| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.3 This is a standard M4 relative velocity question requiring conversion of bearings to vectors, calculation of relative velocity (magnitude and direction), and finding closest approach using perpendicular distance formula. While it involves multiple steps and vector manipulation, it follows a well-established procedure taught explicitly in mechanics modules with no novel problem-solving required. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct velocity triangle drawn | B1 | Correct velocity triangle |
| \(w^2 = 6.3^2 + 10^2 - 2 \times 6.3 \times 10\cos 50°\) | M1 | |
| \(w = 7.66 \text{ ms}^{-1}\) | A1 | |
| \(\frac{\sin\alpha}{6.3} = \frac{\sin 50°}{w}\) | M1 | This mark cannot be earned from work done in part (ii) |
| \(\alpha = 39.04°\) \((\beta = 90.96°)\) | ||
| Bearing is \(205 - \alpha = 166°\) | A1 | Total: 5 |
| OR: \(\begin{pmatrix}6.3\sin 75\\6.3\cos 75\end{pmatrix} - \begin{pmatrix}10\sin 25\\10\cos 25\end{pmatrix} = \begin{pmatrix}1.859\\-7.433\end{pmatrix}\) | M1A1 | |
| \(w = \sqrt{1.859^2 + 7.433^2} = 7.66\) | M1, A1 | Finding magnitude or direction |
| Bearing is \(180 - \tan^{-1}\frac{1.859}{7.433} = 166°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Diagram showing path of \(A\) as viewed from \(B\) | B1 ft | May be implied. Or B1 for correct (ft) expression for \(d^2\) in terms of \(t\) |
| \(d = 2500\sin 14.04\) | M1 | or other complete method |
| \(= 607 \text{ m}\) | A1 | Accept 604.8 to 609. Total: 3 |
| *SR* If \(\beta = 89°\) is used, give A1 for 684.9 to 689.1 |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct velocity triangle drawn | B1 | Correct velocity triangle |
| $w^2 = 6.3^2 + 10^2 - 2 \times 6.3 \times 10\cos 50°$ | M1 | |
| $w = 7.66 \text{ ms}^{-1}$ | A1 | |
| $\frac{\sin\alpha}{6.3} = \frac{\sin 50°}{w}$ | M1 | This mark cannot be earned from work done in part (ii) |
| $\alpha = 39.04°$ $(\beta = 90.96°)$ | | |
| Bearing is $205 - \alpha = 166°$ | A1 | **Total: 5** |
| OR: $\begin{pmatrix}6.3\sin 75\\6.3\cos 75\end{pmatrix} - \begin{pmatrix}10\sin 25\\10\cos 25\end{pmatrix} = \begin{pmatrix}1.859\\-7.433\end{pmatrix}$ | M1A1 | |
| $w = \sqrt{1.859^2 + 7.433^2} = 7.66$ | M1, A1 | Finding magnitude or direction |
| Bearing is $180 - \tan^{-1}\frac{1.859}{7.433} = 166°$ | A1 | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing path of $A$ as viewed from $B$ | B1 ft | May be implied. Or B1 for correct (ft) expression for $d^2$ in terms of $t$ |
| $d = 2500\sin 14.04$ | M1 | or other complete method |
| $= 607 \text{ m}$ | A1 | Accept 604.8 to 609. **Total: 3** |
| | | *SR* If $\beta = 89°$ is used, give A1 for 684.9 to 689.1 |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{a9e010ce-c3a8-4f95-a154-fd16ef3e5e5b-2_823_650_1318_751}
A boat $A$ is travelling with constant speed $6.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a course with bearing $075 ^ { \circ }$. Boat $B$ is travelling with constant speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a course with bearing $025 ^ { \circ }$. At one instant, $A$ is 2500 m due north of $B$ (see diagram).\\
(i) Find the magnitude and bearing of the velocity of $A$ relative to $B$.\\
(ii) Find the shortest distance between $A$ and $B$ in the subsequent motion.
\hfill \mbox{\textit{OCR M4 2008 Q4 [8]}}