A particle \(P\) of mass \(m\) is moving with speed \(u\) on a fixed smooth horizontal surface. The particle strikes a fixed vertical barrier. At the instant of impact the direction of motion of \(P\) makes an angle \(\alpha\) with the barrier. The coefficient of restitution between \(P\) and the barrier is \(e\). As a result of the impact, the direction of motion of \(P\) is turned through \(90^{\circ}\).
(a) Show that \(\tan ^{2} \alpha=\frac{1}{e}\).

The particle \(P\) loses two-thirds of its kinetic energy in the impact.
(b) Find the value of \(\alpha\) and the value of \(e\).