Question

The figure shows a block of mass $M=2m$ having a spherical 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 $\frac{{mv}^2}{R}$
• B. the normal reaction on the ball by the block is $\frac{2}{3}\frac{{mv}^2}{R}$
• 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: (A), (D)

the normal reaction on the ball by the block is $\frac{{mv}^2}{R}$
the acceleration of the block with respect to the ground is $\frac{v^2}{2R}$

velocity of ball respect to the block

$\Rightarrow V_{ball,block}=-v\hat{j}$

From to the diagram, Normal Reaction on the ball

$\Rightarrow F_N$$=\frac{m{v}^2}{R}$

According to third law of Newton's, equal magnitude and opposite direction of force will be reacted on block due to movement of ball downwards

$\Rightarrow a=\frac{F_N}{bloc{k}^{\prime }smass}\Rightarrow a=\frac{F_N}{2m}\Rightarrow a=\frac{\left(\frac{m{v}^2}{R}\right)}{2m}\Rightarrow a=\frac{v^2}{2R}$