| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Apparent wind problems |
| Difficulty | Challenging +1.8 This M3 relative velocity problem requires vector decomposition in two reference frames, solving simultaneous equations to find unknown velocity components, then applying closest approach geometry. While systematic, it demands careful coordinate work, bearing conversions, and multi-step reasoning beyond standard textbook exercises—significantly harder than average but within reach of well-prepared students. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Combined velocity triangles diagram | B1 | B1: For combined velocity triangles (PI) |
| \(\frac{25}{\sin 40°} = \frac{rv_O}{\sin 30°}\) | M1 | M1: Sine rule to find Vel of \(T\) rel to \(Q\) |
| \(rv_O = \frac{25\sin 30°}{\sin 40°}\) \(( = 19.4465)\) | A1 | A1: Correct expression or value |
| \(v_T = \sqrt{10^2 + \left(\frac{25\sin 30°}{\sin 40°}\right)^2 - 2 \times 10 \times \frac{25\sin 30°}{\sin 40°} \times \cos 110°}\) | M1 | M1: Cosine rule to find vel of \(T\) |
| \(v_T = 24.7(222662) \text{ ms}^{-1}\) | A1 | A1: Correct expression or value |
| \(\frac{\sin\theta}{\left(\frac{25\sin 30°}{\sin 40°}\right)} = \frac{\sin 110°}{24.7222662}\) | M1 | M1: Sine rule to find \(\theta\) |
| \(\theta = 47.6601°\) and Bearing: \(318°\) | A1 A1 | A1: Correct \(\theta\) and A1: Correct bearing (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Right-angled velocity triangle | B1 | B1: Right-angled velocity triangle |
| \(\cos \alpha = \frac{10}{24.7}\) | M1 | M1: Using trig to find \(\alpha\) |
| \(\alpha = 66(.11775\ldots)°\) | A1 | A1: Correct equation to find \(\alpha\) |
| Motor cyclist's bearing: \(024°\) | A1 | A1: Correct \(\alpha\) and A1: Correct bearing |
| Answer | Marks | Guidance |
|---|---|---|
| If \(v_T\) makes angle \(\beta\) with the north, then \(\frac{v_T}{\sin 30°} = \frac{15}{\sin(60° - \beta)}\) | M1 | M1: Sine rule to link \(v_T\) and \(\beta\) using 15, 30, 60 |
| \(\frac{v_T}{\sin 110°} = \frac{10}{\sin(\beta - 20°)}\) | M1 | M1: Sine rule to link \(v_T\) and \(\beta\) using 10, 110, 20 |
| \(15\sin 30° \sin(\beta - 20°) = 10\sin 110° \sin(60° - \beta)\) | dM1 | dM1: Correct eqn in \(\beta\) |
| \(\tan \beta = \frac{10\sin 110° \sin 60° + 15\sin 30° \sin 20°}{15\sin 30° \cos 20° + 10\sin 110° \cos 60°}\) | dM1 | dM1: Finding \(\tan\beta\) |
| \(\beta = 42.3°\) | A1 | A1: Correct \(\beta\) \(\square\) |
| Bearing: \(318°\) | A1 | A1: Correct bearing (AG) |
| \(\frac{v_T}{\sin 30°} = \frac{15}{\sin(60° - 42.3°)}\) and \(v_T = 24.7 \text{ ms}^{-1}\) | A1 | A1: Correct vel of \(T\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_T = \left(\frac{a}{b}\right)\) and \(rv_O = \left(\frac{a+10}{b}\right)\) and \(rv_C = \left(\frac{a}{b}, 15\right)\) | B1 B1 | B1: Correct vector for \(v_T\). B1: Correct vector |
| \(\frac{a+10}{b} = -\tan 20°\) OE | M1 | M1: Correct ratio for \(\tan 20°\) |
| \(\frac{a-15}{b} = -\tan 60°\) OE | M1 | M1: Correct ratio for \(\tan 60°\) |
| \(a = -16.65\ldots, \quad b = 18.27\ldots\) | A1 | A1: Both \(a\) and \(b\) correct |
| Bearing: \(270° + \tan^{-1}\left(\frac{18.27}{16.65}\right) = 318°\) | A1 | A1: Correct expression for bearing |
| \(v_T = \sqrt{(16.65\ldots)^2 + (18.27\ldots)^2}\) and \(v_T = 24.7 \text{ ms}^{-1}\) | A1 A1 | A1: CAO (AG). A1: Correct vel of \(T\) |
**(a)**
| Combined velocity triangles diagram | B1 | B1: For combined velocity triangles (PI) |
| $\frac{25}{\sin 40°} = \frac{rv_O}{\sin 30°}$ | M1 | M1: Sine rule to find Vel of $T$ rel to $Q$ |
| $rv_O = \frac{25\sin 30°}{\sin 40°}$ $( = 19.4465)$ | A1 | A1: Correct expression or value |
| $v_T = \sqrt{10^2 + \left(\frac{25\sin 30°}{\sin 40°}\right)^2 - 2 \times 10 \times \frac{25\sin 30°}{\sin 40°} \times \cos 110°}$ | M1 | M1: Cosine rule to find vel of $T$ |
| $v_T = 24.7(222662) \text{ ms}^{-1}$ | A1 | A1: Correct expression or value |
| $\frac{\sin\theta}{\left(\frac{25\sin 30°}{\sin 40°}\right)} = \frac{\sin 110°}{24.7222662}$ | M1 | M1: Sine rule to find $\theta$ |
| $\theta = 47.6601°$ and Bearing: $318°$ | A1 A1 | A1: Correct $\theta$ and A1: Correct bearing (AG) |
**(b)**
| Right-angled velocity triangle | B1 | B1: Right-angled velocity triangle |
| $\cos \alpha = \frac{10}{24.7}$ | M1 | M1: Using trig to find $\alpha$ |
| $\alpha = 66(.11775\ldots)°$ | A1 | A1: Correct equation to find $\alpha$ |
| Motor cyclist's bearing: $024°$ | A1 | A1: Correct $\alpha$ and A1: Correct bearing |
**Total: 13 marks**
---
## Question 7(a) - Alternative 1
| If $v_T$ makes angle $\beta$ with the north, then $\frac{v_T}{\sin 30°} = \frac{15}{\sin(60° - \beta)}$ | M1 | M1: Sine rule to link $v_T$ and $\beta$ using 15, 30, 60 |
| $\frac{v_T}{\sin 110°} = \frac{10}{\sin(\beta - 20°)}$ | M1 | M1: Sine rule to link $v_T$ and $\beta$ using 10, 110, 20 |
| $15\sin 30° \sin(\beta - 20°) = 10\sin 110° \sin(60° - \beta)$ | dM1 | dM1: Correct eqn in $\beta$ |
| $\tan \beta = \frac{10\sin 110° \sin 60° + 15\sin 30° \sin 20°}{15\sin 30° \cos 20° + 10\sin 110° \cos 60°}$ | dM1 | dM1: Finding $\tan\beta$ |
| $\beta = 42.3°$ | A1 | A1: Correct $\beta$ $\square$ |
| Bearing: $318°$ | A1 | A1: Correct bearing (AG) |
| $\frac{v_T}{\sin 30°} = \frac{15}{\sin(60° - 42.3°)}$ and $v_T = 24.7 \text{ ms}^{-1}$ | A1 | A1: Correct vel of $T$ |
---
## Question 7(a) - Alternative 2
| $v_T = \left(\frac{a}{b}\right)$ and $rv_O = \left(\frac{a+10}{b}\right)$ and $rv_C = \left(\frac{a}{b}, 15\right)$ | B1 B1 | B1: Correct vector for $v_T$. B1: Correct vector |
| $\frac{a+10}{b} = -\tan 20°$ OE | M1 | M1: Correct ratio for $\tan 20°$ |
| $\frac{a-15}{b} = -\tan 60°$ OE | M1 | M1: Correct ratio for $\tan 60°$ |
| $a = -16.65\ldots, \quad b = 18.27\ldots$ | A1 | A1: Both $a$ and $b$ correct |
| Bearing: $270° + \tan^{-1}\left(\frac{18.27}{16.65}\right) = 318°$ | A1 | A1: Correct expression for bearing |
| $v_T = \sqrt{(16.65\ldots)^2 + (18.27\ldots)^2}$ and $v_T = 24.7 \text{ ms}^{-1}$ | A1 A1 | A1: CAO (AG). A1: Correct vel of $T$ |
---A quad-bike, a truck and a car are moving on a large, open, horizontal surface in a desert plain. Relative to the quad-bike, which is travelling due west at its maximum speed of $10 \text{ m s}^{-1}$, the truck is moving on a bearing of $340°$. Relative to the car, which is travelling due east at a speed of $15 \text{ m s}^{-1}$, the truck is moving on a bearing of $300°$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the truck is approximately $24.7 \text{ m s}^{-1}$ and that it is moving on a bearing of $318°$, correct to the nearest degree. [8 marks]
\item At the instant when the truck is at a distance of $400$ metres from the quad-bike, the bearing of the truck from the quad-bike is $060°$. The truck continues to move with the same velocity as in part (a). The quad-bike continues to move at a speed of $10 \text{ m s}^{-1}$.
Find the bearing, to the nearest degree, on which the quad-bike should travel in order to approach the truck as closely as possible. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2016 Q7 [13]}}