| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on vector interception requiring position vector formation, equating components to find collision time, and calculating distance after velocity change. The multi-part structure and 6 marks indicate moderate length, but all steps follow routine procedures: position = initial + velocity×time, equate components, substitute. Part (c) adds one extra calculation step but remains straightforward. Slightly easier than average A-level due to being a textbook-style mechanics problem with clear methodology. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement3.02h Motion under gravity: vector form |
| Answer | Marks |
|---|---|
| Cargo ship travels \((9\hat{i} - 6\hat{j})\) km in \(t\) hours | M1 |
| Pos\(^s\) vector after \(t\) hours is \([(\hat{i} + 5\hat{j}) + (9\hat{i} - 6\hat{j})]\) km | A1 |
| \(= [(7 + 9t)\hat{i} + (5 - 6t)\hat{j}]\) km | A1 |
| Pos\(^s\) vector of ferry after \(t\) hours is \((12t + 18\hat{j})\) km |
| Answer | Marks |
|---|---|
| They will collide if coeffs. of \(\hat{i}\) and \(\hat{j}\) in pos\(^s\) vectors are equal | B1 |
| \(7 + 9t = 12t\) and \(56 - 6t = 18t\) are both satisfied when \(t = \frac{7}{3}\) | M1 A1 |
| Collision after \(\frac{7}{3}\) hrs or 2 hrs 20 mins i.e. at 8:20 a.m. | A1 |
| Pos\(^s\) vector = \(12(\frac{7}{3})\hat{i} + 18(\frac{7}{3})\hat{j} = (28\hat{i} + 42\hat{j})\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At 8 a.m. ferry at \((24\hat{i} + 36\hat{j})\) | M2 A1 | |
| \(\frac{1}{4}\) hr at \(21\hat{i} + 6\hat{j} = 7\hat{i} + 2\hat{j}\) so at 8:20 a.m. ferry is at \(3\hat{i} + 38\hat{j}\) | M2 A1 | |
| At 8:20 a.m. cargo ship is at \((28\hat{i} + 42\hat{j})\) | ||
| Dist. between = \(\sqrt{3^2 + 4^2} = 5\) km | M1 A1 | (14) |
**Part (a)**
Cargo ship travels $(9\hat{i} - 6\hat{j})$ km in $t$ hours | M1 |
Pos$^s$ vector after $t$ hours is $[(\hat{i} + 5\hat{j}) + (9\hat{i} - 6\hat{j})]$ km | A1 |
$= [(7 + 9t)\hat{i} + (5 - 6t)\hat{j}]$ km | A1 |
Pos$^s$ vector of ferry after $t$ hours is $(12t + 18\hat{j})$ km | |
**Part (b)**
They will collide if coeffs. of $\hat{i}$ and $\hat{j}$ in pos$^s$ vectors are equal | B1 |
$7 + 9t = 12t$ and $56 - 6t = 18t$ are both satisfied when $t = \frac{7}{3}$ | M1 A1 |
Collision after $\frac{7}{3}$ hrs or 2 hrs 20 mins i.e. at 8:20 a.m. | A1 |
Pos$^s$ vector = $12(\frac{7}{3})\hat{i} + 18(\frac{7}{3})\hat{j} = (28\hat{i} + 42\hat{j})$ | M1 A1 |
**Part (c)**
At 8 a.m. ferry at $(24\hat{i} + 36\hat{j})$ | M2 A1 |
$\frac{1}{4}$ hr at $21\hat{i} + 6\hat{j} = 7\hat{i} + 2\hat{j}$ so at 8:20 a.m. ferry is at $3\hat{i} + 38\hat{j}$ | M2 A1 |
At 8:20 a.m. cargo ship is at $(28\hat{i} + 42\hat{j})$ | |
Dist. between = $\sqrt{3^2 + 4^2} = 5$ km | M1 A1 | (14)
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**Total: 75 marks**7. At 6 a.m. a cargo ship has position vector $( 7 \mathbf { i } + 56 \mathbf { j } ) \mathrm { km }$ relative to a fixed origin $O$ on the coast and moves with constant velocity $( 9 \mathbf { i } - 6 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }$.
A ferry sails from $O$ at 6 a.m. and moves with constant velocity $( 12 \mathbf { i } + 18 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed due east and due north respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the position vector of the cargo ship $t$ hours after 6 a.m. is given by
$$[ ( 7 + 9 t ) \mathbf { i } + ( 56 - 6 t ) \mathbf { j } ] \mathrm { km }$$
and find the position vector of the ferry in terms of $t$.
\item Show that if both vessels maintain their course and speed, they will collide and find the time and position vector at which this occurs.\\
(6 marks)\\
At 8 a.m. the captain of the ferry realises that a collision is imminent and changes course so that the ferry now has velocity $( 21 \mathbf { i } + 6 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }$.
\item Find the distance between the two ships at the time when they would have collided.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [14]}}