Answer to Question #96837 in Mechanics | Relativity for Nash addae

Question

A 60-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 24.0 m/s. If the player was moving upward with a speed of 4.28 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards? Assume the velocity of the player directly after the collision is upwards with a magnitude of 3.87 m/s.

Answer 1

momentum conservation law:

$m_s\vec{v}_{s1}+m_b\vec{v}_{b1}=m_s\vec{v}_{s2}+m_b\vec{v}_{b2}\\ m_s{v}_{s1}-m_b{v}_{b1}=-m_s{v}_{s2}+m_b{v}_{b2}=m_s({v}_{s1}+\Delta v)+m_b{v}_{b2}\\ v_{b2}=\frac{m_s{v}_{s1}-m_b{v}_{b1}-m_s({v}_{s1}+\Delta v)}{m_b}=\\ =\frac{-m_b{v}_{b1}-m_s\Delta v}{m_b} = \\ =\frac{-0.45\cdot24-60\cdot3.87}{0.45}=540 m/s$