| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| 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 -0.3 This is a standard M1 relative velocity problem requiring vector addition and basic trigonometry. Part (a) uses sin⁻¹(0.8/2) to find the angle, part (b) applies Pythagoras for resultant speed and simple time = distance/speed. All steps are routine applications of well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(2\cos\alpha=0.8\) | M1 | Use of cos or sin to find \(\alpha\) with 2 and 0.8 |
| \(\cos\alpha=\frac{0.8}{2}\) | A1 | Correct equation |
| \(\alpha=\cos^{-1}\!\left(\frac{0.8}{2}\right)=66.4°\) | A1 | Correct \(\alpha\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v=\sqrt{2^2-0.8^2}=1.83\text{ ms}^{-1}\) or \(v=2\sin66.4°=1.83\text{ ms}^{-1}\) | M1 | Use of Pythagoras with 2 and 0.8 or trigonometry with angle from above |
| A1 | Correct velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(t=\frac{14}{1.83}=7.64\text{ s}\); allow 7.65 s | M1 | Use of distance over speed from previous |
| A1 | Correct time |
## Question 3:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $2\cos\alpha=0.8$ | M1 | Use of cos or sin to find $\alpha$ with 2 and 0.8 |
| $\cos\alpha=\frac{0.8}{2}$ | A1 | Correct equation |
| $\alpha=\cos^{-1}\!\left(\frac{0.8}{2}\right)=66.4°$ | A1 | Correct $\alpha$ from correct working |
**Part (b)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $v=\sqrt{2^2-0.8^2}=1.83\text{ ms}^{-1}$ or $v=2\sin66.4°=1.83\text{ ms}^{-1}$ | M1 | Use of Pythagoras with 2 and 0.8 or trigonometry with angle from above |
| | A1 | Correct velocity |
**Part (b)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $t=\frac{14}{1.83}=7.64\text{ s}$; allow 7.65 s | M1 | Use of distance over speed from previous |
| | A1 | Correct time |
---3 A boat can travel at a speed of $2 \mathrm {~ms} ^ { - 1 }$ in still water. The boat is to cross a river in which a current flows at a speed of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The angle between the direction in which the boat is pointing and the bank is $\alpha$. The boat travels so that the resultant velocity of the boat is perpendicular to the bank.\\
\includegraphics[max width=\textwidth, alt={}, center]{7e0585ea-062a-487c-8e39-37a4ed414ff8-3_264_1040_575_493}
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha = 66.4 ^ { \circ }$ correct to three significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Find the magnitude of the resultant velocity of the boat.
\item The width of the river is 14 metres. Find the time that it takes for the boat to cross the river.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q3 [7]}}