| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.3 This is a standard M2 differential equations question involving resistance proportional to v². Part (a) is straightforward application of F=ma with given values. Part (b) requires separating variables and integrating 1/v² - a routine technique taught explicitly in M2. The algebra is mechanical with no conceptual challenges beyond following the standard method. |
| 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 |
|---|---|---|
| \(F = ma\): \(-0.08v^2 = 0.05\frac{dv}{dt}\) | M1 | Apply Newton's second law |
| \(\frac{dv}{dt} = \frac{-0.08}{0.05}v^2 = -1.6v^2\) | A1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dt} = -1.6v^2 \Rightarrow \frac{1}{v^2}\frac{dv}{dt} = -1.6\) | M1 | Separate variables |
| \(\int v^{-2}\,dv = \int -1.6\,dt\) | M1 | Correct integration attempt |
| \(-\frac{1}{v} = -1.6t + c\) | A1 | Correct integration |
| When \(t=0\), \(v=3\): \(c = -\frac{1}{3}\) | M1 | Apply initial condition |
| \(-\frac{1}{v} = -1.6t - \frac{1}{3}\) | ||
| \(\frac{1}{v} = 1.6t + \frac{1}{3} = \frac{4.8t + 1}{3} = \frac{5 + 24t}{15}\) | A1 | Correctly shown |
| \(v = \frac{15}{5+24t}\) |
# Question 8:
## Part (a):
| $F = ma$: $-0.08v^2 = 0.05\frac{dv}{dt}$ | M1 | Apply Newton's second law |
|---|---|---|
| $\frac{dv}{dt} = \frac{-0.08}{0.05}v^2 = -1.6v^2$ | A1 | Correctly shown |
## Part (b):
| $\frac{dv}{dt} = -1.6v^2 \Rightarrow \frac{1}{v^2}\frac{dv}{dt} = -1.6$ | M1 | Separate variables |
|---|---|---|
| $\int v^{-2}\,dv = \int -1.6\,dt$ | M1 | Correct integration attempt |
| $-\frac{1}{v} = -1.6t + c$ | A1 | Correct integration |
| When $t=0$, $v=3$: $c = -\frac{1}{3}$ | M1 | Apply initial condition |
| $-\frac{1}{v} = -1.6t - \frac{1}{3}$ | | |
| $\frac{1}{v} = 1.6t + \frac{1}{3} = \frac{4.8t + 1}{3} = \frac{5 + 24t}{15}$ | A1 | Correctly shown |
| $v = \frac{15}{5+24t}$ | | |
---8 A stone, of mass 0.05 kg , is moving along the smooth horizontal floor of a tank, which is filled with oil. At time $t$, the stone has speed $v$. As the stone moves, it experiences a resistance force of magnitude $0.08 v ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 1.6 v ^ { 2 }$$
(2 marks)
\item The initial speed of the stone is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Show that
$$v = \frac { 15 } { 5 + 24 t }$$
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q8 [7]}}