Question

The figure shows a block of mass $M=2\text{m}$ 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:

• A. The normal reaction on the ball by the block is $\frac{m{v}^2}{R}$
• B. The normal reaction on the ball by the block is $\frac{2}{3}\frac{m{v}^2}{R}$
• C. The acceleration of the block with respect to the ground is $\frac{v^2}{3R}$
  
• D. The acceleration of the block with respect to the ground is $\frac{v^2}{2R}$

Answer 1

Answer: (B), (C)

The correct options are
B The normal reaction on the ball by the block is $\frac{2}{3}\frac{m{v}^2}{R}$
C The acceleration of the block with respect to the ground is $\frac{v^2}{3R}$



Velocity of ball with respect to the block
${\overrightarrow{v}}_{\text{ball, block}}=-v\hat{j}$
From the diagram, normal reaction on the ball
$N+ma=\frac{m{v}^2}{R}$.....(i)
According to the third law of Newton, equal magnitude and opposite direction of force will be reacted on block due to movement of ball downwards.
$\Rightarrow a=\frac{N}{bloc{k}^{\prime }smass}$

Putting this value of $a$ in (i)
$2ma+ma=\frac{m{v}^2}{R}$
$\Rightarrow a=\frac{v^2}{3R}$
and,
$N=\frac{2}{3}\frac{m{v}^2}{R}$