| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| 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 closest approach problem requiring students to find the position vectors as functions of time, form the separation vector, minimize using calculus (setting derivative to zero), then calculate the minimum distance. While it involves multiple steps and vector manipulation, it follows a well-established procedure taught explicitly in M4 with no novel insight required. |
| Spec | 1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
Particles P and Q move in a plane with constant velocities. At time $t = 0$ the position vectors of P and Q, relative to a fixed point O in the plane, are $(16\mathbf{i} - 12\mathbf{j})$ m and $(-5\mathbf{i} + 4\mathbf{j})$ m respectively. The velocity of P is $(\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$ and the velocity of Q is $(2\mathbf{i} + \mathbf{j})$ m s$^{-1}$
Find the shortest distance between P and Q in the subsequent motion.
(7 marks)
---\begin{enumerate}
\item Particles $P$ and $Q$ move in a plane with constant velocities. At time $t = 0$ the position vectors of $P$ and $Q$, relative to a fixed point $O$ in the plane, are $( 16 \mathbf { i } - 12 \mathbf { j } ) \mathrm { m }$ and $( - 5 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m }$ respectively. The velocity of $P$ is $( \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $Q$ is $( 2 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
\end{enumerate}
Find the shortest distance between $P$ and $Q$ in the subsequent motion.
\hfill \mbox{\textit{Edexcel M4 2015 Q1 [7]}}