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 the collision waterfalls dead, the rate of flow of water in the nozzle is equal to:
A.$\dfrac{2mg}{V}$
B.$\dfrac{mV}{g}$
C.$\dfrac{mg}{V}$
D. None of these

Answer 1

Hint: Momentum, for any moving body, is just the product of its mass and its velocity. Traditionally, symbolized as ‘p ‘, it could be formulated as$p=mv$ and is also the first derivative of force with respect to time i.e. Force can also be equated as the rate of change of momentum.

Complete step by step answer:
Initial velocity of water =V
Final velocity of water = 0
$\therefore $ The force exerted by the water on ping pong ball can be obtained by:
${{F}_{w}}=rate\text{ }of\text{ }change\text{ }of\text{ }momentum$
${{F}_{w}}=\dfrac{dp}{dt}$
${{F}_{w}}=\dfrac{dm}{dt}({{v}_{f}}-{{v}_{i}})$
${{F}_{w}}=\dfrac{dm}{dt}(v-0)$
${{F}_{w}}=v\dfrac{dm}{dt}$
In order for the ping pong ball to float in air, force due to gravity should cancel out the force exerted by the water on the ball
i.e. ${{F}_{w}}={{F}_{g}}$
 $\Rightarrow v\dfrac{dm}{dt}=mg$
$\Rightarrow \dfrac{dm}{dt}=\dfrac{mg}{v}$
Therefore, the correct option would be option (C) $\dfrac{mg}{V}$

Note:
One must take note of and remember that when an object s in equilibrium, all the forces acting on the body must cancel each other just like in this question where for a point when ping pong ball as in air, the force due to gravity is equated to force exerted by the water so that the ball stays in equilibrium.