a 150g cricket ball moving horizontally at 20m/s was hit straight back to the bowler at 12m/s. if the contact with the bat lasted for 1/25 sec, the average force exerted by the bat on the ball is Share with your friends Share 9 Ravi Gupta answered this change in momentum = mv-(-mu)= m(v+u) = 1501000×20+12 = 4.8 kg m/sforce = change in momentum/time = 4.81/25 = 120 N 23 View Full Answer