| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Bead on straight wire vector force |
| Difficulty | Standard +0.8 This M5 question requires multiple sophisticated steps: finding the component of force along the wire direction, applying work-energy theorem with vector methods, and solving for position using parametric equations. It combines vector geometry, mechanics principles, and algebraic manipulation in a non-routine way that goes beyond standard textbook exercises. |
| Spec | 1.10f Distance between points: using position vectors6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Direction of wire: \((-2\mathbf{i} + 3\mathbf{j})\), unit vector \(\hat{u} = \frac{1}{\sqrt{13}}(-2\mathbf{i} + 3\mathbf{j})\) | B1 | Direction identified |
| Component of force along wire: \((0.5\mathbf{i} + \mathbf{j}) \cdot \frac{1}{\sqrt{13}}(-2\mathbf{i} + 3\mathbf{j}) = \frac{-1+3}{\sqrt{13}} = \frac{2}{\sqrt{13}}\) | M1 A1 | Dot product with unit vector |
| Work-energy theorem: \(F \cdot d = \frac{1}{2}mv^2 - 0\) so \(\frac{2}{\sqrt{13}} \cdot d = \frac{1}{2}(0.4)(25) = 5\) | M1 A1 | |
| Distance: \(d = \frac{5\sqrt{13}}{2}\) | A1 | |
| Position of B: \(\overrightarrow{OB} = (-\mathbf{i}+5\mathbf{j}) + \frac{5\sqrt{13}}{2} \cdot \frac{(-2\mathbf{i}+3\mathbf{j})}{\sqrt{13}} = (-\mathbf{i}+5\mathbf{j}) + 5(-\mathbf{i}+\frac{3}{2}\mathbf{j}) = -6\mathbf{i} + \frac{25}{2}\mathbf{j}\) | M1 A1 | cao |
## Question 1:
**Direction of wire:** $(-2\mathbf{i} + 3\mathbf{j})$, unit vector $\hat{u} = \frac{1}{\sqrt{13}}(-2\mathbf{i} + 3\mathbf{j})$ | B1 | Direction identified
**Component of force along wire:** $(0.5\mathbf{i} + \mathbf{j}) \cdot \frac{1}{\sqrt{13}}(-2\mathbf{i} + 3\mathbf{j}) = \frac{-1+3}{\sqrt{13}} = \frac{2}{\sqrt{13}}$ | M1 A1 | Dot product with unit vector
**Work-energy theorem:** $F \cdot d = \frac{1}{2}mv^2 - 0$ so $\frac{2}{\sqrt{13}} \cdot d = \frac{1}{2}(0.4)(25) = 5$ | M1 A1 |
**Distance:** $d = \frac{5\sqrt{13}}{2}$ | A1 |
**Position of B:** $\overrightarrow{OB} = (-\mathbf{i}+5\mathbf{j}) + \frac{5\sqrt{13}}{2} \cdot \frac{(-2\mathbf{i}+3\mathbf{j})}{\sqrt{13}} = (-\mathbf{i}+5\mathbf{j}) + 5(-\mathbf{i}+\frac{3}{2}\mathbf{j}) = -6\mathbf{i} + \frac{25}{2}\mathbf{j}$ | M1 A1 | cao
---\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}
A bead $P$ of mass 0.4 kg is threaded on a smooth straight horizontal wire. The wire lies along the line with vector equation $\mathbf { r } = ( \mathbf { i } + 2 \mathbf { j } ) + \lambda ( - 2 \mathbf { i } + 3 \mathbf { j } )$. The bead is initially at rest at the point $A$ with position vector $( - \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }$. A constant horizontal force $( 0.5 \mathbf { i } + \mathbf { j } ) \mathrm { N }$ acts on $P$ and moves it along the wire to the point $B$. At $B$ the speed of $P$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Find the position vector of $B$.\\
\hfill \mbox{\textit{Edexcel M5 2016 Q1 [7]}}