| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Challenging +1.2 This is a standard M4 pursuit/closest approach problem requiring vector methods to find when dP/dt · PQ = 0. While it involves multiple steps (setting up position vectors, finding velocity of approach, minimizing distance), the technique is well-established and practiced extensively in M4. The calculations are straightforward once the method is recognized, making it moderately above average difficulty but routine for this module. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02h Motion under gravity: vector form |
| Answer | Marks |
|---|---|
| M1 M1 A1 |
| Answer | Marks |
|---|---|
| M1A1ft A1 |
| Answer | Marks |
|---|---|
| B1 M1 A1 DM1 A1 |
**Part (a)**
$\cos \alpha = \frac{12}{20}$
Bearing is $180° + \alpha = 233°$ (nearest degree)
| M1 M1 A1 |
**Part (b)**
$PN = 2000\cos(135° - \alpha) = 200\sqrt{2}$ m or decimal equivalent
| M1A1ft A1 |
**Part (c)**
Time to closest approach = $\frac{\sqrt{20^2 - 12^2}}{QN} = \frac{\sqrt{20^2 - 12^2}}{2000\sin(135° - \alpha)} = \frac{2000\sin(135° - \alpha)}{16}$
Distance moved by Q = their $t \times 12 = 1050\sqrt{2}$ m or decimal equivalent
| B1 M1 A1 DM1 A1 |
**Total: [12]**
---\begin{enumerate}
\item At noon a motorboat $P$ is 2 km north-west of another motorboat $Q$. The motorboat $P$ is moving due south at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The motorboat $Q$ is pursuing motorboat $P$ at a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and sets a course in order to get as close to motorboat $P$ as possible.\\
(a) Find the course set by $Q$, giving your answer as a bearing to the nearest degree.\\
(b) Find the shortest distance between $P$ and $Q$.\\
(c) Find the distance travelled by $Q$ from its position at noon to the point of closest approach.\\
\end{enumerate}
\section*{June 2009}
\hfill \mbox{\textit{Edexcel M4 2009 Q3 [12]}}