| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2009 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.2 This is a standard M4 potential energy and equilibrium question requiring systematic application of geometric relationships, PE formulation, differentiation, and stability analysis. While it involves multiple steps and careful geometry (finding heights of both masses), the techniques are routine for Further Maths M4 students and the question provides clear guidance through its parts. The algebra is moderately involved but straightforward, placing it slightly above average difficulty. |
| 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 |
|---|---|
| M1 B1 A1 |
| Answer | Marks |
|---|---|
| M1 A1 DM1 DM1 A1 DM1 A1 DM1 |
| Answer | Marks |
|---|---|
| M1 A1 M1 A1 cso |
**Part (a)**
$V = -mg2a\sin 2\theta - \frac{7}{20}mg(L - 4a\sin \theta)$
$= \frac{1}{5}mga(7\sin \theta - 10\sin 2\theta) - \frac{7}{20}mgL$
| M1 B1 A1 |
**Part (b)**
$\frac{dV}{d\theta} = \frac{1}{5}mga(7\cos \theta - 20\cos 2\theta)$
$\frac{1}{5}mga(7\cos \theta - 20\cos 2\theta) = 0$
$7\cos \theta - 20(2\cos^2 \theta - 1) = 0$
$40\cos^2 \theta - 7\cos \theta - 20 = 0$
$(5\cos \theta - 4)(8\cos \theta + 5) = 0$
$\cos \theta = \frac{4}{5}$ or $(\cos \theta = -\frac{5}{8} \Rightarrow 2\theta > 180°)$
| M1 A1 DM1 DM1 A1 DM1 A1 DM1 |
**Part (c)**
$\frac{d^2V}{d\theta^2} = \frac{1}{5}mga(-7\sin \theta + 40\sin 2\theta)$
$= \frac{1}{5}mga(-7\sin \theta + 80\sin \theta \cos \theta)$
When $\cos \theta = \frac{4}{5}$:
$\frac{d^2V}{d\theta^2} = \frac{1}{5}mga\left(-\frac{21}{5} + 80 \times \frac{3}{5} \times \frac{4}{5}\right) = \frac{171}{125}mga$
$> 0$ therefore stable
| M1 A1 M1 A1 cso |
**Total: [16]**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f4c33171-597e-4ef3-9f21-3e2271d48f30-07_479_807_246_571}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light inextensible string of length $2 a$ has one end attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of mass $m$. A second light inextensible string of length $L$, where $L > \frac { 12 a } { 5 }$, has one of its ends attached to $P$ and passes over a small smooth peg fixed at a point $B$. The line $A B$ is horizontal and $A B = 2 a$. The other end of the second string is attached to a particle of mass $\frac { 7 } { 20 } m$, which hangs vertically below $B$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system, when the angle $P A B = 2 \theta$, is
$$\frac { 1 } { 5 } m g a ( 7 \sin \theta - 10 \sin 2 \theta ) + \text { constant. }$$
\item Show that there is only one value of $\cos \theta$ for which the system is in equilibrium and find this value.
\item Determine the stability of the position of equilibrium.\\
\section*{June 2009}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2009 Q4 [16]}}