| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | When is one object due north/east/west/south of another |
| Difficulty | Moderate -0.8 This is a straightforward M1 vectors question requiring only standard kinematics formulas (position = initial position + velocity × time) and basic vector operations. Part (a) is direct substitution, part (b) requires finding a displacement vector and its magnitude, and part (c) involves setting the i-component of relative position to zero. All techniques are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 10j\) | B1 | |
| \(q = (6i + 12j) + (-8i + 6j)t\) | M1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 3\): \(p = 30j\), \(q = -18i + 30j\) | M1 A1 | |
| \(\Rightarrow\) dist. apart = 18 km | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{PQ} = q - p = (6 - 8t)i + (12 - 4t)j\) | M1 | |
| \(t = 3\): \(\overrightarrow{PQ} = -18i + 0j\) | A1 | |
| \( | \overrightarrow{PQ} | ^2 = (6 - 8t)^2 + (12 - 4t)^2\) or \(t = 3 \Rightarrow |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q\) north of \(P \Rightarrow 6 - 8t = 0\) | M1 | |
| \(t = \frac{3}{4}\) | A1 | (2 marks) |
## Part (a)
$p = 10j$ | B1 |
$q = (6i + 12j) + (-8i + 6j)t$ | M1 A1 | (3 marks)
## Part (b)
$t = 3$: $p = 30j$, $q = -18i + 30j$ | M1 A1 |
$\Rightarrow$ dist. apart = 18 km | A1 | (3 marks)
### Alternative (b)
$\overrightarrow{PQ} = q - p = (6 - 8t)i + (12 - 4t)j$ | M1 |
$t = 3$: $\overrightarrow{PQ} = -18i + 0j$ | A1 |
$|\overrightarrow{PQ}|^2 = (6 - 8t)^2 + (12 - 4t)^2$ or $t = 3 \Rightarrow |\overrightarrow{PQ}| = 18$ | A1 |
## Part (c)
$Q$ north of $P \Rightarrow 6 - 8t = 0$ | M1 |
$t = \frac{3}{4}$ | A1 | (2 marks)
**Total: 8 marks**
---Two ships $P$ and $Q$ are moving along straight lines with constant velocities. Initially $P$ is at a point $O$ and the position vector of $Q$ relative to $O$ is $(6\mathbf{i} + 12\mathbf{j})$ km, where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors directed due east and due north respectively. The ship $P$ is moving with velocity $10\mathbf{j}$ km h$^{-1}$ and $Q$ is moving with velocity $(-8\mathbf{i} + 6\mathbf{j})$ km h$^{-1}$. At time $t$ hours the position vectors of $P$ and $Q$ relative to $O$ are $\mathbf{p}$ km and $\mathbf{q}$ km respectively.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf{p}$ and $\mathbf{q}$ in terms of $t$. [3]
\item Calculate the distance of $Q$ from $P$ when $t = 3$. [3]
\item Calculate the value of $t$ when $Q$ is due north of $P$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q4 [8]}}