| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Standard +0.3 This is a straightforward M1 vector kinematics question requiring standard SUVAT equations applied to i-j components. Part (a) is routine substitution into s = ut + ½at², part (b) repeats this with different time, and part (c) requires recognizing that parallel to i means j-component of velocity equals zero. All steps are mechanical applications of standard formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(75\mathbf{i} = (5\mathbf{i} - 2\mathbf{j}) \times 10 + \frac{1}{2}\mathbf{a} \times 10^2\) | M1 | Equation to find \(\mathbf{a}\) from \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) |
| A1 | Correct expression | |
| \(\mathbf{a} = \frac{75\mathbf{i} - 50\mathbf{i} + 20\mathbf{j}}{50} = 0.5\mathbf{i} + 0.4\mathbf{j}\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = (5\mathbf{i} - 2\mathbf{j}) \times 8 + \frac{1}{2}(0.5\mathbf{i} + 0.4\mathbf{j}) \times 8^2\) | M1 | Expression for \(\mathbf{r}\) using \(t=8\) with no extra terms |
| A1 | Correct expressions | |
| \(= 56\mathbf{i} - 3.2\mathbf{j}\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{v} = (5 + 0.5t)\mathbf{i} + (0.4t - 2)\mathbf{j}\) | M1A1 | Expression for \(\mathbf{v}\). Correct expression |
| \(0.4t - 2 = 0\) | dM1 | \(\mathbf{j}\) component equal to zero |
| \(t = \frac{2}{0.4} = 5\) | A1 | Correct \(t\) |
| \(\mathbf{r} = (5\mathbf{i} - 2\mathbf{j}) \times 5 + \frac{1}{2}(0.5\mathbf{i} + 0.4\mathbf{j}) \times 5^2\) | dM1 | Expression for \(\mathbf{r}\) using \(t\) from \(\mathbf{j}\) component equal to zero |
| \(= 31.25\mathbf{i} - 5\mathbf{j} = 31.3\mathbf{i} - 5\mathbf{j}\) | A1 | Total: 6 |
## Question 8:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $75\mathbf{i} = (5\mathbf{i} - 2\mathbf{j}) \times 10 + \frac{1}{2}\mathbf{a} \times 10^2$ | M1 | Equation to find $\mathbf{a}$ from $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ |
| | A1 | Correct expression |
| $\mathbf{a} = \frac{75\mathbf{i} - 50\mathbf{i} + 20\mathbf{j}}{50} = 0.5\mathbf{i} + 0.4\mathbf{j}$ | A1 | **Total: 3** | **AG** Correct $\mathbf{a}$ from correct working |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = (5\mathbf{i} - 2\mathbf{j}) \times 8 + \frac{1}{2}(0.5\mathbf{i} + 0.4\mathbf{j}) \times 8^2$ | M1 | Expression for $\mathbf{r}$ using $t=8$ with no extra terms |
| | A1 | Correct expressions |
| $= 56\mathbf{i} - 3.2\mathbf{j}$ | A1 | **Total: 3** | Correct position vector |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{v} = (5 + 0.5t)\mathbf{i} + (0.4t - 2)\mathbf{j}$ | M1A1 | Expression for $\mathbf{v}$. Correct expression |
| $0.4t - 2 = 0$ | dM1 | $\mathbf{j}$ component equal to zero |
| $t = \frac{2}{0.4} = 5$ | A1 | Correct $t$ |
| $\mathbf{r} = (5\mathbf{i} - 2\mathbf{j}) \times 5 + \frac{1}{2}(0.5\mathbf{i} + 0.4\mathbf{j}) \times 5^2$ | dM1 | Expression for $\mathbf{r}$ using $t$ from $\mathbf{j}$ component equal to zero |
| $= 31.25\mathbf{i} - 5\mathbf{j} = 31.3\mathbf{i} - 5\mathbf{j}$ | A1 | **Total: 6** | Correct position vector |8 A particle is initially at the origin, where it has velocity $( 5 \mathbf { i } - 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. It moves with a constant acceleration $\mathbf { a } \mathrm { ms } ^ { - 2 }$ for 10 seconds to the point with position vector $75 \mathbf { i }$ metres.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { a } = 0.5 \mathbf { i } + 0.4 \mathbf { j }$.
\item Find the position vector of the particle 8 seconds after it has left the origin.
\item Find the position vector of the particle when it is travelling parallel to the unit vector $\mathbf { i }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q8 [12]}}