| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 Newton's second law with air resistance. Part (a) requires setting up F=ma with weight and resistance forces (routine), part (b) involves separating variables and integrating a linear DE (standard technique), and part (c) asks for a sketch showing exponential decay to terminal velocity (straightforward). While it requires multiple steps, all techniques are core M2 syllabus with no novel insight needed. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation of motion (downward positive): \(72g - 240v = 72\frac{dv}{dt}\) | M1 | Applying Newton's second law with correct forces |
| \(72(9.8) - 240v = 72\frac{dv}{dt}\) | ||
| \(705.6 - 240v = 72\frac{dv}{dt}\) | ||
| Dividing by \(-240\): \(-\frac{72}{240}\frac{dv}{dt} = v - \frac{705.6}{240}\) | ||
| \(-\frac{3}{10}\frac{dv}{dt} = v - 2.94\) | A1 | Shown correctly (QED) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Separating variables: \(\int \frac{1}{v - 2.94}dv = \int -\frac{10}{3}dt\) | M1 | Separating variables correctly |
| \(\ln | v - 2.94 | = -\frac{10}{3}t + c\) |
| At \(t = 0\), \(v = 30\): \(c = \ln(27.06)\) | M1 | Using initial condition |
| \(v - 2.94 = 27.06e^{-\frac{10}{3}t}\) | A1 | |
| \(v = 2.94 + 27.06e^{-\frac{10}{3}t}\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Graph starting at \(v = 30\) when \(t = 0\) | B1 | Correct starting point marked |
| Curve decreasing asymptotically toward \(v = 2.94\) | B1 | Correct shape with asymptote indicated at \(v = 2.94\) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of motion (downward positive): $72g - 240v = 72\frac{dv}{dt}$ | M1 | Applying Newton's second law with correct forces |
| $72(9.8) - 240v = 72\frac{dv}{dt}$ | | |
| $705.6 - 240v = 72\frac{dv}{dt}$ | | |
| Dividing by $-240$: $-\frac{72}{240}\frac{dv}{dt} = v - \frac{705.6}{240}$ | | |
| $-\frac{3}{10}\frac{dv}{dt} = v - 2.94$ | A1 | Shown correctly (QED) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Separating variables: $\int \frac{1}{v - 2.94}dv = \int -\frac{10}{3}dt$ | M1 | Separating variables correctly |
| $\ln|v - 2.94| = -\frac{10}{3}t + c$ | A1 | Correct integration both sides |
| At $t = 0$, $v = 30$: $c = \ln(27.06)$ | M1 | Using initial condition |
| $v - 2.94 = 27.06e^{-\frac{10}{3}t}$ | A1 | |
| $v = 2.94 + 27.06e^{-\frac{10}{3}t}$ | A1 | Correct final answer |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph starting at $v = 30$ when $t = 0$ | B1 | Correct starting point marked |
| Curve decreasing asymptotically toward $v = 2.94$ | B1 | Correct shape with asymptote indicated at $v = 2.94$ |
---7 A parachutist, of mass 72 kg , is falling vertically. He opens his parachute at time $t = 0$ when his speed is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. He then experiences an air resistance force of magnitude $240 v$ newtons, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is his speed at time $t$ seconds.
\begin{enumerate}[label=(\alph*)]
\item When $t > 0$, show that $- \frac { 3 } { 10 } \frac { \mathrm {~d} v } { \mathrm {~d} t } = v - 2.94$.
\item Find $v$ in terms of $t$.
\item Sketch a graph to show how, for $t \geqslant 0$, the parachutist's speed varies with time.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2015 Q7 [9]}}