| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question involving vector motion and distance minimization. Part (a) requires basic position vector formula r = r₀ + vt; part (b) involves vector subtraction and magnitude calculation (routine algebra); part (c) uses completing the square or differentiation to find minimum; part (d) substitutes back. All techniques are standard textbook exercises with no novel insight required, though the multi-part structure and algebraic manipulation make it slightly more substantial than trivial recall questions. |
| Spec | 1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{OP} = t(5\mathbf{i} + 2\mathbf{j}) \text{ m}, \quad \overrightarrow{OQ} = (4t\mathbf{i} + 6\mathbf{j}) \text{ m}\) | B1 M1 A1 | 3 marks |
| (b) \(\overrightarrow{PQ} = -t\mathbf{i} + (6-2t)\mathbf{j}\) with \(d^2 = (-t)^2 + (6-2t)^2 = 5t^2 - 24t + 36\) | M1 A1 M1 A1 A1 | 5 marks |
| (c) \(\frac{d}{dt}(d^2) = 10t - 24 = 0 \text{ for minimum}\) giving \(t = 2.4\) | M1 A1 A1 | 3 marks |
| (d) When \(t = 2.4\), \(PQ = \sqrt{7.2} = 2.68 \text{ m}\) | M1 A1 | 2 marks |
| Then \(\overrightarrow{OP} = (12\mathbf{i} + 4.8\mathbf{j}) \text{ m}, \quad \overrightarrow{OQ} = (9.6\mathbf{i} + 6\mathbf{j}) \text{ m}\) | B1 B1 | total: 15 |
(a) $\overrightarrow{OP} = t(5\mathbf{i} + 2\mathbf{j}) \text{ m}, \quad \overrightarrow{OQ} = (4t\mathbf{i} + 6\mathbf{j}) \text{ m}$ | B1 M1 A1 | 3 marks
(b) $\overrightarrow{PQ} = -t\mathbf{i} + (6-2t)\mathbf{j}$ with $d^2 = (-t)^2 + (6-2t)^2 = 5t^2 - 24t + 36$ | M1 A1 M1 A1 A1 | 5 marks
(c) $\frac{d}{dt}(d^2) = 10t - 24 = 0 \text{ for minimum}$ giving $t = 2.4$ | M1 A1 A1 | 3 marks
(d) When $t = 2.4$, $PQ = \sqrt{7.2} = 2.68 \text{ m}$ | M1 A1 | 2 marks
Then $\overrightarrow{OP} = (12\mathbf{i} + 4.8\mathbf{j}) \text{ m}, \quad \overrightarrow{OQ} = (9.6\mathbf{i} + 6\mathbf{j}) \text{ m}$ | B1 B1 | total: 15$\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors. The point $A$ has position vector $6\mathbf{j}$ m relative to an origin $O$. At time $t = 0$ a particle $P$ starts from $O$ and moves with constant velocity $(5\mathbf{i} + 2\mathbf{j})$ ms$^{-1}$. At the same instant a particle $Q$ starts from $A$ and moves with constant velocity $4\mathbf{i}$ ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Write down the position vectors of $P$ and of $Q$ at time $t$ seconds. [3 marks]
\item Show that the distance $d$ m between $P$ and $Q$ at time $t$ seconds is such that
$$d^2 = 5t^2 - 24t + 36.$$ [5 marks]
\item Find the value of $t$ for which $d^2$ is a minimum. [3 marks]
\item Hence find the minimum distance between $P$ and $Q$, and state the position vector of each particle when they are closest together. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q5 [15]}}