11 Two uniform smooth spheres A and B have equal radii and are moving on a smooth horizontal surface. The mass of \(A\) is 0.2 kg and the mass of \(B\) is 0.6 kg . The spheres collide obliquely. When the spheres collide the line joining their centres is parallel to \(\mathbf { i }\). Immediately before the collision the velocity of A is \(\mathbf { u } _ { \mathrm { A } } \mathrm { ms } ^ { - 1 }\) and the velocity of B is \(\mathbf { u } _ { \mathrm { B } } \mathrm { ms } ^ { - 1 }\). The coefficient of restitution between A and B is 0.5. Immediately after the collision the velocity of A is \(( - 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) and the velocity of B is \(( 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\).

1. Find \(\mathbf { u } _ { \mathrm { A } }\) and \(\mathbf { u } _ { \mathrm { B } }\). After the collision B collides with a smooth vertical wall which is parallel to \(\mathbf { j }\).
   The loss in kinetic energy of B caused by the collision with the wall is 1.152 J .
2. Find the coefficient of restitution between B and the wall.
3. Find the angle through which the direction of motion of B is deflected as a result of the collision with the wall. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a69c3e7a-dfc4-438b-84ba-e88c15d421ea-08_730_476_264_251} \captionsetup{labelformat=empty} \caption{Fig. 12}
   \end{figure} The ends of a light inextensible string are fixed to two points A and B in the same vertical line, with A above B. The string passes through a small smooth ring of mass \(m\). The ring is fastened to the string at a point P . When the string is taut the angle APB is a right angle, the angle BAP is \(\theta\) and the perpendicular distance of P from AB is \(r\). The ring moves in a horizontal circle with constant angular velocity \(\omega\) and the string taut as shown in Fig. 12.
4. By resolving horizontally and vertically, show that the tension in the part of the string BP is \(m \left( r \omega ^ { 2 } \cos \theta - g \sin \theta \right)\).
5. Find a similar expression, in terms of \(r , \omega , m , g\) and \(\theta\), for the tension in the part of the string AP. It is given that \(\mathrm { AB } = 5 a\) and \(\mathrm { AP } = 4 a\).
6. Show that \(16 a \omega ^ { 2 } > 5 g\). The ring is now free to move on the string but remains in the same position on the string as before. The string remains taut and the ring continues to move in a horizontal circle.
7. Find the period of the motion of the ring, giving your answer in terms of \(a , g\) and \(\pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a69c3e7a-dfc4-438b-84ba-e88c15d421ea-09_838_1132_280_248} \captionsetup{labelformat=empty} \caption{Fig. 13}
   \end{figure} A step-ladder has two sides AB and AC , each of length \(4 a\). Side AB has weight \(W\) and its centre of mass is at the half-way point; side AC is light. The step-ladder is smoothly hinged at A and the two parts of the step-ladder, AB and AC , are connected by a light taut rope DE , where D is on \(\mathrm { AB } , \mathrm { E }\) is on AC and \(\mathrm { AD } = \mathrm { AE } = a\). A man of weight \(4 W\) stands at a point F on AB , where \(\mathrm { BF } = x\).
   The system is in equilibrium with B and C on a smooth horizontal floor and the sides AB and AC are each at an angle \(\theta\) to the vertical, as shown in Fig. 13.
8. By taking moments about A for side AB of the step-ladder and then for side AC of the step-ladder show that the tension in the rope is $$W \left( 1 + \frac { 2 x } { a } \right) \tan \theta .$$ The rope is elastic with natural length \(\frac { 1 } { 4 } a\) and modulus of elasticity \(W\).
9. Show that the condition for equilibrium is that $$x = \frac { 1 } { 2 } a ( 8 \cos \theta - \cot \theta - 1 ) .$$ \section*{In this question you must show detailed reasoning.}
10. Hence determine, in terms of \(a\), the maximum value of \(x\) for which equilibrium is possible.