| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Challenging +1.2 This is a standard M4 relative velocity/closest approach problem requiring vector setup, differentiation or perpendicularity condition, and straightforward calculation. While it involves multiple parts and vector manipulation, the technique is well-practiced in mechanics modules and follows a predictable solution path without requiring novel insight. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Velocity of \(A\) relative to \(B\) = velocity of \(A\) – velocity of \(B\) | M1 | Setting up relative velocity |
| \(\mathbf{v}_A = 6\mathbf{i}\), \(\mathbf{v}_B = 5(\sin\alpha\mathbf{i} + \cos\alpha\mathbf{j})\) | ||
| For closest approach, relative velocity must be perpendicular to initial displacement | M1 | Key condition stated |
| Initial displacement of \(A\) from \(B\) is \(4\mathbf{j}\) | ||
| Relative velocity \((6 - 5\sin\alpha)\mathbf{i} + (-5\cos\alpha)\mathbf{j}\) must be perpendicular to \(\mathbf{j}\) | ||
| So \(-5\cos\alpha = 0\)... or use dot product condition with \(4\mathbf{j}\) | ||
| \(\sin\alpha = \frac{6}{5}\)... direction found using \(\sin\theta = \frac{6}{5}\) — need \(B\) heading so relative vel. \(\parallel \mathbf{j}\) | M1 | Correct method |
| \(\sin\alpha = \frac{6}{5}\), so bearing = \(\arcsin\left(\frac{6}{5}\right)\)... correct: \(B\) moves so \(v_{AB}\) is due north | ||
| Bearing = \(360° - \arcsin\left(\frac{6}{5}\right)\)... | ||
| \(\sin(\text{angle east of south}) = \frac{6}{5}\)... bearing \(\approx 307°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Least distance = component of displacement perpendicular to relative velocity | M1 | |
| \(= 4\cos\alpha = 4 \times \frac{\sqrt{61}}{5} \cdot \frac{1}{\sqrt{61}}\)... \(= 4\sin\beta\) where \(\cos\beta = \frac{6}{5}\)... | ||
| \(= 4 \times \frac{\sqrt{61}}{5}\)... \(= \frac{4\sqrt{61}}{5}\) ... using \(\cos\alpha = \frac{\sqrt{61}}{5}\): least distance \(= \frac{4\sqrt{61}}{5} \approx 6.26\) km | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Relative speed \(= \lvert \mathbf{v}_{AB} \rvert\) in direction along AB \(= \sqrt{61} - 0\)... relative velocity magnitude | M1 | |
| \(\lvert \mathbf{v}_{AB} \rvert = \sqrt{(6-5\sin\alpha)^2 + (5\cos\alpha)^2}\); with \(\sin\alpha = \frac{6}{5}\)... \(= 5\cos\alpha \cdot\)... | ||
| Relative speed along AB \(= \sqrt{61}\) ... distance along AB / relative speed | M1 | |
| Distance from \(B\) to foot of perpendicular \(= 4\sin\alpha \cdot\)... \(= 4 \times \frac{6}{5} \cdot\)... | M1 | |
| Time \(= \frac{4 \times \frac{6}{5}}{\sqrt{61}/5} \cdot\)... \(= \frac{24}{\sqrt{61}}\) hours | ||
| Time after 10 a.m. \(= \frac{24}{\sqrt{61}} \approx 3.07\) hours \(\approx\) 3 hours 4 minutes | A1 | cao — approximately 1:04 p.m. |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Velocity of $A$ relative to $B$ = velocity of $A$ – velocity of $B$ | M1 | Setting up relative velocity |
| $\mathbf{v}_A = 6\mathbf{i}$, $\mathbf{v}_B = 5(\sin\alpha\mathbf{i} + \cos\alpha\mathbf{j})$ | | |
| For closest approach, relative velocity must be perpendicular to initial displacement | M1 | Key condition stated |
| Initial displacement of $A$ from $B$ is $4\mathbf{j}$ | | |
| Relative velocity $(6 - 5\sin\alpha)\mathbf{i} + (-5\cos\alpha)\mathbf{j}$ must be perpendicular to $\mathbf{j}$ | | |
| So $-5\cos\alpha = 0$... or use dot product condition with $4\mathbf{j}$ | | |
| $\sin\alpha = \frac{6}{5}$... direction found using $\sin\theta = \frac{6}{5}$ — need $B$ heading so relative vel. $\parallel \mathbf{j}$ | M1 | Correct method |
| $\sin\alpha = \frac{6}{5}$, so bearing = $\arcsin\left(\frac{6}{5}\right)$... correct: $B$ moves so $v_{AB}$ is due north | | |
| Bearing = $360° - \arcsin\left(\frac{6}{5}\right)$... | | |
| $\sin(\text{angle east of south}) = \frac{6}{5}$... bearing $\approx 307°$ | A1 | cao |
*(Note: The direction of B's motion: relative velocity of A w.r.t. B must be parallel to displacement AB (due north). So $6 - 5\sin\alpha = 0$, giving $\sin\alpha = \frac{6}{5}$... The bearing is $360° - \arcsin\!\tfrac{6}{5}$... bearing ≈ 307°)*
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Least distance = component of displacement perpendicular to relative velocity | M1 | |
| $= 4\cos\alpha = 4 \times \frac{\sqrt{61}}{5} \cdot \frac{1}{\sqrt{61}}$... $= 4\sin\beta$ where $\cos\beta = \frac{6}{5}$... | | |
| $= 4 \times \frac{\sqrt{61}}{5}$... $= \frac{4\sqrt{61}}{5}$ ... using $\cos\alpha = \frac{\sqrt{61}}{5}$: least distance $= \frac{4\sqrt{61}}{5} \approx 6.26$ km | A1 | cao |
*(least distance $= \frac{4\sqrt{61}}{5}$ km)*
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Relative speed $= \lvert \mathbf{v}_{AB} \rvert$ in direction along AB $= \sqrt{61} - 0$... relative velocity magnitude | M1 | |
| $\lvert \mathbf{v}_{AB} \rvert = \sqrt{(6-5\sin\alpha)^2 + (5\cos\alpha)^2}$; with $\sin\alpha = \frac{6}{5}$... $= 5\cos\alpha \cdot$... | | |
| Relative speed along AB $= \sqrt{61}$ ... distance along AB / relative speed | M1 | |
| Distance from $B$ to foot of perpendicular $= 4\sin\alpha \cdot$... $= 4 \times \frac{6}{5} \cdot$... | M1 | |
| Time $= \frac{4 \times \frac{6}{5}}{\sqrt{61}/5} \cdot$... $= \frac{24}{\sqrt{61}}$ hours | | |
| Time after 10 a.m. $= \frac{24}{\sqrt{61}} \approx 3.07$ hours $\approx$ 3 hours 4 minutes | A1 | cao — approximately 1:04 p.m. |
---\begin{enumerate}
\item At 10 a.m. two walkers $A$ and $B$ are 4 km apart with $A$ due north of $B$. Walker $A$ is moving due east at a constant speed of $6 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Walker $B$ is moving with constant speed $5 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ and walks in the straight line which allows him to pass as close as possible to $A$.
\end{enumerate}
Find\\
(a) the direction of motion of $B$, giving your answer as a bearing,\\
(b) the least distance between $A$ and $B$,\\
(c) the time when the distance between $A$ and $B$ is least.
\hfill \mbox{\textit{Edexcel M4 2013 Q4 [10]}}