Edexcel M4 2003 June — Question 3 9 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2003
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeApparent wind problems
DifficultyChallenging +1.2 This is a standard M4 apparent wind problem requiring vector addition and trigonometry. While it involves multiple steps (setting up two vector equations, resolving components, solving simultaneous equations), the approach is methodical and follows a well-established technique taught in mechanics modules. The symmetry of the 30° angles provides helpful structure, making it more accessible than arbitrary angle problems.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form

3. A man walks due north at a constant speed \(u\) and the wind seems to him to be blowing from the direction \(30 ^ { \circ }\) east of north. On his return journey, when he is walking at the same speed \(u\) due south, the wind seems to him to be blowing from the direction \(30 ^ { \circ }\) south of east. Assuming that the velocity, \(\mathbf { w }\), of the wind relative to the earth is constant, find
  1. the magnitude of \(\mathbf { w }\), in terms of \(u\),
  2. the direction of \(\mathbf { w }\).

Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{v}_W = {_W}\mathbf{v}_M + \mathbf{v}_M\) (used)M1
One vector triangle drawn correctlyM1
Both triangles correctA1 A1
Combining diagramsM1
\(P\hat{Q}R = 90°\)M1
\(QR = 2u\sin 30° = u\)M1
\(\Rightarrow\) triangle \(OQR\) is equilateral
\(\Rightarrow OQ = \mathbf{v}_W = u\)A1
also \(\Rightarrow Q\hat{O}R = 60°\)
Hence direction is from N \(60°\) EA1 (9)
# Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{v}_W = {_W}\mathbf{v}_M + \mathbf{v}_M$ (used) | M1 | |
| One vector triangle drawn correctly | M1 | |
| Both triangles correct | A1 A1 | |
| Combining diagrams | M1 | |
| $P\hat{Q}R = 90°$ | M1 | |
| $QR = 2u\sin 30° = u$ | M1 | |
| $\Rightarrow$ triangle $OQR$ is equilateral | | |
| $\Rightarrow OQ = \mathbf{v}_W = u$ | A1 | |
| also $\Rightarrow Q\hat{O}R = 60°$ | | |
| Hence direction is from N $60°$ E | A1 | (9) |

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3. A man walks due north at a constant speed $u$ and the wind seems to him to be blowing from the direction $30 ^ { \circ }$ east of north. On his return journey, when he is walking at the same speed $u$ due south, the wind seems to him to be blowing from the direction $30 ^ { \circ }$ south of east. Assuming that the velocity, $\mathbf { w }$, of the wind relative to the earth is constant, find\\
(i) the magnitude of $\mathbf { w }$, in terms of $u$,\\
(ii) the direction of $\mathbf { w }$.\\

\hfill \mbox{\textit{Edexcel M4 2003 Q3 [9]}}