The problem is as follows:
The diagram from below shows a sphere labeled $A$ and is moving with a horizontal speed of $v=4\sqrt{10}\,\frac{m}{s}$ over a frictionless table. After the collision the cable makes an angle of $53^{\circ}$ with the vertical and the sphere $A$ ends at rest. Find the $COR$ (coefficient of restitution) and the relationship between the masses.
The alternatives given are as follows:
$\begin{array}{ll} 1.&\frac{1}{2},\, \frac{1}{2}\\ 2.&\frac{1}{3},\, \frac{1}{6}\\ 3.&\frac{1}{2},\, \frac{1}{3}\\ 4.&\frac{1}{8},\, \frac{1}{2}\\ 5.&\frac{1}{2},\, \frac{1}{6}\\ \end{array}$
When it mentions the COR for the collision I'm assuming that it is an inelastic collision. Therefore if treated as such then
$p_1=p_2$
$m_1 v_1=m_2 v_2$
$v_2=4\sqrt{10} \frac{m_1}{m_2}$
But that's how far I went. I'm stuck with this problem, can someone help me here?.
Initial momentum of $A$ is $4\sqrt{10}m_A$. This is equal to the momentum of $B$ after the collision. The kinetic energy of $B$ after the collision is then $p^{2}/2m_B = 80m_{A}^{2}\big/m_B$.
As has been noted, $B$ is raised by $2\text{m}$ by the collision, giving it a potential energy of $20m_{B} = \frac{1}{2}m_{B}v_{B}^{2}$. Note therefore that $v_{B} = \sqrt{40} = 2\sqrt{10}$.
Therefore $80m_{A}^{2} = 20m_{B}^{2}$, meaning $m_{A}/m_{B} = 1/2$.
The COR is then $$\frac{v_{B}}{v_A} = \frac{2\sqrt{10}}{4\sqrt{10}} = 1/2.$$