| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 11 |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.8 This is a standard mechanics with resistance problem requiring two differential equations (F=ma in different forms) and their integration. While it involves multiple steps and careful algebraic manipulation of exponentials/logarithms, the setup is routine for Further Maths students and follows a well-practiced template. The 11 marks reflect length rather than exceptional difficulty. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| \(-(mg + kv) = m\frac{dv}{dt}\) | M1 | |
| \(\int dt = -\frac{1}{k}\int\frac{mk}{mg+kv}dv\) | M1 | |
| \(t = -\frac{m}{k}\ln(mg+kv) + c\) | A1 | |
| \(v = u\) when \(t = 0 \Rightarrow c = \frac{m}{k}\ln(mg+ku)\) | M1 | |
| \(\therefore t = \frac{m}{k}\ln\left(\frac{mg+ku}{mg+kv}\right)\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-(mg+kv) = mv\frac{dv}{dx}\) | M1 | |
| \(\int dx = -\int\frac{mv}{mg+kv}dv = -\frac{m}{k}\int\left(1-\frac{mg}{mg+kv}\right)dv\) | M1, A1 | |
| \(x = -\frac{m}{k}\left(v - \frac{mg}{k}\ln(mg+kv)\right) + c\) | A1 | |
| \(x = 0\) when \(v = u \Rightarrow c = \frac{m}{k}\left(u - \frac{mg}{k}\ln(mg+ku)\right)\) | M1 | |
| \(x = \frac{m}{k}(u-v) + \frac{m^2g}{k^2}\ln\left(\frac{mg+kv}{mg+ku}\right)\) | A1 | 6 |
### (i)
$-(mg + kv) = m\frac{dv}{dt}$ | M1 |
$\int dt = -\frac{1}{k}\int\frac{mk}{mg+kv}dv$ | M1 |
$t = -\frac{m}{k}\ln(mg+kv) + c$ | A1 |
$v = u$ when $t = 0 \Rightarrow c = \frac{m}{k}\ln(mg+ku)$ | M1 |
$\therefore t = \frac{m}{k}\ln\left(\frac{mg+ku}{mg+kv}\right)$ | A1 | **5**
### (ii)
$-(mg+kv) = mv\frac{dv}{dx}$ | M1 |
$\int dx = -\int\frac{mv}{mg+kv}dv = -\frac{m}{k}\int\left(1-\frac{mg}{mg+kv}\right)dv$ | M1, A1 |
$x = -\frac{m}{k}\left(v - \frac{mg}{k}\ln(mg+kv)\right) + c$ | A1 |
$x = 0$ when $v = u \Rightarrow c = \frac{m}{k}\left(u - \frac{mg}{k}\ln(mg+ku)\right)$ | M1 |
$x = \frac{m}{k}(u-v) + \frac{m^2g}{k^2}\ln\left(\frac{mg+kv}{mg+ku}\right)$ | A1 | **6**A stone of mass $m$ is projected vertically upwards with initial velocity $u$. At time $t$, the height risen above the point of projection is $x$ and the resistance to motion is $kv$ when the velocity of the stone is $v$.
\begin{enumerate}[label=(\roman*)]
\item Write down a first-order differential equation relating $v$ and $t$ and hence find $t$ in terms of $v$. [5]
\item Write down a first-order differential equation relating $v$ and $x$ and hence find $x$ in terms of $v$. [6]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q3 [11]}}