Collisions in Two Dimensions

The Core Idea: Why Two Dimensions Changes Everything

In the previous section, every collision happened along a single straight line. A ball hits a wall and bounces straight back; two gliders on an air track meet head-on. Real collisions are rarely that tidy. A car hits another at an intersection; a billiard ball is struck at an angle. In these cases, the velocities before and after the collision are vectors in a plane, not scalars on a line.

The fundamental laws — conservation of linear momentum and conservation of kinetic energy (for elastic collisions) — still hold. But now they give us two component equations (one for the $x$-axis, one for the $y$-axis) instead of one. That extra equation is what makes two-dimensional collisions richer and, in many cases, solvable only when we know enough of the final conditions.

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Setting Up the Problem

Consider two particles, $A$ and $B$, of masses $m_1$ and $m_2$. Particle $A$ moves with initial velocity $\mathbf{v}_{1i}$; particle $B$ is initially at rest ($\mathbf{v}_{2i} = 0$). This is the standard textbook scenario — one target stationary — and it captures the essential physics.

After the collision, both particles move off with velocities $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$. The collision is elastic, so both momentum and kinetic energy are conserved.

We choose a coordinate system that simplifies the algebra: let the initial velocity of $A$ define the $x$-axis. So $\mathbf{v}_{1i} = v_{1i}\,\hat{\mathbf{i}}$. After the collision, $A$ moves at an angle $\theta_1$ above the $x$-axis, and $B$ moves at an angle $\theta_2$ below the $x$-axis (or vice versa — the signs will come from the equations).

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The Two Conservation Laws, Component by Component

Conservation of linear momentum gives two scalar equations:

• $x$-component: $m_1 v_{1i} = m_1 v_{1f} \cos\theta_1 + m_2 v_{2f} \cos\theta_2$
• $y$-component: $0 = m_1 v_{1f} \sin\theta_1 - m_2 v_{2f} \sin\theta_2$

The minus sign in the $y$-equation appears because we have chosen $\theta_2$ to be measured below the $x$-axis, so its $y$-component is negative.

Conservation of kinetic energy (elastic collision) gives the scalar equation:

These three equations contain four unknowns: $v_{1f}$, $v_{2f}$, $\theta_1$, $\theta_2$. (The masses and $v_{1i}$ are known.) So the system is underdetermined — we need one more piece of information to get a unique answer. That extra information usually comes from the geometry of the impact: for example, the angle at which the particles strike each other, or the fact that the collision is "glancing" rather than head-on.

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The Special Case of Equal Masses

When $m_1 = m_2 = m$, the equations simplify dramatically. The momentum equations become:

• $x$: $v_{1i} = v_{1f} \cos\theta_1 + v_{2f} \cos\theta_2$
• $y$: $0 = v_{1f} \sin\theta_1 - v_{2f} \sin\theta_2$

And the energy equation becomes:

Now square the two momentum equations and add them:

Using $\cos^2 + \sin^2 = 1$ and the cosine addition formula $\cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2 = \cos(\theta_1 + \theta_2)$, we get:

But the energy equation says $v_{1i}^2 = v_{1f}^2 + v_{2f}^2$. For both to be true simultaneously, we must have:

Since $v_{1f}$ and $v_{2f}$ are not zero (the particles do move after collision), the only possibility is:

Therefore:

This is a beautiful and testable result. If you shoot one billiard ball into a stationary one of the same mass, the two balls will always leave at $90^\circ$ to each other — provided the collision is elastic and not head-on.

›Proof

The derivation above is the complete proof. The key step is squaring and adding the momentum components, then comparing with the energy equation. The cancellation forces $\cos(\theta_1+\theta_2)=0$.

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The General Case: Unequal Masses

When $m_1 \neq m_2$, the algebra is messier but the logic is the same. The three equations (two momentum, one energy) still have four unknowns. The extra condition is often the impact parameter — the perpendicular distance between the line of motion of the incoming particle and the centre of the target particle. This parameter determines the angles $\theta_1$ and $\theta_2$.

The textbook does not derive a closed-form solution for the general case; instead, it emphasises the method:

1. Write the momentum conservation equations in component form.
2. Write the energy conservation equation. …