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.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = - 10 ( v - 0.5 )$.
\item By using $\int \frac { 1 } { v - 0.5 } \mathrm {~d} v = - \int 10 \mathrm {~d} t$, find $v$ in terms of $t$.
\item 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 $v$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2012 Q6 [10]}}