# Homework Help: Conservation of momentum and mass of a ball

1. Nov 26, 2014

### Quantum Fizzics

1. The problem statement, all variables and given/known data
A 0.160 kg ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.
a) what is the mass of the ball
b) what fraction of the original kinetic energy gets transferred to the second ball.

2. Relevant equations
p=mv

3. The attempt at a solution
I honestly have no idea how to start this. But I assume that this would be a pendelum. So gravity comes in. Since its asking for the mass of the 2nd ball I was thinking of using the equation above. "p" as 10m/s & of course "m" as 0.160kg. Since it says on the question the 2nd ball moves off with half the original speed of the first ball. I would divide it by 2 then use whatever the V of ball 2 & use the same equation except this time I calculate the mass. Idk what do u guys think? I'll do it right now

2. Nov 26, 2014

### Miles Whitmore

Assuming the balls are moving horizontally, gravity will not come into play here.
I'm not sure where you're getting this number and it does not have the correct units for momentum.

Total momentum is conserved in all collisions, so start with that. Using your relevant equation, write down an expression for the momenta of each ball before and after the collision. Since total momentum is conserved, the sum of momenta before the collision equals the sum of momenta after the collision. That gives you one equation, but you will still have too many unknowns to solve for the unknown mass and velocity.

But you are also told this is an elastic collision, so by definition another quantity is also conserved and you can write another conservation equation. Then solve for unknowns.

3. Nov 26, 2014

### haruspex

Gravity doesn't enter into it regardless of direction. The question only concerns events in an arbitrarily short interval of time. The momentum change contributed by gravitational forces is therefore arbitrarily small.

4. Nov 26, 2014

### Miles Whitmore

Good point

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