Question

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???

Answer 1

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.

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     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.

Answer 2

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.