| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.3 This is a standard M1 closest approach problem with clear scaffolding through parts (a)-(d). Part (a) is direct application of r = r₀ + vt, part (b) is given as 'show that', part (c) requires minimizing |AB|² by differentiation (standard technique), and part (d) is basic trigonometry. The question follows a familiar template with no novel insights required, making it slightly easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r} = (13\mathbf{i} + 5\mathbf{j}) + t(3\mathbf{i} - 10\mathbf{j})\) | M1A1 | Expression with correct structure; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{s} = (3\mathbf{i} - 5\mathbf{j}) + t(15\mathbf{i} + 14\mathbf{j})\) | M1A1 | Expression with correct structure; cao |
| \(\overrightarrow{AB} = \mathbf{s} - \mathbf{r}\) | M1 | Allow difference in either order |
| \(\overrightarrow{AB} = (12t - 10)\mathbf{i} + (24t - 10)\mathbf{j}\) km | A1* | Correct given expression correctly obtained; N.B. \(\overrightarrow{AB} = (-10+12t)\mathbf{i} + (-10+24t)\mathbf{j}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = (12t-10)^2 + (24t-10)^2\) \((= 720t^2 - 720t + 200)\) | M1 | Correct expression (with or without square root) |
| Differentiate and equate to 0 OR complete the square OR use \(t = \frac{-b}{2a}\) | M1 | Attempt to differentiate (at least one power decreasing by 1) or complete the square |
| \(1440t - 720 = 0\) oe OR \(720(t - \frac{1}{2})^2 + 20\) | A1 | Correct equation or expression |
| Solve for \(t\); use \((t-\frac{1}{2})^2 \geq 0\); \(t = \frac{720}{2\times720}\) | DM1 | Dependent on previous M; for completing square method, for 'ignoring' the \((t-\frac{1}{2})^2\) term |
| Substitute their \(t\) into their \(AB\) expression | M1 | Must be non-zero; must have square root |
| \(\sqrt{20}\) oe (km), \(4.5\) or better | A1 | cao |
| OR alternative last 5 marks: \(720t^2 - 720t + 200 = D^2\) | M1A1 | Complete method |
| For real \(t\): \(720^2 \geq 4 \times 720(200 - D^2)\) | DM1 | |
| Solve for \(D\), \((D \geq \sqrt{20})\) | M1 | |
| \(\sqrt{20}\) oe (km), \(4.5\) or better | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\overrightarrow{AB} = -4\mathbf{i} + 2\mathbf{j}\) at \(t = \frac{1}{2}\) to obtain a relevant angle, e.g. \(26.56°\); allow e.g. \(\tan\alpha = \frac{1}{2}\) or \(\tan^{-1}\frac{1}{2}\) | M1 | Using their \(t\) value to obtain \(\overrightarrow{AB}\) and a relevant angle |
| Bearing is \(297°\) or better | A1 | cao |
## Question 8:
### Part 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = (13\mathbf{i} + 5\mathbf{j}) + t(3\mathbf{i} - 10\mathbf{j})$ | M1A1 | Expression with correct structure; cao |
### Part 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{s} = (3\mathbf{i} - 5\mathbf{j}) + t(15\mathbf{i} + 14\mathbf{j})$ | M1A1 | Expression with correct structure; cao |
| $\overrightarrow{AB} = \mathbf{s} - \mathbf{r}$ | M1 | Allow difference in either order |
| $\overrightarrow{AB} = (12t - 10)\mathbf{i} + (24t - 10)\mathbf{j}$ km | A1* | Correct given expression correctly obtained; N.B. $\overrightarrow{AB} = (-10+12t)\mathbf{i} + (-10+24t)\mathbf{j}$ is A0 |
### Part 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = (12t-10)^2 + (24t-10)^2$ $(= 720t^2 - 720t + 200)$ | M1 | Correct expression (with or without square root) |
| Differentiate and equate to 0 **OR** complete the square **OR** use $t = \frac{-b}{2a}$ | M1 | Attempt to differentiate (at least one power decreasing by 1) or complete the square |
| $1440t - 720 = 0$ oe **OR** $720(t - \frac{1}{2})^2 + 20$ | A1 | Correct equation or expression |
| Solve for $t$; use $(t-\frac{1}{2})^2 \geq 0$; $t = \frac{720}{2\times720}$ | DM1 | Dependent on previous M; for completing square method, for 'ignoring' the $(t-\frac{1}{2})^2$ term |
| Substitute their $t$ into their $AB$ expression | M1 | Must be non-zero; must have square root |
| $\sqrt{20}$ oe (km), $4.5$ or better | A1 | cao |
| **OR alternative last 5 marks:** $720t^2 - 720t + 200 = D^2$ | M1A1 | Complete method |
| For real $t$: $720^2 \geq 4 \times 720(200 - D^2)$ | DM1 | |
| Solve for $D$, $(D \geq \sqrt{20})$ | M1 | |
| $\sqrt{20}$ oe (km), $4.5$ or better | A1 | |
### Part 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\overrightarrow{AB} = -4\mathbf{i} + 2\mathbf{j}$ at $t = \frac{1}{2}$ to obtain a relevant angle, e.g. $26.56°$; allow e.g. $\tan\alpha = \frac{1}{2}$ or $\tan^{-1}\frac{1}{2}$ | M1 | Using their $t$ value to obtain $\overrightarrow{AB}$ and a relevant angle |
| Bearing is $297°$ or better | A1 | cao |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.]
A ship $A$ moves with constant velocity $( 3 \mathbf { i } - 10 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }$\\
At time $t$ hours, the position vector of $A$ is $\mathbf { r } \mathrm { km }$.\\
At time $t = 0 , A$ is at the point with position vector $( 13 \mathbf { i } + 5 \mathbf { j } ) \mathrm { km }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { r }$ in terms of $t$.
Another ship $B$ moves with constant velocity $( 15 \mathbf { i } + 14 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$\\
At time $t = 0 , B$ is at the point with position vector $( 3 \mathbf { i } - 5 \mathbf { j } ) \mathrm { km }$.
\item Show that, at time $t$ hours,
$$\overrightarrow { A B } = [ ( 12 t - 10 ) \mathbf { i } + ( 24 t - 10 ) \mathbf { j } ] \mathrm { km }$$
Given that the two ships do not change course,
\item find the shortest distance between the two ships,
\item find the bearing of ship $B$ from ship $A$ when the ships are closest.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f1bdc84b-c8a1-4e7c-a2ba-48b40c6a6d36-28_2820_1967_102_100}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q8 [14]}}