What is the final height reached by block 1 after a 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]

In every collision... momentum

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.