| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| 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 standard M1 relative velocity problem requiring vector addition of two velocities at specific angles (north-east and west), followed by magnitude and direction calculations using Pythagoras and trigonometry. The working is straightforward with no conceptual challenges, making it slightly easier than average for A-level, though the multi-step nature and need for careful angle work keeps it close to typical difficulty. |
| 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 |
| Triangle/diagram showing \(v\), 5, 2 at \(45°\) | B1 | Forming a triangle or diagram to find \(v\); do not penalise if sides not in proportion |
| \(v^2 = 2^2 + 5^2 - 2 \times 2 \times 5\cos 45°\) | M1 | Using cosine rule with 2, 5 and any angle; equation must contain a negative sign and a cosine |
| Correct equation | A1 | Note that implied B1 not awarded at this stage |
| \(v = 3.85459 = 3.85 \text{ ms}^{-1}\) (to 3sf) | A1 | Correct velocity from correct working with intermediate calculation shown or final value from value with more than 3sf |
| OR \(v_1 = 5 - 2\cos 45° (= 3.5858)\), \(v_2 = 2\cos 45° (= 1.414)\) | (M1)(A1) | Two perpendicular equations with 2, 5 and \(\sin 45°\) or \(\cos 45°\); both components with correct magnitude; implied B1 can be awarded at this stage |
| \(v = \sqrt{(5-2\cos 45°)^2 + (2\cos 45°)^2} = 3.85 \text{ ms}^{-1}\) | (A1) | Correct velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sin\theta}{2} = \frac{\sin 45°}{3.855}\) | M1 | Use of sine rule with 2, 3.855 or 3.85 or awrt 3.85 and any angle |
| Correct expression | A1 | |
| \(\theta = 21.5°\) | A1 | Correct angle; awrt \(21°\) or \(22°\) |
| Bearing \(= 270 + 21.5 = 292°\) | A1 | Correct bearing; do not penalise candidates who include decimals; accept \(291°\) |
| OR \(\frac{\sin\theta}{5} = \frac{\sin 45°}{3.855}\), \(\theta = 113°\) | (M1)(A1)(A1) | Use of sine rule with 5, 3.855; correct angle awrt \(113°\) or \(114°\); also allow awrt \(66°\) or \(67°\) |
| Bearing \(= 360 - (113.3 - 45) = 292°\) | (A1) | Correct bearing; accept \(291°\) |
| OR \(\tan\theta = \frac{2\cos 45°}{5 - 2\cos 45°}\), \(\theta = 21.5°\) | (M1)(A1)(A1) | Consideration of perpendicular components; correct expression; correct positive angle awrt \(21°\) or \(22°\); also allow method leading to awrt \(68°\) or \(69°\) |
| Bearing \(= 270 + 21.5 = 292°\) | (A1) | Correct bearing; accept \(291°\) |
| OR \(\cos\theta = \frac{3.855^2 + 5^2 - 2^2}{2 \times 5 \times 3.855}\), \(\theta = 21.5°\) | (M1)(A1)(A1) | Use of cosine rule with 2, 3.855 and 5; correct expression; correct angle awrt \(21°\) or \(22°\) |
| Bearing \(= 270 + 21.5 = 292°\) | (A1) | Correct bearing; accept \(291°\) |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Triangle/diagram showing $v$, 5, 2 at $45°$ | B1 | Forming a triangle or diagram to find $v$; do not penalise if sides not in proportion |
| $v^2 = 2^2 + 5^2 - 2 \times 2 \times 5\cos 45°$ | M1 | Using cosine rule with 2, 5 and any angle; equation must contain a negative sign and a cosine |
| Correct equation | A1 | Note that implied B1 not awarded at this stage |
| $v = 3.85459 = 3.85 \text{ ms}^{-1}$ (to 3sf) | A1 | Correct velocity from correct working with intermediate calculation shown or final value from value with more than 3sf |
| **OR** $v_1 = 5 - 2\cos 45° (= 3.5858)$, $v_2 = 2\cos 45° (= 1.414)$ | (M1)(A1) | Two perpendicular equations with 2, 5 and $\sin 45°$ or $\cos 45°$; both components with correct magnitude; implied B1 can be awarded at this stage |
| $v = \sqrt{(5-2\cos 45°)^2 + (2\cos 45°)^2} = 3.85 \text{ ms}^{-1}$ | (A1) | Correct velocity |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sin\theta}{2} = \frac{\sin 45°}{3.855}$ | M1 | Use of sine rule with 2, 3.855 or 3.85 or awrt 3.85 and any angle |
| Correct expression | A1 | |
| $\theta = 21.5°$ | A1 | Correct angle; awrt $21°$ or $22°$ |
| Bearing $= 270 + 21.5 = 292°$ | A1 | Correct bearing; do not penalise candidates who include decimals; accept $291°$ |
| **OR** $\frac{\sin\theta}{5} = \frac{\sin 45°}{3.855}$, $\theta = 113°$ | (M1)(A1)(A1) | Use of sine rule with 5, 3.855; correct angle awrt $113°$ or $114°$; also allow awrt $66°$ or $67°$ |
| Bearing $= 360 - (113.3 - 45) = 292°$ | (A1) | Correct bearing; accept $291°$ |
| **OR** $\tan\theta = \frac{2\cos 45°}{5 - 2\cos 45°}$, $\theta = 21.5°$ | (M1)(A1)(A1) | Consideration of perpendicular components; correct expression; correct positive angle awrt $21°$ or $22°$; also allow method leading to awrt $68°$ or $69°$ |
| Bearing $= 270 + 21.5 = 292°$ | (A1) | Correct bearing; accept $291°$ |
| **OR** $\cos\theta = \frac{3.855^2 + 5^2 - 2^2}{2 \times 5 \times 3.855}$, $\theta = 21.5°$ | (M1)(A1)(A1) | Use of cosine rule with 2, 3.855 and 5; correct expression; correct angle awrt $21°$ or $22°$ |
| Bearing $= 270 + 21.5 = 292°$ | (A1) | Correct bearing; accept $291°$ |7 A boat is travelling in water that is moving north-east at a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The velocity of the boat relative to the water is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ due west.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8c6f9ac0-c24f-48d0-9fb2-883651e791d7-5_275_349_415_504}
\captionsetup{labelformat=empty}
\caption{Velocity of the water}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8c6f9ac0-c24f-48d0-9fb2-883651e791d7-5_81_293_534_1181}
\captionsetup{labelformat=empty}
\caption{Velocity of the boat relative to the water}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the resultant velocity of the boat is $3.85 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to three significant figures.
\item Find the bearing on which the boat is travelling, giving your answer to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q7 [8]}}