A ball of 150g is thrown with velocity of 20 m/s towards the batsman. He hits the ball in the direction opposite to its initial direction of motion with velocity 25m/s. If the ball is hit in 0.01s , then find out the change in momentum of the ball and force applied by the batsman to the ball.

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Solution

Dear Student
$m = 150 g = 0.15 g u = 20 m / s v = - 25 m / s (N e g a t i v e s i g n a s i t s d i r e c t i o n i s o p p o s i t e t o t h a t o f i n i t i a l v e l o c i t y) I n i t i a l M o m e n t u m = m u = 0.15 \times 20 = 3.00 k g . m / s F i n i a l M o m e n t u m = m v = 0.15 \times (- 25) = - 3.75 k g . m / s C h a n g e i n m o m e n t u m = F i n i a l M o m e n t u m - I n i t i a l M o m e n t u m d P = - 3.75 - 3.00 = - 6.75 k g . m / s F o r c e = \frac{C h a n g e i n m o m e n t u m}{T i m e} = \frac{- 6.75}{0.01} = 675 N$
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