Question

After a 0.300-kg rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor
and rebounds to a height of 1.50 m. (a) Determine the magnitude and direction of the impulse
delivered to the ball by the floor. (b) Suppose the ball is in contact with the floor for 0.05 s.
Calculate the average force the floor exerts on the ball.

Answer 1

The impulse is the change of momentum of the ball
coming down:
(1/2) m v^2 = m g h
h = 1.75 meters
v down = -sqrt(2 g h) = -sqrt(2*9.81*1.75)
now v up
v up = +sqrt (2*9.81*1.50)
change in momentum = + m sqrt [ 2 * 9.81 (1.50 +1.75) ]
= impulse
then F * time = impulse
F = impulse / 0.05