| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: find bearing/direction to intercept (exact intercept) |
| Difficulty | Challenging +1.2 This is a standard relative velocity interception problem requiring vector resolution and solving a triangle using the cosine rule. While it involves multiple steps (setting up velocity vectors, applying cosine/sine rules, converting to bearings), these are well-practiced M4 techniques with no novel insight required. The 11 marks reflect computational work rather than conceptual difficulty, placing it moderately above average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Using velocity diagram: \(\frac{\sin \theta}{\sin 45°} = \frac{1500}{2000}\) | M1A1, M1A1, M1A1 | (6 marks) |
| Answer | Marks |
|---|---|
| - (i) \(v^2 = 1500^2 + 2000^2 - 2(1500)(2000)\cos(90 + 13°)^2\) | M1A1 |
| - (ii) \(v \cos 45° = 2000 \cos 13°\) | A1 |
| - (iii) \(\frac{\sin 45°}{2000} = \frac{\sin 103°}{v}\) | M1A1 (5 marks) |
| Answer | Marks |
|---|---|
| Time = \(\frac{100}{v}\) h = 131 s | M1A1 |
| Answer | Marks |
|---|---|
| (i) In the case below \(\alpha\) is bearing; but other relevant angle may be used. One equation in \(t\) and \(\alpha\): e.g. \(2000t \sin \alpha = 50\sqrt{2} - 1500t\) | M1A1, M1A1, M1A1, M1A1√ |
| Second equation in \(t\) and \(\alpha\): e.g. \(2000t \cos \alpha = 50\sqrt{2}\) | A1 |
| Equation in one variable: e.g. \(4 \cos \alpha - 4 \sin \alpha = 3\) | M1A1 |
| Reducing to simple equation e.g. \(4\sqrt{2} \cos(\alpha + 45°) = 3\) | M1A1 |
| Bearing = (0)13° | A1 |
| Substituting for \(\alpha\) to find \(t\): \(t = 131\) s | M1A1 |
| (ii) Using cosine rule: \((2000v)^2 = (1500t)^2 + 100^2 - 2(100)(1500t)\cos 45°\) | M2A1A1 |
| Quadratic form: \(175t^2 + 15\sqrt{2}t - 1 = 0\) | M1A1 |
| Solving: \(t = 131\)s | M1A1 |
| Equation in \(t\) and \(\alpha\) | M1A1 |
| Bearing = (0)13° | A1 |
## Part (a)
Using velocity diagram: $\frac{\sin \theta}{\sin 45°} = \frac{1500}{2000}$ | M1A1, M1A1, M1A1 | (6 marks)
$\theta = 32°$ (32.03°)
Bearing = $90° - (45° + \theta) = 013°$
## Part (b)
Method for $v$:
- (i) $v^2 = 1500^2 + 2000^2 - 2(1500)(2000)\cos(90 + 13°)^2$ | M1A1
- (ii) $v \cos 45° = 2000 \cos 13°$ | A1
- (iii) $\frac{\sin 45°}{2000} = \frac{\sin 103°}{v}$ | M1A1 (5 marks)
$v = 2756$ km h$^{-1}$
Time = $\frac{100}{v}$ h = 131 s | M1A1
[Time = $\frac{100 \cos 45°}{2000 \cos 13°}$ gains M1M1A1 immediately, correct answer gains A2]
**Using displacement method (several variations)**
(i) In the case below $\alpha$ is bearing; but other relevant angle may be used. One equation in $t$ and $\alpha$: e.g. $2000t \sin \alpha = 50\sqrt{2} - 1500t$ | M1A1, M1A1, M1A1, M1A1√
Second equation in $t$ and $\alpha$: e.g. $2000t \cos \alpha = 50\sqrt{2}$ | A1
Equation in one variable: e.g. $4 \cos \alpha - 4 \sin \alpha = 3$ | M1A1
Reducing to simple equation e.g. $4\sqrt{2} \cos(\alpha + 45°) = 3$ | M1A1
Bearing = (0)13° | A1
Substituting for $\alpha$ to find $t$: $t = 131$ s | M1A1
(ii) Using cosine rule: $(2000v)^2 = (1500t)^2 + 100^2 - 2(100)(1500t)\cos 45°$ | M2A1A1
Quadratic form: $175t^2 + 15\sqrt{2}t - 1 = 0$ | M1A1
Solving: $t = 131$s | M1A1
Equation in $t$ and $\alpha$ | M1A1
Bearing = (0)13° | A1
---A pilot flying an aircraft at a constant speed of 2000 kmh$^{-1}$ detects an enemy aircraft 100 km away on a bearing of 045°. The enemy aircraft is flying at a constant velocity of 1500 kmh$^{-1}$ due west. Find
\begin{enumerate}[label=(\roman*)]
\item the course, as a bearing to the nearest degree, that the pilot should set up in order to intercept the enemy aircraft,
\item the time, to the nearest s, that the pilot will take to reach the enemy aircraft. [11]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q4 [11]}}