A ping-pong ball of mass m is floating in air | vedclass

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Question

The impact force,

$F=\frac{\Delta p}{\Delta t}$

Where, $\Delta \mathrm{P}=$ change of momentum of water of mass $\Delta \mathrm{m}$ striking the pen

Vedclass · Accepted Answer

$\frac {mg}{v}$

Vedclass · Answer

The impact force,

$F=\frac{\Delta p}{\Delta t}$

Where, $\Delta \mathrm{P}=$ change of momentum of water of mass $\Delta \mathrm{m}$ striking the pendulum with a speed v during time $\Delta t$

since water falls dead after collision with the ping-pong ball,

$\Delta \mathrm{p}=\Delta \mathrm{mv}$

$\mathbf{O r}$        ${\mathrm{F}=\mathrm{v} \frac{\Delta \mathrm{m}}{\Delta \mathrm{t}}}$

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

since, the ball is in equilibrium,

${F-\mathrm{mg}=0 	ext { or } \mathrm{F}=\mathrm{mg}}$

${\mathrm{v} \frac{\Delta \mathrm{m}}{\Delta \mathrm{t}}=\mathrm{mg}}$

or          ${\frac{\Delta \mathrm{m}}{\Delta \mathrm{t}}=\frac{\mathrm{mg}}{\mathrm{v}}}$

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 mass flow rate of water in the nozzle is equal to