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 this instant :

Options

A. The normal reaction on the ball by the block is `(mv^(2))/R`
B. The normal reaction on the ball by the block is `2/3 (mv^(2))/R`
C. The acceleration of the block with respect to the ground is `(v^(2))/(3R)`
D. The acceleration of the block with respect to the ground is `(v^(2))/(2R)`

Explanation

Let the normal force between the block and the ball be N
For the block, from Newton's `II^(nd)` law, we have N=Ma=2ma
For ball (with respect to the block), from Newton's `II^(nd)` law, we have `N+ma=(mv^(2))/R`
Solve the two equations.