| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | December |
| 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, but follows a standard template for velocity-dependent forces. Part (a) is scaffolded to guide students to the differential equation, parts (b) and (c) involve routine integration techniques (sinh substitution or recognition). While Further Mechanics content is inherently harder than core A-level, this particular question is relatively straightforward within that context with clear signposting and standard methods. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = k\sqrt{9 + 1.25^2} = 13\) | M1 [3.3] | Substituting \(v = 1.25\) and \(F = 13\) |
| \(k = 4\) | A1 [1.1] | |
| \(4\sqrt{9 + v^2} = 8\frac{dv}{dt} \Rightarrow \frac{dv}{\sqrt{9 + v^2}} = \frac{1}{2}\) (AG) | A1 [3.3] | Clear use of \(F = ma\) leading to AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{1}{\sqrt{9 + v^2}} dv = \frac{1}{2}t + c\) | M1 [3.4] | Separating the variables and correctly integrating to obtain bt |
| \(\sinh^{-1}\left(\frac{v}{3}\right) = \frac{1}{2}t + c\) | M1 [3.4] | Integrating other side to obtain \(p\sinh^{-1}qv\) |
| \(c = 0\) | B1 [1.1] | Substituting \(t = 0, v = 0\) into a solution of the DE to obtain \(c\) |
| \(v = 3\sinh\left(\frac{1}{2}t\right)\) | A1 [3.4] | |
| [4] | Condone omission of \(c\) for these two M marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = 3\sinh\left(\frac{1}{2}t\right) \Rightarrow x = C + 6\cosh\left(\frac{1}{2}t\right)\) | M1 [3.4] | Replacing \(v\) with \(\frac{dx}{dt}\) and integrating to obtain \(r\cosh st\) |
| \(t = 0, x = 0 \Rightarrow C = -6 \Rightarrow x = 6\left(\cosh\left(\frac{1}{2}t\right) - 1\right)\) oe | A1 [2.2a] | |
| [2] | Condone omission of \(C\) (or use of 'c' again) |
### (a)
$F = k\sqrt{9 + 1.25^2} = 13$ | M1 [3.3] | Substituting $v = 1.25$ and $F = 13$
$k = 4$ | A1 [1.1]
$4\sqrt{9 + v^2} = 8\frac{dv}{dt} \Rightarrow \frac{dv}{\sqrt{9 + v^2}} = \frac{1}{2}$ (AG) | A1 [3.3] | Clear use of $F = ma$ leading to AG
| [3]
### (b)
$\int\frac{1}{\sqrt{9 + v^2}} dv = \frac{1}{2}t + c$ | M1 [3.4] | Separating the variables and correctly integrating to obtain bt
$\sinh^{-1}\left(\frac{v}{3}\right) = \frac{1}{2}t + c$ | M1 [3.4] | Integrating other side to obtain $p\sinh^{-1}qv$
$c = 0$ | B1 [1.1] | Substituting $t = 0, v = 0$ into a solution of the DE to obtain $c$ | Or using 0 as limits on integrals
$v = 3\sinh\left(\frac{1}{2}t\right)$ | A1 [3.4]
| [4] | Condone omission of $c$ for these two M marks
### (c)
$\frac{dx}{dt} = 3\sinh\left(\frac{1}{2}t\right) \Rightarrow x = C + 6\cosh\left(\frac{1}{2}t\right)$ | M1 [3.4] | Replacing $v$ with $\frac{dx}{dt}$ and integrating to obtain $r\cosh st$
$t = 0, x = 0 \Rightarrow C = -6 \Rightarrow x = 6\left(\cosh\left(\frac{1}{2}t\right) - 1\right)$ oe | A1 [2.2a]
| [2] | Condone omission of $C$ (or use of 'c' again)
---4 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 2018 Q4 [9]}}