6 Alice places a toy, of mass 0.4 kg , on a slope. The toy is set in motion with an initial velocity of \(1 \mathrm {~ms} ^ { - 1 }\) down the slope. The resultant force acting on the toy is \(( 2 - 4 v )\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the toy's velocity at time \(t\) seconds after it is set in motion.

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 10 ( v - 0.5 )\).
2. By using \(\int \frac { 1 } { v - 0.5 } \mathrm {~d} v = - \int 10 \mathrm {~d} t\), find \(v\) in terms of \(t\).
3. Find the time taken for the toy's velocity to reduce to \(0.55 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
   \(7 \quad\) A small bead, of mass \(m\), is suspended from a fixed point \(O\) by a light inextensible string of length \(a\). With the string taut, the bead is at the point \(B\), vertically below \(O\), when it is set into vertical circular motion with an initial horizontal velocity \(u\), as shown in the diagram.
   \includegraphics[max width=\textwidth, alt={}, center]{06c3e260-8167-4616-97d4-0f360a376a0f-5_616_613_520_733} The string does not become slack in the subsequent motion. The velocity of the bead at the point \(A\), where \(A\) is vertically above \(O\), is \(v\).