| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.8 This is a straightforward M1 question testing basic vector addition, magnitude calculation, Newton's second law (F=ma), and angle finding using arctangent. All parts are routine applications of standard formulas with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resultant \(= (9+5-7)\mathbf{i} + (-3+8+3)\mathbf{j}\) | M1 | Adding components |
| \(= 7\mathbf{i} + 8\mathbf{j}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | F | = \sqrt{7^2 + 8^2}\) |
| \(= \sqrt{113} \approx 10.6\) N | A1 | ft their (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = ma \Rightarrow a = \frac{\sqrt{113}}{5}\) | M1 | Use of \(F=ma\) |
| \(\approx 2.12 \text{ ms}^{-2}\) | A1 | ft their (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{8}{7}\) | M1 | Correct trig ratio using components |
| \(\theta = \arctan\left(\frac{8}{7}\right)\) | M1 | Correct method |
| \(\theta \approx 48.8°\) | A1 | cao |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resultant $= (9+5-7)\mathbf{i} + (-3+8+3)\mathbf{j}$ | M1 | Adding components |
| $= 7\mathbf{i} + 8\mathbf{j}$ | A1 | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|F| = \sqrt{7^2 + 8^2}$ | M1 | Correct use of Pythagoras on their resultant |
| $= \sqrt{113} \approx 10.6$ N | A1 | ft their (a) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = ma \Rightarrow a = \frac{\sqrt{113}}{5}$ | M1 | Use of $F=ma$ |
| $\approx 2.12 \text{ ms}^{-2}$ | A1 | ft their (b) |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{8}{7}$ | M1 | Correct trig ratio using components |
| $\theta = \arctan\left(\frac{8}{7}\right)$ | M1 | Correct method |
| $\theta \approx 48.8°$ | A1 | cao |
---2 Three forces act on a particle. These forces are ( $9 \mathbf { i } - 3 \mathbf { j }$ ) newtons, ( $5 \mathbf { i } + 8 \mathbf { j }$ ) newtons and ( $- 7 \mathbf { i } + 3 \mathbf { j }$ ) newtons. The vectors $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Find the resultant of these forces.
\item Find the magnitude of the resultant force.
\item Given that the particle has mass 5 kg , find the magnitude of the acceleration of the particle.
\item Find the angle between the resultant force and the unit vector $\mathbf { i }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q2 [9]}}