| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position from velocity and initial conditions |
| Difficulty | Moderate -0.8 This is a straightforward M1 kinematics question using constant acceleration formulae in vector form. Parts (a) and (b) require direct application of v = u + at and r = r₀ + ut + ½at², while part (c) needs recognizing that 'north-westerly' means the i-component of velocity is zero, then solving for t and finding the distance. All steps are standard textbook procedures with no novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = (6\mathbf{i} + 2.4\mathbf{j}) + (-0.8\mathbf{i} + 0.1\mathbf{j})t\) | M1 | Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) |
| \(\mathbf{v} = (6 - 0.8t)\mathbf{i} + (2.4 + 0.1t)\mathbf{j}\) ms⁻¹ | A1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r} = 13.6\mathbf{i} + (6\mathbf{i} + 2.4\mathbf{j})t + \frac{1}{2}(-0.8\mathbf{i} + 0.1\mathbf{j})t^2\) | M1 | Using \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) |
| \(\mathbf{r} = (13.6 + 6t - 0.4t^2)\mathbf{i} + (2.4t + 0.05t^2)\mathbf{j}\) | A1 A1 | A1 for each correct component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| North-westerly means equal north and west components, so \(\mathbf{i}\) component of velocity \(= -\mathbf{j}\) component of velocity | M1 | Correct interpretation of north-westerly |
| \(6 - 0.8t = -(2.4 + 0.1t)\) | M1 | Setting up equation |
| \(6 - 0.8t = -2.4 - 0.1t\) | A1 | Correct equation |
| \(8.4 = 0.7t\), so \(t = 12\) | A1 | Correct value of \(t\) |
| Position: \(\mathbf{r} = (13.6 + 72 - 57.6)\mathbf{i} + (28.8 + 7.2)\mathbf{j} = 28\mathbf{i} + 36\mathbf{j}\) | M1 A1 | Substituting \(t=12\) into position vector |
| Distance \(= \sqrt{28^2 + 36^2} = \sqrt{784 + 1296} = \sqrt{2080} \approx 45.6\) m | M1 A1 | Using Pythagoras; correct final answer |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = (6\mathbf{i} + 2.4\mathbf{j}) + (-0.8\mathbf{i} + 0.1\mathbf{j})t$ | M1 | Using $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ |
| $\mathbf{v} = (6 - 0.8t)\mathbf{i} + (2.4 + 0.1t)\mathbf{j}$ ms⁻¹ | A1 | Correct expression |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = 13.6\mathbf{i} + (6\mathbf{i} + 2.4\mathbf{j})t + \frac{1}{2}(-0.8\mathbf{i} + 0.1\mathbf{j})t^2$ | M1 | Using $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ |
| $\mathbf{r} = (13.6 + 6t - 0.4t^2)\mathbf{i} + (2.4t + 0.05t^2)\mathbf{j}$ | A1 A1 | A1 for each correct component |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| North-westerly means equal north and west components, so $\mathbf{i}$ component of velocity $= -\mathbf{j}$ component of velocity | M1 | Correct interpretation of north-westerly |
| $6 - 0.8t = -(2.4 + 0.1t)$ | M1 | Setting up equation |
| $6 - 0.8t = -2.4 - 0.1t$ | A1 | Correct equation |
| $8.4 = 0.7t$, so $t = 12$ | A1 | Correct value of $t$ |
| Position: $\mathbf{r} = (13.6 + 72 - 57.6)\mathbf{i} + (28.8 + 7.2)\mathbf{j} = 28\mathbf{i} + 36\mathbf{j}$ | M1 A1 | Substituting $t=12$ into position vector |
| Distance $= \sqrt{28^2 + 36^2} = \sqrt{784 + 1296} = \sqrt{2080} \approx 45.6$ m | M1 A1 | Using Pythagoras; correct final answer |
I can see these are answer space pages (pages 17-20) from an AQA mechanics exam paper (P58718/Jan13/MM1B). These pages contain only blank lined answer spaces for questions 7 and 8 — there is no mark scheme content visible on these pages.
To get the mark scheme content for these questions, you would need the separate **AQA Mark Scheme document** for this paper (January 2013, MM1B).
However, I can solve **Question 8** from the question paper shown:
---7 A particle is initially at the point $A$, which has position vector $13.6 \mathbf { i }$ metres, with respect to an origin $O$. At the point $A$, the particle has velocity $( 6 \mathbf { i } + 2.4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, and in its subsequent motion, it has a constant acceleration of $( - 0.8 \mathbf { i } + 0.1 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle $t$ seconds after it leaves $A$.
\item Find an expression for the position vector of the particle, with respect to the origin $O$, $t$ seconds after it leaves $A$.
\item Find the distance of the particle from the origin $O$ when it is travelling in a north-westerly direction.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-17_2486_1709_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q7 [12]}}