Question

A 0.5 kg ball is dropped from rest at point 1.2 m above the floor. The ball rebounce straight upwards at the hight of 0.70 m . What is the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor

Answer 1

Solution: Recall that force is given by the equation: F~ = ∆~p/∆t. Plugging our values into this

gives:

F = ∆p/∆t

= 4.28 kg $\frac{m}{s}$

$5(10^{-3}) s$

= 855 N