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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 :A. `(2mg)/(v)`B. `(mv)/(g)`C. `(mg)/(v)`D. None of these |
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Answer» The impact force `F=(Deltap//Deltat)`, where `Deltap=` change of momentum of water of mass `Deltam` striking the ball with a speed `v` during time `Delta t`. Since water falls dead after collision with the ping-pong ball `Deltap=DeltamvrArrF=v(Deltam)/(Deltat)` where `(Deltam)/(Deltat)=` rate of flow of water in the nozzle. Since the ball is in equilibrium `F-mg=0rArrF=mg` `rArr(vDeltam)/(Deltat)=mg` `rArr(Deltam)/(Deltat)=(mg)/(v)` |
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