| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 13 |
| 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 vectors question requiring position vectors at time t, collision verification, and relative position calculations. All parts use routine techniques (r = r₀ + vt, vector subtraction, magnitude calculation) with straightforward algebra. The multi-part structure and 13 marks indicate moderate length, but no step requires novel insight—it's textbook application of kinematics vectors, making it slightly easier than average A-level. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) At time \(t\): \(\mathbf{r}_A = (-5 + 2t)\mathbf{i} + (10 + 2t)\mathbf{j}\), \(\mathbf{r}_B = (3 - 2t)\mathbf{i} + (4 + 5t)\mathbf{j}\) i components equal when \(-5 + 2t = 3 - 2t \Rightarrow t = 2\) h \(t = 2\): \(\mathbf{r}_A = -\mathbf{i} + 14\mathbf{j}\), \(\mathbf{r}_B = -\mathbf{i} + 14\mathbf{j} \Rightarrow\) collide | B1, B1, M1 A1, M1 A1 | 6 marks |
| (b) New \(\mathbf{r}_A = (-5 + t)\mathbf{i} + (10 + t)\mathbf{j}\) \(\Rightarrow \overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = (8 - 3t)\mathbf{i} + (-6 + 4t)\mathbf{j}\) | M1 A1 | 2 marks |
| (c) \(t = 2\): \(\overrightarrow{AB} = 2\mathbf{i} + 2\mathbf{j}\), \(\Rightarrow \text{dist.} = \sqrt{2^2 + 2^2} \approx 2.83 \text{ km}\) | M1 M1 A1 | 3 marks |
| (d) B north of A \(\Rightarrow 8 - 3t = 0 \Rightarrow t = 8/3 \Rightarrow\) time 1440 hours | M1 A1 | 2 marks |
**(a)** At time $t$: $\mathbf{r}_A = (-5 + 2t)\mathbf{i} + (10 + 2t)\mathbf{j}$, $\mathbf{r}_B = (3 - 2t)\mathbf{i} + (4 + 5t)\mathbf{j}$ i components equal when $-5 + 2t = 3 - 2t \Rightarrow t = 2$ h $t = 2$: $\mathbf{r}_A = -\mathbf{i} + 14\mathbf{j}$, $\mathbf{r}_B = -\mathbf{i} + 14\mathbf{j} \Rightarrow$ collide | B1, B1, M1 A1, M1 A1 | 6 marks
**(b)** New $\mathbf{r}_A = (-5 + t)\mathbf{i} + (10 + t)\mathbf{j}$ $\Rightarrow \overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = (8 - 3t)\mathbf{i} + (-6 + 4t)\mathbf{j}$ | M1 A1 | 2 marks
**(c)** $t = 2$: $\overrightarrow{AB} = 2\mathbf{i} + 2\mathbf{j}$, $\Rightarrow \text{dist.} = \sqrt{2^2 + 2^2} \approx 2.83 \text{ km}$ | M1 M1 A1 | 3 marks
**(d)** B north of A $\Rightarrow 8 - 3t = 0 \Rightarrow t = 8/3 \Rightarrow$ time 1440 hours | M1 A1 | 2 marks
---[In this question, the horizontal unit vectors $\mathbf{i}$ and $\mathbf{j}$ are directed due East and North respectively.]
A coastguard station $O$ monitors the movements of ships in a channel. At noon, the station's radar records two ships moving with constant speed. Ship $A$ is at the point with position vector $(-5\mathbf{i} + 10\mathbf{j})$ km relative to $O$ and has velocity $(2\mathbf{i} + 2\mathbf{j})$ km h$^{-1}$. Ship $B$ is at the point with position vector $(3\mathbf{i} + 4\mathbf{j})$ km and has velocity $(-2\mathbf{i} + 5\mathbf{j})$ km h$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Given that the two ships maintain these velocities, show that they collide. [6]
\end{enumerate}
The coast guard radios ship $A$ and orders it to reduce its speed to move with velocity $(\mathbf{i} + \mathbf{j})$ km h$^{-1}$.
Given that $A$ obeys this order and maintains this new constant velocity,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find an expression for the vector $\overrightarrow{AB}$ at time $t$ hours after noon. [2]
\item find, to 3 significant figures, the distance between $A$ and $B$ at 1400 hours, [3]
\item Find the time at which $B$ will be due north of $A$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [13]}}