| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| 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 question with straightforward vector addition. Part (a) is a 'show that' using basic trigonometry (tan α = 3/4), and part (b) requires applying Pythagoras theorem to find the resultant velocity. Both parts follow routine procedures with no novel problem-solving required, making it easier than average. |
| 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/Working | Mark | Guidance |
| \(\tan \alpha = \frac{4}{3}\) (or equivalent ratio from the velocity triangle) | M1 | Using the correct ratio from the velocity triangle; resultant velocity is combination of 3 ms⁻¹ parallel and 4 ms⁻¹ perpendicular |
| \(\alpha = 53.1°\) (shown, correct to 3 s.f.) | A1 | Correct conclusion shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Path from \(B\) to \(A\) is at angle \(53.1°\) below the bank (same line, opposite direction) | B1 | Identifying the direction of travel on return |
| Velocity of boat relative to water is \(4\) ms⁻¹ along \(BA\) direction; setting up velocity components | M1 | Resolving boat's velocity relative to water in component form |
| Component of boat velocity parallel to bank: \(4\sin(53.1°) = 3.2\) ms⁻¹ (in direction opposing current) | A1 | Correct parallel component |
| Component perpendicular to bank: \(4\cos(53.1°) = 2.4\) ms⁻¹ | A1 | Correct perpendicular component |
| Resultant parallel component: \(3 - 3.2 = -0.2\) ms⁻¹ (or \(3.2 - 3 = 0.2\) ms⁻¹) | M1 | Adding water velocity and boat velocity components correctly |
| Resultant velocity \(= \sqrt{(0.2)^2 + (2.4)^2}\) | DM1 | Using Pythagoras on resultant components |
| \(= \sqrt{0.04 + 5.76} = \sqrt{5.8} \approx 2.41\) ms⁻¹ | A1 | Correct final answer |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan \alpha = \frac{4}{3}$ (or equivalent ratio from the velocity triangle) | M1 | Using the correct ratio from the velocity triangle; resultant velocity is combination of 3 ms⁻¹ parallel and 4 ms⁻¹ perpendicular |
| $\alpha = 53.1°$ (shown, correct to 3 s.f.) | A1 | Correct conclusion shown |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Path from $B$ to $A$ is at angle $53.1°$ below the bank (same line, opposite direction) | B1 | Identifying the direction of travel on return |
| Velocity of boat relative to water is $4$ ms⁻¹ along $BA$ direction; setting up velocity components | M1 | Resolving boat's velocity relative to water in component form |
| Component of boat velocity parallel to bank: $4\sin(53.1°) = 3.2$ ms⁻¹ (in direction opposing current) | A1 | Correct parallel component |
| Component perpendicular to bank: $4\cos(53.1°) = 2.4$ ms⁻¹ | A1 | Correct perpendicular component |
| Resultant parallel component: $3 - 3.2 = -0.2$ ms⁻¹ (or $3.2 - 3 = 0.2$ ms⁻¹) | M1 | Adding water velocity and boat velocity components correctly |
| Resultant velocity $= \sqrt{(0.2)^2 + (2.4)^2}$ | DM1 | Using Pythagoras on resultant components |
| $= \sqrt{0.04 + 5.76} = \sqrt{5.8} \approx 2.41$ ms⁻¹ | A1 | Correct final answer |
---6 A river has straight parallel banks. The water in the river is flowing at a constant velocity of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ parallel to the banks. A boat crosses the river, from the point $A$ to the point $B$, so that its path is at an angle $\alpha$ to the bank. The velocity of the boat relative to the water is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ perpendicular to the bank. The diagram shows these velocities and the path of the boat.\\
\includegraphics[max width=\textwidth, alt={}, center]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-12_467_988_568_532}
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha = 53.1 ^ { \circ }$, correct to three significant figures.
\item The boat returns along the same straight path from $B$ to $A$. Given that the speed of the boat relative to the water is still $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the magnitude of the resultant velocity of the boat on the return journey.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q6 [8]}}