| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.3 This is a standard M4 relative velocity problem with closest approach. It requires converting bearings to vectors, finding relative velocity (routine vector subtraction), then using perpendicularity for closest approach. The calculations are straightforward with clear methodology, making it slightly easier than average for A-level but typical for M4 content. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity triangle diagram | B1 | If wrong triangle used: B0 M1A0 M1A0 for equivalent work; Full ft in (ii) and (iii) |
| \(v^2 = 240^2 + 185^2 - 2 \times 240 \times 185\cos 75\) | M1 | |
| \(v = 262 \text{ m s}^{-1}\) | A1 | |
| \(\frac{\sin\phi}{185} = \frac{\sin 75}{262.4}\), \(\quad \phi = 42.9°\) | M1 | |
| Bearing is \(25 + \phi = 067.9°\) | A1 | Accept \(68°\), \(67.8°\); Allow other clearly stated descriptions of direction Total: 5 |
| OR \(\mathbf{v} = \begin{pmatrix}240\sin 25\\240\cos 25\end{pmatrix} - \begin{pmatrix}185\sin 310\\185\cos 310\end{pmatrix}\) | M1 | |
| \(= \begin{pmatrix}243\\98.6\end{pmatrix}\) | A1 | |
| \(v = \sqrt{243^2 + 98.6^2} = 262\) | M1, A1 | Finding magnitude or angle |
| Bearing is \(\tan^{-1}\frac{243}{98.6} = 67.9°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Relative displacement diagram | M1 | |
| \(d = 4500\cos 67.9\) | M1 | |
| \(= 1690 \text{ m}\) | A1 ft | Ft from bearing in (i) Total: 3 |
| OR \(\overrightarrow{BA} = \begin{pmatrix}243t - 4500\\98.6t\end{pmatrix}\) | M1 | |
| \(BA^2 = (243t-4500)^2 + (98.6t)^2\) is minimum when \(t = \frac{4500 \times 243}{243^2 + 98.6^2} \quad (=15.9)\) | M1 | |
| Then \(BA = 1690\) | A1 ft | Ft from relative velocity in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Bearing is \(270 + 67.9 = 338°\) | M1, A1 ft | Ft from bearing in (i); SR B1 ft for \(158°\) Total: 2 |
| OR \(\overrightarrow{BA} = \begin{pmatrix}-636\\1570\end{pmatrix}\); Bearing is \(360 - \tan^{-1}\frac{636}{1570} = 338°\) | M1, A1 ft | Ft from relative velocity in (i) |
# Question 5(i):
Velocity triangle diagram | B1 | If wrong triangle used: B0 M1A0 M1A0 for equivalent work; Full ft in (ii) and (iii)
$v^2 = 240^2 + 185^2 - 2 \times 240 \times 185\cos 75$ | M1 |
$v = 262 \text{ m s}^{-1}$ | A1 |
$\frac{\sin\phi}{185} = \frac{\sin 75}{262.4}$, $\quad \phi = 42.9°$ | M1 |
Bearing is $25 + \phi = 067.9°$ | A1 | Accept $68°$, $67.8°$; Allow other clearly stated descriptions of direction **Total: 5**
OR $\mathbf{v} = \begin{pmatrix}240\sin 25\\240\cos 25\end{pmatrix} - \begin{pmatrix}185\sin 310\\185\cos 310\end{pmatrix}$ | M1 |
$= \begin{pmatrix}243\\98.6\end{pmatrix}$ | A1 |
$v = \sqrt{243^2 + 98.6^2} = 262$ | M1, A1 | Finding magnitude or angle
Bearing is $\tan^{-1}\frac{243}{98.6} = 67.9°$ | A1 |
## Question 5(ii):
Relative displacement diagram | M1 |
$d = 4500\cos 67.9$ | M1 |
$= 1690 \text{ m}$ | A1 ft | Ft from bearing in (i) **Total: 3**
OR $\overrightarrow{BA} = \begin{pmatrix}243t - 4500\\98.6t\end{pmatrix}$ | M1 |
$BA^2 = (243t-4500)^2 + (98.6t)^2$ is minimum when $t = \frac{4500 \times 243}{243^2 + 98.6^2} \quad (=15.9)$ | M1 |
Then $BA = 1690$ | A1 ft | Ft from relative velocity in (i)
## Question 5(iii):
Bearing is $270 + 67.9 = 338°$ | M1, A1 ft | Ft from bearing in (i); SR B1 ft for $158°$ **Total: 2**
OR $\overrightarrow{BA} = \begin{pmatrix}-636\\1570\end{pmatrix}$; Bearing is $360 - \tan^{-1}\frac{636}{1570} = 338°$ | M1, A1 ft | Ft from relative velocity in (i)
---5 Two aircraft $A$ and $B$ are flying horizontally at the same height. $A$ has constant velocity $240 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction with bearing $025 ^ { \circ }$, and $B$ has constant velocity $185 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction with bearing $310 ^ { \circ }$.\\
(i) Find the magnitude and direction of the velocity of $A$ relative to $B$.
Initially $A$ is 4500 m due west of $B$. For the instant during the subsequent motion when $A$ and $B$ are closest together, find\\
(ii) the distance between $A$ and $B$,\\
(iii) the bearing of $A$ from $B$.
\hfill \mbox{\textit{OCR M4 2004 Q5 [10]}}