How Do You Solve an Elastic Collision Problem Involving a Sloped Track?

[okaymeka]

Okay I have been working on this problem forever and I don't know what else to try. Can anyone help me?...

A 0.005-kg block is released from rest at the top of a frictionless track 2.50 m above the top of a table. It then collides elastically with a 1.00-kg object that is initially at rest on the table. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the 1.00-kg object land, given that the table is 2.00 m height? (d) How far away from the bottom of the table does the 0.500-kg object eventually land?

 

[tonyh]

Okay I have been working on this problem forever and I don't know what else to try. Can anyone help me?...

A 0.005-kg block is released from rest at the top of a frictionless track 2.50 m above the top of a table. It then collides elastically with a 1.00-kg object that is initially at rest on the table. (a) Determine the velocities of the two objects just after the collision. (b) How high up the track does the 0.500-kg object travel back after the collision? (c) How far away from the bottom of the table does the 1.00-kg object land, given that the table is 2.00 m height? (d) How far away from the bottom of the table does the 0.500-kg object eventually land?

Hmmm, sounds like a homework problem. You need to think about conservation of momentum and conservation of mechanical energy.

 

[physixguru]

Concepts involved:

a.Conservation of momentum
b.Impulse
c.Projectile Motion
d.Projectile motion

I would have loves to do the question for you but the guidelines prevent me from doing so.Show some attempt.

 

[vkmphysics]

This question is incomplete because set of information provided does not support the question asked. if the motion is linearly vertical then why would the block at table will move horizontally, if the block is released vertically. so to solve this problem you must have a clear figure or the situation must describe more about the kinemetics of the question. else if you have figure either post it or describe it and if figure is avaliable with you then it is very easy then just apply conservation of momentum and mechanical energy.

 

[okaymeka]

Sorry guys for not showing my original effort. It was late. I tried finding vf with this equation: mgh + 1/2mv^2 = mgh + 1/2mv^2 and it seems like no matter what numbers I plug in I can't come up with the correct answer. I will try ya'll suggestions to see if I can do something differently. Thanks for replying.

 

[okaymeka]

By the way, there is a figure provided with the question. There is a table that is 2.00 m off the ground with a slope on top of the table that is 2.50 m high with m1 at the top of the slope and m2 at the bottom of the slope. And x is on the ground out from the table.