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
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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
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