Collisions in Two Dimensions | 二维碰撞

📚 Collisions in Two Dimensions | 二维碰撞

In A-Level Physics, most collision problems in one dimension are solved by applying conservation of momentum with simple positive and negative signs. However, real collisions rarely happen along a straight line. When two objects collide at an angle, or when a ball strikes a wall obliquely, the velocities change direction as well as magnitude. These are collisions in two dimensions, and they require us to treat momentum as a vector quantity by resolving it into perpendicular components.

在A-Level物理中，一维碰撞问题通常通过应用动量守恒并使用正负号即可解决。然而，真实碰撞很少沿直线发生。当两个物体以一定角度碰撞，或当球斜撞墙壁时，速度的方向和大小都会同时改变。这就是二维碰撞，我们需要将动量视为矢量，通过将其分解为垂直分量来处理。

1. The Vector Nature of Momentum | 动量的矢量性

Momentum is defined as the product of mass and velocity: p = m v. Since velocity is a vector, momentum is also a vector. It has both magnitude and direction. In one-dimensional problems, we handle direction by assigning positive and negative signs. In two dimensions, a single sign is no longer sufficient.

动量的定义为质量与速度的乘积：p = m v。由于速度是矢量，动量也是矢量，既有大小也有方向。在一维问题中，我们通过赋予正负号来处理方向。而在二维问题中，单一的正负号已不再足够。

Suppose a ball of mass m moves with speed v at an angle θ above the horizontal axis. Its momentum components are:

设一个质量为 m 的球以速度 v 沿与水平轴成 θ 角的方向运动，其动量分量为：

pₓ = m v cos θ, p_y = m v sin θ

Here pₓ and p_y are the horizontal and vertical momentum components respectively. The magnitude of the total momentum is found using Pythagoras’ theorem: p = √(pₓ² + p_y²). This decomposition is the foundation of every two-dimensional collision calculation.

其中 pₓ 和 p_y 分别是水平方向和竖直方向的动量分量。总动量的大小可通过勾股定理求得：p = √(pₓ² + p_y²)。这种分解是处理一切二维碰撞计算的基础。

It is essential to remember that momentum is conserved only when there is no net external force acting on the system. During a collision, the internal forces between the colliding objects are large and act over a very short time interval, so the impulse due to external forces such as friction or gravity is negligible. Therefore, total momentum is conserved during the collision itself.

必须牢记：只有在系统不受合外力作用时，动量才守恒。碰撞过程中，碰撞物体之间的内力很大且作用时间极短，因此摩擦、重力等外力的冲量可以忽略不计。因此，在碰撞过程中总动量守恒。

2. Conservation of Momentum in Two Dimensions | 二维动量守恒

The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. In two dimensions, this principle applies independently to each perpendicular direction. This is because momentum components along a given axis are conserved separately when no external force acts along that axis.

动量守恒定律指出：孤立系统的总动量在碰撞前后保持不变。在二维问题中，该定律分别独立地适用于每个垂直方向。这是因为当某轴方向上没有外力作用时，该方向上的动量分量单独守恒。

For a collision between two objects A and B, we write two separate equations:

对于物体A和B之间的碰撞，我们写出两个独立的方程：

Along x-axis: m_A u_Aₓ + m_B u_Bₓ = m_A v_Aₓ + m_B v_Bₓ

Along y-axis: m_A u_A_y + m_B u_B_y = m_A v_A_y + m_B v_B_y

where u represents the initial velocity components and v represents the final velocity components. Note that u and v are velocity components, not speeds. For example, if object A initially moves with speed u_A at angle α to the x-axis, then u_Aₓ = u_A cos α and u_A_y = u_A sin α.

其中 u 表示初速度分量，v 表示末速度分量。注意 u 和 v 是速度分量而非速率。例如，若物体A以速率 u_A 沿与x轴成 α 角的方向运动，则 u_Aₓ = u_A cos α，u_A_y = u_A sin α。

Because we have two independent equations, a two-dimensional collision problem may involve up to two unknown quantities. Typically, these unknowns are the final speed and direction of one of the objects, or the two final speed components of a single object. The exam often provides enough data to solve for these using simultaneous equations.

由于我们有两个独立的方程，二维碰撞问题最多可以包含两个未知量。通常，未知量是某个物体的末速率和方向，或单个物体末速度的两个分量。考试通常会提供足够的数据，以便通过联立方程求解这些量。

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

Collisions are classified as elastic or inelastic according to whether kinetic energy is conserved. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not — some of it is transformed into heat, sound, or deformation energy. A perfectly inelastic collision is one in which the two objects stick together and move with a common velocity.

碰撞根据动能是否守恒分为弹性碰撞和非弹性碰撞。在弹性碰撞中，动量和动能均守恒。在非弹性碰撞中，动量守恒但动能不守恒——部分动能转化为热能、声能或形变能。完全非弹性碰撞是指两物体碰撞后粘在一起，以共同速度运动。

In two dimensions, the same distinction applies. To test whether a collision is elastic, calculate the total kinetic energy before and after the collision. The kinetic energy of an object is a scalar:

在二维碰撞中，同样的区分依然适用。要判断碰撞是否为弹性碰撞，需计算碰撞前后系统的总动能。物体的动能是标量：

Eₖ = ½ m v²

where v is the speed — the magnitude of the velocity vector. Therefore Eₖ = ½ m (vₓ² + v_y²). You must use speeds, not velocity components with signs, because kinetic energy has no direction.

其中 v 是速率，即速度矢量的大小。因此 Eₖ = ½ m (vₓ² + v_y²)。你必须使用速率而非带符号的速度分量，因为动能没有方向。

A common exam question gives the masses and velocities of two objects before and after a glancing collision and asks whether the collision is elastic. The approach is straightforward: compute the total kinetic energy before the collision, compute the total kinetic energy after the collision, and compare. If they are equal (within experimental precision), the collision is elastic.

常见的考题会给出两个物体在斜碰前后的质量和速度，并要求判断碰撞是否为弹性碰撞。方法很直接：计算碰撞前的总动能，计算碰撞后的总动能，然后进行比较。若二者相等（在实验精度范围内），则为弹性碰撞。

4. Resolving Momentum into Perpendicular Components | 将动量分解为垂直分量

The key technique in two-dimensional collision problems is resolution. Every velocity vector is resolved into two perpendicular components — conventionally along the x-axis and y-axis. The conservation of momentum is then applied separately along each axis.

解决二维碰撞问题的关键技巧是分解。将每个速度矢量分解为两个垂直分量——通常沿x轴和y轴。然后分别沿每个轴应用动量守恒。

The procedure is as follows. First, draw a clear diagram showing the objects before and after the collision, including all velocity arrows and angles. Second, resolve every velocity into x and y components. Third, apply conservation of momentum along the x-axis to obtain one equation. Fourth, apply conservation of momentum along the y-axis to obtain a second equation. Finally, solve the simultaneous equations for the unknown quantities.

步骤如下：首先，画出清晰的示意图，标明碰撞前后各物体的速度箭头和角度。其次，将每个速度分解为x分量和y分量。第三，沿x轴应用动量守恒，得到一个方程。第四，沿y轴应用动量守恒，得到第二个方程。最后，联立求解未知量。

It is crucial to maintain a consistent sign convention. For example, if you take the positive x-direction as the direction of the incident object’s initial motion, then any component pointing in the opposite direction must carry a negative sign. Many students lose marks because they ignore the sign of a component when writing the conservation equation.

保持一致的符号约定至关重要。例如，若取入射物体初始运动方向为正x方向，则任何指向相反方向的分量都必须加负号。许多学生因为在写守恒方程时忽略了分量的符号而失分。

When using a coordinate system, choose axes that simplify the problem. Often, aligning the x-axis with the initial direction of motion of one object eliminates one component — the initial y-component of that object is zero. This reduces the amount of algebra considerably.

使用坐标系时，选择能够简化问题的坐标轴。通常，将x轴与某个物体的初始运动方向对齐，可以消去该物体的一个初始y分量，从而大大减少代数运算量。

5. Oblique Collisions with a Wall | 与墙壁的斜碰撞

A classic two-dimensional collision problem is a ball striking a smooth wall at an angle. When a ball hits a smooth wall, the wall exerts a normal reaction force perpendicular to its surface. Since the wall is smooth, there is no friction, so no force acts parallel to the wall’s surface.

一个经典的二维碰撞问题是球斜撞光滑墙壁。当球撞击光滑墙壁时，墙壁施加以垂直于其表面的法向反作用力。由于墙面光滑，不存在摩擦力，因此没有平行于墙面方向的力。

Consequently, the component of the ball’s momentum parallel to the wall is unchanged during the collision. If the collision with the wall is elastic, the component of velocity perpendicular to the wall is reversed in direction with the same magnitude. If the collision is inelastic, the perpendicular component is reduced by a factor known as the coefficient of restitution.

因此，球的动量在平行于墙壁方向的分量在碰撞过程中保持不变。若球与墙的碰撞是弹性的，则垂直于墙面的速度分量方向反转但大小不变。若是非弹性碰撞，则垂直分量按恢复系数的大小减小。

Consider a ball of mass m moving with speed v striking a wall at an angle θ to the normal. The component of velocity perpendicular to the wall is v cos θ, and the component parallel to the wall is v sin θ. After an elastic rebound, the perpendicular component is −v cos θ, while the parallel component remains v sin θ.

考虑一个质量为 m 的球以速率 v 沿与法线成 θ 角的方向撞击墙壁。垂直于墙面的速度分量为 v cos θ，平行于墙面的速度分量为 v sin θ。弹性反弹后，垂直分量变为 −v cos θ，而平行分量保持 v sin θ 不变。

The change in momentum is therefore double the perpendicular component:

因此动量的变化量等于垂直分量的两倍：

Δp = 2 m v cos θ

This is a very common exam result. Note that the angle in the formula is measured with respect to the normal, not the wall surface. If the angle to the wall is given, you must convert: angle to the normal = 90° − angle to the wall.

这是一个非常常见的考试结论。注意公式中的角度是相对于法线而非墙面测量的。如果题目给出的是与墙面的夹角，你必须换算：与法线的夹角 = 90° − 与墙面的夹角。

6. Collisions Between Two Moving Objects | 两个运动物体之间的碰撞

When two objects collide and then move off in different directions, we must apply conservation of momentum along two perpendicular axes simultaneously. This is the most general type of two-dimensional collision problem in the CIE syllabus.

当两个物体碰撞后沿不同方向运动时，我们必须同时沿两个垂直轴应用动量守恒。这是CIE考纲中最一般的二维碰撞问题类型。

Take the x-axis to be the direction of the first object’s initial motion. Suppose object A of mass m_A moves initially with speed u_A along the x-axis, while object B of mass m_B is initially stationary. After the collision, A moves with speed v_A at angle θ above the axis, and B moves with speed v_B at angle φ below the axis. Conservation of momentum gives:

取x轴为第一个物体的初始运动方向。设质量为 m_A 的物体A以速率 u_A 沿x轴运动，质量为 m_B 的物体B初始静止。碰撞后，A以速率 v_A 沿与x轴上方成 θ 角的方向运动，B以速率 v_B 沿与x轴下方成 φ 角的方向运动。动量守恒给出：

x-axis: m_A u_A = m_A v_A cos θ + m_B v_B cos φ

y-axis: 0 = m_A v_A sin θ − m_B v_B sin φ

The minus sign in the y-axis equation arises because object B moves below the x-axis while object A moves above it. These two equations can be solved for two unknowns — for example, v_B and φ — provided all other quantities are known.

y轴方程中的负号是因为物体B在x轴下方运动而物体A在x轴上方运动。联立这两个方程可以解出两个未知量——例如 v_B 和 φ——前提是其他所有量均为已知。

If both objects are initially moving, the initial momentum components along each axis must both be included. For instance, if object B also has an initial velocity, the x-axis equation becomes m_A u_Aₓ + m_B u_Bₓ = m_A v_Aₓ + m_B v_Bₓ, and similarly for the y-axis.

如果两个物体初始都在运动，则每个轴上的初始动量分量都必须包含在内。例如，若物体B也有初速度，则x轴方程变为 m_A u_Aₓ + m_B u_Bₓ = m_A v_Aₓ + m_B v_Bₓ，y轴方程同理。

7. Worked Example 1: Ball Bouncing Off a Wall | 实例1：球斜撞墙壁反弹

A ball of mass 0.20 kg travels at 5.0 m/s and strikes a smooth vertical wall at an angle of 30° to the normal. It rebounds with the same speed. Calculate the magnitude of the change in momentum of the ball.

一个质量为0.20 kg的球以5.0 m/s的速率运动，沿与法线成30°角的方向撞击光滑竖直墙壁，并以相同速率反弹。求球动量变化量的大小。

Step 1 — Resolve the initial momentum into components. The component perpendicular to the wall is p_perp = m v cos θ = 0.20 × 5.0 × cos 30° = 0.866 N·s. The component parallel to the wall is p_par = m v sin θ = 0.20 × 5.0 × sin 30° = 0.500 N·s.

第一步——将初始动量分解为分量。垂直于墙面的分量为 p_垂直 = m v cos θ = 0.20 × 5.0 × cos 30° = 0.866 N·s。平行于墙面的分量为 p_平行 = m v sin θ = 0.20 × 5.0 × sin 30° = 0.500 N·s。

Step 2 — After the collision, the perpendicular component is reversed: p_perp’ = −0.866 N·s. The parallel component is unchanged: p_par’ = 0.500 N·s.

第二步——碰撞后，垂直分量反向：p_垂直’ = −0.866 N·s。平行分量不变：p_平行’ = 0.500 N·s。

Step 3 — The change in momentum is Δp = p_perp’ − p_perp = −0.866 − 0.866 = −1.732 N·s. The parallel component contributes zero change. Hence the magnitude of the change in momentum is |Δp| = 2 m v cos θ = 2 × 0.20 × 5.0 × cos 30° = 1.73 N·s.

第三步——动量变化量为 Δp = p_垂直’ − p_垂直 = −0.866 − 0.866 = −1.732 N·s。平行分量变化为零。因此动量变化量的大小为 |Δp| = 2 m v cos θ = 2 × 0.20 × 5.0 × cos 30° = 1.73 N·s。

Notice that the total momentum change is in the direction of the normal, perpendicular to the wall. There is no change of momentum in the direction parallel to the wall. This is consistent with the fact that the wall only exerts a normal force on the ball.

注意总动量变化方向沿法线方向，即垂直于墙面。平行于墙面方向没有动量变化。这与墙壁只对球施加法向力的事实一致。

8. Worked Example 2: Glancing Collision of Two Balls | 实例2：两球的斜碰

A ball A of mass 0.50 kg moves at 4.0 m/s along the x-axis and collides with a stationary ball B of mass 0.30 kg. After the collision, ball A moves at 3.0 m/s at an angle of 30° above the x-axis. Calculate the magnitude and direction of the velocity of ball B after the collision.

质量为0.50 kg的球A沿x轴以4.0 m/s运动，与质量为0.30 kg的静止球B碰撞。碰撞后，球A以3.0 m/s的速度沿x轴上方30°角方向运动。求碰撞后球B速度的大小和方向。

Step 1 — Apply conservation of momentum along the x-axis. Before the collision, only ball A has x-momentum: 0.50 × 4.0 = 2.0 N·s. After the collision, ball A has x-momentum 0.50 × 3.0 × cos 30° = 1.299 N·s. Therefore ball B must have x-momentum 2.0 − 1.299 = 0.701 N·s, so v_Bₓ = 0.701 / 0.30 = 2.34 m/s.

第一步——沿x轴应用动量守恒。碰撞前，只有球A具有x方向动量：0.50 × 4.0 = 2.0 N·s。碰撞后，球A的x方向动量为 0.50 × 3.0 × cos 30° = 1.299 N·s。因此球B的x方向动量必须为 2.0 − 1.299 = 0.701 N·s，故 v_Bₓ = 0.701 / 0.30 = 2.34 m/s。

Step 2 — Apply conservation of momentum along the y-axis. Before the collision, the total y-momentum is zero. After the collision, ball A has y-momentum 0.50 × 3.0 × sin 30° = 0.75 N·s (positive). Therefore ball B must have y-momentum −0.75 N·s, so v_B_y = −0.75 / 0.30 = −2.5 m/s. The negative sign indicates that ball B moves below the x-axis.

第二步——沿y轴应用动量守恒。碰撞前，总y方向动量为零。碰撞后，球A的y方向动量为 0.50 × 3.0 × sin 30° = 0.75 N·s（正值）。因此球B的y方向动量必须为 −0.75 N·s，故 v_B_y = −0.75 / 0.30 = −2.5 m/s。负号表示球B在x轴下方运动。

Step 3 — Combine the components to find the magnitude and direction of v_B:

第三步——合成分量，求 v_B 的大小和方向：

v_B = √(v_Bₓ² + v_B_y²) = √(2.34² + 2.5²) = √(5.48 + 6.25) = √11.73 = 3.42 m/s

φ = tan⁻¹ (2.5 / 2.34) = 46.9° below the x-axis

So ball B moves at 3.4 m/s at an angle of approximately 47° below the positive x-axis. This type of calculation — splitting a two-dimensional problem into two independent one-dimensional conservation equations — is exactly what the CIE examiner expects to see in a structured answer.

因此球B以3.4 m/s的速率沿与正x轴下方约47°角的方向运动。这种将二维问题拆分为两个独立的一维守恒方程的计算方式，正是CIE考官期望在规范性解答中看到的过程。

9. Energy Analysis in Two-Dimensional Collisions | 二维碰撞中的能量分析

After solving the momentum equations, it is often useful to check whether the collision is elastic by comparing kinetic energies. Using the previous example, the initial kinetic energy of ball A is Eₖ,initial = ½ × 0.50 × 4.0² = 4.0 J. Ball B is stationary, so its initial kinetic energy is zero.

解出动量方程之后，通常需要通过比较动能来判断碰撞是否为弹性碰撞。沿用上例，球A的初始动能为 Eₖ,初始 = ½ × 0.50 × 4.0² = 4.0 J。球B静止，因此其初始动能为零。

After the collision, ball A has kinetic energy Eₖ,A = ½ × 0.50 × 3.0² = 2.25 J. Ball B has kinetic energy Eₖ,B = ½ × 0.30 × 3.42² = 1.75 J. The total kinetic energy after the collision is 2.25 + 1.75 = 4.00 J.

碰撞后，球A的动能为 Eₖ,A = ½ × 0.50 × 3.0² = 2.25 J。球B的动能为 Eₖ,B = ½ × 0.30 × 3.42² = 1.75 J。碰撞后总动能为 2.25 + 1.75 = 4.00 J。

The total kinetic energy is the same before and after the collision, so this particular collision is perfectly elastic. In general, if the total kinetic energy after the collision is less than before, the collision is inelastic, and the difference represents energy transformed into other forms.

碰撞前后的总动能相同，因此该碰撞为完全弹性碰撞。一般来说，若碰撞后的总动能小于碰撞前，则为非弹性碰撞，差值代表转化为其他形式的能量。

When you perform an energy calculation, be careful to use the speed squared rather than summing velocity components with signs. The kinetic energy of an object moving with components vₓ and v_y is always ½ m (vₓ² + v_y²), which is equivalent to ½ m v².

进行能量计算时，务必使用速率平方，而不是将带符号的速度分量直接相加。具有分量 vₓ 和 v_y 的物体的动能始终为 ½ m (vₓ² + v_y²)，这与 ½ m v² 等价。

10. Common Misconceptions and Exam Tips | 常见误区与考试技巧

One of the most frequent errors in two-dimensional collision questions is treating speed as a vector. Speed must never be substituted directly into a component equation. Always resolve velocity into components using the given angles before applying any momentum equation.

二维碰撞问题中最常见的错误之一是将速率当作矢量处理。速率绝不能直接代入分量方程。在应用任何动量方程之前，务必使用给定角度将速度分解为分量。

A second common mistake is confusing the angle with respect to the normal and the angle with respect to the surface. In a wall-collision problem, if the angle to the wall is given, convert it before using the formula Δp = 2 m v cos θ. Write the angle clearly on your diagram to avoid this error.

第二个常见错误是混淆相对于法线的夹角和相对于表面的夹角。在墙壁碰撞问题中，若给出的是与墙面的夹角，在使用公式 Δp = 2 m v cos θ 之前必须先换算。在图上清楚标出角度以避免此类错误。

Examiners award method marks even when arithmetic goes wrong. Therefore, always show the resolved component equations explicitly. For each axis, write the conservation equation in full symbol form before substituting numbers. This demonstrates your understanding and secures partial credit.

即使计算失误，考官也会给方法分。因此，务必明确写出分解后的分量方程。对每个轴，先写出完整的符号形式守恒方程，再代入数值。这样既展示了你的理解，也能确保获得部分分数。

Finally, always

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Collisions in Two Dimensions | 二维碰撞