| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| 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 problem with straightforward vector addition. Students apply Pythagoras to find resultant speed (√20 m/s), use tan⁻¹(2/4) for the angle, and divide distance by perpendicular component for time. All three parts follow routine procedures with no conceptual challenges beyond basic vector resolution. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resultant velocity \(= \sqrt{4^2 + 2^2}\) | M1 | Correct use of Pythagoras with the two velocity components |
| \(= \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ ms}^{-1}\) | A1 | Accept \(\sqrt{20}\) or \(2\sqrt{5}\) or 4.47 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan \alpha = \frac{2}{4}\) or equivalent trig ratio | M1 | Correct trig statement using 2 and 4 |
| \(\alpha = 26.6°\) | A1 | Accept \(\arctan(\frac{1}{2})\) or 26.57° |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Distance \(XY = \frac{20}{\sin \alpha}\) or equivalent method using components | M1 | Must use perpendicular width of 20 m correctly; e.g. time \(= \frac{20}{4}\) using perpendicular component |
| \(t = \frac{20}{4} = 5 \text{ s}\) | A1 | cao |
## Question 4:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resultant velocity $= \sqrt{4^2 + 2^2}$ | M1 | Correct use of Pythagoras with the two velocity components |
| $= \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ ms}^{-1}$ | A1 | Accept $\sqrt{20}$ or $2\sqrt{5}$ or 4.47 |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan \alpha = \frac{2}{4}$ or equivalent trig ratio | M1 | Correct trig statement using 2 and 4 |
| $\alpha = 26.6°$ | A1 | Accept $\arctan(\frac{1}{2})$ or 26.57° |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Distance $XY = \frac{20}{\sin \alpha}$ or equivalent method using components | M1 | Must use perpendicular width of 20 m correctly; e.g. time $= \frac{20}{4}$ using perpendicular component |
| $t = \frac{20}{4} = 5 \text{ s}$ | A1 | cao |4 A canoe is paddled across a river which has a width of 20 metres. The canoe moves from the point $X$ on one bank of the river to the point $Y$ on the other bank, so that its path is a straight line at an angle $\alpha$ to the banks. The velocity of the canoe relative to the water is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to the banks. The water flows at $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ parallel to the banks.\\
\includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-10_469_1333_543_374}
Model the canoe as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant velocity of the canoe.
\item Find the angle $\alpha$.
\item Find the time that it takes for the canoe to travel from $X$ to $Y$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-11_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q4 [6]}}