| Exam Board | OCR |
|---|---|
| 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 | Challenging +1.3 This is an M4 relative velocity problem requiring vector resolution, closest approach calculations, and bearing conversions. While it involves multiple steps and careful geometric reasoning, it follows a standard mechanics template with guided parts. The 'show that' in part (i) provides scaffolding, and the techniques (relative velocity, perpendicular distance) are well-practiced at this level. Harder than average due to the multi-step nature and M4 content, but not requiring novel insight. |
| 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 |
| [Velocity triangle diagram] | M1 | Velocity triangle with 90° opposite \(v_C\) |
| A1 | Correct velocity triangle | |
| \(\cos\alpha = \frac{12}{15}\), \(\alpha = 36.87°\) (4 sf) | M1 | Finding a relevant angle |
| Bearing of \(v_B\) is \(110 - 36.87 = 073.13 = 073°\) (nearest degree) | A1 ag [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Magnitude is \(\sqrt{15^2 - 12^2} = 9 \text{ ms}^{-1}\) | B1 | Accept 8.95 to 9.05 |
| Direction is 90° from \(v_B\) | M1 | |
| Bearing is \(73.13 + 90 = 163°\) (nearest degree) | A1 [3] | |
| *Alternative:* \(v^2 = 12^2 + 15^2 - 2\times12\times15\cos 37°\), \(v=9\) | B1 | Accept 8.95 to 9.05 |
| \(\frac{\sin\beta}{12} = \frac{\sin 37°}{v}\), \(\beta = 53°\) | M1 | Finding a relevant angle |
| Bearing is \(110 + 53 = 163°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [Diagram showing displacement and relative velocity] | M1 | Diagram indicating initial displacement and relative velocity (may be implied) |
| \(d = 3500\sin 56.87°\) | M1 | |
| Shortest distance is 2930 m (3 sf) | A1 [3] | Accept 2910 to 2950 |
| *Alternative:* \(d^2 = (3500\sin 40° + 2.6...t)^2 + (3500\cos 40° - 8.6...t)^2\) | M1 | |
| Minimum when \(-34432 + 162t = 0\), \(t = 213\) | M1 | Differentiating or completing the square |
| Shortest distance is 2930 m (3 sf) | A1 | Accept 2910 to 2950 |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Velocity triangle diagram] | M1 | Velocity triangle with 90° opposite $v_C$ |
| | A1 | Correct velocity triangle |
| $\cos\alpha = \frac{12}{15}$, $\alpha = 36.87°$ (4 sf) | M1 | Finding a relevant angle |
| Bearing of $v_B$ is $110 - 36.87 = 073.13 = 073°$ (nearest degree) | A1 ag [4] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Magnitude is $\sqrt{15^2 - 12^2} = 9 \text{ ms}^{-1}$ | B1 | Accept 8.95 to 9.05 |
| Direction is 90° from $v_B$ | M1 | |
| Bearing is $73.13 + 90 = 163°$ (nearest degree) | A1 [3] | |
| *Alternative:* $v^2 = 12^2 + 15^2 - 2\times12\times15\cos 37°$, $v=9$ | B1 | Accept 8.95 to 9.05 |
| $\frac{\sin\beta}{12} = \frac{\sin 37°}{v}$, $\beta = 53°$ | M1 | Finding a relevant angle |
| Bearing is $110 + 53 = 163°$ | A1 | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Diagram showing displacement and relative velocity] | M1 | Diagram indicating initial displacement and relative velocity (may be implied) |
| $d = 3500\sin 56.87°$ | M1 | |
| Shortest distance is 2930 m (3 sf) | A1 [3] | Accept 2910 to 2950 |
| *Alternative:* $d^2 = (3500\sin 40° + 2.6...t)^2 + (3500\cos 40° - 8.6...t)^2$ | M1 | |
| Minimum when $-34432 + 162t = 0$, $t = 213$ | M1 | Differentiating or completing the square |
| Shortest distance is 2930 m (3 sf) | A1 | Accept 2910 to 2950 |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{ea62d6d9-ac2f-44e7-8bfb-ae9aeea7109b-2_688_777_1382_683}
From a boat $B$, a cruiser $C$ is observed 3500 m away on a bearing of $040 ^ { \circ }$. The cruiser $C$ is travelling with constant speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a straight line course with bearing $110 ^ { \circ }$ (see diagram). The boat $B$ travels with constant speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a straight line course which takes it as close as possible to the cruiser $C$.\\
(i) Show that the bearing of the course of $B$ is $073 ^ { \circ }$, correct to the nearest degree.\\
(ii) Find the magnitude and the bearing of the velocity of $C$ relative to $B$.\\
(iii) Find the shortest distance between $B$ and $C$ in the subsequent motion.
\hfill \mbox{\textit{OCR M4 2010 Q4 [10]}}