# A ball of mass 250g collides with a bat with velocity 10m/s and returns with the same velocity in 0.01 seconds what is the force applied???

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by ash007

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by ash007

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As the ball is hit.:u=0 v=10m/s ⇒10=0+aX0.01⇒a=1000m/s²

F=ma⇒F=0.25*1000=250N

So the force applied is 250N

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Change in velocity in 0.01 sec = +10m/s - (-10m/s) = 20 m/sec

Change in momentum = 250/1000 kg * 20 m/sec = 5 kg-m/sec

Force = rate of change in momentum = 5 / 0.01 = 500 Newtons

Direction of the force is same as the direction of velocity of the ball after rebounding.

====================

v = u + a t

u and v are in opposite directions. So u is -ve and v is positive.

+10 = -10 + a * 0.01

a = 2000 m/sec²

Force = m a = 250/1000 * 2000 = 500 Newtons, in the direction of v.

Change in momentum = 250/1000 kg * 20 m/sec = 5 kg-m/sec

Force = rate of change in momentum = 5 / 0.01 = 500 Newtons

Direction of the force is same as the direction of velocity of the ball after rebounding.

====================

v = u + a t

u and v are in opposite directions. So u is -ve and v is positive.

+10 = -10 + a * 0.01

a = 2000 m/sec²

Force = m a = 250/1000 * 2000 = 500 Newtons, in the direction of v.