| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 17 |
| 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 on relative motion with constant velocities. Part (a) requires basic trigonometry for bearings, (b) is a straightforward 'show that' using position vectors, (c) repeats bearing calculation, (d) involves setting up a simple equation when the direction condition is met, and (e) requires solving a quadratic from |PQ|=200. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use trig to get an equation in a relevant angle e.g. \(\tan\alpha = 1\) or uses isosceles triangle | M1 | e.g. use of cos or sin or the cosine rule. M0 if not using the velocity of \(Q\) |
| \(135°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{p} = t(15\mathbf{i})\) | M1 A1 | Correct structure for either |
| \(\mathbf{q} = 200\mathbf{j} + t(20\mathbf{i} - 20\mathbf{j})\) | A1 | oe |
| \(\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = 200\mathbf{j} + t(20\mathbf{i} - 20\mathbf{j}) - t(15\mathbf{i})\) | M1 | Allow \(\mathbf{p} - \mathbf{q}\). M0 if they put \(\mathbf{p} = \mathbf{q}\). Need to see at least this line of working |
| \(\overrightarrow{PQ} = 5t\mathbf{i} + (200 - 20t)\mathbf{j}\) (m) | A1* | Correct given answer correctly obtained, allow omission of m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PQ} = 50\mathbf{i}\) at \(t = 10\) (Allow M1 if they find \(\mathbf{p} = 150\mathbf{i}\) and \(\mathbf{q} = 200\mathbf{i}\) at \(t = 10\)) | M1 | Clear attempt to find \(\overrightarrow{PQ}\) at \(t = 10\) or \(\mathbf{p}\) and \(\mathbf{q}\) at \(t = 10\). N.B. This mark could be implied by a correct diagram |
| \(270°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5t = 200 - 20t\) | M1 | Equating components to give an equation in \(t\) only, with no vectors |
| \(t = 8\) | A1 | cao |
| \(\overrightarrow{PQ} = (5\times8)\mathbf{i} + (200 - 20\times8)\mathbf{j}\) | M1 | Substituting their \(t\) value into their \(\overrightarrow{PQ}\) |
| \(PQ = \sqrt{40^2 + 40^2} = 40\sqrt{2}\) (m), \(57\) or better | M1A1 | cao. Finding the magnitude, with square root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((5t)^2 + (200 - 20t)^2 = 200^2\) | M1 | Use of Pythagoras — allow with square root — to obtain an equation in \(t\) |
| \(425t^2 - 8000t = 0\) | A1 | Correct unsimplified 2 term quadratic in \(t\) |
| \(t = 0\) or \(\frac{320}{17} = 18.82\ldots\) Accept 19 or better. Apply isw | A1 | For both answers |
## Question 8:
**Note:** Allow column vectors throughout except for the answer to (b).
---
### Part 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use trig to get an equation in a relevant angle e.g. $\tan\alpha = 1$ or uses isosceles triangle | M1 | e.g. use of cos or sin or the cosine rule. M0 if not using the velocity of $Q$ |
| $135°$ | A1 | cao |
**Total: (2)**
---
### Part 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{p} = t(15\mathbf{i})$ | M1 A1 | Correct structure for either |
| $\mathbf{q} = 200\mathbf{j} + t(20\mathbf{i} - 20\mathbf{j})$ | A1 | oe |
| $\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = 200\mathbf{j} + t(20\mathbf{i} - 20\mathbf{j}) - t(15\mathbf{i})$ | M1 | Allow $\mathbf{p} - \mathbf{q}$. M0 if they put $\mathbf{p} = \mathbf{q}$. Need to see at least this line of working |
| $\overrightarrow{PQ} = 5t\mathbf{i} + (200 - 20t)\mathbf{j}$ (m) | A1* | Correct given answer correctly obtained, allow omission of m |
**Total: (5)**
---
### Part 8(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PQ} = 50\mathbf{i}$ at $t = 10$ (Allow M1 if they find $\mathbf{p} = 150\mathbf{i}$ and $\mathbf{q} = 200\mathbf{i}$ at $t = 10$) | M1 | Clear attempt to find $\overrightarrow{PQ}$ at $t = 10$ or $\mathbf{p}$ and $\mathbf{q}$ at $t = 10$. N.B. This mark could be implied by a correct diagram |
| $270°$ | A1 | cao |
**Total: (2)**
---
### Part 8(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5t = 200 - 20t$ | M1 | Equating components to give an equation in $t$ only, with no vectors |
| $t = 8$ | A1 | cao |
| $\overrightarrow{PQ} = (5\times8)\mathbf{i} + (200 - 20\times8)\mathbf{j}$ | M1 | Substituting their $t$ value into their $\overrightarrow{PQ}$ |
| $PQ = \sqrt{40^2 + 40^2} = 40\sqrt{2}$ (m), $57$ or better | M1A1 | cao. Finding the magnitude, with square root |
**Total: (5)**
---
### Part 8(e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(5t)^2 + (200 - 20t)^2 = 200^2$ | M1 | Use of Pythagoras — allow with square root — to obtain an equation in $t$ |
| $425t^2 - 8000t = 0$ | A1 | Correct unsimplified 2 term quadratic in $t$ |
| $t = 0$ or $\frac{320}{17} = 18.82\ldots$ Accept 19 or better. Apply isw | A1 | For both answers |
**Total: (3)**
**Question Total: (17)**8. [In this question, $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin $O$.]
Two boats, $P$ and $Q$, are moving with constant velocities.\\
The velocity of $P$ is $15 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the velocity of $Q$ is $( 20 \mathbf { i } - 20 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find the direction in which $Q$ is travelling, giving your answer as a bearing.
The boats are modelled as particles.\\
At time $t = 0 , P$ is at the origin $O$ and $Q$ is at the point with position vector $200 \mathbf { j } \mathrm {~m}$. At time $t$ seconds, the position vector of $P$ is $\mathbf { p m }$ and the position vector of $Q$ is $\mathbf { q m }$.
\item Show that
$$\overrightarrow { P Q } = [ 5 t \mathbf { i } + ( 200 - 20 t ) \mathbf { j } ] \mathrm { m }$$
\item Find the bearing of $P$ from $Q$ when $t = 10$
\item Find the distance between $P$ and $Q$ when $Q$ is north east of $P$
\item Find the times when $P$ and $Q$ are 200 m apart.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q8 [17]}}