| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.8 This M4 question requires multiple sophisticated techniques: geometric constraint analysis with pulleys and strings, potential energy formulation for a multi-body system, equilibrium via energy methods (dV/dθ = 0), and stability analysis via second derivative test. While systematic, it demands careful bookkeeping across several connected parts and represents advanced mechanics beyond typical A-level, though the individual calculus steps are standard. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives6.02e Calculate KE and PE: using formulae6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(AP = s - AD - DE = s - L - 2L\sin\theta\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V(\theta) = 2 \times 2mg \times L\cos\theta + \ldots = \ldots + mg(2L\cos\theta - AP)\) | B1, M1 | |
| \(= 4mgL\cos\theta + mg(2L\cos\theta + 2L\sin\theta) + C\) | M1 | |
| \(= 2mgL(3\cos\theta + \sin\theta) + \text{constant}\) | A1 (cao) | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V'(\theta) = 2mgL(-3\sin\theta + \cos\theta) = 0\) | M1, M1 | |
| \(\tan\theta = \frac{1}{3}\) | A1 | |
| \(\theta \approx 18°, 0.32°\) | A1 (awrt 18°, 0.32°) | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V''(\theta) = 2mgL(-3\cos\theta - \sin\theta)\) | M1, A1 | |
| \(V''\left(\arctan\frac{1}{3}\right) = -2\sqrt{10}mgL\) | ||
| \(V''(\theta) < 0\), for any acute \(\theta\) | M1 | |
| Equilibrium is unstable for any acute \(\theta\) | A1 (ft) | (4 marks) |
## Part (a)
$AP = s - AD - DE = s - L - 2L\sin\theta$ | M1, A1 | (2 marks)
## Part (b)
$V(\theta) = 2 \times 2mg \times L\cos\theta + \ldots = \ldots + mg(2L\cos\theta - AP)$ | B1, M1 |
$= 4mgL\cos\theta + mg(2L\cos\theta + 2L\sin\theta) + C$ | M1 |
$= 2mgL(3\cos\theta + \sin\theta) + \text{constant}$ | A1 (cao) | (4 marks)
## Part (c)
$V'(\theta) = 2mgL(-3\sin\theta + \cos\theta) = 0$ | M1, M1 |
$\tan\theta = \frac{1}{3}$ | A1 |
$\theta \approx 18°, 0.32°$ | A1 (awrt 18°, 0.32°) | (4 marks)
## Part (d)
$V''(\theta) = 2mgL(-3\cos\theta - \sin\theta)$ | M1, A1 |
$V''\left(\arctan\frac{1}{3}\right) = -2\sqrt{10}mgL$ | |
$V''(\theta) < 0$, for any acute $\theta$ | M1 |
Equilibrium is **unstable** for any acute $\theta$ | A1 (ft) | (4 marks)
**Total: 14 marks**
---\includegraphics{figure_2}
Two uniform rods $AB$ and $AC$, each of mass $2m$ and length $2L$, are freely jointed at $A$. The mid-points of the rods are $D$ and $E$ respectively. A light inextensible string of length $s$ is fixed to $E$ and passes round small, smooth light pulleys at $D$ and $A$. A particle $P$ of mass $m$ is attached to the other end of the string and hangs vertically. The points $A$, $B$ and $C$ lie in the same vertical plane with $B$ and $C$ on a smooth horizontal surface. The angles $PAB$ and $PAC$ are each equal to $\theta$ ($\theta > 0$), as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Find the length of $AP$ in terms of $s$, $L$ and $\theta$.
[2]
\item Show that the potential energy $V$ of the system is given by
$$V = 2mgL(3\cos\theta + \sin\theta) + \text{constant}.$$
[4]
\item Hence find the value of $\theta$ for which the system is in equilibrium.
[4]
\item Determine whether this position of equilibrium is stable or unstable.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q3 [14]}}