Question

A ball weighing 10 g hits a hard surface with a speed of 5 m/s and rebound with the same speed .the ball remains in contact with the surface for 0.01 s the average force. exerted by the surface on the ball is

Answer 1

Favg = change in movementum /time
F = 2*0.01*5/0.01=10N

Answer 2

Answer:

10 N

Explanation:

Force = change in momentum / time

F = (mv - mu)/t

F = [1/100 ( 5 - (-5) ) ]/ (1/100)       [in SI units]

F = 1/100 * 10 * 100

F = 10 N