A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s.
(a) What impulse acts on the ball during the contact?
(b) If the ball is in contact with the floor for 0.020 s, what is the magnitude of the average force on the floor from the ball?
(a) Impulse on the ball = change in momentum for the ball = 1.2 kg * (10 - (-25)) m/s = ...
Solution is straightforward application of formulas.