| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| 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 +1.2 This is a standard M4 interception problem requiring vector resolution, relative velocity concepts, and optimization. Part (i) is routine interception using sine rule; part (ii) requires minimizing distance using perpendicular relative velocity; part (iii) needs recognizing the critical case where relative velocity is perpendicular to initial position. While multi-part with several techniques, these are well-practiced M4 methods without 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 | Marks | Guidance |
| [Velocity triangle diagram with \(60°\), \(15\), \(18\)] | B1 | Velocity triangle |
| \(\frac{\sin\theta}{15} = \frac{\sin 60°}{18}\) | M1 | Implies previous B1 |
| Bearing is \(046.2°\) (3 sf) | A1 [3] | Accept \(46°\) or \(046°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Velocity triangle with relative velocity perpendicular to \(\mathbf{v}_B\)] | M1 | Relative velocity perpendicular to \(\mathbf{v}_B\) |
| [Correct velocity triangle] | A1 | Correct velocity triangle |
| \(\sin\alpha = \frac{9}{15}\) | M1 | |
| \(\alpha = 36.9°\) | A1 | Or other angle is \(53.1°\) |
| Bearing is \(30° + \alpha = 066.9°\) (3 sf) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Minimum \(V = 15\sin 60°\) | M1 | |
| \(= 13.0\) (3 sf) | A1 [2] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Velocity triangle diagram with $60°$, $15$, $18$] | B1 | Velocity triangle |
| $\frac{\sin\theta}{15} = \frac{\sin 60°}{18}$ | M1 | Implies previous B1 |
| Bearing is $046.2°$ (3 sf) | A1 [3] | Accept $46°$ or $046°$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Velocity triangle with relative velocity perpendicular to $\mathbf{v}_B$] | M1 | Relative velocity perpendicular to $\mathbf{v}_B$ |
| [Correct velocity triangle] | A1 | Correct velocity triangle |
| $\sin\alpha = \frac{9}{15}$ | M1 | |
| $\alpha = 36.9°$ | A1 | Or other angle is $53.1°$ |
| Bearing is $30° + \alpha = 066.9°$ (3 sf) | A1 [5] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Minimum $V = 15\sin 60°$ | M1 | |
| $= 13.0$ (3 sf) | A1 [2] | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-2_460_388_1160_826}
A ship $S$ is travelling with constant velocity $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a course with bearing $120 ^ { \circ }$. A patrol boat $B$ observes the ship when $S$ is due north of $B$. The patrol boat $B$ then moves with constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line (see diagram).\\
(i) Given that $V = 18$, find the bearing of the course of $B$ such that $B$ intercepts $S$.\\
(ii) Given instead that $V = 9$, find the bearing of the course of $B$ such that $B$ passes as close as possible to $S$.\\
(iii) Find the smallest value of $V$ for which it is possible for $B$ to intercept $S$.
\hfill \mbox{\textit{OCR M4 2012 Q3 [10]}}