| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question involving vectors and relative motion. Parts (a)-(d) are routine applications of magnitude formulas, unit vectors, and position vectors requiring straightforward algebraic manipulation. Part (e) involves solving a quadratic inequality, which is slightly more demanding but still follows a standard procedure. The question is well-scaffolded with 'show that' parts guiding students through the solution, making it slightly easier than average for M1. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | dist. between Alison and Bill \(= \sqrt{[(-5)^2 + 12^2]} = 13\) km | M1 A1 |
| (b) | \( | 3i + j |
| \(\therefore v = 2(3i + j) = 6i + 2j\) | M1 A1 | |
| (c) | after \(t\) hours, Alison is at \((2t)i + (5t)j\) | M1 A1 |
| Bill is at \((6t - 5)i + (2t + 12)j\) | A1 | |
| rel to A, pos° vector of B is \((6t - 5 - 2t)i + (2t + 12 - 5t)j\) | M1 | |
| \(= (4t - 5)i + (12 - 3t)j\) km | A1 | |
| (d) | \(d^2 = (4t - 5)^2 + (12 - 3t)^2\) | M1 |
| \(= 16t^2 - 40t + 25 + 144 - 72t + 9t^2 = 25t^2 - 112t + 169\) | A1 | |
| (e) | \(d^2 < 121 \therefore 25t^2 - 112t + 169 < 121\) | M1 |
| \(25t^2 - 112t + 48 < 0\), so \((25t - 12)(t - 4) < 0\) | A1 M1 | |
| suitable method to get \(\frac{12}{25} < t < 4\) | A1 | |
| length of time \(= 3\frac{12}{25}\) hours \(= 3\) hours 31 minutes (to nearest minute) | A1 | (17) |
| Answer | Marks |
|---|---|
| Total | (75) |
(a) | dist. between Alison and Bill $= \sqrt{[(-5)^2 + 12^2]} = 13$ km | M1 A1 |
(b) | $|3i + j| = \sqrt{10}$; speed $= 2\sqrt{10}$ | M1 |
| $\therefore v = 2(3i + j) = 6i + 2j$ | M1 A1 |
(c) | after $t$ hours, Alison is at $(2t)i + (5t)j$ | M1 A1 |
| Bill is at $(6t - 5)i + (2t + 12)j$ | A1 |
| rel to A, pos° vector of B is $(6t - 5 - 2t)i + (2t + 12 - 5t)j$ | M1 |
| $= (4t - 5)i + (12 - 3t)j$ km | A1 |
(d) | $d^2 = (4t - 5)^2 + (12 - 3t)^2$ | M1 |
| $= 16t^2 - 40t + 25 + 144 - 72t + 9t^2 = 25t^2 - 112t + 169$ | A1 |
(e) | $d^2 < 121 \therefore 25t^2 - 112t + 169 < 121$ | M1 |
| $25t^2 - 112t + 48 < 0$, so $(25t - 12)(t - 4) < 0$ | A1 M1 |
| suitable method to get $\frac{12}{25} < t < 4$ | A1 |
| length of time $= 3\frac{12}{25}$ hours $= 3$ hours 31 minutes (to nearest minute) | A1 | (17) |
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**Total** | **(75)** |Two ramblers, Alison and Bill, are out walking. At midday, Alison is at the fixed origin $O$, and Bill is at the point with position vector $(-5\mathbf{i} + 12\mathbf{j})$ km relative to $O$, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular, horizontal unit vectors.
They are both walking with constant velocity – Alison at $(2\mathbf{i} + 5\mathbf{j})$ km h$^{-1}$, and Bill at a speed of $2\sqrt{10}$ km h$^{-1}$ in a direction parallel to the vector $(3\mathbf{i} + \mathbf{j})$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance between the two ramblers at midday. [2 marks]
\item Show that the velocity of Bill is $(6\mathbf{i} + 2\mathbf{j})$ km h$^{-1}$. [3 marks]
\item Show that, at time $t$ hours after midday, the position vector of Bill relative to Alison is
$$[(4t - 5)\mathbf{i} + (12 - 3t)\mathbf{j}] \text{ km.}$$ [5 marks]
\item Show that the distance, $d$ km, between the two ramblers is given by
$$d^2 = 25t^2 - 112t + 169.$$ [2 marks]
\item Using your answer to part $(d)$, find the length of time to the nearest minute for which the distance between the Alison and Bill is less than 11 km. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [17]}}