Pendulum particle collision at lowest point

A particle \(P\), of mass \(m\), is able to move in a vertical circle on the smooth inner surface of a sphere with centre \(O\) and radius \(a\). Points \(A\) and \(B\) are on the inner surface of the sphere and \(A O B\) is a horizontal diameter. Initially, \(P\) is projected vertically downwards with speed \(\sqrt { } \left( \frac { 21 } { 2 } a g \right)\) from \(A\) and begins to move in a vertical circle. At the lowest point of its path, vertically below \(O\), the particle \(P\) collides with a stationary particle \(Q\), of mass \(4 m\), and rebounds. The speed acquired by \(Q\), as a result of the collision, is just sufficient for it to reach the point \(B\).

1. Find the speed of \(P\) and the speed of \(Q\) immediately after their collision.
   In its subsequent motion, \(P\) loses contact with the inner surface of the sphere at the point \(D\), where the angle between \(O D\) and the upward vertical through \(O\) is \(\theta\).
2. Find \(\cos \theta\).

4 A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is held vertically above \(O\) with the string taut and then projected horizontally with speed \(\sqrt { } \left( \frac { 13 } { 3 } a g \right)\). It begins to move in a vertical circle with centre \(O\). When \(P\) is at its lowest point, it collides with a stationary particle of mass \(\lambda m\). The two particles coalesce.

1. Show that the speed of the combined particle immediately after the impact is \(\frac { 5 } { \lambda + 1 } \sqrt { } \left( \frac { 1 } { 3 } a g \right)\). In the subsequent motion, the string becomes slack when the combined particle is at a height of \(\frac { 1 } { 3 } a\) above the level of \(O\).
2. Find the value of \(\lambda\).
3. Find, in terms of \(m\) and \(g\), the instantaneous change in the tension in the string as a result of the collision.