   # Elastic collisions and snooker

Although you would not play snooker using a protractor and a calculator to work out your shots the study of momentum is crucial in what actually happens!

### 1. Collisions in a straight line

We will first consider the more simple case where the two balls collide along a line joining their centres. We have to assume that the two balls are not only exactly the same mass (m) but that the collision between them is perfectly elastic - remember that this means that no kinetic energy is lost in a collision. First think of just one stationary red being hit head on by the white. After the collision the white ball stops dead and the red ball moves off with the velocity that the white ball had before the collision. (This is only exactly true if the two balls are sliding and not rolling!)
We can prove this as follows:

Momentum before collision = momentum after collision muA = mvA + mvB
But because the collision is elastic kinetic energy is also conserved and so:

˝ muA2 = ˝ mvA2 + ˝ mvB2

The masses can be cancelled giving:- uA = vA +vB and uA2 = vA2 + vB2

Using a little algebra to combine these two equations will show you that:

vA = 0     and     VB = uA

In fact it can be proved that when two equal masses collide elastically their velocities are swapped over. If mass A had a velocity u before collision and mass B a velocity v, then after collision A moves with velocity v and B with velocity u.

Example problem
A ball of mass 0.5 kg travelling from left to right at 1.5 ms-1 collides with a ball of equal mass travelling in the same direction at 1.0 ms-1. Prove that they swap velocities after impact.

Momentum before collision = 0.5x1.5 + 0.5x1.0 = 1.25 Ns
Momentum after impact = 1.25 Ns
Kinetic energy before impact = kinetic energy after impact = 0.8125 J
Therefore:
vA + vB = 1.25/0.5 = 2.5 ms-1
vA2 + vB2 = 0.8125x4 = 3.25 = vA2 + (2.5 - vA)2 = vA2 + 6.25 - 5vA + vA2 = 3.25
and so 2vA2 - 5vA + 3 = 0 and when solved this gives vA = 1ms-1 or 1.5 ms-1 but since ball A cannot move through ball B the velocity of ball A must be 1 ms-1.

They have swapped velocities!

### 2. Oblique or angled collisions.

Both balls move off at an angle to the original direction of the white. The interesting thing is that the angle between the red and the white balls is exactly 90o, in theory. The big problem here is spin - if you put some "side" (spin) on the white the angle will be different.

Assuming no spin and that the collision is elastic:

˝ muA2 = ˝ mvA2 + ˝ mvB2 using a vector diagram for the momenta of the two balls before and after the collision we see that:

muA = mvA + mvB

and since from the equation for kinetic energy conservation

uA2 = vA2 + vB2

The vector triangle for velocity must be right angled (Pythagoras). Remember that velocity and therefore momentum is a vector but kinetic energy is a scalar and so we can simply add the kinetic energies of the balls after impact without allowing for their different directions of motion. ## schoolphysics oblique collisions animation

To see an animation of an oblique collision click on the animation link.

## Elastic collisions between particles of unequal mass

### 1. Collisions in a straight line

If a body A with a mass m collides with a second body B with a mass M then the kinetic energy lost by A is:

(a) maximum when m = M

(b) zero when M = infinity

### 2. Oblique collisions

This type of collision can often be seen in cloud chamber and bubble chamber photographs and is very useful in comparing the masses of the colliding particles.
It can be proved that if a mass m collides with a mass M then
(a) if m << M the angle between their paths after collision is > 90o
(b) if m = M the angle between their paths after collision is 90o
(c) if m > M the angle between their paths after collision is < 90o