Question

A cricket ball of mass 150 g moving at $36 m s^{-1}$ strikes a bat and
returns back along the same line at $21 m^{s -1}$ . What is the
change in momentum produced? If the bat remains in contact
with the ball for 1/20 s, what is the average force exerted in
newton.

Answer 1

Mass = 0.15 kg
U= 36mps
V= -21mps [rebound]
So, change in momentum = mv-mu=0.15 [36-(-21)]=0.15*57=8.55
Force = Rate of change of momentum = change in p/t = 8.55 / (1/20) =171 N
Please recheck in case of calculation mistakes