5. A particle \(P\) of mass 3 kg is moving under the action of a constant force \(\mathbf { F }\) newtons. At \(t = 0\), \(P\) has velocity \(( 3 \mathbf { i } - 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). At \(t = 4 \mathrm {~s}\), the velocity of \(P\) is \(( - 5 \mathbf { i } + 11 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find

1. the acceleration of \(P\), in terms of \(\mathbf { i }\) and \(\mathbf { j }\).
2. the magnitude of \(\mathbf { F }\). At \(t = 6 \mathrm {~s} , P\) is at the point \(A\) with position vector ( \(6 \mathbf { i } - 29 \mathbf { j }\) ) m relative to a fixed origin \(O\). At this instant the force \(\mathbf { F }\) newtons is removed and \(P\) then moves with constant velocity. Three seconds after the force has been removed, \(P\) is at the point \(B\).
3. Calculate the distance of \(B\) from \(O\).
   (6) \section*{6.} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{ebbf20a5-63f8-4c80-a8c4-477e1ea1eee7-4_298_1221_358_411}
   \end{figure} A non-uniform rod \(A B\) has length 5 m and weight 200 N . The rod rests horizontally in equilibrium on two smooth supports \(C\) and \(D\), where \(A C = 1.5 \mathrm {~m}\) and \(D B = 1 \mathrm {~m}\), as shown in Fig. 2. The centre of mass of \(A B\) is \(x\) metres from \(A\). A particle of weight \(W\) newtons is placed on the rod at \(A\). The rod remains in equilibrium and the magnitude of the reaction of \(C\) on the rod is 160 N .
4. Show that \(50 x - W = 100\). The particle is now removed from \(A\) and placed on the rod at \(B\). The rod remains in equilibrium and the reaction of \(C\) on the rod now has magnitude 50 N .
5. Obtain another equation connecting \(W\) and \(x\).
6. Calculate the value of \(x\) and the value of \(W\). \section*{7.} \section*{Figure 3}
   \includegraphics[max width=\textwidth, alt={}]{ebbf20a5-63f8-4c80-a8c4-477e1ea1eee7-5_688_1477_379_328}
   Figure 3 shows two particles \(A\) and \(B\), of mass \(m \mathrm {~kg}\) and 0.4 kg respectively, connected by a light inextensible string. Initially \(A\) is held at rest on a fixed smooth plane inclined at \(30 ^ { \circ }\) to the horizontal. The string passes over a small light smooth pulley \(P\) fixed at the top of the plane. The section of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane. The particle \(B\) hangs freely below \(P\). The system is released from rest with the string taut and \(B\) descends with acceleration \(\frac { 1 } { 5 } g\).
7. Write down an equation of motion for \(B\).
8. Find the tension in the string.
9. Prove that \(m = \frac { 16 } { 35 }\).
10. State where in the calculations you have used the information that \(P\) is a light smooth pulley. On release, \(B\) is at a height of one metre above the ground and \(A P = 1.4 \mathrm {~m}\). The particle \(B\) strikes the ground and does not rebound.
11. Calculate the speed of \(B\) as it reaches the ground.
12. Show that \(A\) comes to rest as it reaches \(P\). \section*{END}