perfectly elastic collision

bearhug

2 blocks are free to slide along the frictionless wooden track. The block of mass m1=5.00 kg is released from A, while the block of mass m2= 10.0 kg initially sits @bottom of ramp. The blocks collide @ position Bin a perfectly elastic collision. To what height does m1 rise after collision?

Originally I thought of using Ki + Ui = Kf + Uf where U=mgh
so 1/2m1vi^2 1/2m2vi^2 + (mgh)i = 1/2m1vf^2 1/2m2vf^2+ (mgh)f
However I'm having a hard time figuring this out because I don't know what the velocities of either block is after collision. I do know that the initial of block 2 is 0 m/s. Can someone help me set this problem up? Any help is appreciated.

Doc Al

What else is conserved in every collision?

bearhug

In every collision... momentum

bearhug

In elastic collisios kinetic energy

Doc Al

Right. You have to use that fact to solve this problem. (You just used conservation of energy--but that's not enough.)

bearhug

Does gravitational potential energy have anything to do with this problem?

bearhug

OK so I set up the problem beginning with (m1v1 + m2v2)i = (m1v1 + m2v2)f . Should I start this problem with block 1 m=5.0 kg w/ initial velocity after collision or before. If it's before than initial would be 0 but other wise it wouldn't. Since the question asks for the height after collision I was wondering if this needs to be considered in terms of what's initial and what's final. Any feedback please.

Doc Al

Assuming I understand the problem correctly, here's how to approach it. First figure out the speed of block 1 just before it collides with block 2. Then analyze the collision to determine the speed of block 1 just after the collision. (That involves conservation of momentum and energy.) Once you know the speed of block 1 after the collision, figure out how high it goes.