A small cube of mass m slides down a circular path of radius

A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. The block of mass M rests on a table, and both the cube and the block move without friction and are initially at rest. The cube of mass m starts from the top of the path. Find the velocity v of the cube relative to the ground as it leaves the block.

Solution

We would solve this problem using conservation of momentum as well as conservation of energy.

As the smaller block slides down the velocity it will gain and will cause the larger block to gain due to the \'fall\' must result in a KE which is equal to the initial PE

That is:

0.5 [mv^2 + Mu^2] = mgR

Also, mv = Mu [As the linear momentum must remain conseved]

or u = mv/M
Using this in the above equation for energy to get:
mv^2 + (mv)^2/M = 2mgR

or v^2[1 + 1/M] = 2gR

Hence, v = sqrt[2MgR/(M + 1)]