| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | When is one object due north/east/west/south of another |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving position vectors and relative motion. Part (a) requires finding a bearing from a velocity vector using basic trigonometry. Part (b) requires finding positions at t=4, interpreting 'southwest' as equal displacement in south and west directions, solving a simple linear equation for p, then calculating speed. All techniques are routine for M1 with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{6}{7}\) | M1 | Any trig ratio using 6 and 7 |
| \(\theta = 40.60°...\) | A1 | Correct angle from correct equation e.g. \(49°, 41°, 139°, 131°\)... |
| Bearing is \(360° - 40.60° = 319°\) nearest degree | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r}_A = (20\mathbf{i} - 17\mathbf{j}) + 4(-6\mathbf{i} + 7\mathbf{j}) = (-4\mathbf{i} + 11\mathbf{j})\) | M1 A1 | First M1 for attempt at \(\mathbf{r}_4 = \mathbf{r}_0 + 4\mathbf{v}\) for either \(A\) or \(B\); A1 for \((-4\mathbf{i}+11\mathbf{j})\); \(\mathbf{i}\)'s and \(\mathbf{j}\)'s must be collected |
| \(\mathbf{r}_B = (-8\mathbf{i} + 9\mathbf{j}) + 4(p\mathbf{i} + 2p\mathbf{j}) = (-8+4p)\mathbf{i} + (9+8p)\mathbf{j}\) | A1 | \(\mathbf{i}\)'s and \(\mathbf{j}\)'s must be collected |
| \(\mathbf{r}_A - \mathbf{r}_B = (4-4p)\mathbf{i} + (2-8p)\mathbf{j}\) | DM1 | Dependent on first M1; must subtract both \(\mathbf{i}\) and \(\mathbf{j}\) components |
| \(-8 + 4p - {-4} = 9 + 8p - 11\) | M1 A1 | Third M1 for equating \(\mathbf{i}\) and \(\mathbf{j}\) components of difference (M0 if no difference) to give equation in \(p\) only |
| \(p = -0.5\) | A1 | Correct value of \(p\) |
| \(\mathbf{v}_B = (-0.5\mathbf{i} - \mathbf{j})\) | M1 | Using \(p\) value to obtain velocity vector for \(B\) |
| \( | \mathbf{v}_B | = \sqrt{(-0.5)^2 + (-1)^2}\) |
| \(= \frac{\sqrt{5}}{2} = 1.1 \text{ ms}^{-1}\) or better | A1 | \(\frac{\sqrt{5}}{2}\) oe or 1.1 or better |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{6}{7}$ | M1 | Any trig ratio using 6 and 7 |
| $\theta = 40.60°...$ | A1 | Correct angle from correct equation e.g. $49°, 41°, 139°, 131°$... |
| Bearing is $360° - 40.60° = 319°$ nearest degree | A1 | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r}_A = (20\mathbf{i} - 17\mathbf{j}) + 4(-6\mathbf{i} + 7\mathbf{j}) = (-4\mathbf{i} + 11\mathbf{j})$ | M1 A1 | First M1 for attempt at $\mathbf{r}_4 = \mathbf{r}_0 + 4\mathbf{v}$ for either $A$ or $B$; A1 for $(-4\mathbf{i}+11\mathbf{j})$; $\mathbf{i}$'s and $\mathbf{j}$'s must be collected |
| $\mathbf{r}_B = (-8\mathbf{i} + 9\mathbf{j}) + 4(p\mathbf{i} + 2p\mathbf{j}) = (-8+4p)\mathbf{i} + (9+8p)\mathbf{j}$ | A1 | $\mathbf{i}$'s and $\mathbf{j}$'s must be collected |
| $\mathbf{r}_A - \mathbf{r}_B = (4-4p)\mathbf{i} + (2-8p)\mathbf{j}$ | DM1 | Dependent on first M1; must subtract both $\mathbf{i}$ and $\mathbf{j}$ components |
| $-8 + 4p - {-4} = 9 + 8p - 11$ | M1 A1 | Third M1 for equating $\mathbf{i}$ and $\mathbf{j}$ components of difference (M0 if no difference) to give equation in $p$ only |
| $p = -0.5$ | A1 | Correct value of $p$ |
| $\mathbf{v}_B = (-0.5\mathbf{i} - \mathbf{j})$ | M1 | Using $p$ value to obtain velocity vector for $B$ |
| $|\mathbf{v}_B| = \sqrt{(-0.5)^2 + (-1)^2}$ | M1 | Finding magnitude (available even if $\mathbf{v}_B$ not correct form) |
| $= \frac{\sqrt{5}}{2} = 1.1 \text{ ms}^{-1}$ or better | A1 | $\frac{\sqrt{5}}{2}$ oe or 1.1 or better |
---2. [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin $O$.]
At time $t = 0$, a bird $A$ leaves its nest, that is located at the point with position vector $( 20 \mathbf { i } - 17 \mathbf { j } ) \mathrm { m }$, and flies with constant velocity $( - 6 \mathbf { i } + 7 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. At the same time a second bird $B$ leaves its nest which is located at the point with position vector $( - 8 \mathbf { i } + 9 \mathbf { j } ) \mathrm { m }$ and flies with constant velocity ( $p \mathbf { i } + 2 p \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$, where $p$ is a constant. At time $t = 4 \mathrm {~s}$, bird $B$ is south west of bird $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the direction of motion of $A$, giving your answer as a bearing to the nearest degree.
\item Find the speed of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2019 Q2 [13]}}