| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Bearing and speed from velocity vector |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring basic vector operations: (a) finding magnitude of a velocity vector, (b) using dot product to find angle between two vectors, (c) using distance-speed-time with vector geometry. All techniques are standard and the question provides significant scaffolding (showing the angle is 60°, stating it's equilateral). Slightly easier than average due to the structured parts and routine calculations. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed \(= \sqrt{(3)^2 + (\sqrt{3})^2} = 2\sqrt{3}\) m s\(^{-1}\) | B1 | AO 1.1b; AWRT 3.46; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(\tan^{-1}\dfrac{\sqrt{3}}{3}\) or \(\tan^{-1}\dfrac{3}{\sqrt{3}}\) to find angle between one of the velocity vectors relative to i or j direction | M1 | AO 3.1a; sight of sine/cosine rule using magnitude of \(AC\) scores M0 R0 |
| Angle between \(AB\) and i direction \(= \tan^{-1}\dfrac{\sqrt{3}}{3} = 30°\); Angle between \(BC\) and i direction \(= \tan^{-1}\dfrac{\sqrt{3}}{3} = 30°\); Angle \(ABC = 30° + 30° = 60°\) | R1 | AO 2.1; must include clear reference to angle \(ABC\) or indicate it on diagram; when using trig ratios, i and j must not be included |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time taken from \(B\) to \(C = \dfrac{9}{3} = 3\) seconds | R1 | AO 2.2a |
| \(\overrightarrow{OC} = \begin{bmatrix}1\\0\end{bmatrix} + 3\begin{bmatrix}-3\\\sqrt{3}\end{bmatrix}\) | M1 | AO 3.1a; displacement from \(B\) to \(C\) of form \(t\begin{bmatrix}-3\\\sqrt{3}\end{bmatrix}\) where \(1 < t \leq 9\) |
| \(= \begin{bmatrix}-8\\3\sqrt{3}\end{bmatrix}\) | A1 | AO 1.1b |
## Question 18(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed $= \sqrt{(3)^2 + (\sqrt{3})^2} = 2\sqrt{3}$ m s$^{-1}$ | B1 | AO 1.1b; AWRT 3.46; condone missing units |
**Subtotal: 1 mark**
---
## Question 18(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\tan^{-1}\dfrac{\sqrt{3}}{3}$ or $\tan^{-1}\dfrac{3}{\sqrt{3}}$ to find angle between one of the velocity vectors relative to **i** or **j** direction | M1 | AO 3.1a; sight of sine/cosine rule using magnitude of $AC$ scores M0 R0 |
| Angle between $AB$ and **i** direction $= \tan^{-1}\dfrac{\sqrt{3}}{3} = 30°$; Angle between $BC$ and **i** direction $= \tan^{-1}\dfrac{\sqrt{3}}{3} = 30°$; Angle $ABC = 30° + 30° = 60°$ | R1 | AO 2.1; must include clear reference to angle $ABC$ or indicate it on diagram; when using trig ratios, **i** and **j** must not be included |
**Subtotal: 2 marks**
---
## Question 18(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time taken from $B$ to $C = \dfrac{9}{3} = 3$ seconds | R1 | AO 2.2a |
| $\overrightarrow{OC} = \begin{bmatrix}1\\0\end{bmatrix} + 3\begin{bmatrix}-3\\\sqrt{3}\end{bmatrix}$ | M1 | AO 3.1a; displacement from $B$ to $C$ of form $t\begin{bmatrix}-3\\\sqrt{3}\end{bmatrix}$ where $1 < t \leq 9$ |
| $= \begin{bmatrix}-8\\3\sqrt{3}\end{bmatrix}$ | A1 | AO 1.1b |
**Subtotal: 3 marks**
---18 In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors representing due east and due north respectively.
A particle, $T$, is moving on a plane at a constant speed.\\
The path followed by $T$ makes the exact shape of a triangle $A B C$.\\
$T$ moves around $A B C$ in an anticlockwise direction as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-28_447_366_671_925}
On its journey from $A$ to $B$ the velocity vector of $T$ is $( 3 \mathbf { i } + \sqrt { 3 } \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
18
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $T$ as it moves from $A$ to $B$
18
\item On its journey from $B$ to $C$ the velocity vector of $T$ is $( - 3 \mathbf { i } + \sqrt { 3 } \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ Show that the acute angle $A B C = 60 ^ { \circ }$\\
18
\item It is given that $A B C$ is an equilateral triangle.\\
$T$ returns to its initial position after 9 seconds.\\
Vertex $B$ lies at position vector $\left[ \begin{array} { l } 1 \\ 0 \end{array} \right]$ metres with respect to a fixed origin $O$\\
Find the position vector of $C$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q18 [6]}}