| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Bearing and speed from velocity vector |
| Difficulty | Standard +0.3 This is a standard M1 vectors question with routine steps: finding a bearing (basic trigonometry), writing position vectors using r = r₀ + vt (direct formula application), vector subtraction, and solving a quadratic from |QP| = 10. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{2}{9}\), \(\theta = 12.5°\), bearing \(103°\) | M1 A1 A1 (3) | M1 for \(\tan\theta = \pm\frac{2}{9}\) or \(\pm\frac{9}{2}\) or use \(\sin\theta\) or \(\cos\theta\); First A1 for \(\theta = \pm13°\) or \(\pm77°\) or \(\pm12.5°\) or \(\pm77.5°\) or better; Second A1 for \(103°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(\mathbf{p} = (9\mathbf{i} + 10\mathbf{j}) + t(9\mathbf{i} - 2\mathbf{j})\) | M1 A1 | M1 for clear attempt; First A1 for \(\mathbf{p}\) oe |
| (ii) \(\mathbf{q} = (\mathbf{i} + 4\mathbf{j}) + t(4\mathbf{i} + 8\mathbf{j})\) | A1 (3) | Second A1 for \(\mathbf{q}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{QP} = (8 + 5t)\mathbf{i} + (6 - 10t)\mathbf{j}\) | M1 A1 (2) | M1 for \(\mathbf{p} - \mathbf{q}\) or \(\mathbf{q} - \mathbf{p}\) with substituted values; A1 correct answer (identical coefficients of \(\mathbf{i}\) and \(\mathbf{j}\), allow column vectors) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(D^2 = (8+5t)^2 + (6-10t)^2\) | M1 | First M1 for attempt to find \(QP\) or \(QP^2\) in terms of \(t\) only |
| \(= 125t^2 - 40t + 100\) | A1 | First A1 for correct expression |
| \(100 = 125t^2 - 40t + 100\) | M1 | Second M1 for \(\sqrt{\text{(3 term quadratic)}} = 10\) or (3 term quadratic) \(= 100\) |
| \(0 = 5t(25t - 8)\) | M1 | Third M1 for quadratic \(= 0\) and attempt to solve |
| \(t = 0\) or \(0.32\) | A1 A1 (6) | Second A1 for \(t=0\); Third A1 for \(t = 0.32\) oe |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{2}{9}$, $\theta = 12.5°$, bearing $103°$ | M1 A1 A1 (3) | M1 for $\tan\theta = \pm\frac{2}{9}$ or $\pm\frac{9}{2}$ or use $\sin\theta$ or $\cos\theta$; First A1 for $\theta = \pm13°$ or $\pm77°$ or $\pm12.5°$ or $\pm77.5°$ or better; Second A1 for $103°$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (i) $\mathbf{p} = (9\mathbf{i} + 10\mathbf{j}) + t(9\mathbf{i} - 2\mathbf{j})$ | M1 A1 | M1 for clear attempt; First A1 for $\mathbf{p}$ oe |
| (ii) $\mathbf{q} = (\mathbf{i} + 4\mathbf{j}) + t(4\mathbf{i} + 8\mathbf{j})$ | A1 (3) | Second A1 for $\mathbf{q}$ oe |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{QP} = (8 + 5t)\mathbf{i} + (6 - 10t)\mathbf{j}$ | M1 A1 (2) | M1 for $\mathbf{p} - \mathbf{q}$ or $\mathbf{q} - \mathbf{p}$ with substituted values; A1 correct answer (identical coefficients of $\mathbf{i}$ and $\mathbf{j}$, allow column vectors) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $D^2 = (8+5t)^2 + (6-10t)^2$ | M1 | First M1 for attempt to find $QP$ or $QP^2$ in terms of $t$ only |
| $= 125t^2 - 40t + 100$ | A1 | First A1 for correct expression |
| $100 = 125t^2 - 40t + 100$ | M1 | Second M1 for $\sqrt{\text{(3 term quadratic)}} = 10$ or (3 term quadratic) $= 100$ |
| $0 = 5t(25t - 8)$ | M1 | Third M1 for quadratic $= 0$ and attempt to solve |
| $t = 0$ or $0.32$ | A1 A1 (6) | Second A1 for $t=0$; Third A1 for $t = 0.32$ oe |
---7. [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 $O$.]
Two ships, $P$ and $Q$, are moving with constant velocities.\\
The velocity of $P$ is $( 9 \mathbf { i } - 2 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ and the velocity of $Q$ is $( 4 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find the direction of motion of $P$, giving your answer as a bearing to the nearest degree.
When $t = 0$, the position vector of $P$ is $( 9 \mathbf { i } + 10 \mathbf { j } ) \mathrm { km }$ and the position vector of $Q$ is $( \mathbf { i } + 4 \mathbf { j } ) \mathrm { km }$. At time $t$ hours, the position vectors of $P$ and $Q$ are $\mathbf { p } \mathrm { km }$ and $\mathbf { q } \mathrm { km }$ respectively.
\item Find an expression for
\begin{enumerate}[label=(\roman*)]
\item $\mathbf { p }$ in terms of $t$,
\item $\mathbf { q }$ in terms of $t$.
\end{enumerate}\item Hence show that, at time $t$ hours,
$$\overrightarrow { Q P } = ( 8 + 5 t ) \mathbf { i } + ( 6 - 10 t ) \mathbf { j }$$
\item Find the values of $t$ when the ships are 10 km apart.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q7 [14]}}