| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| 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 velocities at 45°, followed by magnitude calculation using Pythagoras and direction using trigonometry. While it involves multiple steps, the techniques are standard M1 mechanics with no conceptual difficulty or novel insight required—slightly easier than average due to the convenient 45° angle. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing 50, 135°, 180° angles | B1 | Diagram (may be implied). The shape is sufficient, but 50 and 180 must be seen. The 135° may be replaced by 45° or be absent. |
| \(v^2 = 50^2 + 180^2 - 2 \times 50 \times 180 \cos 135°\) | M1, A1 | Use of cosine rule with 50, 180 and either 135° or 45°. Correct equation. |
| \(v = 218\) ms\(^{-1}\) | A1 | Correct result for \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin \alpha}{\sin 135°} = \frac{50}{218.24}\) | M1, A1F | Use of the sine rule with 50, 135° or 45° and AWRT 218 or candidate's answer to part (a) to at least 3SF. Correct equation (must have 135° not 45°). |
| \(\alpha = 9.3°\) | A1 | Correct angle. |
| Bearing is 351° | A1 | Three figure bearing. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \alpha = \frac{35.36}{215.36}\) | M1, A1 | Use of trig to find angle. Correct equation. |
| \(\alpha = 9.3°\) | A1 | Correct angle. |
| Bearing is 351° | A1 | Three figure bearing. |
**4(a)**
| Diagram showing 50, 135°, 180° angles | B1 | Diagram (may be implied). The shape is sufficient, but 50 and 180 must be seen. The 135° may be replaced by 45° or be absent. |
| $v^2 = 50^2 + 180^2 - 2 \times 50 \times 180 \cos 135°$ | M1, A1 | Use of cosine rule with 50, 180 and either 135° or 45°. Correct equation. |
| $v = 218$ ms$^{-1}$ | A1 | Correct result for $v$ |
**4(b) - Primary Solution**
| $\frac{\sin \alpha}{\sin 135°} = \frac{50}{218.24}$ | M1, A1F | Use of the sine rule with 50, 135° or 45° and AWRT 218 or candidate's answer to part (a) to at least 3SF. Correct equation (must have 135° not 45°). |
| $\alpha = 9.3°$ | A1 | Correct angle. |
| Bearing is 351° | A1 | Three figure bearing. |
**4(b) - Alternative Solution**
| $\tan \alpha = \frac{35.36}{215.36}$ | M1, A1 | Use of trig to find angle. Correct equation. |
| $\alpha = 9.3°$ | A1 | Correct angle. |
| Bearing is 351° | A1 | Three figure bearing. |
Note: The cosine rule could be used instead of the sine rule here. Apply mark scheme as for sine rule.4 An aeroplane is travelling due north at $180 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the air. The air is moving north-west at $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant velocity of the aeroplane.
\item Find the direction of the resultant velocity, giving your answer as a three-figure bearing to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q4 [8]}}