| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question on position vectors with constant velocity. Parts (i)-(iv) involve basic vector arithmetic and plotting. Part (v) requires differentiation to find velocity and bearing calculation. Part (vi) involves finding maximum distance using calculus, but all techniques are standard for this topic. The multi-part structure and mark allocation make it slightly above routine drill exercises, but it requires no novel insight—just systematic application of learned 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 displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A: \(t=0\), \(\mathbf{r}=\begin{pmatrix}3\\2\end{pmatrix}\), B: \(t=2\), \(\mathbf{r}=\begin{pmatrix}15\\18\end{pmatrix}\) | B1 | Award this mark automatically if the displacement is correct |
| \(\begin{pmatrix}15\\18\end{pmatrix}-\begin{pmatrix}3\\2\end{pmatrix}=\begin{pmatrix}12\\16\end{pmatrix}\) | B1 | Finding the displacement. Follow through from position vectors for A and B |
| \(\sqrt{12^2+16^2}=20\) The distance AB is 20 km. | B1 | Cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | 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. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Points A and B plotted correctly, line segment AB drawn for *Rosemary* | B1 | Points A and B plotted correctly, with no FT from part (i), and the line segment AB for the *Rosemary*. No extra lines or curves. |
| A curve between A and B for the *Sage* | B1 | For the *Sage*, a curve between A and B. B0 for two line segments. Nothing extra. No FT from part (i). |
| Passes through \((9,6)\) | B1 | Condone no labels |
| [3] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A: $t=0$, $\mathbf{r}=\begin{pmatrix}3\\2\end{pmatrix}$, B: $t=2$, $\mathbf{r}=\begin{pmatrix}15\\18\end{pmatrix}$ | B1 | Award this mark automatically if the displacement is correct |
| $\begin{pmatrix}15\\18\end{pmatrix}-\begin{pmatrix}3\\2\end{pmatrix}=\begin{pmatrix}12\\16\end{pmatrix}$ | B1 | Finding the displacement. Follow through from position vectors for A and B |
| $\sqrt{12^2+16^2}=20$ The distance AB is 20 km. | B1 | Cao |
| **[3]** | | |
### Part (ii):
| Answer/Working | Mark | 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. |
| **[1]** | | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Points A and B plotted correctly, line segment AB drawn for *Rosemary* | B1 | Points A and B plotted correctly, with no FT from part (i), and the line segment AB for the *Rosemary*. No extra lines or curves. |
| A curve between A and B for the *Sage* | B1 | For the *Sage*, a curve between A and B. B0 for two line segments. Nothing extra. No FT from part (i). |
| Passes through $(9,6)$ | B1 | Condone no labels |
| **[3]** | | |5 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 Q5 [18]}}