| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 10 |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring differential equations (separating variables with v dv/dx = F/m), exponential integration, work-energy principles, and limiting behavior analysis. While the individual techniques are standard for Further Maths, the multi-step derivation with resistive forces proportional to v² and the need to apply initial conditions correctly makes this moderately challenging, though still within expected Further Maths territory. |
| Spec | 6.02a Work done: concept and definition6.02c Work by variable force: using integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - kv^2 = m\frac{dv}{dx}\) | M1 | N11 with 3 terms and \(a = v\frac{dv}{dx}\) used; Condone sign errors and/or missing m |
| \(m \int \frac{-2kv}{T - kv^2} dv = \int -2k \, dx\) | M1 | Separating the variables |
| \(m\ln(T - kv^2) = -2kx + c\) | M1 | Correctly integrating both sides |
| \(\ln(T - kv^2) = \frac{-2kx + c}{m} \Rightarrow T - kv^2 = e^{\frac{-2kx+c}{m}}\) | M1 | Exponentiation, giving \(f(v) = e^{g(x)}\) |
| \(\Rightarrow T - kv^2 = e^{-\frac{2k}{m}}e^m \Rightarrow v^2 = \frac{1}{k}\left(T - Ae^{-\frac{2k}{m}}\right)\) | A1 | AG so intermediate step needed; oe, e.g. stating \(A = e^{\frac{c}{m}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Work done by constant force is \(T(l - 0) = T\) | B1 | |
| \(x = 0, v = 0\) gives \(v^2 = \frac{T}{k}\left(1 - e^{-\frac{2k}{m}}\right)\) | M1 | Use of initial conditions to find \(A\) |
| KE gain is \(\frac{1}{2}mv_1^2 = \frac{mT}{2k}\left(1 - e^{-\frac{2k}{m}}\right)\) | M1 | Use of initial conditions to find \(A\) |
| So work done against resistance \(= T - \frac{mT}{2k}\left(1 - e^{-\frac{2k}{m}}\right)\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 0, v = 0\) gives \(v^2 = \frac{T}{k}\left(1 - e^{-\frac{2k}{m}}\right)\) | M1 | Use of initial conditions to find \(A\) |
| WD against resistance \(= \int_0^l kv^2 \, dx = T \int_0^l \left(1 - e^{-\frac{2k}{m}}\right) dx\) | M1 | |
| \(= T\left[x + \frac{m}{2k}e^{-\frac{2k}{m}}\right]_0^l\) | M1 | |
| \(= T\left(1 + \frac{m}{2k}e^{-\frac{2k}{m}} - \frac{m}{2k}\right)\) | A1 | oe |
| Answer | Marks |
|---|---|
| The limiting velocity is \(\sqrt{\frac{T}{k}}\) | B1 |
## (i)
$T - kv^2 = m\frac{dv}{dx}$ | M1 | N11 with 3 terms and $a = v\frac{dv}{dx}$ used; Condone sign errors and/or missing m
$m \int \frac{-2kv}{T - kv^2} dv = \int -2k \, dx$ | M1 | Separating the variables
$m\ln(T - kv^2) = -2kx + c$ | M1 | Correctly integrating both sides
$\ln(T - kv^2) = \frac{-2kx + c}{m} \Rightarrow T - kv^2 = e^{\frac{-2kx+c}{m}}$ | M1 | Exponentiation, giving $f(v) = e^{g(x)}$
$\Rightarrow T - kv^2 = e^{-\frac{2k}{m}}e^m \Rightarrow v^2 = \frac{1}{k}\left(T - Ae^{-\frac{2k}{m}}\right)$ | A1 | AG so intermediate step needed; oe, e.g. stating $A = e^{\frac{c}{m}}$
## (ii)
Work done by constant force is $T(l - 0) = T$ | B1 |
$x = 0, v = 0$ gives $v^2 = \frac{T}{k}\left(1 - e^{-\frac{2k}{m}}\right)$ | M1 | Use of initial conditions to find $A$
KE gain is $\frac{1}{2}mv_1^2 = \frac{mT}{2k}\left(1 - e^{-\frac{2k}{m}}\right)$ | M1 | Use of initial conditions to find $A$
So work done against resistance $= T - \frac{mT}{2k}\left(1 - e^{-\frac{2k}{m}}\right)$ | A1 | oe
**Alternative solution**
$x = 0, v = 0$ gives $v^2 = \frac{T}{k}\left(1 - e^{-\frac{2k}{m}}\right)$ | M1 | Use of initial conditions to find $A$
WD against resistance $= \int_0^l kv^2 \, dx = T \int_0^l \left(1 - e^{-\frac{2k}{m}}\right) dx$ | M1 |
$= T\left[x + \frac{m}{2k}e^{-\frac{2k}{m}}\right]_0^l$ | M1 |
$= T\left(1 + \frac{m}{2k}e^{-\frac{2k}{m}} - \frac{m}{2k}\right)$ | A1 | oe
## (iii)
The limiting velocity is $\sqrt{\frac{T}{k}}$ | B1 |
---A particle $P$ of mass $m$ moves along the positive $x$-axis. When its displacement from the origin $O$ is $x$ its velocity is $v$, where $v \geqslant 0$. It is subject to two forces: a constant force $T$ in the positive $x$ direction, and a resistive force which is proportional to $v^2$.
\begin{enumerate}[label=(\roman*)]
\item Show that $v^2 = \frac{1}{k}\left(T - Ae^{-\frac{2kx}{m}}\right)$ where $A$ and $k$ are constants. [5]
\end{enumerate}
$P$ starts from rest at $O$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find an expression for the work done against the resistance to motion as $P$ moves from $O$ to the point where $x = 1$. [4]
\item Find an expression for the limiting value of the velocity of $P$ as $x$ increases. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q6 [10]}}