Inelastic collision: block moving down a frictionless ramp

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Homework Help Overview

The problem involves two suitcases on a frictionless ramp, where one suitcase slides down and collides inelastically with another suitcase at rest. The discussion centers around the conservation of energy and momentum in the context of inelastic collisions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of mechanical energy and the application of momentum conservation in inelastic collisions. Questions are raised about the validity of the proposed approach and the assumptions regarding energy conversion and momentum transfer.

Discussion Status

Some participants affirm the original poster's approach, while others suggest reconsidering the method used for momentum transfer. There is an acknowledgment of different methods being presented, but no consensus has been reached on a single approach.

Contextual Notes

Participants note the importance of not providing complete solutions and emphasize the need for hints and guidance rather than full methods. The discussion reflects a focus on understanding the principles involved rather than arriving at a definitive answer.

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Homework Statement


In the figure provided, two suitcases are on a 6.36 m high ramp to passengers waiting in a baggage terminal. The top suitcase is released from rest, and it slides down the ramp and hits the second suitcase. If the suitcase at the top has a mass of 11.8 kg and the other suitcase has a mass of 23.6 kg, what is their combined speed if an inelastic collision is achieved? Assume the ramp to be frictionless.

Homework Equations


KE = (1/2)*mv2
PE = mgh
m1v1+m2v2 = (m1+m2)v

The Attempt at a Solution


I just had a question about whether this is the approach to solve this question or not.
So since the ramp is frictionless, total mechanical energy is conserved.
You use mgh = (1/2)*mv2 , isolate v and solve for v for the first suitcase.
Then with the v you plug it into the inelastic collision formula to solve for the new combined velocity of the two suitcases.

m1v1+m2v2= (m1+m2)v
Isolate for v and v2 is 0?
 
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I think it looks right to me. All the potential energy of the first suitcase gets converted to kinetic energy. That suitcase collides with the 2nd suitcase which is at rest. Since the collision is inelastic, there is conservation of momentum, but not conservation of energy.
 
Inelastic collision so momentum transfer during collision is given by the equation Δp = μΔv where μ = reduced mass 11.8 x 23.6 /(11.8 + 23.6) and Δv = √(2gh).

Using g = 9.8 m/s2 we obtain Δp = 87.83 Ns and dividing by mass of suitcase 2, velocity is 3.72 m/s.

Now check on suitcase 1:

11.8 x √(2gh) - 87.83 = 43.92 and dividing by mass of suitcase 1 , velocity is also 3.72 m/s
 
neilparker62 said:
Inelastic collision so momentum transfer during collision is given by the equation Δp = μΔv where μ = reduced mass 11.8 x 23.6 /(11.8 + 23.6) and Δv = √(2gh).

Using g = 9.8 m/s2 we obtain Δp = 87.83 Ns and dividing by mass of suitcase 2, velocity is 3.72 m/s.

Now check on suitcase 1:

11.8 x √(2gh) - 87.83 = 43.92 and dividing by mass of suitcase 1 , velocity is also 3.72 m/s
This is a homework forum. Please do not post complete solutions (unless it has already been solved and you are just showing a better way). Just point out errors in the posted attempt and offer hints on how to proceed.
In the present case, a simple "yes" would have been appropriate.
 
Apologies - I was showing another method but that wasn't what was asked by the OP.
 

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