| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Challenging +1.8 This is a Further Maths M4 question requiring energy methods for equilibrium. It involves calculating gravitational PE for a composite rigid body system and elastic PE, then using dV/dθ=0 for equilibrium and d²V/dθ² for stability. The geometry requires careful coordinate work with trigonometry, and the algebraic manipulation to reach the given form is non-trivial. While systematic, it demands multiple advanced techniques and extended reasoning beyond standard A-level. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(BR = 2 \times 2a\cos\theta = 4a\cos\theta\) | B1 | |
| \(\text{EPE} = 3mg\frac{(4a\cos\theta)^2}{2 \times 2a}\) | M1 | |
| \(= 12mga\cos^2\theta = 6mga + 6mga\cos 2\theta\) | A1 | |
| GPE: taking AR as zero level; \(\text{GPE} = \text{GPE of } AB + \text{GPE of } BC\) | M1+M1 | |
| \(= 4mg \times a\sin 2\theta + 2mg(2a\sin 2\theta - a/2\cos 2\theta)\) | A1 | |
| \(= 8mga\sin 2\theta - mga\cos 2\theta\) | ||
| \(\Rightarrow\) Total \(V = 8mga\sin 2\theta + 5mga\cos 2\theta + \text{constant}\) ** | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{d\theta} = 16mga\cos 2\theta - 10mga\sin 2\theta\) | M1A1 | |
| \(\frac{dV}{d\theta} = 0 \Rightarrow 10\sin 2\theta = 16\cos 2\theta\) | M1 | |
| \(\tan 2\theta = \frac{8}{5} \Rightarrow \theta = 0.51 \ \text{radians} \ (29.0°)\) | A1 | (4) |
| Or: \(8mga\sin 2\theta + 5mga\cos 2\theta = \sqrt{89}mga\cos(2\theta - \alpha),\ \tan\alpha = \frac{8}{5}\) | M1A1 | |
| Turning pts when \(2\theta - \alpha = n\pi \Rightarrow \theta = 0.51 \ \text{rads}\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2V}{d\theta^2} = -32mga\sin 2\theta - 20mga\cos 2\theta\) | M1 | |
| \(\theta = 0.51 \Rightarrow \frac{d^2V}{d\theta^2} < 0\), equilibrium is unstable | M1A1 | cso (3) Total: 14 |
| Or: \(2\theta - \alpha = 0 \Rightarrow \cos(2\theta - \alpha) = 1\); Max value \(\Rightarrow\) equilibrium is unstable |
## Question 7:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $BR = 2 \times 2a\cos\theta = 4a\cos\theta$ | B1 | |
| $\text{EPE} = 3mg\frac{(4a\cos\theta)^2}{2 \times 2a}$ | M1 | |
| $= 12mga\cos^2\theta = 6mga + 6mga\cos 2\theta$ | A1 | |
| GPE: taking AR as zero level; $\text{GPE} = \text{GPE of } AB + \text{GPE of } BC$ | M1+M1 | |
| $= 4mg \times a\sin 2\theta + 2mg(2a\sin 2\theta - a/2\cos 2\theta)$ | A1 | |
| $= 8mga\sin 2\theta - mga\cos 2\theta$ | | |
| $\Rightarrow$ Total $V = 8mga\sin 2\theta + 5mga\cos 2\theta + \text{constant}$ ** | A1 | (7) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{d\theta} = 16mga\cos 2\theta - 10mga\sin 2\theta$ | M1A1 | |
| $\frac{dV}{d\theta} = 0 \Rightarrow 10\sin 2\theta = 16\cos 2\theta$ | M1 | |
| $\tan 2\theta = \frac{8}{5} \Rightarrow \theta = 0.51 \ \text{radians} \ (29.0°)$ | A1 | (4) |
| Or: $8mga\sin 2\theta + 5mga\cos 2\theta = \sqrt{89}mga\cos(2\theta - \alpha),\ \tan\alpha = \frac{8}{5}$ | M1A1 | |
| Turning pts when $2\theta - \alpha = n\pi \Rightarrow \theta = 0.51 \ \text{rads}$ | M1A1 | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2V}{d\theta^2} = -32mga\sin 2\theta - 20mga\cos 2\theta$ | M1 | |
| $\theta = 0.51 \Rightarrow \frac{d^2V}{d\theta^2} < 0$, equilibrium is unstable | M1A1 | cso (3) **Total: 14** |
| Or: $2\theta - \alpha = 0 \Rightarrow \cos(2\theta - \alpha) = 1$; Max value $\Rightarrow$ equilibrium is unstable | | |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2b891a9c-3abe-4e88-ba94-b6abcb37b4c3-13_451_1077_315_370}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a framework $A B C$, consisting of two uniform rods rigidly joined together at $B$ so that $\angle A B C = 90 ^ { \circ }$. The rod $A B$ has length $2 a$ and mass $4 m$, and the rod $B C$ has length $a$ and mass $2 m$. The framework is smoothly hinged at $A$ to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length $2 a$ and modulus of elasticity $3 m g$, is attached to $A$. The string passes through a small smooth ring $R$ fixed at a distance $2 a$ from $A$, on the same horizontal level as $A$ and in the same vertical plane as the framework. The other end of the string is attached to $B$. The angle $A R B$ is $\theta$, where $0 < \theta < \frac { \pi } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy $V$ of the system is given by
$$V = 8 a m g \sin 2 \theta + 5 a m g \cos 2 \theta + \text { constant }$$
\item Find the value of $\theta$ for which the system is in equilibrium.
\item Determine the stability of this position of equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2011 Q7 [14]}}