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The figure shows a block of mass M = 2m having a spherical smooth cavity of radius R placed on a smooth horizontal surface. There is a small ball of mass m moving at an instant vertically downward with a velocity v with respect to the block at the instant shown in figure. At this instant:

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The acceleration of the block with respect to the ground is $\frac{v^{2}}{2 R}$

The acceleration of the block with respect to the ground is $\frac{v^{2}}{3 R}$

The normal reaction on the ball by the block is $\frac{m v^{2}}{R}$

The normal reaction on the ball by the block is $\frac{2}{3} \frac{m v^{2}}{R}$

answer is B, C.

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Detailed Solution

Let the normal force between the block and the ball be N.

For the block, from Newton’s II^nd law, we have $N = M a = 2 m a$ .
For ball (with respect to the block), from Newton’s II^nd law, we have $N + m a = \frac{m v^{2}}{R}$ .

Solve the two equations.

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