| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | October |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Moderate -0.3 This is a standard M1 vectors question requiring routine application of speed formula, relative velocity concepts, and bearing calculations. Part (a) is straightforward Pythagoras, part (b) involves setting up position vectors and comparing coefficients (though multi-step), and part (c) requires solving a quadratic and applying trigonometry. While it has multiple parts worth 16 marks total, each component uses well-practiced techniques without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors3.02a Kinematics language: position, displacement, velocity, acceleration3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | 3 2 + 1 2 2 | M1 |
| 1 5 3 , 3 1 7 , 12 or better ( k m h − 1 ) | A1 |
| Answer | Marks |
|---|---|
| 8(a) | M1 Use of Pythagoras with square root |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b) | ( − 9 i + 6 j ) + t ( 3 i + 1 2 j ) | M1 |
| ( 1 6 i + 6 j ) + t ( p i + q j ) | A1 | |
| A B = b − a = (1 6 i + 6 j ) + t ( p i + q j ) − ( ( − 9 i + 6 j ) + t ( 3 i + 1 2 j ) ) | M1 A1 |
| Answer | Marks |
|---|---|
| cancel. | M1 |
| p=−9 , q=3 | A1 |
| Answer | Marks |
|---|---|
| 8(b) | M1 Correct structure for either |
| Answer | Marks | Guidance |
|---|---|---|
| 8(c) | ( 2 5 − 1 2 t ) 2 + ( − 9 t ) 2 = 1 5 2 (225t2 −600t+400=0) | M1A1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| ( 9 i − 1 2 j ) Note that this a method mark. | DM1 |
| Answer | Marks |
|---|---|
| 1 2 | M1 |
| =37o | A1 |
| Bearing is 323o to nearest degree | A1 |
| Answer | Marks |
|---|---|
| 8(c) | M1 Use of Pythagoras to give an equation in t only |
Question 8:
--- 8(a) ---
8(a) | 3 2 + 1 2 2 | M1
1 5 3 , 3 1 7 , 12 or better ( k m h − 1 ) | A1
(2)
8(a) | M1 Use of Pythagoras with square root
A1 cao
--- 8(b) ---
8(b) | ( − 9 i + 6 j ) + t ( 3 i + 1 2 j ) | M1 | A1
( 1 6 i + 6 j ) + t ( p i + q j ) | A1
A B = b − a = (1 6 i + 6 j ) + t ( p i + q j ) − ( ( − 9 i + 6 j ) + t ( 3 i + 1 2 j ) ) | M1 A1
= 25+t(p−3)]i+t(q−12)j
Compare with: ( 2 5 − 1 2 t ) i − 9 t j or e.g. use b = A B + a to obtain an
equation in p only and an equation in q only. May be implied by correct
answers only.
(−12= p−3 and −9=q−12)
N.B. This mark may not be available if they go wrong and the t’s don’t
cancel. | M1
p=−9 , q=3 | A1
(7)
8(b) | M1 Correct structure for either
A1 cao
A1 cao
M1 Allow a − b
A1 for a correct unsimplified expression for either b – a or a – b
M1 for an equation in p only and an equation in q only
A1 cao
--- 8(c) ---
8(c) | ( 2 5 − 1 2 t ) 2 + ( − 9 t ) 2 = 1 5 2 (225t2 −600t+400=0) | M1A1
4
t =
3 | A1
( 9 i − 1 2 j ) Note that this a method mark. | DM1
9
t a n =
1 2 | M1
=37o | A1
Bearing is 323o to nearest degree | A1
(7)
(16)
Notes for question 8
8(c) | M1 Use of Pythagoras to give an equation in t only
Allow with a square root.
A1 Correct unsimplified quadratic equation
A1 t = 1.3 or better
DM1 Use of their t to find A B o r B A , dependent on previous M. May
be implied. Allow if they use one of their two incorrect t values.
M1 For an equation in a relevant angle for their AB. Could be implied
by a relevant angle seen on a diagram which could need checking with a
calculator
A1 Correct relevant angle e.g 3 7 o , 5 3 o , 1 2 7 o , e t c or better
A1 cao
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom[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 ships, $A$ and $B$, are moving with constant velocities.
The velocity of $A$ is $(3\mathbf{i} + 12\mathbf{j})\text{ kmh}^{-1}$ and the velocity of $B$ is $(p\mathbf{i} + q\mathbf{j})\text{ kmh}^{-1}$
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $A$. [2]
The ships are modelled as particles.
At 12 noon, $A$ is at the point with position vector $(-9\mathbf{i} + 6\mathbf{j})$ km and $B$ is at the point with position vector $(16\mathbf{i} + 6\mathbf{j})$ km.
At time $t$ hours after 12 noon,
$$\overrightarrow{AB} = [(25 - 12t)\mathbf{i} - 9t\mathbf{j}] \text{ km}$$
\item Find the value of $p$ and the value of $q$. [7]
\item Find the bearing of $A$ from $B$ when the ships are 15 km apart, giving your answer to the nearest degree. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q8 [16]}}