Question

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. $\frac{2mg}{v}$
• B. $\frac{mv}{g}$
• C. $\frac{mg}{v}$
• D. None of these.

Answer 1

Answer: (C)

$\frac{mg}{v}$

Answer 2

The impact force F = (∆p/∆t) where ∆p = change of momentum of water of mass ∆m striking the ball a speed v during time ∆t. Since water falls dead after collision with the ping-pong ball

∆p = ∆m v ⇒ F = v$\frac{\Delta m}{\Delta t}$ upward on the ball

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

Since the ball is in equilibrium

F – mg = 0 ⇒ F = mg

⇒ $\frac{v\Delta m}{\Delta t}$ = mg ⇒ $\frac{\Delta m}{\Delta t}$= $\frac{mg}{v}$.

Hence, (3) is the correct choice.