| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question with vectors requiring routine application of position-velocity relationships and collision conditions. Parts (a)-(c) involve straightforward calculations with given formulas, while part (d) requires finding positions at a specific time and computing distance—all standard textbook exercises with no novel problem-solving required. |
| Spec | 1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{(20\mathbf{i}+34\mathbf{j})-(15\mathbf{i}+36\mathbf{j})}{0.5}\) | M1 | Complete method to find velocity expression. Allow if 30 minutes used. M1A0 if missing brackets |
| \((10\mathbf{i}-4\mathbf{j})\)* | A1* | Reaches given answer from fully correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((15\mathbf{i}+36\mathbf{j})+t(10\mathbf{i}-4\mathbf{j})\) | M1 A1 | Finds expression for \(\mathbf{p}\) in terms of \(t\) with correct structure; correct answer in terms of \(\mathbf{i}\), \(\mathbf{j}\) and \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Verify using \(t=1.5\) in \(\mathbf{p}\) or \(\mathbf{q}\) | M1 | Substitutes \(t=1.5\) into given \(\mathbf{q}\) or their \(\mathbf{p}\) |
| \(\mathbf{p} = (15\mathbf{i}+36\mathbf{j})+1.5(10\mathbf{i}-4\mathbf{j}) = 30\mathbf{i}+30\mathbf{j}\) | A1 | \(P\) equation correct |
| \(\mathbf{q} = (42-8\times1.5)\mathbf{i}+(9+14\times1.5)\mathbf{j} = 30\mathbf{i}+30\mathbf{j}\) | A1 | \(Q\) equation correct. \(\mathbf{p}\) or \(\mathbf{q} = 30\mathbf{i}+30\mathbf{j}\) alone can imply correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equate \(\mathbf{i}\): \(15+10t = 42-8t \rightarrow t=1.5\) | M1 A1 | Equates coefficients of \(\mathbf{i}\) or \(\mathbf{j}\) using given \(\mathbf{q}\) and their \(\mathbf{p}\) |
| Equate \(\mathbf{j}\): \(36-4t = 9+14t \rightarrow t=1.5\) | A1 | Allow both A marks if \(t=1.5\) written only once |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{15+10t}{36-4t} = \frac{42-8t}{9+14t}\) | M1 | Uses ratio of components |
| \(\rightarrow t = 1.5\) or \(-8.5\) | A1 | Two \(t\) values |
| Verifies components both 30 at \(t=1.5\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(30\mathbf{i}+30\mathbf{j}\) | A1(B1) | The A mark is now treated as a B mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Position of \(P\) at 14:30 is \(40\mathbf{i}+26\mathbf{j}\) | B1 | Position of \(P\) at 14:30 |
| \(\mathbf{q} = (42-8\times0.5)\mathbf{i}+(9+14\times0.5)\mathbf{j}\) \(= 38\mathbf{i}+16\mathbf{j}\) | M1 | Use \(t=0.5\) to find new position of \(Q\) at 12:30 |
| \(15\mathbf{j}\) seen or implied | B1 | Correct expression seen for new velocity of \(Q\) |
| New position of \(Q\) at 14:30: \(\mathbf{q} = (38\mathbf{i}+16\mathbf{j})+2(15\mathbf{j})\) N.B. M0 if 2.5 used | M1 | Complete method to find new position of \(Q\) at 14:30, using new \(\mathbf{v}\) for \(Q\) |
| \(\mathbf{q} = 38\mathbf{i}+46\mathbf{j}\) | A1 | Correct position |
| \( | PQ | = \sqrt{(40-38)^2+(26-46)^2}\) |
| \(= \sqrt{404}\) or \(2\sqrt{101}\) (km) | A1 | Correct surd answer |
## Question 6:
### Part 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{(20\mathbf{i}+34\mathbf{j})-(15\mathbf{i}+36\mathbf{j})}{0.5}$ | M1 | Complete method to find velocity expression. Allow if 30 minutes used. M1A0 if missing brackets |
| $(10\mathbf{i}-4\mathbf{j})$* | A1* | Reaches given answer from fully correct working |
### Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(15\mathbf{i}+36\mathbf{j})+t(10\mathbf{i}-4\mathbf{j})$ | M1 A1 | Finds expression for $\mathbf{p}$ in terms of $t$ with correct structure; correct answer in terms of $\mathbf{i}$, $\mathbf{j}$ and $t$ |
### Part 6(c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Verify using $t=1.5$ in $\mathbf{p}$ or $\mathbf{q}$ | M1 | Substitutes $t=1.5$ into given $\mathbf{q}$ or their $\mathbf{p}$ |
| $\mathbf{p} = (15\mathbf{i}+36\mathbf{j})+1.5(10\mathbf{i}-4\mathbf{j}) = 30\mathbf{i}+30\mathbf{j}$ | A1 | $P$ equation correct |
| $\mathbf{q} = (42-8\times1.5)\mathbf{i}+(9+14\times1.5)\mathbf{j} = 30\mathbf{i}+30\mathbf{j}$ | A1 | $Q$ equation correct. $\mathbf{p}$ or $\mathbf{q} = 30\mathbf{i}+30\mathbf{j}$ alone can imply correct equation |
**ALT1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equate $\mathbf{i}$: $15+10t = 42-8t \rightarrow t=1.5$ | M1 A1 | Equates coefficients of $\mathbf{i}$ or $\mathbf{j}$ using given $\mathbf{q}$ and their $\mathbf{p}$ |
| Equate $\mathbf{j}$: $36-4t = 9+14t \rightarrow t=1.5$ | A1 | Allow both A marks if $t=1.5$ written only once |
**ALT2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{15+10t}{36-4t} = \frac{42-8t}{9+14t}$ | M1 | Uses ratio of components |
| $\rightarrow t = 1.5$ or $-8.5$ | A1 | Two $t$ values |
| Verifies components both 30 at $t=1.5$ | A1 | — |
### Part 6(c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $30\mathbf{i}+30\mathbf{j}$ | A1(B1) | The A mark is now treated as a B mark |
### Part 6(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Position of $P$ at 14:30 is $40\mathbf{i}+26\mathbf{j}$ | B1 | Position of $P$ at 14:30 |
| $\mathbf{q} = (42-8\times0.5)\mathbf{i}+(9+14\times0.5)\mathbf{j}$ $= 38\mathbf{i}+16\mathbf{j}$ | M1 | Use $t=0.5$ to find new position of $Q$ at 12:30 |
| $15\mathbf{j}$ seen or implied | B1 | Correct expression seen for new velocity of $Q$ |
| New position of $Q$ at 14:30: $\mathbf{q} = (38\mathbf{i}+16\mathbf{j})+2(15\mathbf{j})$ N.B. M0 if 2.5 used | M1 | Complete method to find new position of $Q$ at 14:30, using new $\mathbf{v}$ for $Q$ |
| $\mathbf{q} = 38\mathbf{i}+46\mathbf{j}$ | A1 | Correct position |
| $|PQ| = \sqrt{(40-38)^2+(26-46)^2}$ | dM1 | Use of Pythagoras. Dependent on both previous M marks |
| $= \sqrt{404}$ or $2\sqrt{101}$ (km) | A1 | Correct surd answer |
---\begin{enumerate}
\item \hspace{0pt} [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 $O$.]
\end{enumerate}
At 12:00, a ship $P$ sets sail from a harbour with position vector $( 15 \mathbf { i } + 36 \mathbf { j } ) \mathrm { km }$. At 12:30, $P$ is at the point with position vector $( 20 \mathbf { i } + 34 \mathbf { j } ) \mathrm { km }$.
Given that $P$ moves with constant velocity,\\
(a) show that the velocity of $P$ is $( 10 \mathbf { i } - 4 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }$
At time $t$ hours after 12:00, the position vector of $P$ is $\mathbf { p } \mathrm { km }$.\\
(b) Find an expression for $\mathbf { p }$ in terms of $\mathbf { i } , \mathbf { j }$ and $t$.
A second ship $Q$ is also travelling at a constant velocity.\\
At time $t$ hours after 12:00, the position vector of $Q$ is given by $\mathbf { q } \mathrm { km }$, where
$$\mathbf { q } = ( 42 - 8 t ) \mathbf { i } + ( 9 + 14 t ) \mathbf { j }$$
Ships $P$ and $Q$ are modelled as particles.\\
If both ships maintained their course,\\
(c) (i) verify that they would collide at 13:30\\
(ii) find the position vector of the point at which the collision would occur.
At 12:30 $Q$ changes speed and direction to avoid the collision.\\
Ship $Q$ now travels due north with a constant speed of $15 \mathrm { kmh } ^ { - 1 }$\\
Ship $P$ maintains the same constant velocity throughout.\\
(d) Find the exact distance between $P$ and $Q$ at 14:30
\hfill \mbox{\textit{Edexcel M1 2023 Q6 [15]}}