| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Standard +0.3 This is a standard M2 differential equation problem with air resistance proportional to √v. Part (a) requires separating variables and integrating (∫v^(-1/2)dv), then applying initial conditions—routine for M2 students. Part (b) is trivial once (a) is complete. The non-standard power (√v rather than v or v²) adds slight novelty but the method is well-practiced, making this 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 |
|---|---|---|
| (a) \(20\frac{dv}{dt} = -10\sqrt{v}\) and \(\frac{dv}{dt} = \frac{\sqrt{v}}{2}\) | M1, A1, dM1 | — |
| \(\int\frac{1}{\sqrt{v}}dv = \int-\frac{1}{2}dt\) (AG) and \(2\sqrt{v} = -\frac{t}{2} + c\) | dM1, A1, dM1 | — |
| \(t = 0, v = 25 \Rightarrow c = 10\), so \(v = \left(\frac{20 - t}{4}\right)^2\) | A1 | 7 marks |
| (b) \(t = 20\) | B1 | 1 mark |
**(a)** $20\frac{dv}{dt} = -10\sqrt{v}$ and $\frac{dv}{dt} = \frac{\sqrt{v}}{2}$ | M1, A1, dM1 | — | Applying Newton's second law with $\frac{dv}{dt}$; correct differential equation; separating variables
$\int\frac{1}{\sqrt{v}}dv = \int-\frac{1}{2}dt$ (AG) and $2\sqrt{v} = -\frac{t}{2} + c$ | dM1, A1, dM1 | — | Integrating; correct integrals with or without $c$; finding the constant of integration
$t = 0, v = 25 \Rightarrow c = 10$, so $v = \left(\frac{20 - t}{4}\right)^2$ | A1 | 7 marks | Correct final result from correct working
**(b)** $t = 20$ | B1 | 1 mark | Correct time
**Total for Question 7: 8 marks**
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## TOTAL: 75 marks7 A particle of mass 20 kg moves along a straight horizontal line. At time $t$ seconds the velocity of the particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resistance force of magnitude $10 \sqrt { v }$ newtons acts on the particle while it is moving. At time $t = 0$ the velocity of the particle is $25 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at time $t$
$$v = \left( \frac { 20 - t } { 4 } \right) ^ { 2 }$$
\item State the value of $t$ when the particle comes to rest.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q7 [8]}}