| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.8 This M4 relative velocity problem requires vector decomposition, bearing conversions, and solving simultaneous equations from velocity triangles. Part (i) involves standard relative motion calculation, part (ii) requires finding closest approach using perpendicular distance, while parts (iii-iv) demand working backwards from given information to find unknown velocities—more sophisticated than typical mechanics questions but within expected M4 scope. |
| 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 |
| Diagram showing relative velocity | M1 | Suitable diagram showing relative velocity. *May be implied* |
| \(x^2 = 80^2 + 36^2 - 2\times80\times36\cos 40°\) | M1 | |
| \(x = 57.30\) | ||
| Relative velocity has magnitude \(\frac{x}{3} = 19.1\) km h\(^{-1}\) | A1 ag | |
| \(\frac{\sin\alpha}{36} = \frac{\sin 40°}{57.30}\), \(\alpha = 23.82°\) | M1 | Or other valid method for finding a relevant angle |
| Relative velocity has bearing \(40 + \alpha = 063.8°\) | A1 ag | |
| [5] | ||
| *OR, using components:* | ||
| Diagram | M1 | Implied by both components correct |
| East \(\frac{80\sin 40°}{3}\) \((= 17.14)\) | M1 | |
| North \(\frac{80\cos 40° - 36}{3}\) \((= 8.428)\) | M1 | |
| Speed \(\sqrt{17.14^2 + 8.428^2} = 19.1\) | A1 ag | |
| Bearing \(\tan^{-1}\frac{17.14}{8.428} = 063.8°\) | A1 ag |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Shortest distance \(d = 80\sin 23.82°\) | M1 | or \(36\sin 63.8°\) |
| \(= 32.3\) km (3 sf) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Velocity diagram with 28 opposite a known angle | M1 | *May be implied* (28 opposite a known angle between sides with positive and negative slopes) |
| \(\frac{\sin\beta}{19.10} = \frac{\sin 41.18°}{28}\), \(\beta = 26.69°\) | M1 | |
| Bearing of \(P\) is \(105 + \beta = 131.7°\) (1 dp) | A1 | *Using components for (iii) and (iv): M2A1 for \(\theta = 131.7°\) or \(v = 39.4\); M1A1 for other quantity* |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{v_Q}{\sin 112.13°} = \frac{28}{\sin 41.18°}\) | M1 | Or other valid method for finding speed |
| Speed of \(Q\) is 39.4 km h\(^{-1}\) (3 sf) | A1 | |
| [2] |
# Question 6:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing relative velocity | M1 | Suitable diagram showing relative velocity. *May be implied* |
| $x^2 = 80^2 + 36^2 - 2\times80\times36\cos 40°$ | M1 | |
| $x = 57.30$ | | |
| Relative velocity has magnitude $\frac{x}{3} = 19.1$ km h$^{-1}$ | A1 ag | |
| $\frac{\sin\alpha}{36} = \frac{\sin 40°}{57.30}$, $\alpha = 23.82°$ | M1 | Or other valid method for finding a relevant angle |
| Relative velocity has bearing $40 + \alpha = 063.8°$ | A1 ag | |
| | [5] | |
| *OR, using components:* | | |
| Diagram | M1 | Implied by both components correct |
| East $\frac{80\sin 40°}{3}$ $(= 17.14)$ | M1 | |
| North $\frac{80\cos 40° - 36}{3}$ $(= 8.428)$ | M1 | |
| Speed $\sqrt{17.14^2 + 8.428^2} = 19.1$ | A1 ag | |
| Bearing $\tan^{-1}\frac{17.14}{8.428} = 063.8°$ | A1 ag | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Shortest distance $d = 80\sin 23.82°$ | M1 | or $36\sin 63.8°$ |
| $= 32.3$ km (3 sf) | A1 | |
| | [2] | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Velocity diagram with 28 opposite a known angle | M1 | *May be implied* (28 opposite a known angle between sides with positive and negative slopes) |
| $\frac{\sin\beta}{19.10} = \frac{\sin 41.18°}{28}$, $\beta = 26.69°$ | M1 | |
| Bearing of $P$ is $105 + \beta = 131.7°$ (1 dp) | A1 | *Using components for (iii) and (iv): M2A1 for $\theta = 131.7°$ or $v = 39.4$; M1A1 for other quantity* |
| | [3] | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{v_Q}{\sin 112.13°} = \frac{28}{\sin 41.18°}$ | M1 | Or other valid method for finding speed |
| Speed of $Q$ is 39.4 km h$^{-1}$ (3 sf) | A1 | |
| | [2] | |6 Two ships $P$ and $Q$ are moving on straight courses with constant speeds. At one instant $Q$ is 80 km from $P$ on a bearing of $220 ^ { \circ }$. Three hours later, $Q$ is 36 km due south of $P$.\\
(i) Show that the velocity of $Q$ relative to $P$ is $19.1 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ in the direction with bearing $063.8 ^ { \circ }$ (both correct to 3 significant figures).\\
(ii) Find the shortest distance between the two ships in the subsequent motion.
Given that the speed of $P$ is $28 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ and $Q$ is travelling in the direction with bearing $105 ^ { \circ }$, find\\
(iii) the bearing of the direction in which $P$ is travelling,\\
(iv) the speed of $Q$.
\hfill \mbox{\textit{OCR M4 2011 Q6 [12]}}