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Perfectly elastic collision 
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#1
Oct906, 09:08 AM

P: 79

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. 


#3
Oct906, 11:52 AM

P: 79

In every collision... momentum



#4
Oct906, 11:54 AM

P: 79

Perfectly elastic collision
In elastic collisios kinetic energy



#5
Oct906, 11:58 AM

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P: 41,303




#6
Oct906, 03:38 PM

P: 79

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



#7
Oct906, 05:53 PM

P: 79

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.



#8
Oct1006, 11:48 AM

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P: 41,303

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.



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