| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Challenging +1.2 This is a standard M4 elastic string equilibrium problem requiring energy methods and stability analysis. While it involves multiple components (gravitational PE, elastic PE, and calculus for equilibrium), the techniques are routine for Further Maths students: applying standard PE formulas, differentiating to find equilibrium, and using the second derivative test for stability. The algebra is moderately involved but follows predictable patterns for this topic. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(OQ = 4a\sin\theta\) | B1 | |
| \(V = (-)\,mga\sin 2\theta + \frac{mg}{2\sqrt{3}\cdot 2a}(4a\sin\theta - a)^2 + C\) | B1; M1 A1 | |
| \(= -mga\sin 2\theta + \frac{mga^2}{4a\sqrt{3}}(16\sin^2\theta - 8\sin\theta + 1) + C\) | M1 | |
| \(= -mga\sin 2\theta + \frac{mga}{4\sqrt{3}}(8(1-\cos 2\theta) - 8\sin\theta) + C\) | M1 | |
| \(V = -\frac{mga}{\sqrt{3}}(2\cos 2\theta + \sqrt{3}\sin 2\theta + 2\sin\theta) + C\) | A1 c.s.o | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(V'(\theta) = -\frac{mga}{\sqrt{3}}(-4\sin 2\theta + 2\sqrt{3}\cos 2\theta + 2\cos\theta)\) | M1 A1 | |
| \(V'\!\left(\frac{\pi}{6}\right) = -\frac{mga}{\sqrt{3}}\left(-2\sqrt{3} + 2\sqrt{3}\cdot\frac{1}{2} + 2\cdot\frac{\sqrt{3}}{2}\right) = 0\) | M1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(V''(\theta) = \frac{mga}{\sqrt{3}}(+8\cos 2\theta + 4\sqrt{3}\sin 2\theta + 2\sin\theta)\) | M1 A1 | |
| \(V''\!\left(\frac{\pi}{6}\right) = \frac{11mga}{\sqrt{3}} > 0 \therefore\) stable | M1 A1 c.s.o | (4 marks) |
# Question 4:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| $OQ = 4a\sin\theta$ | B1 | |
| $V = (-)\,mga\sin 2\theta + \frac{mg}{2\sqrt{3}\cdot 2a}(4a\sin\theta - a)^2 + C$ | B1; M1 A1 | |
| $= -mga\sin 2\theta + \frac{mga^2}{4a\sqrt{3}}(16\sin^2\theta - 8\sin\theta + 1) + C$ | M1 | |
| $= -mga\sin 2\theta + \frac{mga}{4\sqrt{3}}(8(1-\cos 2\theta) - 8\sin\theta) + C$ | M1 | |
| $V = -\frac{mga}{\sqrt{3}}(2\cos 2\theta + \sqrt{3}\sin 2\theta + 2\sin\theta) + C$ | A1 c.s.o | (7 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $V'(\theta) = -\frac{mga}{\sqrt{3}}(-4\sin 2\theta + 2\sqrt{3}\cos 2\theta + 2\cos\theta)$ | M1 A1 | |
| $V'\!\left(\frac{\pi}{6}\right) = -\frac{mga}{\sqrt{3}}\left(-2\sqrt{3} + 2\sqrt{3}\cdot\frac{1}{2} + 2\cdot\frac{\sqrt{3}}{2}\right) = 0$ | M1 A1 | (4 marks) |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| $V''(\theta) = \frac{mga}{\sqrt{3}}(+8\cos 2\theta + 4\sqrt{3}\sin 2\theta + 2\sin\theta)$ | M1 A1 | |
| $V''\!\left(\frac{\pi}{6}\right) = \frac{11mga}{\sqrt{3}} > 0 \therefore$ stable | M1 A1 c.s.o | (4 marks) |
**Total: 15 marks**
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{4eb9c38d-66f4-40ba-b7cf-2c2bd19ad087-3_506_967_339_608}
\end{center}
\end{figure}
A uniform rod $P Q$, of length $2 a$ and mass $m$, is free to rotate in a vertical plane about a fixed smooth horizontal axis through the end $P$. The end $Q$ is attached to one end of a light elastic string, of natural length $a$ and modulus of elasticity $\frac { m g } { 2 \sqrt { 3 } }$. The other end of the string is attached to a fixed point $O$, where $O P$ is horizontal and $O P = 2 a$, as shown in Fig. 2. $\angle O P Q$ is denoted by $2 \theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that, when the string is taut, the potential energy of the system is
$$- \frac { m g a } { \sqrt { 3 } } ( 2 \cos 2 \theta + \sqrt { 3 } \sin 2 \theta + 2 \sin \theta ) + \text { constant } .$$
\item Verify that there is a position of equilibrium at $\theta = \frac { \pi } { 6 }$.
\item Determine whether this is a position of stable equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q4 [15]}}