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a cricket ball of mass 150g is moving with a velocity of 12m/s and is hit by a bat so that it turns back with a velocity of 20m/s. The force of blow acts for a time of 0.01s. then find the change in momentum of ball and also find the average force exerted by the bat on the ball.
Mass of the ball, m = 150 g = 0.15 kgInitial speed of the ball, u = -12 m/sFinal speed of the bal after being hit, v = 20 m/sImpulse = change in momentum = final momentum initial momentum= Impulse = 0.15 × 20 0.15 × (-12) = 4.8 Ns
Mass of the ball, m = 150 g = 0.15 kg
Initial speed of the ball, u = -12 m/s
Final speed of the bal after being hit, v = 20 m/s
Impulse = change in momentum = final momentum initial momentum
= Impulse = 0.15 × 20 0.15 × (-12) = 4.8 Ns
We know that Force = Impulse/ Time Impulse = mass * change in velocityTime = 0.01 sMass m = 150 g = 1.5 x 10 ̄ ¹ kgChange in velocity = Final - intial velocity = 20 - (-12) = 32 m/sTherefore Force = mass * change in velocity / time = 1.5 x 10 ̄ ¹ x 32 / 0.01= 15 x 32 = 480 NAverage force on the ball is 480 N
Dear student,Please find the attached solution to your problem. Change in momentum of the ball = mv – mu = 0.15 x (20 – ( – 12)) = 0.15 x 32 = 4.8 NsFrom impulse momentum theorem, we know,Impulse = change in Linear momentumhence, F.Δt = 4.8hence, F = 4.8/0.01 = 480 N Hope it helps.Thanks and regards,Kushagra
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