| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Particle motion - velocity/time (dv/dt = f(v,t)) |
| Difficulty | Moderate -0.3 This is a standard M2 differential equations question with heavily scaffolded steps. Parts (a)(i) and (a)(ii) are 'show that' questions requiring simple force equation setup and Newton's second law application. Part (b) guides students through separation of variables with the integral already set up. The techniques are routine for M2 students, though it requires careful algebraic manipulation and understanding of the exponential solution form. Slightly easier than average due to extensive scaffolding. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resultant downward force = \(65 \times 9.8 - 260v\) | B1 | Must show \(65 \times 9.8 - 260v = 65(9.8 - 4v)\) |
| \(= 65(9.8 - 4v)\) newtons | Given answer, must show working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(65\frac{\mathrm{d}v}{\mathrm{d}t} = -65(4v - 9.8)\) | M1 | Apply \(F = ma\) with correct mass |
| \(\frac{\mathrm{d}v}{\mathrm{d}t} = -(4v - 9.8) = -4(v - 2.45)\) | A1 | Given answer, must show working clearly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int \frac{1}{v-2.45}\,\mathrm{d}v = -\int 4\,\mathrm{d}t\) shown by separating variables | M1 | Must show the separation step |
| \(\ln\ | v - 2.45\ | = -4t + c\) |
| Using \(t = 0\), \(v = 19.6\): \(c = \ln(17.15)\) | M1 | Substituting initial conditions |
| \(\ln\ | v - 2.45\ | = -4t + \ln(17.15)\) |
| \(v = 2.45 + 17.15e^{-4t}\) | A1 | Correct final expression for \(v\) |
## Question 8:
**(a)(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Resultant downward force = $65 \times 9.8 - 260v$ | B1 | Must show $65 \times 9.8 - 260v = 65(9.8 - 4v)$ |
| $= 65(9.8 - 4v)$ newtons | | Given answer, must show working |
**(a)(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $65\frac{\mathrm{d}v}{\mathrm{d}t} = -65(4v - 9.8)$ | M1 | Apply $F = ma$ with correct mass |
| $\frac{\mathrm{d}v}{\mathrm{d}t} = -(4v - 9.8) = -4(v - 2.45)$ | A1 | Given answer, must show working clearly |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \frac{1}{v-2.45}\,\mathrm{d}v = -\int 4\,\mathrm{d}t$ shown by separating variables | M1 | Must show the separation step |
| $\ln\|v - 2.45\| = -4t + c$ | A1 | Correct integration both sides |
| Using $t = 0$, $v = 19.6$: $c = \ln(17.15)$ | M1 | Substituting initial conditions |
| $\ln\|v - 2.45\| = -4t + \ln(17.15)$ | A1 | Correct equation with $c$ evaluated |
| $v = 2.45 + 17.15e^{-4t}$ | A1 | Correct final expression for $v$ |
These pages (22, 23, and 24) from the AQA January 2011 MM2B paper contain **no questions or mark scheme content** — they are blank continuation/overflow pages with only the standard notice:
> "There are no questions printed on this page"
> "DO NOT WRITE ON THIS PAGE — ANSWER IN THE SPACES PROVIDED"
There is no mark scheme content to extract from these pages.8 Vicky has mass 65 kg and is skydiving. She steps out of a helicopter and falls vertically. She then waits a short period of time before opening her parachute. The parachute opens at time $t = 0$ when her speed is $19.6 \mathrm {~ms} ^ { - 1 }$, and she then experiences an air resistance force of magnitude $260 v$ newtons, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is her speed at time $t$ seconds.
\begin{enumerate}[label=(\alph*)]
\item When $t > 0$ :
\begin{enumerate}[label=(\roman*)]
\item show that the resultant downward force acting on Vicky is
65(9.8-4v) newtons
\item show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = - 4 ( v - 2.45 )$.
\end{enumerate}\item By showing that $\int \frac { 1 } { v - 2.45 } \mathrm {~d} v = - \int 4 \mathrm {~d} t$, find $v$ in terms of $t$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-21_2349_1707_221_153}
\end{center}
DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED
DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q8 [8]}}