Answer:
[tex]v_0 = 5.89 m/s[/tex]
Explanation:
As we know that the ball swings out to maximum angle of 50 degree
so we will have use energy conservation to find the initial speed of the ball
[tex]\frac{1}{2}mv^2 = mgL(1 - cos\theta)[/tex]
[tex]v = \sqrt{2gL(1 - cos\theta)}[/tex]
so we have
[tex]v = \sqrt{2(9.81)(1)(1 - cos50)}[/tex]
[tex]v = 2.65 m/s[/tex]
now we can use momentum conservation for this collision
[tex]m_1v_0 = m_1v_{1f} + m_2v_{2f}[/tex]
[tex]29 v_0 = 29v_{1f} + 100(2.65)[/tex]
[tex]v_0 = v_{1f} + 9.14[/tex]
also we know that it is elastic collision
so we will have
[tex]2.65 - v_{1f} = v_0[/tex]
now from above two equations we have
[tex]v_0 = 5.89 m/s[/tex]