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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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Question

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.

Answer in Brief
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Solution

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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Chapter 2: Force and Laws of Motion - Very Short Answers 2 [Page 75]

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Lakhmir Singh Physics Part 1 [English] Class 9
Chapter 2 Force and Laws of Motion
Very Short Answers 2 | Q 37.2 | Page 75

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