| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward M5 work-energy problem requiring calculation of work done (force dot product with displacement) and application of the work-energy principle. The vector arithmetic is routine and the method is standard, making it slightly easier than average for A-level mechanics. |
| Spec | 6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction of wire: \((13\mathbf{i} + 5\mathbf{j}) - (4\mathbf{i} - \mathbf{j}) = 9\mathbf{i} + 6\mathbf{j}\) | B1 | Finding displacement vector \(\overrightarrow{AB}\) |
| Unit vector along wire: \(\frac{1}{\sqrt{117}}(9\mathbf{i} + 6\mathbf{j})\) or \(\frac{1}{\sqrt{13}}(3\mathbf{i} + 2\mathbf{j})\) | M1 | Attempt unit vector along wire |
| Component of force along wire: \((0.2\mathbf{i} + 0.3\mathbf{j}) \cdot \frac{1}{\sqrt{13}}(3\mathbf{i} + 2\mathbf{j})\) | M1 | Dot product of force with unit vector |
| \(= \frac{0.6 + 0.6}{\sqrt{13}} = \frac{1.2}{\sqrt{13}}\) | A1 | Correct value |
| Distance \(AB = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13}\) | B1 | Correct distance |
| Work-energy: \(\frac{1}{2}(0.2)v^2 = \frac{1.2}{\sqrt{13}} \times 3\sqrt{13}\) | M1 | Applying work-energy theorem |
| \(0.1v^2 = 3.6\) | A1 | |
| \(v = 6 \text{ m s}^{-1}\) | A1 | cao |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction of wire: $(13\mathbf{i} + 5\mathbf{j}) - (4\mathbf{i} - \mathbf{j}) = 9\mathbf{i} + 6\mathbf{j}$ | B1 | Finding displacement vector $\overrightarrow{AB}$ |
| Unit vector along wire: $\frac{1}{\sqrt{117}}(9\mathbf{i} + 6\mathbf{j})$ or $\frac{1}{\sqrt{13}}(3\mathbf{i} + 2\mathbf{j})$ | M1 | Attempt unit vector along wire |
| Component of force along wire: $(0.2\mathbf{i} + 0.3\mathbf{j}) \cdot \frac{1}{\sqrt{13}}(3\mathbf{i} + 2\mathbf{j})$ | M1 | Dot product of force with unit vector |
| $= \frac{0.6 + 0.6}{\sqrt{13}} = \frac{1.2}{\sqrt{13}}$ | A1 | Correct value |
| Distance $AB = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13}$ | B1 | Correct distance |
| Work-energy: $\frac{1}{2}(0.2)v^2 = \frac{1.2}{\sqrt{13}} \times 3\sqrt{13}$ | M1 | Applying work-energy theorem |
| $0.1v^2 = 3.6$ | A1 | |
| $v = 6 \text{ m s}^{-1}$ | 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 0.2 kg is threaded on a smooth straight horizontal wire. The bead is at rest at the point $A$ with position vector $( 4 \mathbf { i } - \mathbf { j } ) \mathrm { m }$. A force $( 0.2 \mathbf { i } + 0.3 \mathbf { j } ) \mathrm { N }$ acts on $P$ and moves it to the point $B$ with position vector $( 13 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }$.
Find the speed of $P$ at $B$.\\
\hfill \mbox{\textit{Edexcel M5 2014 Q1 [5]}}