| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 9 |
| 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 problem requiring vector subtraction, finding closest approach using perpendicular distance or calculus, and working with bearings. While it involves multiple steps and careful coordinate work, it follows a well-established method taught explicitly in mechanics modules, making it slightly easier than average overall. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Relative velocity magnitude | M1 | \( |
| Relative velocity | A1 | \( |
| Bearing angle calculation | M1 | \(\frac{\sin\theta}{20} = \frac{\sin 115°}{29 \cdot 640...}\); \(\theta = 37 \cdot 7°\) |
| Bearing | A1 | The relative velocity is on bearing \(167°\) |
| Closest distance | M1 | Closest distance \(= 70\sin 12 \cdot 7° = 15 \cdot 4 \text{ km}\) |
| Time to closest approach | M1 | \(t = \frac{70\cos 12 \cdot 7°}{29 \cdot 64} = 2 \cdot 30 \text{ hrs}\) |
| Final answer | A1 | Ships closest together at \(2:18\) am |
**Relative velocity magnitude** | M1 | $|_A\mathbf{v}_B| = \sqrt{20^2 + 15^2 - 2 \times 20 \times 15 \times \cos 115°} = \sqrt{878 \cdot 640...}$
**Relative velocity** | A1 | $|_A\mathbf{v}_B| = 29 \cdot 6 \text{ km h}^{-1}$ (3 s.f.)
**Bearing angle calculation** | M1 | $\frac{\sin\theta}{20} = \frac{\sin 115°}{29 \cdot 640...}$; $\theta = 37 \cdot 7°$
**Bearing** | A1 | The relative velocity is on bearing $167°$
**Closest distance** | M1 | Closest distance $= 70\sin 12 \cdot 7° = 15 \cdot 4 \text{ km}$
**Time to closest approach** | M1 | $t = \frac{70\cos 12 \cdot 7°}{29 \cdot 64} = 2 \cdot 30 \text{ hrs}$
**Final answer** | A1 | Ships closest together at $2:18$ am
---7 At midnight, ship $A$ is 70 km due north of ship $B$. Ship $A$ travels with constant velocity $20 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ in the direction with bearing $140 ^ { \circ }$. Ship $B$ travels with constant velocity $15 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ in the direction with bearing $025 ^ { \circ }$.\\
(i) Find the magnitude and direction of the velocity of $A$ relative to $B$.\\
(ii) Find the distance between the ships when they are at their closest, and find the time when this occurs.
\hfill \mbox{\textit{OCR M4 2002 Q7 [9]}}