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प्रश्न
A 150 g ball, travelling at 30 m/s, strikes the palm of a player’s hand and is stopped in 0.05 second. Find the force exerted by the ball on the hand.
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उत्तर
Mass of the ball, m = 150 g = 0.15 kg
Initial velocity of the ball, u = 30 m/s
Initial momentum of the ball, pi = mu = 4.5 kg.m/s
The player’s hand stops the ball in 0.05 s.
Thus, final momentum of the ball is, pf = 0
So, change in momentum of the ball = pf − pi
= 0 − 4.5
= −4.5 kg.m/s
So, rate of change of momentum is = −4.5/0.05 = −90 N
This is the force exerted by the hand on the ball.
So, the force exerted by the ball on the hand is = 90 N
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