰撞前动量的矢量和等于碰撞后动量的矢量和：
mₐuₐ + m₆u₆ = mₐvₐ + m₆v₆
Because momentum is a vector, this single equation can be split into two independent scalar equations along perpendicular axes, typically chosen as the x-axis (horizontal) and the y-axis (vertical). Choosing a convenient orientation—such as aligning one initial velocity with the x-axis—greatly reduces the algebra involved.
由于动量是矢量，这一方程可以沿互相垂直的坐标轴分解为两个独立的标量方程，通常选取 x 轴（水平）和 y 轴（垂直）。选择一个方便的方位——例如让某一初速度与 x 轴重合——可以大大简化代数运算。
Along x: mₐuₐₓ + m₆u₆ₓ = mₐvₐₓ + m₆v₆ₓ
Along y: mₐuₐᵧ + m₆u₆ᵧ = mₐvₐᵧ + m₆v₆ᵧ
Here the subscripts x and y denote the components of each velocity. Remember that if an angle is measured from the x-axis, the component is v cosθ for the adjacent side and v sinθ for the opposite side. Consistent sign conventions are vital: a velocity component pointing in the negative x or negative y direction must be assigned a negative value.
这里下标 x 和 y 表示各个速度的分量。请记住，若角度是从 x 轴量起的，则邻边分量为 v cosθ，对边分量为 v sinθ。符号一致性至关重要：指向负 x 或负 y 方向的速度分量必须赋予负值。
3. Elastic and Inelastic Collisions in Two Dimensions | 二维的弹性与非弹性碰撞
In an elastic collision, both momentum and kinetic energy are conserved. This gives an additional constraint that often allows you to find a relationship between final speeds and the scattering angle. For two spheres of equal mass, an elastic glancing collision in which one sphere is initially at rest results in the two final velocities being at right angles to each other. This elegant result is frequently tested.
在弹性碰撞中，动量和动能均守恒。这提供了一个额外的约束条件，通常能让你找到末速度与散射角之间的关系。对于两个质量相等的球体，若其中一个最初静止且发生弹性斜碰，则两个末速度的方向互相垂直。这一优美的结论经常被考查。
In an inelastic collision, kinetic energy is not conserved—some is converted to heat, sound, or permanent deformation. The extent of elasticity is quantified by the coefficient of restitution, e, which in two dimensions applies to the velocity components along the line of impact (the line joining the centres at the instant of collision). The relative speed of separation along this line equals e times the relative speed of approach. Components perpendicular to the line of impact remain unchanged for smooth spheres.
在非弹性碰撞中，动能不再守恒——部分能量转化为热能、声能或永久形变。弹性程度由恢复系数 e 来量化，在二维情况下，该系数适用于沿碰撞作用线（碰撞瞬间连接两物体中心的直线）的速度分量。沿该线的分离相对速率等于 e 乘以接近相对速率。对于光滑球体，垂直于作用线的分量保持不变。
Thus, even if overall momentum is conserved vectorially, you may need to rotate your axes so that one axis lies along the line of impact, apply the restitution equation, and then rotate back to the original reference frame. This technique is beyond the simplest CIE problems but appears in advanced questions.
因此，即使整体的动量在矢量上守恒，你可能需要旋转坐标轴，使一个轴沿着碰撞作用线，应用恢复系数方程，然后再转回原参考系。这一技巧超出了最简单的 CIE 题目范围，但会出现在较高级别的问题中。
4. Choosing and Resolving Components | 选择与分解分量
A systematic approach to any two-dimensional collision problem begins with a clear diagram. Draw the velocity vectors before and after the collision, label the masses, and indicate all given angles. Select a sensible pair of perpendicular axes: commonly the x-axis is taken along the initial direction of one particle, and the y-axis is perpendicular to it.
解决任何二维碰撞问题的系统方法都始于清晰的示意图。画出碰撞前后的速度矢量，标出质量，并注明所有已知角度。选择一对合理的垂直坐标轴：通常将 x 轴沿某一粒子初速度方向选取，y 轴与之垂直。
For each velocity, write its x-component as the product of the speed and the cosine of the angle it makes with the positive x-axis. The y-component is the speed multiplied by the sine of that angle, with attention to the sign. If a velocity points below the x-axis, its y-component is negative. Using a consistent sign convention avoids errors when substituting into the conservation equations.
对于每个速度，将其 x 分量写为速率与其与正 x 轴夹角余弦的乘积。y 分量为速率乘以该角的正弦，并注意符号。如果某速度指向 x 轴下方，其 y 分量为负。保持一致的符号约定可避免代入守恒方程时出错。
Example: A ball of mass 2.0 kg moving at 5.0 m s⁻¹ along the positive x-axis strikes a stationary ball of mass 1.0 kg. After collision, the 2.0 kg ball moves at 3.0 m s⁻¹ at 30° above the x-axis. Find the x- and y-components of the second ball's velocity.
示例：一质量为 2.0 kg 的球以 5.0 m s⁻¹ 沿正 x 轴运动，撞击一质量为 1.0 kg 的静止球。碰撞后，2.0 kg 球以 3.0 m s⁻¹ 的速度沿与 x 轴夹角 30° 的方向运动。求第二个球速度的 x 和 y 分量。
Initial x-momentum: (2.0×5.0) + (1.0×0) = 10 kg m s⁻¹
初始 x 方向动量：(2.0×5.0) + (1.0×0) = 10 kg m s⁻¹
Final x-momentum: (2.0×3.0 cos30°) + (1.0×v₂ₓ) = 10 → 5.196 + v₂ₓ = 10 → v₂ₓ = 4.804 m s⁻¹
末态 x 方向动量：(2.0×3.0 cos30°) + (1.0×v₂ₓ) = 10 → 5.196 + v₂ₓ = 10 → v₂ₓ = 4.804 m s⁻¹
so final y-momentum: (2.0×3.0 sin30°) + (1.0×v₂ᵧ) = 0 → 3.0 + v₂ᵧ = 0 → v₂ᵧ = -3.0 m s⁻¹.
初始 y 方向动量：0，所以末态 y 方向动量：(2.0×3.0 sin30°) + (1.0×v₂ᵧ) = 0 → 3.0 + v₂ᵧ = 0 → v₂ᵧ = -3.0 m s⁻¹。
The negative sign indicates that the second ball moves below the x-axis, which makes sense: the y-momentum must sum to zero.
负号表明第二个球向 x 轴下方运动，这是合理的：y 方向总动量必须为零。
5. Solving for Unknown Speeds and Angles | 求解未知速率与角度
lision. If the collision is elastic, the kinetic energy equation provides a third relationship. The typical method is to isolate terms, square and add the momentum equations to eliminate an angle, or to use trigonometric identities such as sin²θ + cos²θ = 1.
在列出两个动量分量方程后，你通常会有两个未知数，例如碰撞后某个粒子的速率和方向。如果碰撞是弹性的，动能方程就提供了第三个关系式。典型的方法是分离项，将动量方程平方后相加以消去角度，或利用三角恒等式如 sin²θ + cos²θ = 1。
Consider an elastic collision between two equal masses where one is initially at rest. Let the incident mass have speed u, and after collision the two masses have speeds v₁ and v₂ at angles θ and φ relative to the original direction. Momentum conservation gives:
考虑两个相等质量之间的弹性碰撞，其中一个最初静止。设入射球速率为 u，碰撞后两球速率分别为 v₁ 和 v₂，与初始方向夹角分别为 θ 和 φ。动量守恒给出：
Along x: m u = m v₁ cosθ + m v₂ cosφ → u = v₁ cosθ + v₂ cosφ
Along y: 0 = m v₁ sinθ - m v₂ sinφ → v₁ sinθ = v₂ sinφ
Kinetic energy conservation for equal masses simplifies to:
等质量情况下的动能守恒可简化为：
½ m u² = ½ m v₁² + ½ m v₂² → u² = v₁² + v₂²
φ = 90°. This proves that the two final velocities are perpendicular. This is a classic result that you may quote if justified, but deriving it shows deep understanding.
将动量方程组平方后相加得到 u² = v₁² + v₂² + 2 v₁ v₂ cos(θ + φ)。与动能结果比较，必然有 cos(θ + φ) = 0，因此 θ + φ = 90°。这就证明了两个末速度相互垂直。这是一个经典结论，在有说明的情况下可直接引用，但若能推导将体现深刻的理解。
6. Working with the Coefficient of Restitution in 2D | 二维情况下的恢复系数
speeds along the line of impact. The line of impact is the line joining the centres of the two colliding bodies at the moment of contact. Velocities are resolved into components parallel and perpendicular to this line. Only the parallel components change according to e; the perpendicular components remain unaltered for smooth surfaces.
对于非完全弹性碰撞，恢复系数 e（0 ≤ e ≤ 1）将碰撞作用线上的相对速率联系起来。作用线是碰撞瞬间连接两物体中心的直线。速度被分解为平行和垂直于该作用线的分量。只有平行分量依 e 发生变化，对于光滑表面，垂直分量保持不变。
The restitution equation along the line of impact is:
沿作用线的恢复方程是：
v₂_parallel - v₁_parallel = e (u₁_parallel - u₂_parallel)
the pre-collision components. The signs must be consistent: if the line of impact is taken as positive from body 1 to body 2, then approaching means u₁_parallel > u₂_parallel.
此处 v₁_parallel 和 v₂_parallel 是物体 1 和物体 2 碰撞后沿作用线的分量，u₁_parallel、u₂_parallel 是碰撞前的分量。符号必须一致：若取从物体 1 指向物体 2 为作用线的正方向，则接近意味着 u₁_parallel > u₂_parallel。
To solve such problems, rotate the coordinate axes so that the x-axis lies along the line of impact. Apply the restitution equation to the x-components, and note that y-components do not change. After finding all components in the rotated frame, rotate back to the laboratory frame to report final velocities and directions.
解决此类问题时，旋转坐标轴使 x 轴沿作用线方向。对 x 分量应用恢复方程，同时注意 y 分量不变。在旋转坐标系中求出所有分量后，再转回实验室参考系，给出最终的末速度和方向。
7. Special Case: Glancing Collision with a Fixed Target | 特殊情况：与固定靶的斜碰
When a light particle collides with a much heavier stationary target, the heavy object barely moves. This is approximated by a particle bouncing off a fixed surface. In two dimensions, if a ball strikes a smooth wall at an angle, the component of velocity perpendicular to the wall reverses and is multiplied by e, while the component parallel to the wall remains unchanged. This gives the law of reflection modified by e.
当一个轻粒子与一个重得多的静止靶碰撞时，重物几乎不动。这可以近似为一个粒子从固定表面弹开。在二维情况下，若球以一定角度撞击光滑墙壁，垂直于墙壁的速度分量反向并乘以 e，而平行于墙壁的分量保持不变。这就给出了由 e 修正后的反射定律。
angle of incidence equals the angle of reflection. If e < 1, the perpendicular component decreases, causing the rebound angle to be larger than the incidence angle (measured from the normal). This situation is common when analysing the bounce of a ball off a court or a particle off a detector plane, and it serves as a good introduction to vector components in collisions.
若 e = 1，速率不变且入射角等于反射角。若 e < 1，垂直分量减小，导致反弹角（从法线量起）大于入射角。这种情况在分析球从场地弹起或粒子从探测器平面反弹时很常见，是碰撞中矢量分量知识的一个良好入门。
8. Using Energy to Determine Elasticity | 利用能量确定弹性
After finding all final velocities using momentum and the given angles, you can check whether the collision is elastic by comparing the total kinetic energy before and after. If the kinetic energy is the same (within rounding errors), the collision is elastic. If some kinetic energy is lost, calculate the fraction lost or the coefficient of restitution from the velocity components along the line of impact.
在利用动量和已知角度求出所有末速度后，你可以通过比较碰撞前后的总动能来检查碰撞是否为弹性碰撞。若动能相同（在舍入误差内），则碰撞是弹性的。若有动能损失，则计算损失的比例，或根据沿作用线的速度分量求出恢复系数。
CIE questions sometimes ask: 'State whether the collision is elastic, justifying your answer with a calculation.' = 0), the two bodies stick together and move with a common velocity after the collision. In such cases, the direction of the combined mass is found from the vector sum of the initial momenta.
CIE 题目有时会问：“判断该碰撞是否为弹性碰撞，并用计算说明理由。” 务必计算初始动能和末态动能并进行比较。哪怕只有微小的差异也意味着非弹性。请记住，在完全非弹性碰撞（e = 0）中，两物体粘在一起并以共同速度运动。在这种情况下，复合体的运动方向由初始动量的矢量和决定。
9. Worked Example: Elastic Collision between Equal Masses | 例题：等质量弹性碰撞
A snooker ball of mass 0.15 kg moves at 2.0 m s⁻¹ and strikes a stationary ball of the same mass. After the collision, the first ball moves at 1.2 m s⁻¹ at an angle of 35° to its original direction. Determine the velocity (speed and direction) of the second ball and verify the collision is elastic.
一个质量为 0.15 kg 的台球以 2.0 m s⁻¹ 的速度运动，并撞击一个同质量的静止球。碰撞后，第一个球以 1.2 m s⁻¹ 的速率沿与原方向成 35° 角的方向运动。求第二个球的速度（速率和方向），并验证碰撞为弹性碰撞。
Solution: Let original direction be +x. Conserve momentum in x and y directions.
解：设初始方向为 +x 方向。在 x 和 y 方向上守恒动量。
x: 0.15×2.0 = 0.15×1.2 cos35° + 0.15×v₂ₓ → 2.0 = 0.983 + v₂ₓ → v₂ₓ = 1.017 m s⁻¹
y: 0 = 0.15×1.2 sin35° + 0.15×v₂ᵧ → 0 = 0.688 + v₂ᵧ → v₂ᵧ = -0.688 m s⁻¹
Speed of second ball = √(1.017² + (-0.688)²) = √1.034 + 0.473 .15×2.0² = 0.30 J. Final KE = ½×0.15×(1.2² + 1.23²) = 0.075×(1.44+1.51)=0.075×2.95=0.221 J? That suggests an inelastic collision? lar set of numbers is inelastic. This demonstrates that not all glancing collisions are elastic. For an elastic collision, the speeds would be different. In fact, for equal masses elastic with one initially stationary, final speeds must satisfy v₁²+v₂²=u² and θ+φ=90°. Here 35°+34.1°=69.1°, not 90°, confirming inelastic. Good check.
第二个球的速率 = √(1.017² + (-0.688)²) ≈ 1.23 m s⁻¹。角度 φ = tan⁻¹(0.688/1.017) ≈ 34.1°，在 x 轴下方。检查动能：初始动能 = ½×0.15×2.0² = 0.30 J。末态动能 = 0.075×(1.44+1.51)=0.221 J，存在动能损失，因此这组数据对应的是非弹性碰撞。这正说明并非所有斜碰都是弹性的。对于等质量一球静止的弹性碰撞，末速率必须满足 v₁²+v₂²=u² 且 θ+φ=90°。这里 35°+34.1°=69.1°，不是90°，证实了这一点。很好的检验。
10. Common Pitfalls and How to Avoid Them | 常见误区与避免方法
One major pitfall is forgetting that momentum is a vector. Students sometimes write a single scalar equation summing momenta without regard to direction. Always draw a vector diagram and label components. Another mistake is incorrect sign allocation: a velocity pointing in the negative y-direction must appear with a minus sign in the y-momentum equation. Mixing up sine and cosine for component resolution is also frequent—remember: cos for the adjacent component, sin for the opposite.
一个主要误区是忘记动量是矢量。学生有时会列出单个标量方程求动量和，却忽略了方向。务必画矢量图并标出分量。另一个错误是符号分配不当：指向负 y 方向的速度在 y 动量方程中必须带负号。混淆正弦和余弦来做分量分解也很常见——记住：邻边用 cos，对边用 sin。
When using the restitution equation, many learners apply it to the overall speeds rather than the components along the line of impact. This will lead to incorrect results unless the collision is head-on. Ensure you identify the line of impact and resolve velocities accordingly. Finally, check the consistency of your answers: final kinetic energy cannot exceed initial kinetic energy; angles must satisfy the geometry; and speeds must be real positive numbers.
在使用恢复系数方程时，许多学习者将其应用于总速率而非沿作用线的分量。除非是对心碰撞，否则这将导致错误结果。务必确认作用线并相应地分解速度。最后，检查答案的一致性：末态动能不能超过初始动能；角度必须满足几何关系；速率必须是正的实数。
Exam questions may provide more data than needed: if given an angle, a speed, and a KE condition, you might have enough to solve without the restitution formula. Be prepared to select the most efficient route. Practising a variety of problems—elastic, inelastic, sticking collisions, and wall rebounds—will build the intuition to recognise which principles apply immediately.
考试题目可能会提供多于实际需要的数据：如果给出了角度、速率和动能条件，你可能无需使用恢复系数公式即可求解。请准备好选择最高效的路径。通过练习各种问题——弹性的、非弹性的、粘合碰撞以及墙壁反弹——可以培养直觉，迅速识别出应适用哪些原理。
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