2
\includegraphics[max width=\textwidth, alt={}, center]{184020e1-7ff2-4172-8d33-baff963afa76-2_531_760_927_696} Two smooth vertical walls each with their base on a smooth horizontal surface intersect at an angle of \(60 ^ { \circ }\). A small smooth sphere \(P\) is moving on the horizontal surface with speed \(u\) when it collides with the first vertical wall at the point \(D\). The angle between the direction of motion of \(P\) and the wall is \(\alpha ^ { \circ }\) before the collision and \(75 ^ { \circ }\) after the collision. The speed of \(P\) after this collision is \(v\) and the coefficient of restitution between \(P\) and the first wall is \(e\). Sphere \(P\) then collides with the second vertical wall at the point \(E\). The speed of \(P\) after this second collision is \(\frac { 1 } { 4 } u\) (see diagram). The coefficient of restitution between \(P\) and the second wall is \(\frac { 3 } { 4 }\).

1. By considering the collision at \(E\), show that \(v = \frac { \sqrt { } 2 } { 5 } u\).
2. Find the value of \(\alpha\) and the value of \(e\).