A ball whose mass is 0.2kg hits the floor with a speed of 8m/s and rebounds upward with a speed of 7 m/s. The ball was in contact with the floor for 0.5 ms.

Question

(a) What was the average magnitude of the force exerted on the ball by the floor?
(b) Calculate the magnitude of the gravitational force that the Earth exerts on the ball.
(c) In a collision for a brief time, there are forces between the colliding objects that are much greater than external forces.

Answer 1

You know the third law, right?
Net force = rate of change of momentum = m a
so find the change of momentum P
Pdown = .2 * -.8
Pup = .2 * .7
dP = change in P = .2 (.7 - -.8) = .2 * 1.5 = .3 kg m/s
dT = .5 * 10^-3 s
so
F = dP/dt = (.3/.5)10^3 = 600 N

gravity force = m g = .3 * 9.81 N, about 3 N, much less

Answer 2

I mean second law !