| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question testing basic vector concepts: finding distance using Pythagoras, showing constant velocity by differentiating, plotting paths, finding speed/direction from velocity components, and finding maximum separation. All techniques are standard and clearly signposted, though part (vi) requires completing the square which adds slight challenge. Slightly easier than average due to clear structure and routine methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A: \(t=0,\ \mathbf{r} = \begin{pmatrix}3\\2\end{pmatrix}\); B: \(t=2,\ \mathbf{r} = \begin{pmatrix}15\\18\end{pmatrix}\) | B1 | Awarded automatically if displacement is correct |
| \(\begin{pmatrix}15\\18\end{pmatrix} - \begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}12\\16\end{pmatrix}\) | B1 | Finding displacement; follow through from position vectors for A and B |
| \(\sqrt{12^2 + 16^2} = 20\) — distance AB is 20 km | B1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = \begin{pmatrix}6\\8\end{pmatrix}\) which is constant | B1 | Any valid argument. Accept \(\begin{pmatrix}6\\8\end{pmatrix}\) with no comment. Do not accept \(a=0\) without explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Points A and B plotted correctly with line segment AB for the *Rosemary* | B1 | No FT from part (i). No extra lines or curves |
| A curve between A and B for the *Sage* | B1 | B0 for two line segments. No FT from part (i) |
| Curve passes through \((9, 6)\) | B1 | Condone no labels |
# Question 8:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| A: $t=0,\ \mathbf{r} = \begin{pmatrix}3\\2\end{pmatrix}$; B: $t=2,\ \mathbf{r} = \begin{pmatrix}15\\18\end{pmatrix}$ | B1 | Awarded automatically if displacement is correct |
| $\begin{pmatrix}15\\18\end{pmatrix} - \begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}12\\16\end{pmatrix}$ | B1 | Finding displacement; follow through from position vectors for A and B |
| $\sqrt{12^2 + 16^2} = 20$ — distance AB is 20 km | B1 | Cao |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = \begin{pmatrix}6\\8\end{pmatrix}$ which is constant | B1 | Any valid argument. Accept $\begin{pmatrix}6\\8\end{pmatrix}$ with no comment. Do not accept $a=0$ without explanation |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Points A and B plotted correctly with line segment AB for the *Rosemary* | B1 | No FT from part (i). No extra lines or curves |
| A curve between A and B for the *Sage* | B1 | B0 for two line segments. No FT from part (i) |
| Curve passes through $(9, 6)$ | B1 | Condone no labels |8 In this question, positions are given relative to a fixed origin, O. The $x$-direction is east and the $y$-direction north; distances are measured in kilometres.
Two boats, the Rosemary and the Sage, are having a race between two points A and B.\\
The position vector of the Rosemary at time $t$ hours after the start is given by
$$\mathbf { r } = \binom { 3 } { 2 } + \binom { 6 } { 8 } t , \text { where } 0 \leqslant t \leqslant 2 .$$
The Rosemary is at point A when $t = 0$, and at point B when $t = 2$.\\
(i) Find the distance AB .\\
(ii) Show that the Rosemary travels at constant velocity.
The position vector of the Sage is given by
$$\mathbf { r } = \binom { 3 ( 2 t + 1 ) } { 2 \left( 2 t ^ { 2 } + 1 \right) }$$
(iii) Plot the points A and B .
Draw the paths of the two boats for $0 \leqslant t \leqslant 2$.\\
(iv) What can you say about the result of the race?\\
(v) Find the speed of the Sage when $t = 2$. Find also the direction in which it is travelling, giving your answer as a compass bearing, to the nearest degree.\\
(vi) Find the displacement of the Rosemary from the Sage at time $t$ and hence calculate the greatest distance between the boats during the race.
\hfill \mbox{\textit{OCR MEI M1 2012 Q8 [18]}}