| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Challenging +1.8 This M4 question requires setting up elastic potential energy using Hooke's law, gravitational PE for a rotating rod, then finding equilibrium via differentiation. The geometry (finding string extension using cosine rule) and algebraic manipulation of the derivative are non-trivial, making this significantly harder than average but standard for M4 level. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| GPE: (from a fixed point) e.g. \(mga\cos\theta\) (+C) | M1, B1 | |
| EPE: \(\frac{1}{2}mg\frac{(ext)^2}{4a}\) | M1A1 | |
| \(BC^2 = (4a)^2 + (2a)^2 - 2(4a)(2a)\cos\theta = 20a^2 - 16a^2\cos\theta\) | M1A1 | |
| \(\Rightarrow\) EPE = \(\frac{1}{2}mga[5 - 4\cos\theta - 2\sqrt{\{5 - 4\cos\theta\}} + 1]\) | M1 | |
| Applied: | A1√ | |
| \(V = \) GPE + EPE (+C) | A1 | (9 marks) |
| Answer | Marks |
|---|---|
| \(= mga\{-\cos\theta - \sqrt{\{5 - 4\cos\theta\}}\} + \text{constant}\) | A1 (no errors seen) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{d\theta} = mga\left\{\sin\theta - \frac{4\sin\theta}{2\sqrt{5 - 4\cos\theta}}\right\}\) | M1A1 | |
| \(\frac{dV}{d\theta} = 0\); \([\sin\theta\{\sqrt{(5 - 4\cos\theta)} - 2\} = 0]\) | M1 | |
| \(\Rightarrow \sin\theta = 0\) or \(\sqrt{(5 - 4\cos\theta)} = 2\) | A1 | |
| \(\Rightarrow \theta = 0\) or \(\pi\) (\(0°\) or \(180°\)) | M1A1 | (6 marks) |
| \(\Rightarrow \theta = \cos^{-1}(\frac{1}{4}) = 1.32\) (75.5°) | M1A1 |
## Part (a)
GPE: (from a fixed point) e.g. $mga\cos\theta$ (+C) | M1, B1
EPE: $\frac{1}{2}mg\frac{(ext)^2}{4a}$ | M1A1
$BC^2 = (4a)^2 + (2a)^2 - 2(4a)(2a)\cos\theta = 20a^2 - 16a^2\cos\theta$ | M1A1
$\Rightarrow$ EPE = $\frac{1}{2}mga[5 - 4\cos\theta - 2\sqrt{\{5 - 4\cos\theta\}} + 1]$ | M1
Applied: | A1√
$V = $ GPE + EPE (+C) | A1 | (9 marks)
$= mga\{-\cos\theta - \sqrt{(5 - 4\cos\theta)} + 3\} + C$ ($\surd$ dep. on all Ms)
$= mga\{-\cos\theta - \sqrt{\{5 - 4\cos\theta\}}\} + \text{constant}$ | A1 (no errors seen)
## Part (b)
$\frac{dV}{d\theta} = mga\left\{\sin\theta - \frac{4\sin\theta}{2\sqrt{5 - 4\cos\theta}}\right\}$ | M1A1
$\frac{dV}{d\theta} = 0$; $[\sin\theta\{\sqrt{(5 - 4\cos\theta)} - 2\} = 0]$ | M1
$\Rightarrow \sin\theta = 0$ or $\sqrt{(5 - 4\cos\theta)} = 2$ | A1
$\Rightarrow \theta = 0$ or $\pi$ ($0°$ or $180°$) | M1A1 | (6 marks)
$\Rightarrow \theta = \cos^{-1}(\frac{1}{4}) = 1.32$ (75.5°) | M1A1\includegraphics{figure_3}
A uniform rod $AB$, of mass $m$ and length $2a$, can rotate freely in a vertical plane about a fixed smooth horizontal axis through $A$. The fixed point $C$ is vertically above $A$ and $AC = 4a$. A light elastic string, of natural length $2a$ and modulus of elasticity $\frac{1}{4}mg$, joins $B$ to $C$. The rod $AB$ makes an angle $\theta$ with the upward vertical at $A$, as shown in Fig. 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system is
$$-mga[\cos \theta + \sqrt{(5 - 4 \cos \theta)}] + \text{constant}.$$ [9]
\item Hence determine the values of $\theta$ for which the system is in equilibrium. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q7 [15]}}