| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: find bearing/direction to intercept (exact intercept) |
| Difficulty | Standard +0.8 This M3 interception problem requires vector resolution with bearings, relative velocity calculation, and solving a geometric interception scenario with time constraints. While the individual techniques are standard (bearing conversions, relative velocity, triangle solving), the multi-stage problem-solving with several parts and the need to set up the interception geometry correctly makes this moderately challenging, above average for A-level but within typical M3 scope. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(_pv_F = \sqrt{4^2 + 2^2}\) | M1 | |
| \(= 4.47\) m s\(^{-1}\) or \(2\sqrt{5}\) ms\(^{-1}\) or \(\sqrt{20}\) ms\(^{-1}\) | A1 | |
| \(\theta = \tan^{-1}\frac{2}{4}\) | M1 | |
| \(\theta = 26.6°\) | A1F | |
| Bearing \(= 40° + 180° - 26.6° = 193°\) | A1F | |
| Alternative: Comp. due west \(= 4\sin40° - 2\sin50° = 1.04\) ms\(^{-1}\) | (M1) | OE; resolving in two directions |
| Comp. due south \(= 2\cos50° + 4\cos40° = 4.35\) ms\(^{-1}\) | ||
| \(_pv_F = \sqrt{1.04^2 + 4.35^2} = 4.47\) ms\(^{-1}\) | (A1) | |
| \(\theta = \tan^{-1}\frac{1.04}{4.35}\) or \(\tan^{-1}\frac{4.35}{1.04}\) | (M1) | |
| \(\theta = 13.4°\) or \(76.6°\) | (A1F) | |
| Bearing \(= 13.4° + 180°\) or \(270° - 76.6° = 193°\) | (A1F) | |
| Alternative: Correct triangle | (M1) | Any orientation |
| \(_pv_F = \sqrt{1.04^2 + 4.35^2} = 4.47\) ms\(^{-1}\) | (A1) | |
| Rel. Vel. Triangle angle \(26.6°\) or \(63.4°\) | (A1) | |
| Bearing \(= 40° + 180° - 26.6°\) or \(63.4° + 40° + 90° = 193°\) | (M1)(A1F) | |
| Subtotal | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v_F = v_p + {}_pv_F\) | ||
| \(\frac{\sin\alpha}{2} = \frac{\sin140°}{4}\) | M1A1 | |
| \(\alpha = 18.7°\) | A1F | |
| Bearing \(= 90° + 18.7° = 109°\) | A1F | |
| Alternative: \(2\sin40° = 4\sin\alpha\) | (M1) | |
| \(\alpha = \sin^{-1}\left(\frac{1}{2}\sin40°\right)\) | (A1) | |
| \(\alpha = 18.7°\); Bearing \(= 109°\) | (A1F)(A1F) | |
| Subtotal | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\beta = 180° - (140° + 18.7°) = 21.3°\) | B1F | |
| \(\frac{{}_pv_F}{\sin21.3°} = \frac{4}{\sin140°}\) | M1 | |
| \(_pv_F = 2.2568\) ms\(^{-1}\) | A1F | |
| \(t = \frac{1500}{2.2568} = 665\) sec | A1F | |
| Alternative: \(_Fv_p = 4\cos18.7° - 2\cos40° = 2.2568\) | (M1)(A2,1,0) | o.e. resolving in two directions |
| \(t = \frac{1500}{2.2568} = 665\) sec | (A1F) | |
| Subtotal | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| No cross wind, calm lake, instantaneous change of direction by the patrol boat | B1 | Any sensible assumption |
| Subtotal | 1 | |
| Total | 14 |
## Question 3:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $_pv_F = \sqrt{4^2 + 2^2}$ | M1 | |
| $= 4.47$ m s$^{-1}$ or $2\sqrt{5}$ ms$^{-1}$ or $\sqrt{20}$ ms$^{-1}$ | A1 | |
| $\theta = \tan^{-1}\frac{2}{4}$ | M1 | |
| $\theta = 26.6°$ | A1F | |
| Bearing $= 40° + 180° - 26.6° = 193°$ | A1F | |
| **Alternative:** Comp. due west $= 4\sin40° - 2\sin50° = 1.04$ ms$^{-1}$ | (M1) | OE; resolving in two directions |
| Comp. due south $= 2\cos50° + 4\cos40° = 4.35$ ms$^{-1}$ | | |
| $_pv_F = \sqrt{1.04^2 + 4.35^2} = 4.47$ ms$^{-1}$ | (A1) | |
| $\theta = \tan^{-1}\frac{1.04}{4.35}$ or $\tan^{-1}\frac{4.35}{1.04}$ | (M1) | |
| $\theta = 13.4°$ or $76.6°$ | (A1F) | |
| Bearing $= 13.4° + 180°$ or $270° - 76.6° = 193°$ | (A1F) | |
| **Alternative:** Correct triangle | (M1) | Any orientation |
| $_pv_F = \sqrt{1.04^2 + 4.35^2} = 4.47$ ms$^{-1}$ | (A1) | |
| Rel. Vel. Triangle angle $26.6°$ or $63.4°$ | (A1) | |
| Bearing $= 40° + 180° - 26.6°$ or $63.4° + 40° + 90° = 193°$ | (M1)(A1F) | |
| **Subtotal** | **5** | |
### Part (b)(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $v_F = v_p + {}_pv_F$ | | |
| $\frac{\sin\alpha}{2} = \frac{\sin140°}{4}$ | M1A1 | |
| $\alpha = 18.7°$ | A1F | |
| Bearing $= 90° + 18.7° = 109°$ | A1F | |
| **Alternative:** $2\sin40° = 4\sin\alpha$ | (M1) | |
| $\alpha = \sin^{-1}\left(\frac{1}{2}\sin40°\right)$ | (A1) | |
| $\alpha = 18.7°$; Bearing $= 109°$ | (A1F)(A1F) | |
| **Subtotal** | **4** | |
### Part (b)(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\beta = 180° - (140° + 18.7°) = 21.3°$ | B1F | |
| $\frac{{}_pv_F}{\sin21.3°} = \frac{4}{\sin140°}$ | M1 | |
| $_pv_F = 2.2568$ ms$^{-1}$ | A1F | |
| $t = \frac{1500}{2.2568} = 665$ sec | A1F | |
| **Alternative:** $_Fv_p = 4\cos18.7° - 2\cos40° = 2.2568$ | (M1)(A2,1,0) | o.e. resolving in two directions |
| $t = \frac{1500}{2.2568} = 665$ sec | (A1F) | |
| **Subtotal** | **4** | |
### Part (b)(iii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| No cross wind, calm lake, instantaneous change of direction by the patrol boat | B1 | Any sensible assumption |
| **Subtotal** | **1** | |
| **Total** | **14** | |
---3 A fishing boat is travelling between two ports, $A$ and $B$, on the shore of a lake. The bearing of $B$ from $A$ is $130 ^ { \circ }$. The fishing boat leaves $A$ and travels directly towards $B$ with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A patrol boat on the lake is travelling with speed $4 \mathrm {~ms} ^ { - 1 }$ on a bearing of $040 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{719b82f7-2ab5-48db-9b2a-98284096a78a-3_713_1319_443_406}
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of the fishing boat relative to the patrol boat, giving your answer as a speed together with a bearing.
\item When the patrol boat is 1500 m due west of the fishing boat, it changes direction in order to intercept the fishing boat in the shortest possible time.
\begin{enumerate}[label=(\roman*)]
\item Find the bearing on which the patrol boat should travel in order to intercept the fishing boat.
\item Given that the patrol boat intercepts the fishing boat before it reaches $B$, find the time, in seconds, that it takes the patrol boat to intercept the fishing boat after changing direction.
\item State a modelling assumption necessary for answering this question, other than the boats being particles.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2009 Q3 [14]}}