| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Standard +0.8 This M4 mechanics question requires setting up position vectors, finding the relative velocity, minimizing distance using calculus or vector projection, and working backwards to find the bearing. It demands multi-step problem-solving with vectors and optimization, significantly above average difficulty but standard for M4 level. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Vector triangle drawn correctly | M1 A1 | |
| \(\cos\theta = \frac{12}{13}\) | M1 | |
| \((\theta = 22.6°)\) | ||
| Course is \(360° - 22.6°\) | ||
| \(= 337°\) (AWRT) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(v = \sqrt{13^2 - 12^2} = 5\) | ||
| \(t = \frac{6\cos\theta}{5} = 1.107\) | ||
| Time is 1.06 p.m. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(d = 6\sin\theta = 6 \times \frac{5}{13} = 2.31\) km | M1 A1 | AWRT 2.3 km, (2 marks) |
# Question 3:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| Vector triangle drawn correctly | M1 A1 | |
| $\cos\theta = \frac{12}{13}$ | M1 | |
| $(\theta = 22.6°)$ | | |
| Course is $360° - 22.6°$ | | |
| $= 337°$ (AWRT) | A1 | (4 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $v = \sqrt{13^2 - 12^2} = 5$ | | |
| $t = \frac{6\cos\theta}{5} = 1.107$ | | |
| Time is 1.06 p.m. | | |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| $d = 6\sin\theta = 6 \times \frac{5}{13} = 2.31$ km | M1 A1 | AWRT 2.3 km, (2 marks) |
**Total: 11 marks**
---3. At noon, two boats $A$ and $B$ are 6 km apart with $A$ due east of $B$. Boat $B$ is moving due north at a constant speed of $13 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Boat $A$ is moving with constant speed $12 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ and sets a course so as to pass as close as possible to boat $B$. Find
\begin{enumerate}[label=(\alph*)]
\item the direction of motion of $A$, giving your answer as a bearing,
\item the time when the boats are closest,
\item the shortest distance between the boats.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q3 [11]}}