| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector from magnitude and bearing |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question testing basic vector concepts: sketching a 2D vector, calculating magnitude using Pythagoras (√(16+25)), finding bearing from arctan(5/4), and scalar multiplication. All steps are routine applications of standard formulas with no problem-solving insight required, making it easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch: O, \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{r}\)] | B1 [1] | Only require correct quadrant. Vectors must have arrows. Need not label \(\mathbf{r}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{4^2 + (-5)^2}\) | M1 | Accept \(\sqrt{4^2 - 5^2}\) |
| \(= \sqrt{41}\) or \(6.4031...\) so \(6.40\) (3 s.f.) | A1 | |
| Need \(180 - \arctan\left(\frac{4}{5}\right)\) | M1 | Or equivalent. Award for \(\arctan\left(\pm\frac{4}{5}\right)\) or \(\arctan\left(\pm\frac{5}{4}\right)\) or equivalent seen without 180 or 90 |
| \(141.340\) so \(141°\) | A1 [4] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(12\mathbf{i} - 15\mathbf{j}\) or \(\begin{pmatrix}12\\-15\end{pmatrix}\) | B1 [1] | Do not award for magnitude given as the answer. Penalise spurious notation by 1 mark at most once in paper |
## Question 5:
### Part (i)
[Sketch: O, $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{r}$] | B1 [1] | Only require correct quadrant. Vectors must have arrows. Need not label $\mathbf{r}$
### Part (ii)
$\sqrt{4^2 + (-5)^2}$ | M1 | Accept $\sqrt{4^2 - 5^2}$
$= \sqrt{41}$ or $6.4031...$ so $6.40$ (3 s.f.) | A1 |
Need $180 - \arctan\left(\frac{4}{5}\right)$ | M1 | Or equivalent. Award for $\arctan\left(\pm\frac{4}{5}\right)$ or $\arctan\left(\pm\frac{5}{4}\right)$ or equivalent seen without 180 or 90
$141.340$ so $141°$ | A1 [4] | cao
### Part (iii)
$12\mathbf{i} - 15\mathbf{j}$ or $\begin{pmatrix}12\\-15\end{pmatrix}$ | B1 [1] | Do not award for magnitude given as the answer. Penalise spurious notation by 1 mark at most once in paper
---5 A particle has a position vector $\mathbf { r }$, where $\mathbf { r } = 4 \mathbf { i } - 5 \mathbf { j }$ and $\mathbf { i }$ and $\mathbf { j }$ are unit vectors in the directions east and north respectively.\\
(i) Sketch $\mathbf { r }$ on a diagram showing $\mathbf { i }$ and $\mathbf { j }$ and the origin O .\\
(ii) Calculate the magnitude of $\mathbf { r }$ and its direction as a bearing.\\
(iii) Write down the vector that has the same direction as $\mathbf { r }$ and three times its magnitude.
\hfill \mbox{\textit{OCR MEI M1 Q5 [6]}}