| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 8 |
| Topic | Power and driving force |
| Type | Power from force and derived speed (non-equilibrium) |
| Difficulty | Standard +0.3 This is a straightforward Further Mechanics question requiring standard techniques: KE calculation from velocity components, integration of force to find velocity (using F=ma), and solving P=F·v=0. While it involves vectors and calculus, each step follows directly from definitions with no novel insight required. Slightly above average difficulty due to being Further Maths content and multi-step nature. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| \(KE = \frac{1}{2} \times 2 \times \left(\frac{-19.5}{-60}\right) \cdot \left(\frac{-19.5}{-60}\right) = 19.5^2 + 60^2\) awrt 3980 J | M1, A1 [2] | Using \(KE = \frac{1}{2}mv.v\) or \(\frac{1}{2}mv^2\) and attempting to find dot product or magnitude |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{4t}{-2}\right) = 2a = 2\frac{dv}{dt}\) | M1 [1.1] | Using \(F = ma\) with \(a = \frac{dv}{dt}\) soi |
| \(\Rightarrow v = \left(\frac{t^2}{-t}\right) + u\) | M1 [1.1] | Integrating to find v (u may be missing or may look like a scalar at this stage) |
| \(t = 0 \Rightarrow u = \left(\frac{-19.5}{-60}\right)\) so \(v = \left(\frac{t^2 - 19.5}{-t - 60}\right)\) oe | A1 [1.1] | |
| \(P = F.v = \left(\frac{4t}{-2}\right) \cdot \left(\frac{t^2 - 19.5}{-t - 60}\right) = 4t^3 - 76t + 120\) | M1 [1.1] | Using \(P = F.v\) and attempt at dot product |
| \(4t^3 - 7t + 120 = 0\) | A1 [1.1] | |
| \(t = 2, 3\) | A1 [2.2a] | BC If \(t = -5\) included then A0 |
### (a)
$KE = \frac{1}{2} \times 2 \times \left(\frac{-19.5}{-60}\right) \cdot \left(\frac{-19.5}{-60}\right) = 19.5^2 + 60^2$ awrt 3980 J | M1, A1 [2] | Using $KE = \frac{1}{2}mv.v$ or $\frac{1}{2}mv^2$ and attempting to find dot product or magnitude
### (b)
$\left(\frac{4t}{-2}\right) = 2a = 2\frac{dv}{dt}$ | M1 [1.1] | Using $F = ma$ with $a = \frac{dv}{dt}$ soi
$\Rightarrow v = \left(\frac{t^2}{-t}\right) + u$ | M1 [1.1] | Integrating to find **v** (u may be missing or may look like a scalar at this stage) | Must be reasonable attempt at integration
$t = 0 \Rightarrow u = \left(\frac{-19.5}{-60}\right)$ so $v = \left(\frac{t^2 - 19.5}{-t - 60}\right)$ oe | A1 [1.1]
$P = F.v = \left(\frac{4t}{-2}\right) \cdot \left(\frac{t^2 - 19.5}{-t - 60}\right) = 4t^3 - 76t + 120$ | M1 [1.1] | Using $P = F.v$ and attempt at dot product
$4t^3 - 7t + 120 = 0$ | A1 [1.1]
$t = 2, 3$ | A1 [2.2a] | BC If $t = -5$ included then A0
---1 A particle, $P$, of mass 2 kg moves in two dimensions. Its initial velocity is $\binom { - 19.5 } { - 60 } \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the initial kinetic energy of $P$.
For $t \geqslant 0 , P$ is acted upon only by a variable force $\mathbf { F } = \binom { 4 t } { - 2 } \mathrm {~N}$, where $t$ is the time in seconds.
\item Find
\begin{itemize}
\item the velocity of $P$ in terms of $t$,
\item the times when the power generated by $\mathbf { F }$ is zero.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q1 [8]}}