| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Variable Force |
| Type | Force depends on velocity v |
| Difficulty | Challenging +1.2 This is a Further Mechanics differential equation problem requiring Newton's second law with time-dependent forces, solving a first-order linear ODE with integrating factor, then integrating velocity to find displacement. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are standard for FM students: the ODE form is given in part (i), and the solution method is routine application of integrating factors. The numerical integration in part (iii) is straightforward once v(t) is found. |
| Spec | 4.10c Integrating factor: first order equations6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \frac{6}{t^2} - \frac{3v}{t} = -1.5a = 1.5\frac{dv}{dt} = \frac{dv}{dt} + \frac{2v}{t} = -\frac{4}{t^2}\) | B1 [1] | AG; Using Newton II with \(a = \frac{dv}{dt}\) and rearranging |
| Answer | Marks | Guidance |
|---|---|---|
| Integrating factor is \(e^{\int \frac{2}{t}dt} = e^{2\ln t} = t^2\) | M1 | |
| \(DE\) is \(\frac{d}{dt}(vt^2) = 4\) | A1 | |
| \(vt^2 = 4t + c\) | M1 | Using their integrating factor |
| \(v = 0\) when \(t = 2\) gives \(0 = 8 + c\) | M1 | Use of \(v = 0\), \(t = 2\) in their equation to find a constant of integration |
| \(vt^2 = 4t - 8\) | A1 | |
| when \(t = 20\), speed of the piston is 0.18 m s\(^{-1}\) | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{4}{t} - \frac{8}{t^2} \Rightarrow\) distance = \(\int_2^{20} \left(\frac{4}{t} - \frac{8}{t^2}\right) dt\) | M1 | Use \(v = \frac{dx}{dt}\) and expression of distance moved as a definite integral |
| \(= 5.6103...\) so distance is 5.61 m correct to 3sf | A1 | Correct integrand and limits; BC; AG so justification for 3sf accuracy is needed |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{4}{t} - \frac{8}{t^2} \Rightarrow x = 4 \ln t + \frac{8}{t} + C\) | M1 | Use \(v = \frac{dx}{dt}\) and attempt to integrate |
| \(x = 0\) when \(t = 2\) gives \(C = -4 \ln 2 - 4\) | A1 | |
| so \(t = 20\) gives \(x = 4 \ln 20 + 0.4 - 4 \ln 2 - 4 = 5.61\) (3sf) | A1 [3] | \((C = -6.77...)\); \((5.61034...)\); AG |
## (i)
$F = \frac{6}{t^2} - \frac{3v}{t} = -1.5a = 1.5\frac{dv}{dt} = \frac{dv}{dt} + \frac{2v}{t} = -\frac{4}{t^2}$ | B1 [1] | AG; Using Newton II with $a = \frac{dv}{dt}$ and rearranging | Initial $F = ma$ equation must be clear
## (ii)
Integrating factor is $e^{\int \frac{2}{t}dt} = e^{2\ln t} = t^2$ | M1 |
$DE$ is $\frac{d}{dt}(vt^2) = 4$ | A1 |
$vt^2 = 4t + c$ | M1 | Using their integrating factor
$v = 0$ when $t = 2$ gives $0 = 8 + c$ | M1 | Use of $v = 0$, $t = 2$ in their equation to find a constant of integration
$vt^2 = 4t - 8$ | A1 |
when $t = 20$, speed of the piston is 0.18 m s$^{-1}$ | A1 [7] |
## (iii)
$\frac{dx}{dt} = \frac{4}{t} - \frac{8}{t^2} \Rightarrow$ distance = $\int_2^{20} \left(\frac{4}{t} - \frac{8}{t^2}\right) dt$ | M1 | Use $v = \frac{dx}{dt}$ and expression of distance moved as a definite integral
$= 5.6103...$ so distance is 5.61 m correct to 3sf | A1 | Correct integrand and limits; BC; AG so justification for 3sf accuracy is needed
**Alternative solution:**
$\frac{dx}{dt} = \frac{4}{t} - \frac{8}{t^2} \Rightarrow x = 4 \ln t + \frac{8}{t} + C$ | M1 | Use $v = \frac{dx}{dt}$ and attempt to integrate
$x = 0$ when $t = 2$ gives $C = -4 \ln 2 - 4$ | A1 |
so $t = 20$ gives $x = 4 \ln 20 + 0.4 - 4 \ln 2 - 4 = 5.61$ (3sf) | A1 [3] | $(C = -6.77...)$; $(5.61034...)$; AG
---8 A piston of mass 1.5 kg moves in a straight line inside a long straight horizontal cylinder. At time $t \mathrm {~s}$ the displacement of the piston from its initial position at one end of the cylinder is $x \mathrm {~m}$ and its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{a8c9d007-e67f-4637-9e74-630ba9a91442-5_168_805_1726_630}
The piston starts moving when $t = 2$ and is brought to rest when it reaches the other end of the cylinder. While the piston is in motion it is acted on by a force of magnitude $\frac { 6 } { t ^ { 2 } } \mathrm {~N}$ in the positive $x$ direction, and also by a force of magnitude $\frac { 3 v } { t } \mathrm {~N}$ resisting the motion.\\
(i) Show that, while the piston is in motion, $\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 2 v } { t } = \frac { 4 } { t ^ { 2 } }$.
The piston reaches the other end of the cylinder when $t = 20$.\\
(ii) Find the speed of the piston immediately before it is brought to rest.\\
(iii) Show that the piston travels a distance of 5.61 m , correct to 3 significant figures.
\section*{OCR}
\section*{Oxford Cambridge and RSA}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q8 [11]}}