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

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(C) The impact force $F$ exerted by the water on the ball is given by the rate of change of momentum of the water.$F = \frac{\Delta p}{\Delta t}$Here,

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$\frac{mg}{v}$

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(C) The impact force $F$ exerted by the water on the ball is given by the rate of change of momentum of the water.$F = \frac{\Delta p}{\Delta t}$Here,$\Delta p$ is the change in momentum of water of mass $\Delta m$ striking the ball with speed $v$ during time interval $\Delta t$. Since the water falls dead after the collision,its final velocity is $0$.$\Delta p = \Delta m \cdot v - \Delta m \cdot 0 = \Delta m \cdot v$Therefore,$F = v \frac{\Delta m}{\Delta t}$,where $\frac{\Delta m}{\Delta t}$ is the mass flow rate of water.Since the ball is in equilibrium,the upward force exerted by the water must balance the weight of the ball:$F = mg$Substituting the expression for $F$:$v \frac{\Delta m}{\Delta t} = mg$$\frac{\Delta m}{\Delta t} = \frac{mg}{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 the water falls dead,the mass flow rate of water in the nozzle is equal to

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