| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Verifying given motion properties |
| Difficulty | Moderate -0.5 This is a straightforward M2 mechanics question involving Newton's second law and separable differential equations. Part (a) is direct application of F=ma, part (b) is standard separation of variables with given initial conditions, and part (c) is simple substitution and logarithm manipulation. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(F = ma\): \(-0.05mv = m\frac{dv}{dt}\), so \(\frac{dv}{dt} = -0.05v\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{dv}{v} = -\int 0.05dt\), \(\ln v = -0.05t + c\). When \(t = 0, v = 20\), so \(C = 20\). \(v = 20e^{-0.05t}\) | B1 M1 M1 A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| When \(v = 10\): \(10 = 20e^{-0.05t}\), \(e^{0.05t} = 2\), \(t = \frac{1}{0.05}\ln 2 = 13.9\) | M1 A1 A1 | 3 marks |
**Part (a)**
Using $F = ma$: $-0.05mv = m\frac{dv}{dt}$, so $\frac{dv}{dt} = -0.05v$ | B1 | 1 mark | Need to see m terms
**Part (b)**
$\int\frac{dv}{v} = -\int 0.05dt$, $\ln v = -0.05t + c$. When $t = 0, v = 20$, so $C = 20$. $v = 20e^{-0.05t}$ | B1 M1 M1 A1 | 4 marks | Need first 2 terms. Fully correct solutions
**Part (c)**
When $v = 10$: $10 = 20e^{-0.05t}$, $e^{0.05t} = 2$, $t = \frac{1}{0.05}\ln 2 = 13.9$ | M1 A1 A1 | 3 marks | Accept $20\ln 2$
**Total: 8 marks**
---6 A car, of mass $m$, is moving along a straight smooth horizontal road. At time $t$, the car has speed $v$. As the car moves, it experiences a resistance force of magnitude $0.05 m v$. No other horizontal force acts on the car.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.05 v$$
\item When $t = 0$, the speed of the car is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Show that $v = 20 \mathrm { e } ^ { - 0.05 t }$.
\item Find the time taken for the speed of the car to reduce to $10 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q6 [8]}}