| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.8 This M4 relative velocity question requires understanding of vector interception conditions, solving for unknown velocity components using perpendicularity, and finding closest approach distance. While the individual techniques are standard for M4, the multi-part structure requiring conceptual understanding of interception (relative velocity parallel to relative position) and closest approach geometry makes this moderately challenging, above average difficulty. |
| 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 multiplication3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Velocity of C relative to S \(= (8\mathbf{i}+u\mathbf{j})-(12\mathbf{i}+16\mathbf{j})\) | M1 | |
| \(= (-4\mathbf{i}+(u-16)\mathbf{j})\) m s\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| C intercepts S \(\Rightarrow\) relative velocity is parallel to \(\mathbf{i}\), so \(u - 16 = 0\), \(u = 16\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 10 km at 4 km h\(^{-1}\) takes 2.5 hours, so 2.30pm | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 8\), relative velocity \(= -4\mathbf{i} - 8\mathbf{j}\) | B1 | |
| Correct distance identified | B1 | |
| \(\tan\theta = \dfrac{8}{4} = 2 \Rightarrow \sin\theta = \dfrac{2}{\sqrt{5}}\) | ||
| \(\sin\theta = \dfrac{d}{10} = \dfrac{2}{\sqrt{5}}\) | M1A1 | |
| \(d = \dfrac{20}{\sqrt{5}} = 4\sqrt{5} = 8.9\) (km) | A1 |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Velocity of C relative to S $= (8\mathbf{i}+u\mathbf{j})-(12\mathbf{i}+16\mathbf{j})$ | M1 | |
| $= (-4\mathbf{i}+(u-16)\mathbf{j})$ m s$^{-1}$ | A1 | |
**(2 marks)**
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| C intercepts S $\Rightarrow$ relative velocity is parallel to $\mathbf{i}$, so $u - 16 = 0$, $u = 16$ | M1A1 | |
**(2 marks)**
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| 10 km at 4 km h$^{-1}$ takes 2.5 hours, so 2.30pm | M1A1 | |
**(2 marks)**
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 8$, relative velocity $= -4\mathbf{i} - 8\mathbf{j}$ | B1 | |
| Correct distance identified | B1 | |
| $\tan\theta = \dfrac{8}{4} = 2 \Rightarrow \sin\theta = \dfrac{2}{\sqrt{5}}$ | | |
| $\sin\theta = \dfrac{d}{10} = \dfrac{2}{\sqrt{5}}$ | M1A1 | |
| $d = \dfrac{20}{\sqrt{5}} = 4\sqrt{5} = 8.9$ (km) | A1 | |
**(5 marks)**
**Total: 11 marks**\begin{enumerate}
\item \hspace{0pt} [In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are due east and due north respectively.]
\end{enumerate}
A coastguard patrol boat $C$ is moving with constant velocity $( 8 \mathbf { i } + u \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$. Another ship $S$ is moving with constant velocity $( 12 \mathbf { i } + 16 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$.\\
(a) Find, in terms of $u$, the velocity of $C$ relative to $S$.
At noon, $S$ is 10 km due west of $C$.\\
If $C$ is to intercept $S$,\\
(b) (i) find the value of $u$.\\
(ii) Using this value of $u$, find the time at which $C$ would intercept $S$.
If instead, at noon, $C$ is moving with velocity $( 8 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ and continues at this constant velocity,\\
(c) find the distance of closest approach of $C$ to $S$.\\
\hfill \mbox{\textit{Edexcel M4 2011 Q3 [11]}}