Question

A ball weighing 10g hits a hard surface vertically with a speed of 5 ms–1 and rebounds 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

10 N as average force= ΔP/ Δt where ΔP is the change in momentum, Δt is the time period so we have F = ΔP/ Δt = (mv2-mv1)/0.01 where m is mass of ball of in kg = [0.01*(5-(-5))]/0.01 = 10 N