| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Standard +0.8 This is a relative velocity problem requiring vector decomposition, solving simultaneous equations from components, finding closest approach distance, and time calculations. While M4 relative velocity is a standard topic, the multi-step nature (finding actual velocity from relative velocity, then geometric optimization) and the bearing calculations make this moderately challenging, above average difficulty but not requiring exceptional insight. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \( | \mathbf{v_Q} | ^2 = (10\cos 30°)^2 + (16 - 10\sin 30°)^2 = 75 + 121\) |
| \( | \mathbf{v_Q} | = 14 \text{ ms}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta \approx 51.8° \Rightarrow \text{bearing } 308°\) (nearest degree) | M1, A1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| At nearest approach: \(PN = 20\sin 30 = 10 \text{ km}\) | M1 A1, A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Time} \approx 4.44 \text{ pm}\) (AWRT) | M1 A1, A1 | (3 marks) |
## Part (a)(i)
$\mathbf{v_Q} = \mathbf{v_P} + \mathbf{v_{P}}$
$|\mathbf{v_Q}|^2 = (10\cos 30°)^2 + (16 - 10\sin 30°)^2 = 75 + 121$
$|\mathbf{v_Q}| = 14 \text{ ms}^{-1}$ | M1 A1, M1, A1, A1 | (5 marks)
## Part (a)(ii)
$\tan\theta = \frac{16 - \sin 30}{10\cos 30}$ (o.e.)
$\theta \approx 51.8° \Rightarrow \text{bearing } 308°$ (nearest degree) | M1, A1, A1 | (3 marks)
## Part (b)
At nearest approach: $PN = 20\sin 30 = 10 \text{ km}$ | M1 A1, A1 | (3 marks)
## Part (c)
$\text{Time} = \frac{20\cos 30}{10} \approx 1.732 \text{ hrs}$
$\text{Time} \approx 4.44 \text{ pm}$ (AWRT) | M1 A1, A1 | (3 marks)
**Total: 12 marks**
---Two ships $P$ and $Q$ are moving with constant velocity. At 3 p.m., $P$ is 20 km due north of $Q$ and is moving at 16 km h$^{-1}$ due west. To an observer on ship $P$, ship $Q$ appears to be moving on a bearing of $030°$ at 10 km h$^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the speed of $Q$,
\item the direction in which $Q$ is moving, giving your answer as a bearing to the nearest degree,
\end{enumerate}
[6]
\item the shortest distance between the ships,
[3]
\item the time at which the two ships are closest together.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2006 Q3 [12]}}