Question

A soft ball (m = 0.14 kg) moving horizontally to the right hits a wall with a speed of 16.2 m/s. After the collision the ball bounces back and moves horizontally to the left at a speed of 14.4 m/s. If the collision lasts for 0.0015 s, what is the magnitude of the (average) force exerted on the ball by the wall?

Answer 1

mass = m = 0.14 kg

initial velocity = Vi = 16.2 m/sec

final velocity = Vf = - 14.4 m/sec

time = t = 0.0015 sec

magnitude of force F = m ( Vf - Vi ) / t

                             F = 0.14 * ( -14.4 - 16.2 ) / 0.0015

                             F = 2856 N