| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| 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 +1.8 This M4 question requires coordinate geometry to find string length, energy methods for equilibrium, differentiation of a composite function involving a square root, and algebraic manipulation to reach a cubic equation in sin θ. The stability analysis adds another layer. While systematic, it demands multiple sophisticated techniques and careful algebra across several steps, making it significantly harder than average A-level questions but not requiring exceptional insight. |
| 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 |
| \(\sqrt{4a^2+16a^2-16a^2\sin\theta}\) | M1 A1 | Length of string expression |
| \(V = -4mga\cos\theta - mg(L - \sqrt{4a^2+16a^2-16a^2\sin\theta})\) | M1 A1 | |
| \(= -4mga\cos\theta - mgL + 2mga\sqrt{5-4\sin\theta}\) | ||
| \(= 2mga\{\sqrt{5-4\sin\theta} - 2\cos\theta\} + \text{constant}\) ** | A1 | Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V'(\theta) = 2mga\left\{\dfrac{-2\cos\theta}{\sqrt{5-4\sin\theta}} + 2\sin\theta\right\}\) | M1 A1 | |
| For equilibrium, \(V'(\theta)=0\): \(\left\{\dfrac{-2\cos\theta}{\sqrt{5-4\sin\theta}} + 2\sin\theta\right\} = 0\) | M1 | |
| \(\dfrac{\cos^2\theta}{5-4\sin\theta} = \sin^2\theta\) | ||
| \(1-\sin^2\theta = \sin^2\theta(5-4\sin\theta)\) | DM1 | |
| \(4\sin^3\theta - 6\sin^2\theta + 1 = 0\) ** | A1 | Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V''(\theta) = 2mga\!\left(\dfrac{\sqrt{5-4\sin\theta}\cdot 2\sin\theta - \dfrac{-2\cos\theta\cdot(-4\cos\theta)}{2\sqrt{5-4\sin\theta}}}{(5-4\sin\theta)} + 2\cos\theta\right)\) | M1 A1 A1 | |
| \(V''\!\left(\dfrac{\pi}{6}\right) = 2mga\left\{\dfrac{\sqrt{3} - \dfrac{8\times\frac{3}{4}}{2\sqrt{3}}}{3} + \sqrt{3}\right\} = 2mga\sqrt{3} > 0\) so stable | DM1 A1 | Total: 5 marks |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{4a^2+16a^2-16a^2\sin\theta}$ | M1 A1 | Length of string expression |
| $V = -4mga\cos\theta - mg(L - \sqrt{4a^2+16a^2-16a^2\sin\theta})$ | M1 A1 | |
| $= -4mga\cos\theta - mgL + 2mga\sqrt{5-4\sin\theta}$ | | |
| $= 2mga\{\sqrt{5-4\sin\theta} - 2\cos\theta\} + \text{constant}$ ** | A1 | Total: 5 marks |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V'(\theta) = 2mga\left\{\dfrac{-2\cos\theta}{\sqrt{5-4\sin\theta}} + 2\sin\theta\right\}$ | M1 A1 | |
| For equilibrium, $V'(\theta)=0$: $\left\{\dfrac{-2\cos\theta}{\sqrt{5-4\sin\theta}} + 2\sin\theta\right\} = 0$ | M1 | |
| $\dfrac{\cos^2\theta}{5-4\sin\theta} = \sin^2\theta$ | | |
| $1-\sin^2\theta = \sin^2\theta(5-4\sin\theta)$ | DM1 | |
| $4\sin^3\theta - 6\sin^2\theta + 1 = 0$ ** | A1 | Total: 5 marks |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V''(\theta) = 2mga\!\left(\dfrac{\sqrt{5-4\sin\theta}\cdot 2\sin\theta - \dfrac{-2\cos\theta\cdot(-4\cos\theta)}{2\sqrt{5-4\sin\theta}}}{(5-4\sin\theta)} + 2\cos\theta\right)$ | M1 A1 A1 | |
| $V''\!\left(\dfrac{\pi}{6}\right) = 2mga\left\{\dfrac{\sqrt{3} - \dfrac{8\times\frac{3}{4}}{2\sqrt{3}}}{3} + \sqrt{3}\right\} = 2mga\sqrt{3} > 0$ so stable | DM1 A1 | Total: 5 marks |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{60202547-5d12-405f-bc83-2907419ec354-09_413_1212_262_365}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The end $A$ of a uniform rod $A B$, of length $2 a$ and mass $4 m$, is smoothly hinged to a fixed point. The end $B$ is attached to one end of a light inextensible string which passes over a small smooth pulley, fixed at the same level as $A$. The distance from $A$ to the pulley is $4 a$. The other end of the string carries a particle of mass $m$ which hangs freely, vertically below the pulley, with the string taut. The angle between the rod and the downward vertical is $\theta$, where $0 < \theta < \frac { \pi } { 2 }$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system is
$$2 m g a ( \sqrt { } ( 5 - 4 \sin \theta ) - 2 \cos \theta ) + \text { constant }$$
\item Hence, or otherwise, show that any value of $\theta$ which corresponds to a position of equilibrium of the system satisfies the equation
$$4 \sin ^ { 3 } \theta - 6 \sin ^ { 2 } \theta + 1 = 0 .$$
\item Given that $\theta = \frac { \pi } { 6 }$ corresponds to a position of equilibrium, determine its stability.
\section*{L \\
$\_\_\_\_$}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2010 Q5 [15]}}