| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done by constant force - vector setup |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem requiring students to find the component of force along the wire, calculate work done, apply the work-energy principle, and use the constraint that the point lies on the given line. While it involves vectors and multiple steps (5-6 marks typical), each step follows standard M5 procedures without requiring novel insight or complex problem-solving. |
| Spec | 1.10f Distance between points: using position vectors6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = 2a - 1\) | B1 | |
| \((6\mathbf{i} - 2\mathbf{j}).((5-a)\mathbf{i} + (9-b)\mathbf{j}) = \frac{1}{2} \times 0.08 \times 10^2\) | M1 A2 | M1 for use of work-energy principle with usual rules; A2 for correct equation with dot product evaluated |
| \(-3a + b = -4\) | DM1 | Second DM1 for solving two simultaneous equations for either \(a\) or \(b\) |
| \(a = 3,\ b = 5\) | A1, A1 | Third A1 for \(a=3\); Fourth A1 for \(b=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((6\mathbf{i} - 2\mathbf{j}) \cdot \frac{1}{\sqrt{5}}(\mathbf{i} + 2\mathbf{j}) = 0.08a\) | M1 | M1 for use of \(\mathbf{F} = m\mathbf{a}\) along the wire |
| \(a = 5\sqrt{5}\) | A1 | First A1 for correct \(a\) |
| \(10^2 = 2 \times 5\sqrt{5} \times s\) | DM1 | Second DM1 for using \(v^2 = u^2 + 2as\) along wire |
| \(s = 2\sqrt{5}\) | A1 | Second A1 for \(s = 2\sqrt{5}\) |
| \(\mathbf{s} = 2(\mathbf{i} + 2\mathbf{j})\) | B1 | B1 for use of \(y = 2x - 1\) |
| \(a = 3,\ b = 5\) | A1, A1 | Third A1 for \(a=3\); Fourth A1 for \(b=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda(\mathbf{i} + 2\mathbf{j})\) | B1 | |
| \((6\mathbf{i} - 2\mathbf{j}).\lambda(\mathbf{i} + 2\mathbf{j}) = \frac{1}{2}(0.08)(10^2)\) | M1 A2 | |
| \(6\lambda - 4\lambda = 4 \Rightarrow \lambda = 2\) | DM1 | |
| \((2\mathbf{i} + 4\mathbf{j}) = (5-a)\mathbf{i} + (9-b)\mathbf{j}\) | A1 | |
| \(a = 3,\ b = 5\) | A1 |
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = 2a - 1$ | B1 | |
| $(6\mathbf{i} - 2\mathbf{j}).((5-a)\mathbf{i} + (9-b)\mathbf{j}) = \frac{1}{2} \times 0.08 \times 10^2$ | M1 A2 | M1 for use of work-energy principle with usual rules; A2 for correct equation with dot product evaluated |
| $-3a + b = -4$ | DM1 | Second DM1 for solving two simultaneous equations for either $a$ or $b$ |
| $a = 3,\ b = 5$ | A1, A1 | Third A1 for $a=3$; Fourth A1 for $b=5$ |
**Alternative (force-acceleration):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(6\mathbf{i} - 2\mathbf{j}) \cdot \frac{1}{\sqrt{5}}(\mathbf{i} + 2\mathbf{j}) = 0.08a$ | M1 | M1 for use of $\mathbf{F} = m\mathbf{a}$ along the wire |
| $a = 5\sqrt{5}$ | A1 | First A1 for correct $a$ |
| $10^2 = 2 \times 5\sqrt{5} \times s$ | DM1 | Second DM1 for using $v^2 = u^2 + 2as$ along wire |
| $s = 2\sqrt{5}$ | A1 | Second A1 for $s = 2\sqrt{5}$ |
| $\mathbf{s} = 2(\mathbf{i} + 2\mathbf{j})$ | B1 | B1 for use of $y = 2x - 1$ |
| $a = 3,\ b = 5$ | A1, A1 | Third A1 for $a=3$; Fourth A1 for $b=5$ |
**Alternative (lambda method):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda(\mathbf{i} + 2\mathbf{j})$ | B1 | |
| $(6\mathbf{i} - 2\mathbf{j}).\lambda(\mathbf{i} + 2\mathbf{j}) = \frac{1}{2}(0.08)(10^2)$ | M1 A2 | |
| $6\lambda - 4\lambda = 4 \Rightarrow \lambda = 2$ | DM1 | |
| $(2\mathbf{i} + 4\mathbf{j}) = (5-a)\mathbf{i} + (9-b)\mathbf{j}$ | A1 | |
| $a = 3,\ b = 5$ | A1 | |
---\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal, $x - y$ plane.]
\end{enumerate}
A bead $P$ of mass 0.08 kg is threaded on a smooth straight horizontal wire which lies along the line with equation $y = 2 x - 1$. The unit of length on both axes is the metre. Initially the bead is at rest at the point $( a , b )$. A force $( 6 \mathbf { i } - 2 \mathbf { j } ) \mathrm { N }$ acts on $P$ and moves it along the wire so that $P$ passes through the point $( 5,9 )$ with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Find the value of $a$ and the value of $b$.
\hfill \mbox{\textit{Edexcel M5 2017 Q1 [7]}}