| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Relative velocity: find resultant velocity (magnitude and/or direction) |
| Difficulty | Moderate -0.8 This is a straightforward relative velocity problem requiring basic vector addition. Part (a) uses simple speed = distance/time (V = 240/20 = 12), part (b) applies Pythagoras (√(90² + 12²)), and part (c) uses basic trigonometry for bearing. All steps are routine M1 mechanics with no problem-solving insight needed, making it easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| \(240 = 20V\), \(V = \frac{240}{20} = 12\) | M1, A1 | Correct equation. Correct \(V\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \sqrt{90^2 + 12^2} = 90.8 \text{ m s}^{-1}\) | M1A1 | M1: Equation or expression to find \(v\) or \(v^2\) based on the use of Pythagoras. Must have \(a +\). Allow their value of \(V\) from part (a). A1: Correct velocity. AWRT 90.8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \alpha = \frac{12}{90}\), \(\alpha = 008°\) OR \(\sin \alpha = \frac{12}{\sqrt{8244}}\), \(\alpha = 008°\) OR \(\cos \alpha = \frac{90}{\sqrt{8244}}\), \(\alpha = 008°\) OR \(\tan^{-1}\left(\frac{90}{12}\right) = 82.4\), \(\alpha = 90 - 82 = 008°\) OR \(\sin^{-1}\left(\frac{90}{\sqrt{8244}}\right) = 82.4\), \(\alpha = 90 - 82 = 008°\) OR \(\cos^{-1}\left(\frac{90}{\sqrt{8244}}\right) = 82.4\), \(\alpha = 90 - 82 = 008°\) | M1A1; dM1, A1 | M1: Seeing tan with 12 or their \(V\) from (a) and 90, either way round. A1: Seeing AWRT 8 or 82. A1: Final answer of 008°. CAO. M1: Use of sin or cos with 12 or their \(V\) from (a) or 90 and their value, which may be approximated, from part (b). A1: Seeing AWRT 8 or 82. A1: Final answer of 008°. CAO. If working in radians, do not award final A1 mark unless converted to degrees. Note that intermediate answers of AWRT 0.13 or AWRT 1.4 can be accepted for M1A1. |
**4. (a)**
$240 = 20V$, $V = \frac{240}{20} = 12$ | M1, A1 | Correct equation. Correct $V$. | 2 marks
**4. (b)**
$v = \sqrt{90^2 + 12^2} = 90.8 \text{ m s}^{-1}$ | M1A1 | M1: Equation or expression to find $v$ or $v^2$ based on the use of Pythagoras. Must have $a +$. Allow their value of $V$ from part (a). A1: Correct velocity. AWRT 90.8 | 2 marks
OR (If finding the angle first.) M1: Using 12 or 90 with the sin or cos of their angle. Allow their value of $V$ from part (a). A1: Correct velocity. AWRT 90.8
**4. (c)**
$\tan \alpha = \frac{12}{90}$, $\alpha = 008°$ OR $\sin \alpha = \frac{12}{\sqrt{8244}}$, $\alpha = 008°$ OR $\cos \alpha = \frac{90}{\sqrt{8244}}$, $\alpha = 008°$ OR $\tan^{-1}\left(\frac{90}{12}\right) = 82.4$, $\alpha = 90 - 82 = 008°$ OR $\sin^{-1}\left(\frac{90}{\sqrt{8244}}\right) = 82.4$, $\alpha = 90 - 82 = 008°$ OR $\cos^{-1}\left(\frac{90}{\sqrt{8244}}\right) = 82.4$, $\alpha = 90 - 82 = 008°$ | M1A1; dM1, A1 | M1: Seeing tan with 12 or their $V$ from (a) and 90, either way round. A1: Seeing AWRT 8 or 82. A1: Final answer of 008°. CAO. M1: Use of sin or cos with 12 or their $V$ from (a) or 90 and their value, which may be approximated, from part (b). A1: Seeing AWRT 8 or 82. A1: Final answer of 008°. CAO. If working in radians, do not award final A1 mark unless converted to degrees. Note that intermediate answers of AWRT 0.13 or AWRT 1.4 can be accepted for M1A1. | 3 marks
**Total for Question 4: 7 marks**
---4 An aeroplane is flying in air that is moving due east at $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Relative to the air, the aeroplane has a velocity of $90 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ due north. During a 20 second period, the motion of the air causes the aeroplane to move 240 metres to the east.
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $V$.
\item Find the magnitude of the resultant velocity of the aeroplane.
\item Find the direction of the resultant velocity, giving your answer as a three-figure bearing, correct to the nearest degree.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2016 Q4 [3]}}