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\includegraphics[max width=\textwidth, alt={}, center]{67af8d98-85af-42b1-9e7f-c6380a1f8a3f-4_638_473_260_836} A particle \(P\), of mass 3.5 kg , is in equilibrium suspended from the top \(A\) of a smooth slope inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 40 } { 49 }\), by an elastic rope of natural length 4 m and modulus of elasticity 112 N (see diagram). Another particle \(Q\), of mass 0.5 kg , is released from rest at \(A\) and slides freely downwards until it reaches \(P\) and becomes attached to it.

1. Find the value of \(V ^ { 2 }\), where \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(Q\) immediately before it becomes attached to \(P\), and show that the speed of the combined particles, immediately after \(Q\) becomes attached to \(P\), is \(\frac { 1 } { 2 } \sqrt { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The combined particles slide downwards for a distance of \(X \mathrm {~m}\), before coming instantaneously to rest at \(B\).
2. Show that \(28 X ^ { 2 } - 8 X - 5 = 0\).