| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Relative velocity: find resultant velocity (magnitude and/or direction) |
| Difficulty | Moderate -0.3 This is a straightforward relative velocity problem requiring vector addition of two perpendicular velocities (west and south), followed by finding magnitude using Pythagoras and direction using inverse tangent. It's slightly easier than average as it involves perpendicular vectors with simple numbers and standard M1 techniques, but still requires proper understanding of relative velocity and bearings. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resultant velocity has components: south = 6, west = 2 | M1 | Correct vector addition |
| \(V = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ ms}^{-1}\) | A1 | Accept \(\sqrt{40}\) or \(2\sqrt{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{2}{6}\) | M1 | Correct trigonometric ratio |
| \(\theta = \arctan\left(\frac{2}{6}\right) = 18.43...°\) | A1 | Correct angle from south |
| Bearing \(= 180° + 18° = 198°\) | A1 | Correct bearing to nearest degree |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resultant velocity has components: south = 6, west = 2 | M1 | Correct vector addition |
| $V = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ ms}^{-1}$ | A1 | Accept $\sqrt{40}$ or $2\sqrt{10}$ |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{2}{6}$ | M1 | Correct trigonometric ratio |
| $\theta = \arctan\left(\frac{2}{6}\right) = 18.43...°$ | A1 | Correct angle from south |
| Bearing $= 180° + 18° = 198°$ | A1 | Correct bearing to nearest degree |
---2 A yacht is sailing through water that is flowing due west at $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The velocity of the yacht relative to the water is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ due south. The yacht has a resultant velocity of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a bearing of $\theta$.
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $V$.
\item Find $\theta$, giving your answer to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2015 Q2 [5]}}