| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Standard +0.3 This is a standard M3 interception problem requiring vector methods and relative velocity. Part (a) involves setting up position vectors and solving for interception direction using basic trigonometry. Part (b) requires finding closest approach using perpendicular distance or minimization. While multi-step, these are well-practiced techniques with straightforward application, making it slightly easier than average for A-level. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Position of B relative to A: \((12\sin 65°, 12\cos 65°)\) = \((10.876, 5.071)\) km | M1 | Setting up relative position |
| Velocity of B = \((0, 10)\) km h\(^{-1}\) | ||
| For interception, velocity of A must be directed toward moving B | M1 | Correct method using velocity triangle |
| Using sine rule in velocity triangle: \(\frac{\sin\theta}{10} = \frac{\sin 65°}{18}\) | A1 | Correct equation |
| \(\sin\theta = \frac{10\sin 65°}{18} \Rightarrow \theta \approx 30.2°\) | A1 | |
| Bearing \(= 065° - 30.2° \approx 035°\) | A1 | Accept 034° to 035° |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Velocity of A on bearing 065°: \(\mathbf{v}_A = (18\sin 65°, 18\cos 65°)\) | M1 | |
| \(\mathbf{v}_A = (16.314, 7.607)\) km h\(^{-1}\) | A1 | |
| Velocity of B: \(\mathbf{v}_B = (0, 10)\) | ||
| Relative velocity of A w.r.t. B: \((16.314, -2.393)\) | M1 | |
| Relative position B w.r.t. A: \((10.876, 5.071)\) | M1 | |
| Distance = \(\frac{ | 10.876 \times (-2.393) - 5.071 \times 16.314 | }{\sqrt{16.314^2 + 2.393^2}}\) |
| \(= \frac{ | {-26.027 - 82.769} | }{16.488}\) |
| \(\approx \frac{108.796}{16.488} \approx 6.60\) km | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Time of closest approach: \(t = \frac{\mathbf{r} \cdot \mathbf{v}_{rel}}{ | \mathbf{v}_{rel} | ^2}\) |
| \(t = \frac{10.876 \times 16.314 + 5.071 \times (-2.393)}{16.314^2 + 2.393^2}\) | M1 | |
| \(t = \frac{177.534 - 12.135}{271.747} \approx \frac{165.399}{271.747} \approx 0.609\) h | A1 | |
| Time \(\approx 12{:}37\) (approximately 37 minutes after noon) | Accept equivalent |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Position of B relative to A: $(12\sin 65°, 12\cos 65°)$ = $(10.876, 5.071)$ km | M1 | Setting up relative position |
| Velocity of B = $(0, 10)$ km h$^{-1}$ | | |
| For interception, velocity of A must be directed toward moving B | M1 | Correct method using velocity triangle |
| Using sine rule in velocity triangle: $\frac{\sin\theta}{10} = \frac{\sin 65°}{18}$ | A1 | Correct equation |
| $\sin\theta = \frac{10\sin 65°}{18} \Rightarrow \theta \approx 30.2°$ | A1 | |
| Bearing $= 065° - 30.2° \approx 035°$ | A1 | Accept 034° to 035° |
## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Velocity of A on bearing 065°: $\mathbf{v}_A = (18\sin 65°, 18\cos 65°)$ | M1 | |
| $\mathbf{v}_A = (16.314, 7.607)$ km h$^{-1}$ | A1 | |
| Velocity of B: $\mathbf{v}_B = (0, 10)$ | | |
| Relative velocity of A w.r.t. B: $(16.314, -2.393)$ | M1 | |
| Relative position B w.r.t. A: $(10.876, 5.071)$ | M1 | |
| Distance = $\frac{|10.876 \times (-2.393) - 5.071 \times 16.314|}{\sqrt{16.314^2 + 2.393^2}}$ | M1 | Perpendicular distance method |
| $= \frac{|{-26.027 - 82.769}|}{16.488}$ | A1 | |
| $\approx \frac{108.796}{16.488} \approx 6.60$ km | A1 | |
## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Time of closest approach: $t = \frac{\mathbf{r} \cdot \mathbf{v}_{rel}}{|\mathbf{v}_{rel}|^2}$ | M1 | Dot product method |
| $t = \frac{10.876 \times 16.314 + 5.071 \times (-2.393)}{16.314^2 + 2.393^2}$ | M1 | |
| $t = \frac{177.534 - 12.135}{271.747} \approx \frac{165.399}{271.747} \approx 0.609$ h | A1 | |
| Time $\approx 12{:}37$ (approximately 37 minutes after noon) | | Accept equivalent |6 At noon, two ships, $A$ and $B$, are a distance of 12 km apart, with $B$ on a bearing of $065 ^ { \circ }$ from $A$. The ship $B$ travels due north at a constant speed of $10 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. The ship $A$ travels at a constant speed of $18 \mathrm {~km} \mathrm {~h} ^ { - 1 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a90a2de3-5cc0-4e87-b29a-2562f86eee17-16_492_585_445_738}
\begin{enumerate}[label=(\alph*)]
\item Find the direction in which $A$ should travel in order to intercept $B$. Give your answer as a bearing.
\item In fact, the ship $A$ actually travels on a bearing of $065 ^ { \circ }$.
\begin{enumerate}[label=(\roman*)]
\item Find the distance between the ships when they are closest together.
\item Find the time when the ships are closest together.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2012 Q6 [14]}}