a ball of mass 0.25kg moves on frictinless horizontal floor and hits a vertical wall with speed 5m/s . the ball rebound with speed 40m/s. if the ball was in contact with the ball for 0.150 seconds, then the average force that acts on the ball is

Average force = (change in linear momentum)/timeInitial momentum= 0.25kg× (+5)m/s =1.25kg-m/SFinal momentum=0.25kg×(-40)m/s = -10kg-m/SAverage force= (-10-1.25)kg-m/s/0.15s =-11.25/0.15kg-m/s^2 =-75 Joules