Answer:
Part a)
[tex]m_Av_A = m_Av_A' cos\theta_A' + m_Bv_B'cos\theta_B'[/tex]
Part b)
[tex]0 = m_Av_A' sin\theta_A' - m_Bv_B'sin\theta_B'[/tex]
Part c)
[tex]\theta_B' = 46.9 degree[/tex]
Part d)
[tex]v_B' = 1.15 m/s[/tex]
Explanation:
Part a)
As we know that momentum is conserved as there is no external force on the system of two balls
So here we will have
Along X direction we have
[tex]m_Av_A = m_Av_A' cos\theta_A' + m_Bv_B'cos\theta_B'[/tex]
Part b)
Now momentum conservation along Y direction is given as
[tex]0 = m_Av_A' sin\theta_A' - m_Bv_B'sin\theta_B'[/tex]
Part c)
Now we have
[tex]0.115(2.80) = 0.115(2.10) cos30 + 0.144v_B'cos\theta_B'[/tex]
[tex]v_B'cos\theta_B' = 0.784[/tex]
[tex]0 = 0.115(2.10)sin30 - 0.144v_B'sin\theta_B'[/tex]
[tex]v_B'sin\theta_B' = 0.839[/tex]
now from above two equations we have
[tex]\theta_B' = 46.9 degree[/tex]
Part d)
Now from any of the above equation we have
[tex]v_B' = 1.15 m/s[/tex]