| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Standard +0.8 This is a relative velocity/closest approach problem requiring vector decomposition, minimization of distance, and bearing calculations. While the mechanics are standard M4 content, it demands careful setup of position vectors, understanding of perpendicular approach for minimum distance, and multi-step reasoning beyond routine exercises. Harder than typical A-level but not requiring exceptional insight. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
4 A cruise ship $C$ is sailing due north at a constant speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A boat $B$, initially 2000 m due west of $C$, sails with constant speed $11 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a straight line course which takes it as close as possible to $C$.\\
(i) Find the bearing of the direction in which $B$ sails.\\
(ii) Find the shortest distance between $B$ and $C$ in the subsequent motion.
\hfill \mbox{\textit{OCR M4 2003 Q4 [8]}}