| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Topic | Variable Force |
| Type | Force depends on velocity v |
| Difficulty | Challenging +1.2 This is a Further Mechanics question requiring separation of variables and integration of expressions involving √(9+v²), which leads to sinh/arcsinh functions. While it requires multiple steps and familiarity with hyperbolic functions, the question provides clear guidance through parts (a)-(c), with part (a) essentially giving away the differential equation setup. The integrations are standard for Further Maths students who have covered hyperbolic functions, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 4.10c Integrating factor: first order equations6.06a Variable force: dv/dt or v*dv/dx methods |
3 A particle $P$ of mass 8 kg moves in a straight line on a smooth horizontal plane. At time $t \mathrm {~s}$ the displacement of $P$ from a fixed point $O$ on the line is $x \mathrm {~m}$ and the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$. Initially, $P$ is at rest at $O$.\\
$P$ is acted on by a horizontal force, directed along the line away from $O$, with magnitude proportional to $\sqrt { 9 + v ^ { 2 } }$. When $v = 1.25$, the magnitude of this force is 13 N .
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { 1 } { \sqrt { 9 + v ^ { 2 } } } \frac { \mathrm {~d} v } { \mathrm {~d} t } = \frac { 1 } { 2 }$.
\item Find an expression for $v$ in terms of $t$ for $t \geqslant 0$.
\item Find an expression for $x$ in terms of $t$ for $t \geqslant 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q3 [9]}}