| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Bearing and speed from velocity vector |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic vector operations: substituting t=2.5, calculating magnitude for speed, and finding bearing from components using arctan. All steps are routine applications of standard formulas with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature and bearing calculations. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(t=2.5\): \(\mathbf{v} = \binom{-5}{10} + 2.5\binom{6}{-8} = \binom{10}{-10}\) | M1 A1 | |
| East component = North component in magnitude, East positive, North negative → bearing 135° confirmed | A1 | Must show both components equal and correct signs |
| Speed \(= \sqrt{10^2+10^2} = 10\sqrt{2} \approx 14.1\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Position at \(t=2.5\): \(\mathbf{r} = 2.5\binom{-5}{10} + \frac{1}{2}(2.5)^2\binom{6}{-8}\) | M1 | Integrate velocity |
| \(= \binom{-12.5}{25} + \binom{18.75}{-25} = \binom{6.25}{0}\) | A1 | |
| Bearing from O: displacement is \(\binom{6.25}{0}\), i.e. due East | M1 | |
| Bearing = 090° | A1 |
# Question 6:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $t=2.5$: $\mathbf{v} = \binom{-5}{10} + 2.5\binom{6}{-8} = \binom{10}{-10}$ | M1 A1 | |
| East component = North component in magnitude, East positive, North negative → bearing 135° confirmed | A1 | Must show both components equal and correct signs |
| Speed $= \sqrt{10^2+10^2} = 10\sqrt{2} \approx 14.1$ m s$^{-1}$ | A1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Position at $t=2.5$: $\mathbf{r} = 2.5\binom{-5}{10} + \frac{1}{2}(2.5)^2\binom{6}{-8}$ | M1 | Integrate velocity |
| $= \binom{-12.5}{25} + \binom{18.75}{-25} = \binom{6.25}{0}$ | A1 | |
| Bearing from O: displacement is $\binom{6.25}{0}$, i.e. due East | M1 | |
| Bearing = 090° | A1 | |6 The velocity of a model boat, $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, is given by
$$\mathbf { v } = \binom { - 5 } { 10 } + t \binom { 6 } { - 8 }$$
where $t$ is the time in seconds and the vectors $\binom { 1 } { 0 }$ and $\binom { 0 } { 1 }$ are east and north respectively.\\
(i) Show that when $t = 2.5$ the boat is travelling south-east (i.e. on a bearing of $135 ^ { \circ }$ ). Calculate its speed at this time.
The boat is at a point O when $t = 0$.\\
(ii) Calculate the bearing of the boat from O when $t = 2.5$.
\hfill \mbox{\textit{OCR MEI M1 2007 Q6 [7]}}