Question

Kim and mike are on opposing dodgeball teams. Kim grabs a ball of mass 0.400 kg and throws it with a speed of 1.2m/s at the same time that Kim is throwing her ball mike had a slightly lighter under inflated ball with mass 0.35kg and throws it at Kim with a speed of 1.8m/s, his ball travels at an angle of 75 degrees to kim. As a result of the collision kim’s ball is deflected at an angle of 30 degrees from its original path with a speed of 1.6m/s. Determine the speed and angle of mike’s ball do not assume the collision is elastic.

Answer 1

$m_1v_{x1}-m_2v_{x2}=m_1v_{x1}'-m_2v_{x2}'\to$

$v_{x2}'=\frac{-m_1v_{x1}+m_2v_{x2}+m_1v_{x1}'}{m_2}=\frac{-0.4\cdot 1.2+0.35\cdot 1.8\cdot\cos75°+0.4\cdot 1.6\cdot\cos30°}{0.35}=0.678\ (m/s)$

$m_2v_{y2}=m_1v_{y1}'+m_2v_{y2}'\to$

$v_{y2}'=\frac{m_2v_{y2}-m_1v_{y1}'}{m_2}=\frac{0.35\cdot 1.8\cdot\sin75°-0.4\cdot1.6\cdot\sin30°}{0.35}=0.824\ (m/s)$

$v_2'=\sqrt{0.678^2+0.824^2}=1.067\ (m/s)$ . Answer

$\alpha=\tan^{-1}\frac{v_{y2}'}{v_{x2}'}=\tan^{-1}\frac{0.824}{0.678}=50.54°$ . Answer