# Momentum of mass problem

## Homework Statement

A small object of mass m slides down a wedge of mass 2m and exits smoothly onto a frictionless table. The wedge is initially at rest on the table. If the object is initially at rest at a height h above the table, find the velocity of the wedge when the object leaves it.

mv=mv

## The Attempt at a Solution

So far, I know that the velocity of mass m is (2gh)^.5, but I'm not sure how to find the speed of the wedge while it's sliding down. At first I tried to set m(2gh)^.5 = (2m)v, but that doesn't give me the right answer. Any help would be appreciated!

Hmmm......isnt the mass of the system(wedge) initially supposed to be 2m+m=3m???

So does that mean the velocity is (2/3gh)^.5? (mgh = .5 (3m) v^2) Even after that, I'm not quite sure how to set up the equation after the block leaves the wedge.

First the question asks only what is the velocity when the wedge breaks contact, not what its doing while the two are in contact. Now since it says "exits smoothly" I would assume all the velocity is now horizontal. Conservation of momentum along the x axis? Initially it was zero before the block was released you can assume. Since there have been no external forces other than gravity which is directed along the Y axis, will the momentum on the x axis be conserved?

So then the velocity of the block on the horizontal is (2gh)^.5, which equals the momentum of the wedge? That gives me the equation m (2gh)^.5 = 2m v, which can't be right.

why not? I thought it was, but maybe I'm overlooking something. Like we now have excess energy? so maybe we need to conserve both??

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The answer in the back is -((gh)/3)^.5

so lets apply what I suggested:
mgh=1/2*m*vb^2 plus 1/2*2m*Vinc^2 and using m*vb=-2m*Vinc and solving for
Vinc,