Question

A small cube of mass 'm' slides down a circular path of radius 'R'
cut into a large block of mass 'M'. 'M' rests on a table and both
blocks move without friction. The blocks initially are at rest and
'm' starts from the top of the path. Find the velocity'v' of the cube
as it leaves the block. Initially the line joining m
and the centre is horizontal.​

Answer 1

Answer:

v=√2mgR/[m+M]

Explanation:

As the cube starts from the top of the large block in a semi-circular path, therefore all its potential energy will convert into kinetic energy during its course. Then the equation will be

P.Em = K.Em

mgR = 1/2mv²-------(1)              R=radius of circular path

As the large block is also moving so

P.EM = K.EM

As initially large block is lying on table and at rest so P.E will be=0

K.E of the large block = 1/2 Mv²

0 = 1/2Mv²-------(2)

Add equation (1) and (2)

0+mgR=1/2mv² + 1/2 Mv²

mgR=1/2v²[m+M]

or

v²= 2mgR/[m+M]

or

v=√2mgR/[m+M]

Hope it has answered your question.