| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from two position vectors |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question testing basic vector kinematics. Part (a) requires finding displacement AB and dividing by velocity (standard formula application). Part (b) uses constant acceleration equations with vectors, which is routine bookwork. All steps are direct applications of standard formulas with no problem-solving insight required, making it easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{d} = 3\mathbf{i} - 6\mathbf{j}\) | B1 | Accept \(\pm\mathbf{d}\) or displacements of 3, 6 shown on diagram |
| \(3\mathbf{i} - 6\mathbf{j} = (\mathbf{i} - 2\mathbf{j})t\) | M1 | Or equivalent method for \(t\); accept ratio of vectors leading directly to \(\pm 3\) |
| \(t = 3\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = (\mathbf{i} - 2\mathbf{j}) \times 4 + \frac{1}{2} \times 2\mathbf{j} \times 16\) | M1 | Full method for vector expression giving change in position |
| A1 | For correct subs (gives \(4\mathbf{i} + 8\mathbf{j}\)) | |
| \(+6\mathbf{i} - 4\mathbf{j}\) | M1 | |
| \(= 10\mathbf{i} + 4\mathbf{j}\) | A1F | FT slip provided obtain vector expression (\(\mathbf{u} = 0\) gives \(6\mathbf{i} + 12\mathbf{j}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(A(3,2)\), \(C(10,4)\); \(\mathbf{d} = 7\mathbf{i} + 2\mathbf{j}\) | M1 | Attempt to find vector \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) using candidate's \(C\) |
| \( | \mathbf{d} | = \sqrt{7^2 + 2^2}\) |
| \(AC = \sqrt{53} = 7.28\) | A1F | FT d provided two non-zero components; accept \(\sqrt{53}\) |
## Question 6:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{d} = 3\mathbf{i} - 6\mathbf{j}$ | B1 | Accept $\pm\mathbf{d}$ or displacements of 3, 6 shown on diagram |
| $3\mathbf{i} - 6\mathbf{j} = (\mathbf{i} - 2\mathbf{j})t$ | M1 | Or equivalent method for $t$; accept ratio of vectors leading directly to $\pm 3$ |
| $t = 3$ | A1 | CAO | **Total: 3** |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = (\mathbf{i} - 2\mathbf{j}) \times 4 + \frac{1}{2} \times 2\mathbf{j} \times 16$ | M1 | Full method for vector expression giving change in position |
| | A1 | For correct subs (gives $4\mathbf{i} + 8\mathbf{j}$) |
| $+6\mathbf{i} - 4\mathbf{j}$ | M1 | |
| $= 10\mathbf{i} + 4\mathbf{j}$ | A1F | FT slip provided obtain vector expression ($\mathbf{u} = 0$ gives $6\mathbf{i} + 12\mathbf{j}$) | **Total: 4** |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $A(3,2)$, $C(10,4)$; $\mathbf{d} = 7\mathbf{i} + 2\mathbf{j}$ | M1 | Attempt to find vector $\overrightarrow{AC}$ or $\overrightarrow{CA}$ using candidate's $C$ |
| $|\mathbf{d}| = \sqrt{7^2 + 2^2}$ | | |
| $AC = \sqrt{53} = 7.28$ | A1F | FT **d** provided two non-zero components; accept $\sqrt{53}$ | **Total: 2** |
---6 The points $A$ and $B$ have position vectors $( 3 \mathbf { i } + 2 \mathbf { j } )$ metres and $( 6 \mathbf { i } - 4 \mathbf { j } )$ metres respectively. The vectors $\mathbf { i }$ and $\mathbf { j }$ are in a horizontal plane.
\begin{enumerate}[label=(\alph*)]
\item A particle moves from $A$ to $B$ with constant velocity $( \mathbf { i } - 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$. Calculate the time that the particle takes to move from $A$ to $B$.
\item The particle then moves from $B$ to a point $C$ with a constant acceleration of $2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 2 }$. It takes 4 seconds to move from $B$ to $C$.
\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $C$.
\item Find the distance $A C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q6 [9]}}