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
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
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
And found v(21)=SQRT(2gR)
Since the system began with no momentum and momentum must be conserved:
What I'm trying to find is V2, I know that V2=V21+V1, so substituting V1=V2-V21 into the momentum equation...
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.