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## Homework Statement

If there is a frictionless circular path with radius R cut into a block of mass M1 and a block of mass M2 slides down the path, what is the velocity of mass M2 as it leaves the block?

There is no friction between M1 and the ground.

M2=mass of falling block

M1=mass of

v(21)=velocity of M2 with respect to M1

v(1)=velocity of M1 with respect to the earth

v(2)=velocity of M2 with respect to the earth

## Homework Equations

I know that this is a conservation of momentum problem, so I used the equations. E(initial)=E(final) and P(initial)=P(final)

## The Attempt at a Solution

By using conservation of energy, I found the velocity of the mass M2 with respect to M1 as it leaves the circular path. Since it falls a distance R, I set

(M2)gR=m[v(21)]^2/2

And found v(21)=SQRT(2gR)

Since the system began with no momentum and momentum must be conserved:

0=(M2)(V2)+(M1)(V1)

What I'm trying to find is V2, I know that V2=V21+V1, so substituting V1=V2-V21 into the momentum equation...

0=(M2)(V2)+(M1)(V2-V21)=(M2)(V2)+(M1)(V2-[SQRT(2gR)])

Now the final answer I got was that V2=M1[SQRT(2gR)]/(M1+M2).

But the answer hint was that if M1=M2, that V2=SQRT(gR) which wasn't what my answer would get.