Q.

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:


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a

2mgv

b

 mvg

c

 mgv

d

None of these 

answer is C.

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Detailed Solution

Concept: The change in momentum acts as the force that propels the ball forward when it is being struck by the water jet that is keeping it afloat. The weight of the ping-pong ball balances this force, therefore when it is floating, the ball is immobile due to the balance.
The product of mass and speed in mathematics is called the momentum, which is defined as the result of a body with mass m travelling with a velocity.
p=mv
The rate of change of momentum is directly proportional to the force acting on the body, according to Newton's second law.
If the equation were expressed in differential form,
F= dpdt
We know that p=mv
F= ddt(mv) 
The velocity change between the nozzle's opening and the top of the water jet exiting from the nozzle is constant when we use this equation to describe the case of a water jet flowing out of the nozzle. We must take into account the rate of flow of water as the rate of change of the mass of water because the water is emerging. As a result, while the mass is changing, the velocity is constant.
Consequently, the force that the water exerts when it leaves the nozzle,
F= Vdmdt
The ping-pong ball is in balance since it is floating on the water. This indicates that the weight of the ball balances the force generated by the water flowing through the nozzle.
The ball's weight is W=mg.
Also F=W
Vdmdt=mg
The water flow rate in this instance,
dmdt=mgv
Hence, the correct option is 3.
 
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