| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector mechanics question requiring standard applications of F=ma, constant acceleration equations (v=u+at, s=ut+½at²), and basic vector operations. Part (a) is direct division, parts (b)(i-ii) are routine kinematics, and (b)(iii) requires recognizing that north-east means equal i and j components. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{F} = m\mathbf{a}\): \((9\mathbf{i} + 12\mathbf{j}) = 10\mathbf{a}\) | M1 | Applying Newton's second law |
| \(\mathbf{a} = (0.9\mathbf{i} + 1.2\mathbf{j}) \text{ ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{v} = (2.2\mathbf{i} + \mathbf{j}) + (0.9\mathbf{i} + 1.2\mathbf{j})t\) | M1 | Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) |
| At \(t=5\): \(\mathbf{v} = (6.7\mathbf{i} + 7\mathbf{j})\) | A1 | |
| \(\mathbf{r} = (2.2\mathbf{i} + \mathbf{j})t + \frac{1}{2}(0.9\mathbf{i} + 1.2\mathbf{j})t^2\) | M1 | Using \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) |
| At \(t=5\): \(\mathbf{r} = (11\mathbf{i} + 5\mathbf{j}) + (11.25\mathbf{i} + 15\mathbf{j}) = (22.25\mathbf{i} + 20\mathbf{j})\) | A1 | |
| Distance \(= \sqrt{22.25^2 + 20^2} = \sqrt{495.0625 + 400} = \sqrt{895.0625} \approx 29.9 \text{ m}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{v} = (2.2 + 0.9t)\mathbf{i} + (1 + 1.2t)\mathbf{j}\) | M1 | |
| A1 | Both components correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| North-east means \(\mathbf{i}\) and \(\mathbf{j}\) components equal | M1 | |
| \(2.2 + 0.9t = 1 + 1.2t\) | M1 | Setting components equal |
| \(1.2 = 0.3t\), \(t = 4\) | A1 |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{F} = m\mathbf{a}$: $(9\mathbf{i} + 12\mathbf{j}) = 10\mathbf{a}$ | M1 | Applying Newton's second law |
| $\mathbf{a} = (0.9\mathbf{i} + 1.2\mathbf{j}) \text{ ms}^{-2}$ | A1 | |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{v} = (2.2\mathbf{i} + \mathbf{j}) + (0.9\mathbf{i} + 1.2\mathbf{j})t$ | M1 | Using $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ |
| At $t=5$: $\mathbf{v} = (6.7\mathbf{i} + 7\mathbf{j})$ | A1 | |
| $\mathbf{r} = (2.2\mathbf{i} + \mathbf{j})t + \frac{1}{2}(0.9\mathbf{i} + 1.2\mathbf{j})t^2$ | M1 | Using $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ |
| At $t=5$: $\mathbf{r} = (11\mathbf{i} + 5\mathbf{j}) + (11.25\mathbf{i} + 15\mathbf{j}) = (22.25\mathbf{i} + 20\mathbf{j})$ | A1 | |
| Distance $= \sqrt{22.25^2 + 20^2} = \sqrt{495.0625 + 400} = \sqrt{895.0625} \approx 29.9 \text{ m}$ | A1 | |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{v} = (2.2 + 0.9t)\mathbf{i} + (1 + 1.2t)\mathbf{j}$ | M1 | |
| | A1 | Both components correct |
## Part (b)(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| North-east means $\mathbf{i}$ and $\mathbf{j}$ components equal | M1 | |
| $2.2 + 0.9t = 1 + 1.2t$ | M1 | Setting components equal |
| $1.2 = 0.3t$, $t = 4$ | A1 | |
---7 A particle, of mass 10 kg , moves on a smooth horizontal surface. A single horizontal force, $( 9 \mathbf { i } + 12 \mathbf { j } )$ newtons, acts on the particle.
The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the particle.
\item At time $t$ seconds, the velocity of the particle is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0$, the velocity of the particle is $( 2.2 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the particle is at the origin.
\begin{enumerate}[label=(\roman*)]
\item Find the distance between the particle and the origin when $t = 5$.
\item Express $\mathbf { v }$ in terms of $t$.
\item Find $t$ when the particle is travelling north-east.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-15_2484_1709_223_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q7 [11]}}