A 380 kg steel ball strikes a wall with a speed of 90 ms at

A 3.80 kg steel ball strikes a wall with a speed of 9.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball? (Assume right is the positive direction.)

Solution

Change in momentum of ball = m (vf - vi)

= m 2 v sin60

and change in momentum = F x deltat

2 x 3.80 x 9 x sin60 = F x 0.2

F = 296.2 N