| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic vector operations and bearing calculation. Part (i) involves simple trigonometry (arctan) and bearing conversion. Part (ii) requires equating components of parallel vectors, leading to a simple linear equation. Both parts are routine applications of standard techniques with no problem-solving insight needed. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(270 - \arctan\left(\frac{6}{4}\right)\) | M1 | Award for \(\arctan p\) seen where \(p = \pm\frac{6}{4}\) or \(\frac{4}{6}\), or equivalent |
| \(= 213.69...\) so \(214°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Need \((-4+3k)\mathbf{i} + (-6-2k)\mathbf{j} = \lambda(7\mathbf{i} - 9\mathbf{j})\) | M1 | Attempt to get LHS in the direction of \((7\mathbf{i} - 9\mathbf{j})\). Could be done by finding (tangents of) angles. Accept the use of \(\lambda = 1\). |
| either: \(\frac{-4+3k}{-6-2k} = \frac{7}{-9}\), or equivalent | M1 | Attempt to solve their *. Allow \(= \frac{7}{9}, \frac{9}{7}, -\frac{9}{7}\) |
| A1 | Expression correct | |
| \(k = 6\) | A1 | Award full marks for \(k = 6\) found WWW |
| or: \(-4+3k = 7\lambda\) and \(-6-2k = -9\lambda\) | M1 | Attempt to solve their *. Must have both equations. |
| A1 | Correct equations | |
| \(k = 6\) | A1 | Award full marks for \(k = 6\) found WWW |
| trial and error method | M1 | Any attempt to find the value of \(k\) and 'test'. M1 Systematic attempt in (the equivalent of) their *. Award full marks for \(k = 6\) found WWW |
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $270 - \arctan\left(\frac{6}{4}\right)$ | M1 | Award for $\arctan p$ seen where $p = \pm\frac{6}{4}$ or $\frac{4}{6}$, or equivalent |
| $= 213.69...$ so $214°$ | A1 | cao |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Need $(-4+3k)\mathbf{i} + (-6-2k)\mathbf{j} = \lambda(7\mathbf{i} - 9\mathbf{j})$ | M1 | Attempt to get LHS in the direction of $(7\mathbf{i} - 9\mathbf{j})$. Could be done by finding (tangents of) angles. Accept the use of $\lambda = 1$. |
| **either**: $\frac{-4+3k}{-6-2k} = \frac{7}{-9}$, or equivalent | M1 | Attempt to solve **their** *. Allow $= \frac{7}{9}, \frac{9}{7}, -\frac{9}{7}$ |
| | A1 | Expression correct |
| $k = 6$ | A1 | Award full marks for $k = 6$ found WWW |
| **or**: $-4+3k = 7\lambda$ and $-6-2k = -9\lambda$ | M1 | Attempt to solve **their** *. Must have both equations. |
| | A1 | Correct equations |
| $k = 6$ | A1 | Award full marks for $k = 6$ found WWW |
| **trial and error method** | M1 | Any attempt to find the value of $k$ and 'test'. M1 Systematic attempt in (the equivalent of) their *. Award full marks for $k = 6$ found WWW |
---5 In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are pointing east and north respectively.\\
(i) Calculate the bearing of the vector $- 4 \mathbf { i } - 6 \mathbf { j }$.
The vector $- 4 \mathbf { i } - 6 \mathbf { j } + k ( 3 \mathbf { i } - 2 \mathbf { j } )$ is in the direction $7 \mathbf { i } - 9 \mathbf { j }$.\\
(ii) Find $k$.
\hfill \mbox{\textit{OCR MEI M1 2010 Q5 [6]}}