# Homework Help: Elastic Billiard Ball Collision

1. Feb 27, 2013

### iandelaney

1. The problem statement, all variables and given/known data
A cue ball with speed U hits a stationary red ball of equal mass. The collision is elastic (ie no energy is converted into other forms). After the collision the cue ball is moving at an angle θ to its original path. Find the final speed of the cue ball.

2. Relevant equations

conservation of momentum: mu=mv1mv2

conservation of energy: 1/2mu2 = 1/2mv12 + 1/2mv22

3. The attempt at a solution

As I understand it this should be a fairly standard problem but I have been struggling to find any help on the tinterweb hence my posting :-)

I know that from the conservation of momentum that the velocities of the two balls after the collision will be equal to the velocity of the cue ball initially. I would then like to substitute u=v1+v2 into the conservation of energy to obtain (v1+v2)2 = v12+v22 which is only the case when v1 and v2 are zero. I was wondering if anyone could spot my mistake?

Ian

2. Feb 27, 2013

### clamtrox

The velocities are vectors, remember that. You can get rid of the velocity of the red ball using conservation momentum, plug that into the equation of conservation of energy. You get the angle dependence from the dot product between the two velocities.

3. Feb 27, 2013

### iandelaney

When I do that I obviously find that v2=u-v1 and when I sub this into the conservation of energy equation it still comes out as either v1=u or v1=0, and I am assuming that this can't be the case as it states in the question that the ball continues to move after the collision at an angle θ :s

Last edited: Feb 27, 2013
4. Feb 27, 2013

### Ray Vickson

Such problems are almost always easier to solve if you first go into the CM (center-of-mass) system, then transform back to the lab system to finally finish the calculation. In the CM system, the total momentum = 0 before (and after) the collision, so if the masses of the balls are equal, their speeds are equal in the CM system before and after the collision. If energy is conserved in the lab system it is also conserved in the CM system, and vice-versa.