Coefficient of Restitution and Collisions in Two Dimensions

Beyond Head-On Collisions

Conservation of momentum tells you that the total momentum before a collision equals the total after. But on its own it cannot tell you how two balls will move after they collide: you need to know how "bouncy" the collision is. Real collisions are also rarely head-on: snooker balls glance off each other at angles, and cars collide at junctions. This tutorial covers the coefficient of restitution, which measures bounciness, Newton's experimental law, elastic and inelastic collisions, kinetic energy loss, bouncing balls, and collisions in two dimensions where momentum is conserved in each direction separately. It builds on the tutorial on momentum, impulse and collisions.

The Coefficient of Restitution

Newton found experimentally that, for two given bodies, the speed at which they separate after a collision is a fixed fraction of the speed at which they approach:

e = speed of separation / speed of approach.

For two bodies moving along a line, with velocities u₁ and u₂ before and v₁ and v₂ after:

v₂ − v₁ = −e(u₂ − u₁), or equivalently v₂ − v₁ = e(u₁ − u₂).

The number e is the coefficient of restitution, and 0 ≤ e ≤ 1.

e = 1: perfectly elastic. The bodies separate as fast as they approached and no kinetic energy is lost. Collisions between gas molecules and, very nearly, between steel ball bearings are like this.
0 < e < 1: inelastic. Some kinetic energy becomes heat, sound and deformation. A tennis ball on a court has e of roughly 0.7 to 0.8.
e = 0: perfectly inelastic. The bodies stick together and move with a common velocity.

Momentum is conserved in every collision (if no external force acts), whatever the value of e. Kinetic energy is conserved only when e = 1.

Solving a Collision Using the Coefficient of Restitution

With two unknown final velocities you need two equations: conservation of momentum and Newton's law of restitution.

Example 1. A 2 kg ball moving at 6 m/s hits a stationary 1 kg ball head-on. The coefficient of restitution is 0.5. Find the velocities after impact and the kinetic energy lost.

Momentum: 2 × 6 + 1 × 0 = 2v₁ + 1v₂, so 2v₁ + v₂ = 12.

Restitution: v₂ − v₁ = 0.5 × (6 − 0) = 3.

Subtracting the second equation from the first: 3v₁ = 9, so v₁ = 3 m/s, and v₂ = 3 + 3 = 6 m/s. Both move forward; the lighter ball moves off faster.

Kinetic energy before = ½ × 2 × 36 = 36 J. After = ½ × 2 × 9 + ½ × 1 × 36 = 9 + 18 = 27 J. 9 J (a quarter of the energy) is lost.

With e = 1 the same collision gives v₁ = 2 m/s and v₂ = 8 m/s and no energy loss; with e = 0 both move at 4 m/s, losing 12 J.

Signs Matter

Choose a positive direction and stick to it. A velocity in the opposite direction is negative. If your answer for a final velocity comes out negative, that body is moving backwards after the collision. For example, a light ball hitting a heavy one head-on usually bounces back with a negative velocity.

Bouncing Off a Fixed Surface

When a ball hits a fixed wall or floor, the surface doesn't move, so the rebound speed is simply e times the impact speed: v = eu.

Example 2. A ball is dropped from 5 m onto a floor with e = 0.8. How high does it bounce?

The impact speed is u = √(2gh), so the rebound speed is eu. Since height is proportional to speed squared, the rebound height is e²h = 0.64 × 5 = 3.2 m. The next bounce reaches 0.64 × 3.2 = 2.048 m, and so on, a geometric sequence with ratio e².

This gives a simple way to measure e: drop a ball from a known height h and measure the bounce height h′. Then e = √(h′/h). Sports bodies use this kind of test; football and basketball rules specify how high a ball must bounce when dropped from a set height.

Collisions in Two Dimensions

Momentum is a vector, so in two dimensions it is conserved separately in each direction:

total x-momentum before = total x-momentum after;
total y-momentum before = total y-momentum after.

Example 3 (vehicles that stick together). A 1,000 kg car travelling east at 20 m/s collides at a junction with a 3,000 kg truck travelling north at 10 m/s. They lock together. Find their velocity just after the crash.

x-momentum: 1000 × 20 = 20,000 kg m/s (the truck has none in this direction).
y-momentum: 3000 × 10 = 30,000 kg m/s.
Combined mass 4,000 kg, so the velocity components are 20,000/4,000 = 5 m/s east and 30,000/4,000 = 7.5 m/s north.

Speed = √(5² + 7.5²) = 9.01 m/s, at tan⁻¹(7.5/5) = 56.3° north of east.

Kinetic energy before = ½ × 1000 × 400 + ½ × 3000 × 100 = 350,000 J. After = ½ × 4000 × 9.01² ≈ 162,500 J. About 54% of the kinetic energy was converted into deformation, heat and sound, which is why crumple zones are designed to absorb it. Accident investigators use exactly this method in reverse: from the skid marks after the crash they work out the speeds before it.

Glancing Elastic Collisions of Equal Masses

A special result is very useful in snooker and in nuclear physics. When a moving ball makes an elastic, glancing collision with an identical stationary ball, the two balls move off at right angles to each other.

Why: momentum conservation with equal masses gives u = v₁ + v₂ (as vectors), so the three velocities form a triangle. Kinetic energy conservation gives u² = v₁² + v₂². By the converse of Pythagoras, the angle between v₁ and v₂ is 90°.

Example 4. A snooker ball moving at 4 m/s strikes an identical stationary ball elastically and is deflected through 30°. Find both final speeds.

The second ball moves off at 90° − 30° = 60° to the original direction. In the right-angled triangle of velocities, v₁ = 4 cos 30° = 3.46 m/s and v₂ = 4 cos 60° = 2.00 m/s. Check energy: 3.46² + 2² = 12 + 4 = 16 = 4² ✓.

Oblique Impact with a Wall

When a ball hits a smooth wall at an angle, the wall can only push perpendicular to itself. So:

the velocity component parallel to the wall is unchanged;
the component perpendicular to the wall is reversed and multiplied by e.

A ball hitting a smooth wall at 10 m/s at 60° to the wall has a parallel component of 10 cos 60° = 5 m/s and a perpendicular component of 10 sin 60° = 8.66 m/s. With e = 0.5 it rebounds with components 5 m/s and 4.33 m/s, a speed of 6.61 m/s at tan⁻¹(4.33/5) = 40.9° to the wall. With e < 1, the ball always leaves at a shallower angle than it arrived.

Summary

Coefficient of restitution: e = speed of separation ÷ speed of approach, between 0 and 1.
Solve a head-on collision with momentum conservation plus v₂ − v₁ = e(u₁ − u₂).
Momentum is always conserved; kinetic energy only when e = 1.
A ball dropped from height h rebounds to e²h, so e = √(h′/h).
In two dimensions, conserve momentum separately in x and y; equal masses in an elastic glancing collision separate at 90°.