Q. A spring pong ball of mass $m$ is floating in air by a jet of water emerging out of a nozzle. If the water strikes the ping pong ball with a speed $v$ and just after collision water falls dead, the rate of flow of water in the nozzle is equal to
A) $\frac{2mg}{v}$
B) $\frac{m}{g}$
C) $\frac{mg}{v}$
D) $\frac{2m}{vg}$
Explanation
(c) The impact force $F = \frac{\mathrm{\Delta}p}{\mathrm{\Delta}t} = v\frac{\mathrm{\Delta}m}{\mathrm{\Delta}t}$ where $\frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} =$ rate of flow of water in the nozzle and $v$ the velocity of water jet.
Since the ball is in equilibrium $F = mg$ where $m =$ mass of ping pong ball.
$\Rightarrow \ v\frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} = mg$ or rate of flow of water $\frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} = \frac{mg}{v}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.