Conservation of Momentum and Energy in Elastic Collisions

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In the discussion on the conservation of momentum and energy in elastic collisions, a scenario is presented where a smaller package slides down a frictionless chute and collides with a larger stationary package. The initial calculation determines the common speed after the inelastic collision to be 0.865 m/s. The second part of the discussion addresses the elastic collision, where the user struggles to find the rebound height of the smaller package, consistently arriving at the original height of 2.44 m. Participants emphasize the need to apply both conservation of momentum and conservation of energy equations correctly to resolve the confusion. The conversation highlights the complexities of elastic collisions and the importance of accurate equation setup.
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A package of mass m is released from rest at a warehouse loading dock and slides down a 2.44 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 7.00m, from the bottom of the chute. Suppose the packages stick together. What is their common speed after the collision?

Answer: 0.865 m/s

I solved this easily using kinematics to determine the small mass's velocity upon hitting the large mass. Then using conservation of momentum to isoalte the combined velocity. However, this secodn question confuses me:

Suppose the collision between the packages is elastic. To what height does the package of mass m rebound?

I've tried conservation of energy equations and always end up with a rebound to the original height of 2.44m. Can anyone offer suggestions?
 
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mathewings said:
I've tried conservation of energy equations and always end up with a rebound to the original height of 2.44m. Can anyone offer suggestions?
What makes you think your answer is incorrect?
 
How did you set up your conservation of energy equation for the collision? (You still need conservation of momentum as well.)
 
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