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Question

If n balls hit elastically and normally on a surface per unit time and all the balls are of mass m moving with the same velocity u, then force on the surface is:

A
m×u×n
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B
2×m×u×n
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C
12×mu2n
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D
mu2n
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Solution

The correct option is B 2×m×u×n
Since the ball rebounds with same speed but in opposite direction.
So, change in linear momentum of one ball ΔP=PfPi=mu(mu)=2mu
Number of ball hitting the surface per second is n
So, total change in momentum in one second ΔPT=n×2mu
So, force acting on the surface F=2mnu

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