| 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 relative motion and interception. Part (a) requires finding a unit vector and scaling by speed (routine calculation), part (b) is similar, and part (c) involves equating times using distance/speed. While multi-step, it follows a predictable template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{AB} = -50i + 120j\), which has magnitude 130 | M1 A1 | |
| \(v_C = \frac{2.6}{130}(-50i + 120j) = (-i + 2.4j)\) ms\(^{-1}\) | M1 A1 | |
| (b) \(\overrightarrow{OB} = -20i + 60j\), of magnitude \(\sqrt{4000} = 20\sqrt{10}\) | M1 A1 | |
| \(v_D = \frac{k\sqrt{10}}{20\sqrt{10}}(-20i + 60j) = k(-i + 3j)\) ms\(^{-1}\) | M1 A1 | |
| (c) Position vectors at time \(t\) are \((30 - t)i + (2.4t - 60)j\) and \(kt(-i + 3j)\) | B1 B1 | |
| When these are equal, \(30 - t = -kt\) and \(2.4t - 60 = 3kt\) | M1 | |
| \(-60 + 2.4t = -3(30 - t)\) | ||
| \(t = 50\) | A1 M1 A1 | |
| \(k = 0.4\) | Total: 14 marks |
**(a)** $\overrightarrow{AB} = -50i + 120j$, which has magnitude 130 | M1 A1 |
$v_C = \frac{2.6}{130}(-50i + 120j) = (-i + 2.4j)$ ms$^{-1}$ | M1 A1 |
**(b)** $\overrightarrow{OB} = -20i + 60j$, of magnitude $\sqrt{4000} = 20\sqrt{10}$ | M1 A1 |
$v_D = \frac{k\sqrt{10}}{20\sqrt{10}}(-20i + 60j) = k(-i + 3j)$ ms$^{-1}$ | M1 A1 |
**(c)** Position vectors at time $t$ are $(30 - t)i + (2.4t - 60)j$ and $kt(-i + 3j)$ | B1 B1 |
When these are equal, $30 - t = -kt$ and $2.4t - 60 = 3kt$ | M1 |
$-60 + 2.4t = -3(30 - t)$ | |
$t = 50$ | A1 M1 A1 |
$k = 0.4$ | | **Total: 14 marks**6. The points $A$ and $B$ have position vectors $( 30 \mathbf { i } - 60 \mathbf { j } ) \mathrm { m }$ and $( - 20 \mathbf { i } + 60 \mathbf { j } ) \mathrm { m }$ respectively relative to an origin $O$, where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors. A cyclist, Chris, starts at $A$ and cycles towards $B$ with constant speed $2.6 \mathrm {~ms} ^ { - 1 }$. Another cyclist, Doug, starts at $O$ and cycles towards $B$ with constant speed $k \sqrt { } 10 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that Chris's velocity vector is $( - \mathbf { i } + 2 \cdot 4 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$.
\item Find Doug's velocity vector in the form $k ( a \mathbf { i } + b \mathbf { j } ) \mathrm { ms } ^ { - 1 }$.
Given that Chris and Doug arrive at $B$ at the same time,
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [14]}}