A 65-g handball is dropped to the floor from a height of 1 m, and bounces to a height of 0.80 m. What is the magnitude of the impulse received by the ball from the floor?

Answer:

First case :: while drop

V_i = initial velocity at height 1 m = 0 m/s

V_f = final velocity just before hitting the floor

d = displacement = 1 m

a = acceleration = 9.8

Using the equation

V_f² = V_i² + 2 a d

V_f² = 0² + 2 x 9.8 x 1

V_f = 4.43 m/s

Second case ::

First case :: while going up after hitting the floor

V_i = initial velocity after hitting the floor

V_f = final velocity at height 0.80 m = 0 m/s

d = displacement = 0.80 m

a = acceleration = - 9.8

Using the equation

V_f² = V_i² + 2 a d

0² = V_i² + 2 x (-9.8) x (0.8)

V_i = 3.96 m/s

Change in velocity = 3.96 - (-4.43) = 8.39 m/s

Impulse = change in momentum = m x change in veocity = 0.065 x 8.39 = 0.55 kgm/s