A ping-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 mass flow of water in the nozzle is equal to:

A ping-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 mass flow of water in the nozzle is equal to:

1. A

$\frac{2\mathrm{mg}}{\mathrm{v}}$

2. B

$\frac{\mathrm{mv}}{\mathrm{g}}$

3. C

$\frac{\mathrm{mg}}{\mathrm{v}}$

4. D

None of these

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Solution:

The impact force F = (dp/dt) where dp = change of momentum of water of mass dm striking  the ball a speed v during time dt.

Since water falls dead after collision with the ping-pong ball
$\mathrm{\Delta p}=\mathrm{\Delta mv}⇒\mathrm{F}=\mathrm{v}\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}$upward on the ball

Where $\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}$= rate of flow of water in the nozzle.

Since the ball is in equilibrium
$\begin{array}{l}\mathrm{F}-\mathrm{mg}=0⇒\mathrm{F}=\mathrm{mg}\\ ⇒=\mathrm{mg}⇒\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}=\frac{\mathrm{mg}}{\mathrm{v}}\end{array}$

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