| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Minimum speed to intercept |
| Difficulty | Challenging +1.2 This is a relative velocity interception problem requiring vector methods and optimization. Part (a) involves setting up velocity triangles and applying the constraint that ship A's speed cannot exceed 30 km/h, leading to a geometric inequality. Part (b) requires minimizing time subject to the speed constraint. While this is a multi-step M4 problem requiring careful vector analysis and some geometric insight, it follows standard relative velocity techniques taught in the module. The 12 marks and two-part structure indicate moderate difficulty, but the methods are well-practiced in M4 courses, placing it slightly above average difficulty. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.02e Two-dimensional constant acceleration: with vectors3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| Diagram showing: 70 km, points P, Q, H, angles 30°, 30°, 36°, and note "Minimum speed for interception = \(2\sqrt{c\sqrt{s}-3s} = 18°\) \(\cos\theta = \frac{18}{30}\) (= \(\frac{3}{5}\)) \(\Rightarrow\) tan\(\theta = \frac{4}{3}\) | M1 A1, M1 A1, M1, A1 (7), (guidance: "Explanation") |
| Answer | Marks |
|---|---|
| Time \(= \frac{70}{(18\sqrt{3} + 24)} = 1.27\) km | M1 A2, M1 A1 (5), (12) |
## Part (a)
Diagram showing: 70 km, points P, Q, H, angles 30°, 30°, 36°, and note "Minimum speed for interception = $2\sqrt{c\sqrt{s}-3s} = 18°$ $\cos\theta = \frac{18}{30}$ (= $\frac{3}{5}$) $\Rightarrow$ tan$\theta = \frac{4}{3}$ | M1 A1, M1 A1, M1, A1 (7), (guidance: "Explanation") |
## Part (b)
$A\theta = 36\cos 30° + 30 \sin\theta$ $(18\sqrt{3} + 24)$
Time $= \frac{70}{(18\sqrt{3} + 24)} = 1.27$ km | M1 A2, M1 A1 (5), (12) |
---A ship $A$ has maximum speed 30 km h$^{-1}$. At time $t = 0$, $A$ is 70 km due west of $B$ which is moving at a constant speed of 36 km h$^{-1}$ on a bearing of 300°. Ship $A$ moves on a straight course at a constant speed and intercepts $B$. The course of $A$ makes an angle $\theta$ with due north.
\begin{enumerate}[label=(\alph*)]
\item Show that $-\arctan \frac{4}{3} \leq \theta \leq \arctan \frac{4}{3}$. [7]
\item Find the least time for $A$ to intercept $B$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q6 [12]}}