| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.3 This is a straightforward vector kinematics question using standard SUVAT equations with i-j components. Part (a) requires simple calculation of acceleration from two velocities, part (b) applies standard position formula with given initial conditions, and part (c) involves substitution and magnitude calculation. All steps are routine M1 techniques with no problem-solving 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 |
|---|---|---|
| \(-\mathbf{i}+\mathbf{j} = 2\mathbf{i}-\mathbf{j}+10\mathbf{a}\) | M1, A1 | Use of velocity equation; Correct equation |
| \(\mathbf{a} = -0.3\mathbf{i}+0.2\mathbf{j}\) | A1 (3) | Correct \(\mathbf{a}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = (2\mathbf{i}-\mathbf{j})t + \frac{1}{2}(-0.3\mathbf{i}+0.2\mathbf{j})t^2 + 20\mathbf{i}\) | M1 | Use of constant acceleration equation for position |
| \(= (2t - 0.15t^2 + 20)\mathbf{i} + (-t + 0.1t^2)\mathbf{j}\) | A1, A1ft (3) | Correct \(\mathbf{i}\) component; Correct \(\mathbf{j}\) component (ft incorrect acceleration) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r}(20) = (2\times20 - 0.15\times20^2+20)\mathbf{i}+(-20+0.1\times20^2)\mathbf{j}\) | M1 | Substituting \(t=20\) into expression for \(\mathbf{r}\) |
| \(= 0\mathbf{i}+20\mathbf{j}\), so due north of origin | A1 (2) | Correct conclusion from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v}(20) = 2\mathbf{i}-\mathbf{j}+20(-0.3\mathbf{i}+0.2\mathbf{j})\) | M1, A1 | Finding velocity at \(t=20\); Correct velocity |
| \(= -4\mathbf{i}+3\mathbf{j}\) | m1 | Finding magnitude |
| \( | \mathbf{v}(20) | = \sqrt{4^2+3^2} = 5 \text{ ms}^{-1}\) |
## Question 7:
### Part (a)
$-\mathbf{i}+\mathbf{j} = 2\mathbf{i}-\mathbf{j}+10\mathbf{a}$ | M1, A1 | Use of velocity equation; Correct equation
$\mathbf{a} = -0.3\mathbf{i}+0.2\mathbf{j}$ | A1 (3) | Correct $\mathbf{a}$
### Part (b)
$\mathbf{r} = (2\mathbf{i}-\mathbf{j})t + \frac{1}{2}(-0.3\mathbf{i}+0.2\mathbf{j})t^2 + 20\mathbf{i}$ | M1 | Use of constant acceleration equation for position
$= (2t - 0.15t^2 + 20)\mathbf{i} + (-t + 0.1t^2)\mathbf{j}$ | A1, A1ft (3) | Correct $\mathbf{i}$ component; Correct $\mathbf{j}$ component (ft incorrect acceleration)
### Part (c)(i)
$\mathbf{r}(20) = (2\times20 - 0.15\times20^2+20)\mathbf{i}+(-20+0.1\times20^2)\mathbf{j}$ | M1 | Substituting $t=20$ into expression for $\mathbf{r}$
$= 0\mathbf{i}+20\mathbf{j}$, so due north of origin | A1 (2) | Correct conclusion from correct working
### Part (c)(ii)
$\mathbf{v}(20) = 2\mathbf{i}-\mathbf{j}+20(-0.3\mathbf{i}+0.2\mathbf{j})$ | M1, A1 | Finding velocity at $t=20$; Correct velocity
$= -4\mathbf{i}+3\mathbf{j}$ | m1 | Finding magnitude
$|\mathbf{v}(20)| = \sqrt{4^2+3^2} = 5 \text{ ms}^{-1}$ | A1ft (4) | Correct speed (ft incorrect acceleration)
---7 The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively. A yacht moves with a constant acceleration. At time $t$ seconds the position vector of the yacht is $\mathbf { r }$ metres. When $t = 0$ the velocity of the yacht is $( 2 \mathbf { i } - \mathbf { j } ) \mathrm { ms } ^ { - 1 }$, and when $t = 10$ the velocity of the yacht is $( - \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the yacht.
\item When $t = 0$ the yacht is 20 metres due east of the origin. Find an expression for $\mathbf { r }$ in terms of $t$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that when $t = 20$ the yacht is due north of the origin.
\item Find the speed of the yacht when $t = 20$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q7 [12]}}