Particle-barrier collision with angle

6 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 \(90 ^ { \circ }\).

1. Show that \(\tan ^ { 2 } \alpha = \frac { 1 } { e }\).
   The particle \(P\) loses two-thirds of its kinetic energy in the impact.
2. Find the value of \(\alpha\) and the value of \(e\).

1. \begin{figure}[h] \end{figure} A fixed smooth plane is inclined to the horizontal at an angle of \(45 ^ { \circ }\). A particle \(P\) is moving horizontally and strikes the plane. Immediately before the impact, \(P\) is moving in a vertical plane containing a line of greatest slope of the inclined plane. Immediately after the impact, \(P\) is moving in a direction which makes an angle of \(30 ^ { \circ }\) with the inclined plane, as shown in Figure 1. Find the fraction of the kinetic energy of \(P\) which is lost in the impact.

3. A smooth uniform sphere \(P\) of mass \(m\) is falling vertically and strikes a fixed smooth inclined plane with speed \(u\). The plane is inclined at an angle \(\theta , \theta < 45 ^ { \circ }\), to the horizontal. The coefficient of restitution between \(P\) and the inclined plane is \(e\). Immediately after \(P\) strikes the plane, \(P\) moves horizontally.

1. Show that \(e = \tan ^ { 2 } \theta\).
2. Show that the magnitude of the impulse exerted by \(P\) on the plane is \(m u \sec \theta\).