M4 June 2015 Q1
1. 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)
| Scheme | Marks |
|---|---|
| \(\mathbf{r}_P - \mathbf{r}_Q\) | M1 |
| \(= \begin{pmatrix}21 - t\\-16 + t\end{pmatrix}\) | A1 |
| \(d^2 = (21 - t)^2 + (-16 + t)^2\) | |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}d^2 = -2(21 - t) + 2(-16 + t)\ (= -74 + 4t)\) | M1 |
| M1 | |
| Min when \(t = 18.5\) (s) | A1 |
| Relative position \(\begin{pmatrix}2.5\\2.5\end{pmatrix}\), distance \(\sqrt{2.5^2 + 2.5^2}\) (m) | M1 |
| \(= \sqrt{\dfrac{25}{2}} = 3.54\) (m) | A1 |
| (7) | |
| (7 marks) |
Notes
M1 Find position vector of one particle relative to the other. \(\mathbf{r}_P = \begin{pmatrix}16 + t\\-12 + 2t\end{pmatrix},\ \mathbf{r}_Q = \begin{pmatrix}-5 + 2t\\4 + t\end{pmatrix}\)
A1 Accept +/-
Pythagoras
M1 Differentiate \(d\) or \(d^2\) wrt \(t\)
M1 Set derivative = 0 and solve for \(t\)
M1 Substitute their \(t\) to find \(d\)
alt1
| \(\mathbf{r}_P - \mathbf{r}_Q\) | M1 |
| \(= \begin{pmatrix}21 - t\\-16 + t\end{pmatrix}\) | A1 |
| \(d^2 = (21 - t)^2 + (-16 + t)^2\ \left(= 2t^2 - 74t + 697\right)\) | M1 |
| M1 | |
| \(2(t - 18.5)^2 - 684.5 + 697\) | A1 |
| Min \(d^2 = 697 - 684.5\) | M1 |
| Min. \(d = \sqrt{697 - 684.5} = \sqrt{12.5}\) | A1 |
M1 Position of \(P\) relative to \(Q\)
A1 Accept +/-
M1 Use Pythagoras to express \(d^2\) as a quadratic in \(t\)
M1 Complete the square
M1 Use completed square to find minimum value for their expression
alt2
| \(\mathbf{r}_P - \mathbf{r}_Q\) | M1 |
| \(= \begin{pmatrix}21 - t\\-16 + t\end{pmatrix}\) | A1 |
| Relative velocity \(\begin{pmatrix}-1\\1\end{pmatrix}\) | M1 |
| \(:\ \begin{pmatrix}21 - t\\-16 + t\end{pmatrix}\bullet\begin{pmatrix}-1\\1\end{pmatrix} = -(21 - t) + (-16 + t) = 0\), | M1 |
| \(t = 18.5\) (s) | A1 |
| Relative position \(\begin{pmatrix}2.5\\2.5\end{pmatrix}\), distance \(\sqrt{2.5^2 + 2.5^2}\) (m) | M1 |
| \(= \sqrt{\dfrac{25}{2}} = 3.54\) (m) | A1 |
M1 Position of \(P\) relative to \(Q\)
A1 Accept +/-
M1 Set scalar product of relative position and relative velocity = 0 and solve for \(t\).
M1 Substitute their \(t\) to find \(d\)
alt 3
| \(\mathbf{r}_P - \mathbf{r}_Q\) | M1 |
| \(= \begin{pmatrix}21\\-16\end{pmatrix}\) | A1 |
| Relative velocity \(\begin{pmatrix}-1\\1\end{pmatrix}\) | M1 |
| M1 | |
| \(\cos\theta = \dfrac{-37}{\sqrt{2}\sqrt{697}} \quad (-0.991\ldots)\) | A1 |
| \(d = PQ\sin\theta\) | M1 |
| \(= \sqrt{697} \times \sqrt{1 - \dfrac{37^2}{2 \times 697}} = \dfrac{5}{\sqrt{2}} \approx 3.54\) | A1 |
M1 Initial position of \(P\) relative to \(Q\)
A1 Accept +/-
M1 Use scalar product to find \(\cos\theta\)
A1 Accept +/-
M1 Use trig to find distance
(Corrected from the printed mark scheme: the \(\cos\theta\) line is printed as \(\frac{-37}{\sqrt{2}\sqrt{687}}\ (-0.998\ldots)\); \(|PQ| = \sqrt{21^2 + 16^2} = \sqrt{697}\), as in the next line.)