A steel ball of mass 40.0g is dropped from a height 2.00m, it rebounds 1.60m what is the impluse during the time of contact.
mgh_1 + (1/2)m(v_i^2) = mgh_2 + (1/2)m(v_f^2)
mgh_1 = (1/2)m(v_f^2)
2gh_1 = v_f^2
v_f = sqrt(2gh_1), where h1 = 2.00m
Now impluse is the change in momentum so:
J = p_2 - p_1 -> J = 0 - m*sqrt(2gh_1)
Right, since the ball is stopped at 1.6m the final momentum has to be zero? However, I dont think this is the right answer.