| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector mechanics question requiring standard techniques: vector addition, parallel vectors condition (using scalar multiples), substitution, and converting to magnitude/bearing. All steps are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.03a Force: vector nature and diagrams3.03d Newton's second law: 2D vectors3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((4\mathbf{i}-6\mathbf{j})+(p\mathbf{i}+q\mathbf{j}) = (4+p)\mathbf{i}+(q-6)\mathbf{j}\) | M1 | Adding forces, collecting \(\mathbf{i}\)s and \(\mathbf{j}\)s |
| \(\frac{(4+p)}{(q-6)} = \frac{2}{1}\) or \(-\frac{2}{1}\) or \(\frac{1}{2}\) or \(-\frac{1}{2}\) | DM1 A1 | Equation in \(p\) and \(q\) only |
| \(2q - 12 = 4 + p\) | DM1 A1 | |
| \(p - 2q = -16\) | DM1 A1 | Given answer; (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(q = 3 \Rightarrow p = -10\) | B1 | Seen or implied |
| Either: \(0.5\mathbf{a} = -6\mathbf{i} - 3\mathbf{j}\) Or: \( | \mathbf{R} | = \sqrt{(-6)^2+(-3)^2}\) |
| \(\mathbf{a} = -12\mathbf{i} - 6\mathbf{j}\) or \(= \sqrt{45}\) | A1 | |
| \( | \mathbf{a} | = \sqrt{(-12)^2+(-6)^2}\) or \(0.5a = \sqrt{45}\) |
| \(a = \sqrt{180} = 13.4\ \text{ms}^{-2}\) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| e.g. \(\tan\theta = \frac{12}{6} \Rightarrow \theta = 63.4°\) | M1A1 | Relevant trig ratio from \(\mathbf{a}\) or \(\mathbf{R}\) |
| Bearing \(= 180° + 63.4° = 243°\) (nearest degree) | A1cao | (3 marks) |
# Question 6:
## Part 6(a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $(4\mathbf{i}-6\mathbf{j})+(p\mathbf{i}+q\mathbf{j}) = (4+p)\mathbf{i}+(q-6)\mathbf{j}$ | M1 | Adding forces, collecting $\mathbf{i}$s and $\mathbf{j}$s |
| $\frac{(4+p)}{(q-6)} = \frac{2}{1}$ or $-\frac{2}{1}$ or $\frac{1}{2}$ or $-\frac{1}{2}$ | DM1 A1 | Equation in $p$ and $q$ only |
| $2q - 12 = 4 + p$ | DM1 A1 | |
| $p - 2q = -16$ | DM1 A1 | Given answer; (5 marks) |
## Part 6(b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $q = 3 \Rightarrow p = -10$ | B1 | Seen or implied |
| **Either:** $0.5\mathbf{a} = -6\mathbf{i} - 3\mathbf{j}$ **Or:** $|\mathbf{R}| = \sqrt{(-6)^2+(-3)^2}$ | M1 | Use of $\mathbf{F} = 0.5\mathbf{a}$ with resultant force |
| $\mathbf{a} = -12\mathbf{i} - 6\mathbf{j}$ or $= \sqrt{45}$ | A1 | |
| $|\mathbf{a}| = \sqrt{(-12)^2+(-6)^2}$ or $0.5a = \sqrt{45}$ | M1 | Finding magnitude |
| $a = \sqrt{180} = 13.4\ \text{ms}^{-2}$ | A1 | (5 marks) |
## Part 6(c):
| Working | Marks | Guidance |
|---------|-------|----------|
| e.g. $\tan\theta = \frac{12}{6} \Rightarrow \theta = 63.4°$ | M1A1 | Relevant trig ratio from $\mathbf{a}$ or $\mathbf{R}$ |
| Bearing $= 180° + 63.4° = 243°$ (nearest degree) | A1cao | (3 marks) |
---6. [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively]
Two forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ act on a particle $P$ of mass 0.5 kg .\\
$\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - 6 \mathbf { j } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( p \mathbf { i } + q \mathbf { j } ) \mathrm { N }$.\\
Given that the resultant force of $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ is in the same direction as $- 2 \mathbf { i } - \mathbf { j }$,
\begin{enumerate}[label=(\alph*)]
\item show that $p - 2 q = - 16$
Given that $q = 3$
\item find the magnitude of the acceleration of $P$,
\item find the direction of the acceleration of $P$, giving your answer as a bearing to the nearest degree.
XXXXXXXXXXIXITEINTIIS AREA XX女X女X女X女X DO NOT WIRIE IN THS AREA.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q6 [13]}}