| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Standard +0.8 This is a relative velocity/closest approach problem from M4 requiring vector decomposition, minimization using calculus or geometry, and interpretation of the physical scenario. While systematic, it demands more sophisticated problem-solving than routine mechanics questions and involves multiple connected parts with non-trivial geometric insight. |
| 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 | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Diagram showing triangle with sides 8, 6, 10 and angle \(\theta\) | M1 | Correct velocity triangle |
| \(\cos\theta = \dfrac{6}{10} \Rightarrow \theta = 53.1°\) | M1 A1 | |
| Bearing is \(307°\) | A1 | Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(d = 8\sin\theta\ (= 8 \times 0.8)\) | M1 A1 | |
| \(= 6.4\) km | A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = \dfrac{8\cos\theta}{\sqrt{10^2 - 6^2}}\) | M1 A1 | |
| \(= 0.6\) hrs | A1 | |
| Time is 12:36 pm | A1 | Total: 3 marks |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing triangle with sides 8, 6, 10 and angle $\theta$ | M1 | Correct velocity triangle |
| $\cos\theta = \dfrac{6}{10} \Rightarrow \theta = 53.1°$ | M1 A1 | |
| Bearing is $307°$ | A1 | Total: 4 marks |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $d = 8\sin\theta\ (= 8 \times 0.8)$ | M1 A1 | |
| $= 6.4$ km | A1 | Total: 3 marks |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = \dfrac{8\cos\theta}{\sqrt{10^2 - 6^2}}$ | M1 A1 | |
| $= 0.6$ hrs | A1 | |
| Time is 12:36 pm | A1 | Total: 3 marks |
---\begin{enumerate}
\item At 12 noon, $\operatorname { ship } A$ is 8 km due west of $\operatorname { ship } B$. Ship $A$ is moving due north at a constant speed of $10 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Ship $B$ is moving at a constant speed of $6 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing so that it passes as close to $A$ as possible.\\
(a) Find the bearing on which ship $B$ moves.\\
(b) Find the shortest distance between the two ships.\\
(c) Find the time when the two ships are closest.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2010 Q3 [10]}}