| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Moderate -0.3 This is a straightforward mechanics vectors question requiring magnitude calculation using Pythagoras, bearing from trigonometry, showing parallel vectors by scalar multiplication, and finding an unknown component using proportionality. All techniques are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \( | \mathbf{F} | = 12.5\) so \(12.5\) N |
| bearing is \(90 - \arctan\dfrac{12}{3.5}\) | M1 | Use of arctan with \(3.5\) and \(12\) or equiv |
| \(= (0)16.260\ldots\) so \((0)16.3°\) (3 s.f.) | A1 | May be obtained directly as \(\arctan\dfrac{3.5}{12}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(24/7 = 12/3.5\) or \(\ldots\) | E1 | Accept statement following \(\mathbf{G} = 2\mathbf{F}\) shown. |
| \(\mathbf{G} = 2\mathbf{F}\) so \( | \mathbf{G} | = 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{9+12}{3.5} = \dfrac{-18+q}{12}\) | M1 | Or equivalent or in scalar equations. Accept \(\dfrac{21}{q-18}\) or \(\dfrac{q-18}{21} = \tan(i)\) or \(\tan(90-(i))\) |
| so \(q = 6 \times 12 + 18 = 90\) | A1 | Accept \(90\mathbf{j}\) |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $|\mathbf{F}| = 12.5$ so $12.5$ N | B1 | |
| bearing is $90 - \arctan\dfrac{12}{3.5}$ | M1 | Use of arctan with $3.5$ and $12$ or equiv |
| $= (0)16.260\ldots$ so $(0)16.3°$ (3 s.f.) | A1 | May be obtained directly as $\arctan\dfrac{3.5}{12}$ |
**Subtotal: 3**
---
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $24/7 = 12/3.5$ or $\ldots$ | E1 | Accept **statement** following $\mathbf{G} = 2\mathbf{F}$ shown. |
| $\mathbf{G} = 2\mathbf{F}$ so $|\mathbf{G}| = 2|\mathbf{F}|$ | B1 | Accept equivalent in words. |
**Subtotal: 2**
---
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{9+12}{3.5} = \dfrac{-18+q}{12}$ | M1 | Or equivalent or in scalar equations. Accept $\dfrac{21}{q-18}$ or $\dfrac{q-18}{21} = \tan(i)$ or $\tan(90-(i))$ |
| so $q = 6 \times 12 + 18 = 90$ | A1 | Accept $90\mathbf{j}$ |
**Subtotal: 2**
---
**Total: 7**7 A force $\mathbf { F }$ is given by $\mathbf { F } = ( 3.5 \mathbf { i } + 12 \mathbf { j } ) \mathrm { N }$, where $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors east and north respectively.\\
(i) Calculate the magnitude of $\mathbf { F }$ and also its direction as a bearing.\\
(ii) $\mathbf { G }$ is the force $( 7 \mathbf { i } + 24 \mathbf { j } ) \mathrm { N }$. Show that $\mathbf { G }$ and $\mathbf { F }$ are in the same direction and compare their magnitudes.\\
(iii) Force $\mathbf { F } _ { 1 }$ is $( 9 \mathbf { i } - 18 \mathbf { j } ) \mathrm { N }$ and force $\mathbf { F } _ { 2 }$ is $( 12 \mathbf { i } + q \mathbf { j } ) \mathrm { N }$. Find $q$ so that the sum $\mathbf { F } _ { 1 } + \mathbf { F } _ { 2 }$ is in the direction of $\mathbf { F }$.
\hfill \mbox{\textit{OCR MEI M1 Q7 [7]}}