| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | River crossing: reach point directly opposite (find angle and/or time) |
| Difficulty | Moderate -0.8 This is a standard M1 relative velocity problem with straightforward vector addition. Part (a) is given (show that), parts (b) and (c) require basic Pythagoras and trigonometry after recognizing the velocity triangle, and part (d) is a standard modeling assumption question. The setup is clear, the mathematics is routine, and it's a typical textbook exercise requiring no novel insight. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02e Two-dimensional constant acceleration: with vectors |
| \includegraphics[max width=\textwidth, alt={}]{c022c936-72bc-4cf9-8f98-285f12c1d479-08_72_1689_1617_154} | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resultant velocity \(= \frac{16}{10} = 1.6 \text{ ms}^{-1}\) | B1 | Must show distance/time = 16/10 = 1.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V^2 = 1.6^2 + 1.2^2\) | M1 | Use of Pythagoras with 1.6 and 1.2 |
| \(V^2 = 2.56 + 1.44 = 4\) | A1 | Correct calculation |
| \(V = 2 \text{ ms}^{-1}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan \alpha = \frac{1.2}{1.6}\) or \(\sin \alpha = \frac{1.2}{2}\) | M1 | Correct trig ratio using their V |
| \(\alpha = 36.9°\) or \(36.87°\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The boat is modelled as a particle | B1 | Accept equivalent statements |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Resultant velocity $= \frac{16}{10} = 1.6 \text{ ms}^{-1}$ | B1 | Must show distance/time = 16/10 = 1.6 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $V^2 = 1.6^2 + 1.2^2$ | M1 | Use of Pythagoras with 1.6 and 1.2 |
| $V^2 = 2.56 + 1.44 = 4$ | A1 | Correct calculation |
| $V = 2 \text{ ms}^{-1}$ | A1 | CAO |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan \alpha = \frac{1.2}{1.6}$ or $\sin \alpha = \frac{1.2}{2}$ | M1 | Correct trig ratio using their V |
| $\alpha = 36.9°$ or $36.87°$ | A1 | CAO |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| The boat is modelled as a particle | B1 | Accept equivalent statements |
---4 A river has parallel banks which are 16 metres apart. The water in the river flows at $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ parallel to the banks. A boat sets off from one bank at the point $A$ and travels perpendicular to the bank so that it reaches the point $B$, which is directly opposite the point $A$. It takes the boat 10 seconds to cross the river.
The velocity of the boat relative to the water has magnitude $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is at an angle $\alpha$ to the bank, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-08_400_1011_667_511}
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the resultant velocity of the boat is $1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item $\quad$ Find $V$.
\item Find $\alpha$.
\item State one modelling assumption that you needed to make about the boat.
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.......... $\_\_\_\_$\\
\includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-09_40_118_529_159}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q4 [7]}}