Momentum problem

Question

A 300.0 g ball is dropped from a height of 5.1 m onto a table top. After colliding with the table it bounces straight up a distance of 4.0 m. If the ball is in contact with the table for 0.22 s, what is the force, in Newtons, exerted by the table on the ball?

Answer 1

Dropped from a height of 5.1 m---Use the free-fall equation to get the velocity at impact:

d = 1/2 * a * t^2 = 5.1 (a = 9.81)

t^2 = 5.1 * 2 / 9.81 , t = 1.02

velocity at impact = a*t = 9.81 * 1.02 = 10.0

momentum at impact: 10 * 300 = 3,000

Same procedure to get the rebound velocity:

d = 1/2 * a * t^2 = 4

t^2 = 4 * 2 / 9.81

t = 0.90

v = 0.90 * 9.81 = 8.86

momentum after impact: - 8.86 * 300 = -2,658

Impulse supplied by the table: -3,000 - 2,658 = 5,658

Impulse supplied in 0.22 seconds, so force = 5,658 / 0.22 = 25,718 N