| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| 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 question requiring vector subtraction to find relative velocity (part a is a 'show that' with the answer given), then using the relative velocity framework to find closest approach distance and time. The techniques are routine for M4 students: resolving velocities into components, vector subtraction, and applying the standard closest approach formula. While it requires multiple steps and careful angle work, it follows a well-practiced procedure without requiring novel insight. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
A cyclist $C$ is moving with a constant speed of $10$ m s$^{-1}$ due south. Cyclist $D$ is moving with a constant speed of $16$ m s$^{-1}$ on a bearing of $240°$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the velocity of $C$ relative to $D$ is $14$ m s$^{-1}$.
[3]
\end{enumerate}
At $2$ pm, $D$ is $4$ km due east of $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find
\begin{enumerate}[label=(\roman*)]
\item the shortest distance between $C$ and $D$ during the subsequent motion,
\item the time, to the nearest minute, at which this shortest distance occurs.
\end{enumerate}
[7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2006 Q3 [10]}}