| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Challenging +1.2 This is a multi-part relative velocity problem requiring vector geometry and trigonometry. While it involves several steps (finding relative velocity direction, converting between absolute and relative motion, calculating closest approach), the techniques are standard M4 fare: using the sine rule for the geometry, vector addition for velocities, and basic kinematics. The 'show that' in part (i) provides scaffolding. More challenging than average due to the multi-step nature and spatial reasoning required, but follows predictable M4 patterns without requiring novel insight. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| As viewed from \(P\) | [Diagram shown] | |
| \(\sin \alpha = \frac{1790}{7400}\) | M1 | |
| \(\alpha = 14.0°\) | A1 (ag) | |
| Bearing of relative velocity is \(50 - \alpha = 036°\) or \(50 + \alpha = 064°\) | B1 ft | For 64 or ft \(50 + \alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity diagram | B1 | Correct diagram (may be implied) |
| \(\frac{\sin \beta}{7} = \frac{\sin 106}{10}\) | M1 | Correct triangle must be intended |
| \(\beta = 42.3°\) | A1 | |
| Bearing of \(v_Q\) is \(36 + \beta = 078.3°\) | A1 | Accept \(78°\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{w}{\sin 31.7} = \frac{10}{\sin 106}\) | M1 | If cosine rule is used, M1 also requires an attempt at solving the quadratic |
| \(w = 5.47 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} w \sin 36 \\ w \cos 36 \end{pmatrix} = \begin{pmatrix} 10 \sin \theta \\ 10 \cos \theta \end{pmatrix} - \begin{pmatrix} 7 \sin 110 \\ 7 \cos 110 \end{pmatrix}\) | B1 | |
| Obtaining an equation in \(\theta\) only, and solving it | M1 | e.g. \(10 \sin \theta - 7.2654 \cos \theta = 8.3173\) or A1A1 if another angle found first |
| \(\theta = 78.3°\) | A2 | |
| Obtaining an equation in \(w\) only, and solving it | M1 | |
| \(w = 5.47 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(QC = \sqrt{7400^2 - 1790^2} = 7180 \text{ m}\) | M1 | |
| Time taken is \(\frac{7180}{5.468}\) | M1 | (Or M2 for other complete method for finding the time); For attempt at relative distance ÷ \(w\) (not awarded for \(7400 ÷ w\)) or \(21.9\) minutes; It is \(7180 ÷ w\) |
| \(= 1310 \text{ s}\) | A1 ft |
| Answer | Marks |
|---|---|
| Bearing of \(CP\) is \(90 + 36 = 126°\) | B1 |
## Part (i)
As viewed from $P$ | | [Diagram shown]
$\sin \alpha = \frac{1790}{7400}$ | M1 |
$\alpha = 14.0°$ | A1 (ag) |
Bearing of relative velocity is $50 - \alpha = 036°$ or $50 + \alpha = 064°$ | B1 ft | For 64 or ft $50 + \alpha$
**Total: 3 marks**
---
## Part (ii)
Velocity diagram | B1 | Correct diagram (may be implied)
$\frac{\sin \beta}{7} = \frac{\sin 106}{10}$ | M1 | Correct triangle must be intended
$\beta = 42.3°$ | A1 |
Bearing of $v_Q$ is $36 + \beta = 078.3°$ | A1 | Accept $78°$
**Total: 4 marks**
---
## Part (iii)
$\frac{w}{\sin 31.7} = \frac{10}{\sin 106}$ | M1 | If cosine rule is used, M1 also requires an attempt at solving the quadratic
$w = 5.47 \text{ m s}^{-1}$ | A1 |
**Total: 2 marks**
**Alternative for (ii) and (iii)**
$\begin{pmatrix} w \sin 36 \\ w \cos 36 \end{pmatrix} = \begin{pmatrix} 10 \sin \theta \\ 10 \cos \theta \end{pmatrix} - \begin{pmatrix} 7 \sin 110 \\ 7 \cos 110 \end{pmatrix}$ | B1 |
Obtaining an equation in $\theta$ only, and solving it | M1 | e.g. $10 \sin \theta - 7.2654 \cos \theta = 8.3173$ or A1A1 if another angle found first
$\theta = 78.3°$ | A2 |
Obtaining an equation in $w$ only, and solving it | M1 |
$w = 5.47 \text{ m s}^{-1}$ | A1 |
---
## Part (iv)
$QC = \sqrt{7400^2 - 1790^2} = 7180 \text{ m}$ | M1 |
Time taken is $\frac{7180}{5.468}$ | M1 | (Or M2 for other complete method for finding the time); For attempt at relative distance ÷ $w$ (not awarded for $7400 ÷ w$) or $21.9$ minutes; It is $7180 ÷ w$
$= 1310 \text{ s}$ | A1 ft |
**Total: 3 marks**
---
## Part (v)
Bearing of $CP$ is $90 + 36 = 126°$ | B1 |
**Total: 1 mark**
---\includegraphics{figure_6}
A ship $P$ is moving with constant velocity 7 m s$^{-1}$ in the direction with bearing 110°. A second ship $Q$ is moving with constant speed 10 m s$^{-1}$ in a straight line. At one instant $Q$ is at the point $X$, and $P$ is 7400 m from $Q$ on a bearing of 050° (see diagram). In the subsequent motion, the shortest distance between $P$ and $Q$ is 1790 m.
\begin{enumerate}[label=(\roman*)]
\item Show that one possible direction for the velocity of $Q$ relative to $P$ has bearing 036°, to the nearest degree, and find the bearing of the other possible direction of this relative velocity. [3]
\end{enumerate}
Given that the velocity of $Q$ relative to $P$ has bearing 036°, find
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the bearing of the direction in which $Q$ is moving, [4]
\item the magnitude of the velocity of $Q$ relative to $P$, [2]
\item the time taken for $Q$ to travel from $X$ to the position where the two ships are closest together, [3]
\item the bearing of $P$ from $Q$ when the two ships are closest together. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2006 Q6 [13]}}