Question

marble is dropped from rest onto the floor 2.4 meters below. if the marble bounces straight upward to a height of 1.2 meters, what is the impulse delivered to marble by the floor?

Answer 1

The impulse is equal to change in momentum

$$\Delta p=|m{\bf v}_f-m{\bf v}_i|.$$

The law of conservation of energy gives

$$v_i=\sqrt{2gh_i}, \quad v_f=\sqrt{2gh_f}.$$

Hence

$$\Delta p=m\sqrt{2g}(\sqrt{h_f}+\sqrt{h_i})$$

$$=m\sqrt{2\times 9.8}(\sqrt{1.2}+\sqrt{2.4})=11.7m$$