| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two vector forces (direction/magnitude conditions) |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of dot product or trigonometry to find an angle. Part (b) requires setting up and solving simultaneous equations from vector components, which is a standard M1 technique. The question involves multiple steps but uses routine methods without requiring novel insight or complex problem-solving. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{9}{13}\) | M1 A1 | M1 for \(\tan\theta = 9/13\) or \(13/9\); First A1 for correct equation allowing correct adjustment to angle |
| \(\theta = 34.7°\) | A1 (3) | Second A1 for \(\theta = 35°\) or better or \(325°\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a(2\mathbf{i}-\mathbf{j})+b(\mathbf{i}+3\mathbf{j})=(9\mathbf{i}+13\mathbf{j})\) | M1 A2 | First M1 for \(\mathbf{P}+\mathbf{Q}=9\mathbf{i}+13\mathbf{j}\); A2: treat as B1 for \(a(2\mathbf{i}-\mathbf{j})\) seen, B1 for \(b(\mathbf{i}+3\mathbf{j})\) seen; if same \(a\) and \(b\) used, lose one B mark |
| \(2a+b=9\), \(-a+3b=13\) | M1 | Second M1 for equating \(\mathbf{i}\)-components *and* \(\mathbf{j}\)-components to give two equations in two unknowns |
| \(a=2,\ b=5\) | M1 A1 A1 | Third M1 for eliminating one unknown; A1 for \(a=2\); A1 for \(b=5\) |
| \(\mathbf{P}=(4\mathbf{i}-2\mathbf{j})\ \text{N};\ \mathbf{Q}=(5\mathbf{i}+15\mathbf{j})\ \text{N}\) | A1 A1 (9) | A1 for \(\mathbf{P}=(4\mathbf{i}-2\mathbf{j})\); A1 for \(\mathbf{Q}=(5\mathbf{i}+15\mathbf{j})\); N.B. can score all marks if they 'spot' the answers |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{9}{13}$ | M1 A1 | M1 for $\tan\theta = 9/13$ or $13/9$; First A1 for correct equation allowing correct adjustment to angle |
| $\theta = 34.7°$ | A1 (3) | Second A1 for $\theta = 35°$ or better or $325°$ or better |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a(2\mathbf{i}-\mathbf{j})+b(\mathbf{i}+3\mathbf{j})=(9\mathbf{i}+13\mathbf{j})$ | M1 A2 | First M1 for $\mathbf{P}+\mathbf{Q}=9\mathbf{i}+13\mathbf{j}$; A2: treat as B1 for $a(2\mathbf{i}-\mathbf{j})$ seen, B1 for $b(\mathbf{i}+3\mathbf{j})$ seen; if same $a$ and $b$ used, lose one B mark |
| $2a+b=9$, $-a+3b=13$ | M1 | Second M1 for equating $\mathbf{i}$-components *and* $\mathbf{j}$-components to give two equations in two unknowns |
| $a=2,\ b=5$ | M1 A1 A1 | Third M1 for eliminating one unknown; A1 for $a=2$; A1 for $b=5$ |
| $\mathbf{P}=(4\mathbf{i}-2\mathbf{j})\ \text{N};\ \mathbf{Q}=(5\mathbf{i}+15\mathbf{j})\ \text{N}$ | A1 A1 (9) | A1 for $\mathbf{P}=(4\mathbf{i}-2\mathbf{j})$; A1 for $\mathbf{Q}=(5\mathbf{i}+15\mathbf{j})$; N.B. can score all marks if they 'spot' the answers |
---7. A force $\mathbf { F }$ is given by $\mathbf { F } = ( 9 \mathbf { i } + 13 \mathbf { j } )$ N.
\begin{enumerate}[label=(\alph*)]
\item Find the size of the angle between the direction of $\mathbf { F }$ and the vector $\mathbf { j }$.
The force $\mathbf { F }$ is the resultant of two forces $\mathbf { P }$ and $\mathbf { Q }$. The line of action of $\mathbf { P }$ is parallel to the vector ( $2 \mathbf { i } - \mathbf { j }$ ). The line of action of $\mathbf { Q }$ is parallel to the vector ( $\mathbf { i } + 3 \mathbf { j }$ ).
\item Find, in terms of $\mathbf { i }$ and $\mathbf { j }$,
\begin{enumerate}[label=(\roman*)]
\item the force $\mathbf { P }$,
\item the force $\mathbf { Q }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q7 [12]}}