| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2020 |
| Session | May |
| Marks | 10 |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem using v dv/dx = F/m with straightforward integration. Part (a) requires setting up and solving F = ma with the given force, then using two boundary conditions to find constants—routine for FM students. Part (b) involves separating variables and integrating to find time, which is a well-practiced technique. The algebra is slightly involved but follows standard methods without requiring novel insight. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums6.06a Variable force: dv/dt or v*dv/dx methods |
11.
A particle, $P$, of mass 0.4 kg is moving along the positive $x$-axis, in the positive $x$ direction under the action of a single force. At time $t$ seconds, $t > 0 , P$ is $x$ metres from the origin $O$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The force is acting in the direction of $x$ increasing and has magnitude $\frac { k } { v }$ newtons, where $k$ is a constant.
At $x = 3 , v = 2$ and at $x = 6 , v = 2.5$
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 3 } = \frac { 61 x + 9 } { 24 }$
The time taken for the speed of $P$ to increase from $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is $T$ seconds.
\item Use algebraic integration to show that $T = \frac { 81 } { 61 }$
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2020 Q11 [10]}}