OCR M4 2005 June — Question 4 8 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2005
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeClosest approach when exact intercept not possible
DifficultyChallenging +1.2 This is a relative velocity/closest approach problem requiring vector resolution, forming equations for position vectors, and minimizing distance. While it involves multiple steps (resolving velocities, setting up relative motion, finding closest approach), the techniques are standard M4 material with no novel insight required. Slightly above average difficulty due to the bearing system and optimization component, but well within expected mechanics questions.
Spec1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors

4 A boat \(A\) has constant velocity \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction with bearing \(110 ^ { \circ }\). A boat \(B\), which is initially 250 m due south of \(A\), moves with constant speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction which takes it as close as possible to \(A\).
  1. Find the bearing of the direction in which \(B\) moves.
  2. Find the shortest distance between \(A\) and \(B\) in the subsequent motion.

4 A boat $A$ has constant velocity $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction with bearing $110 ^ { \circ }$. A boat $B$, which is initially 250 m due south of $A$, moves with constant speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction which takes it as close as possible to $A$.\\
(i) Find the bearing of the direction in which $B$ moves.\\
(ii) Find the shortest distance between $A$ and $B$ in the subsequent motion.

\hfill \mbox{\textit{OCR M4 2005 Q4 [8]}}