Q. A ball of mass $m$ falls vertically to the ground from a height $h_{1}$ and rebound to a height $h_{2}$. The change in momentum of the ball on striking the ground is
A) $mg(h_{1} - h_{2})$
B) $mg(\sqrt{2gh_{1}} + \sqrt{2gh_{2}})$
C) $m\sqrt{2g(h_{1} + h_{2})}$
D) $m\sqrt{2g}(h_{1} + h_{2})$
Explanation
(b) When ball falls vertically downward from height $h_{1}$its velocity ${\overrightarrow{v}}_{1} = \sqrt{2gh_{1}}$
And its velocity after collision ${\overrightarrow{v}}_{2} = \sqrt{2gh_{2}}$
Change in momentum
$\mathrm{\Delta}\overrightarrow{P} = m\left( {\overrightarrow{v}}_{2} - {\overrightarrow{v}}_{1} \right) = m(\sqrt{2gh_{1}} + \sqrt{2gh_{2}})$
[Because ${\overrightarrow{v}}_{1}$ and ${\overrightarrow{v}}_{2}$ are opposite in direction]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.