a crane drops a 1.0 kg steel ball onto a steel plate. the ball's speed just before impact is 2.0 m/s. if the average impact force is 300 N, and the ball is in contact with the plate for 0.01 seconds, what is the bounce back speed right after the impact.

Answer 1

F=Δp/Δt
FΔt=Δp =p₂-(-p₁)=mv₁+mv₂
(since p₁ directed ↓and p₂ ↑)
v₂ = {FΔt- mv₁}/m =
={300•0.01 - 1•2}/1 = 1 m/s