| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D constant acceleration |
| Difficulty | Moderate -0.8 This is a straightforward multi-part SUVAT question in 2D requiring only direct application of standard kinematic equations (v = u + at, s = ut + ½at²) with vector components. All parts are routine calculations with no problem-solving insight needed, making it easier than average but not trivial due to the vector notation and multiple parts. |
| Spec | 1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) | M1 | Correct formula used |
| \(\mathbf{v} = (-2\mathbf{i}+2\mathbf{j}) + 20(0.25\mathbf{i}+0.3\mathbf{j})\) | A1 | Correct substitution |
| \(\mathbf{v} = 3\mathbf{i} + 8\mathbf{j}\) m s\(^{-1}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Travelling due north means \(\mathbf{i}\) component \(= 0\) | M1 | Correct condition stated |
| \(-2 + 0.25t = 0\) | A1 | Correct equation |
| \(t = 8\) s | A1 | Correct time |
| \(\mathbf{v} = (2 + 0.3 \times 8)\mathbf{j} = 4.4\mathbf{j}\) m s\(^{-1}\) | A1 | Correct velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) | M1 | Correct formula |
| \(\mathbf{r} = (9\mathbf{i}+7\mathbf{j}) + 20(-2\mathbf{i}+2\mathbf{j}) + \frac{1}{2}(0.25\mathbf{i}+0.3\mathbf{j})(400)\) | A1 | Correct substitution |
| \(\mathbf{r} = (9-40+50)\mathbf{i} + (7+40+60)\mathbf{j} = 19\mathbf{i} + 107\mathbf{j}\) m | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Average velocity \(= \frac{\Delta \mathbf{r}}{\Delta t} = \frac{(19\mathbf{i}+107\mathbf{j})-(9\mathbf{i}+7\mathbf{j})}{20}\) | M1 | Correct method |
| \(= \frac{10\mathbf{i}+100\mathbf{j}}{20} = 0.5\mathbf{i} + 5\mathbf{j}\) m s\(^{-1}\) | A1 | Correct answer |
# Question 7:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ | M1 | Correct formula used |
| $\mathbf{v} = (-2\mathbf{i}+2\mathbf{j}) + 20(0.25\mathbf{i}+0.3\mathbf{j})$ | A1 | Correct substitution |
| $\mathbf{v} = 3\mathbf{i} + 8\mathbf{j}$ m s$^{-1}$ | A1 | Correct answer |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Travelling due north means $\mathbf{i}$ component $= 0$ | M1 | Correct condition stated |
| $-2 + 0.25t = 0$ | A1 | Correct equation |
| $t = 8$ s | A1 | Correct time |
| $\mathbf{v} = (2 + 0.3 \times 8)\mathbf{j} = 4.4\mathbf{j}$ m s$^{-1}$ | A1 | Correct velocity |
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ | M1 | Correct formula |
| $\mathbf{r} = (9\mathbf{i}+7\mathbf{j}) + 20(-2\mathbf{i}+2\mathbf{j}) + \frac{1}{2}(0.25\mathbf{i}+0.3\mathbf{j})(400)$ | A1 | Correct substitution |
| $\mathbf{r} = (9-40+50)\mathbf{i} + (7+40+60)\mathbf{j} = 19\mathbf{i} + 107\mathbf{j}$ m | A1 | Correct answer |
## Part (d)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Average velocity $= \frac{\Delta \mathbf{r}}{\Delta t} = \frac{(19\mathbf{i}+107\mathbf{j})-(9\mathbf{i}+7\mathbf{j})}{20}$ | M1 | Correct method |
| $= \frac{10\mathbf{i}+100\mathbf{j}}{20} = 0.5\mathbf{i} + 5\mathbf{j}$ m s$^{-1}$ | A1 | Correct answer |
---7 A particle moves on a smooth horizontal plane. It is initially at the point $A$, with position vector $( 9 \mathbf { i } + 7 \mathbf { j } ) \mathrm { m }$, and has velocity $( - 2 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. The particle moves with a constant acceleration of $( 0.25 \mathbf { i } + 0.3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$ for 20 seconds until it reaches the point $B$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of the particle at the point $B$.
\item Find the velocity of the particle when it is travelling due north.
\item Find the position vector of the point $B$.
\item Find the average velocity of the particle as it moves from $A$ to $B$.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c022c936-72bc-4cf9-8f98-285f12c1d479-15_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q7 [12]}}