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 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: When the jet of water is hitting the ball, keeping it floating, the change in the momentum translates to the force that keeps the ball floating. This force is balanced by the weight of the ping-pong ball which balances the force and hence, the ball is stationary when it is floating.

Complete step-by-step answer:
The momentum is defined as the effect produced by the body of mass m when it is moving with a velocity and mathematically, it is the product of mass and velocity.
$p = mv$
The Newton’s second law states that –
The rate of change of momentum is directly proportional to the force applied on the body.
If we were to write the equation in the differential form –
$F = \dfrac{{dp}}{{dt}}$
Substituting the value of momentum,
$F = \dfrac{d}{{dt}}\left( {mv} \right)$
If we apply this equation to the case of a water jet flowing out of the nozzle, the velocity change is constant from the opening of the nozzle and the top of the water jet emerging out of the nozzle. Since the water is emerging, we have to consider the rate of flow of water as the rate of change of the mass of water. Thus, the velocity is constant and the mass is changing.
So, the force exerted by the water emerging from the nozzle,
$F = v\dfrac{{dm}}{{dt}}$
Since the ping-pong ball is floating on the water, the ping-pong ball is in equilibrium. This means that the force exerted by the water through the nozzle is balanced by the weight of the ball.
The weight of the ball, $W = mg$
Equating,
$F = W$
$v\dfrac{{dm}}{{dt}} = mg$
Here, the rate of flow of water,
$\dfrac{{dm}}{{dt}} = \dfrac{{mg}}{v}$

Hence, the correct option is C.

Note: The students are aware that Newton’s second law gives the equation, $F = ma$
This equation is derived from the equation that is defined above.
$F = \dfrac{d}{{dt}}\left( {mv} \right)$
With the mass being constant for a body, the velocity changing constantly, we get –
$F = m\dfrac{{dv}}{{dt}}$
The term $\dfrac{{dv}}{{dt}}$ is the rate of change of velocity with time, which is equal to the quantity acceleration.
$a = \dfrac{{dv}}{{dt}}$
Thus,
$F = ma$
So, we see that the famous equation above is derived from the force being equal to the rate of change of momentum.