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

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(C) The impact force exerted by the water on the ball is given by Newton's second law in terms of momentum change:$F = \frac{\Delta p}{\Delta t}$Here,

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$\frac{mg}{v}$

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(C) The impact force exerted by the water on the ball is given by Newton's second law in terms of momentum change:$F = \frac{\Delta p}{\Delta t}$Here,$\Delta p$ is the change in momentum of the water of mass $\Delta m$ striking the ball with speed $v$ during time $\Delta t$. Since the water falls dead after the collision,its final velocity is $0$.Therefore,the change in momentum is $\Delta p = \Delta m \cdot v - 0 = \Delta m \cdot v$.Substituting this into the force equation:$F = \frac{\Delta m \cdot v}{\Delta t} = v \left( \frac{\Delta m}{\Delta t} \right)$Where $\frac{\Delta m}{\Delta t}$ is the rate of flow of water in the nozzle. Since the ball is floating in equilibrium,the upward force exerted by the water must balance the weight of the ball:$F = mg$Equating the two expressions for $F$:$v \left( \frac{\Delta m}{\Delta t} \right) = mg$Solving for the rate of flow:$\frac{\Delta m}{\Delta t} = \frac{mg}{v}$

$A$ 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 rate of flow (in $kg/s$) of water in the nozzle is equal to

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