Inelastic collision: block moving down a frictionless ramp

AI Thread Summary
The discussion revolves around calculating the combined speed of two suitcases after an inelastic collision on a frictionless ramp. The first suitcase, with a mass of 11.8 kg, slides down from a height of 6.36 m, converting potential energy to kinetic energy. The conservation of momentum is applied to find the final velocity after the collision, where the second suitcase, weighing 23.6 kg, is initially at rest. The calculations indicate that both suitcases end up moving at a combined speed of 3.72 m/s post-collision. Participants emphasize the importance of not posting complete solutions in homework forums, focusing instead on guiding the original poster with hints and corrections.
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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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