AQA M1 2010 June — Question 5 5 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeRelative velocity: find resultant velocity (magnitude and/or direction)
DifficultyModerate -0.8 This is a straightforward M1 relative velocity question requiring basic vector addition using Pythagoras (part a) and inverse tangent for bearing (part b). The setup is clear with perpendicular components, making it easier than average A-level content which typically involves more problem-solving or algebraic manipulation.
Spec1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors

5 An aeroplane is travelling along a straight line between two points, \(A\) and \(B\), which are at the same height. The air is moving due east at a speed of \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Relative to the air, the aeroplane travels due north at a speed of \(100 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Find the magnitude of the resultant velocity of the aeroplane.
    (3 marks)
  2. Find the bearing on which the aeroplane is travelling, giving your answer to the nearest degree.
    (2 marks)
    QUESTION PART REFERENCE
    \includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-11_2484_1709_223_153}

Question 5:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
Resultant velocity components: north \(= 100\), east \(= 30\)M1 Correct use of components
\(v = \sqrt{100^2 + 30^2}\)M1 Pythagoras applied correctly
\(v = \sqrt{10900} \approx 104.4 \text{ ms}^{-1}\)A1 cao
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\tan\alpha = \frac{30}{100}\)M1 Correct trig ratio used
Bearing \(= 017°\) (to nearest degree)A1 cao
## Question 5:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Resultant velocity components: north $= 100$, east $= 30$ | M1 | Correct use of components |
| $v = \sqrt{100^2 + 30^2}$ | M1 | Pythagoras applied correctly |
| $v = \sqrt{10900} \approx 104.4 \text{ ms}^{-1}$ | A1 | cao |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{30}{100}$ | M1 | Correct trig ratio used |
| Bearing $= 017°$ (to nearest degree) | A1 | cao |
5 An aeroplane is travelling along a straight line between two points, $A$ and $B$, which are at the same height. The air is moving due east at a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Relative to the air, the aeroplane travels due north at a speed of $100 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant velocity of the aeroplane. \\
(3 marks)
\item Find the bearing on which the aeroplane is travelling, giving your answer to the nearest degree. \\
(2 marks) \\

\begin{center}
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QUESTION PART REFERENCE &  \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-11_2484_1709_223_153}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{AQA M1 2010 Q5 [5]}}