A small cube of mass 'm' slides down a circular path of radius 'R' formed from a large block of mass 'M' as shown in figure 'M' rests on a table and both blocks move without friction. The blocks are initially 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 center is horizontal

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Solution

By conservation of energy. -

$m g h = m g (h - R) + \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2}$

By conservation of moment ion-

$M V = m v$

$v = \sqrt{(\frac{M}{M + m}) 2 g R}$

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A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surfaced of a fixed table. Initially the right edge of the block is at x = 0, in a coordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following option is/are correct ?

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