Question

A ball of mass 200 g is dropped from a height 3 m above the ground onto a hard floor, and rebounds to a height of 2.2 m. If the ball is in contact with the floor for 0.002 s calculate the average force exerted by the floor on the ball

Answer 1

V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8
V = 7.67 m/s.


V^2 = Vo^2 + 2g*h = 0 @ max ht.
Vo^2 - 19.6*2.2 = 0
Vo^2 = 43.12
Vo = 6.57 m/s.

a = (-6.57-7.67)/0.002s = -7120 m/s^2.

F = m*a = 0.20 * (-7120) = -1424 N.

Answer 2

A ball of mass 1kg is dropped from a height of 7m and rebounds to a height of 4.5m. Calculate:
(I) its KE just before impact
(II) its initial rebound velocity and KE
(III) Account for the loss of KE on the impact.