Momentum problem

[psycovic23]

1. The problem statement, all variables and given/known data
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.


2. Relevant equations
mv=mv


3. 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!

[anantchowdhary]

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

[psycovic23]

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.

[denverdoc]

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?

[psycovic23]

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.

[denverdoc]

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??

[psycovic23]

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

[denverdoc]

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,
should lead to the correct answer.