| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | River crossing: perpendicular heading or minimum time (find drift and/or time) |
| Difficulty | Moderate -0.3 This is a standard M1 relative velocity question using vector triangles and basic trigonometry. Part (a) involves straightforward application of tan and cos to find V and u from a right-angled triangle. Part (b) requires the cosine rule on a non-right triangle, which is slightly more involved but still routine for M1. The question is well-scaffolded with clear diagrams and follows a predictable pattern, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v = 12.5\) (12.48) | M1 | \(\frac{12}{\sin \text{ or } \cos \text{ of } 74° \text{ or } 16°}\) |
| A1 | Total: 2 | or Pythagoras with 3.44; SC if Pythagoras used in circular solution; M1 (1st use) A1 A1 each answer (3 max) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\tan 74° = \frac{12}{u}\) | M1A1F | \(\sqrt{}\) incorrect \(v\) if used |
| \(u = 3.44\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\theta = 135°\) | B1 | Alt: \(12\cos\) or \(\sin 45°\) B1, Full method |
| \(v^2 = 3.44^2 + 12^2 - 2 \times 12 \times 3.44\cos 135°\) | M1, A1\(\sqrt{}\) | subs, all correct |
| \(v = 14.6\) | A1\(\sqrt{}\) | Total: 4 |
## Question 4:
### Part (a)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $v = 12.5$ (12.48) | M1 | $\frac{12}{\sin \text{ or } \cos \text{ of } 74° \text{ or } 16°}$ |
| | A1 | Total: 2 | or Pythagoras with 3.44; SC if Pythagoras used in circular solution; M1 (1st use) A1 A1 each answer (3 max) |
### Part (a)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\tan 74° = \frac{12}{u}$ | M1A1F | $\sqrt{}$ incorrect $v$ if used |
| $u = 3.44$ | A1 | Total: 3 | cao |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\theta = 135°$ | B1 | Alt: $12\cos$ or $\sin 45°$ B1, Full method |
| $v^2 = 3.44^2 + 12^2 - 2 \times 12 \times 3.44\cos 135°$ | M1, A1$\sqrt{}$ | subs, all correct | Alt: $v^2 = (12\sin45°)^2 + (3.44 + 12\sin45°)^2$ M1, (8.485)(11.925) A1 |
| $v = 14.6$ | A1$\sqrt{}$ | Total: 4 | $\sqrt{}$ incorrect subtraction $\rightarrow 135°$ |
---4 Water flows in a constant direction at a constant speed of $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A boat travels in the water at a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the water.
\begin{enumerate}[label=(\alph*)]
\item The direction in which the boat travels relative to the water is perpendicular to the direction of motion of the water. The resultant velocity of the boat is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $74 ^ { \circ }$ to the direction of motion of the water, as shown in the diagram.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c220e6c4-2676-4022-8301-7d720dc082b2-3_120_164_662_488}
\captionsetup{labelformat=empty}
\caption{Velocity of the water}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c220e6c4-2676-4022-8301-7d720dc082b2-3_126_186_667_890}
\captionsetup{labelformat=empty}
\caption{Velocity of the boat relative to the water}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find $V$.
\item Show that $u = 3.44$, correct to three significant figures.
\end{enumerate}\item The boat changes course so that it travels relative to the water at an angle of $45 ^ { \circ }$ to the direction of motion of the water. The resultant velocity of the boat is now of magnitude $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The velocity of the water is unchanged, as shown in the diagram below.
$$\xrightarrow { 3.44 \mathrm {~m} \mathrm {~s} ^ { - 1 } }$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c220e6c4-2676-4022-8301-7d720dc082b2-3_132_273_1493_895}
\end{center}
Velocity of the boat relative to the water\\
\includegraphics[max width=\textwidth, alt={}, center]{c220e6c4-2676-4022-8301-7d720dc082b2-3_232_355_1498_1384}
Find the value of $v$.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q4 [7]}}