| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2006 |
| Session | January |
| Marks | 17 |
| 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 setting up potential energy for a constrained system with multiple masses, finding equilibrium via differentiation, and stability analysis via second derivative. While methodical, it demands careful geometric reasoning (relating string lengths to angle θ), algebraic manipulation of trigonometric expressions, and understanding of energy methods—significantly beyond routine mechanics but follows standard M4 equilibrium patterns. |
| 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 |
|---|---|
| PE of R = \(-\sqrt{2}mga\cos 2\theta\) (+c) | (1) |
| PE of LH mass = \(-\frac{3}{2}mg(2a - 2a\sin(45 + \theta))\) (+c) | (2) |
| PE of RH mass = \(-\frac{3}{2}mg(2a - 2a\sin(45 - \theta))\) (+c) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \sqrt{2}mga(3\cos\theta - \cos 2\theta) + \text{constant}\) (*) | B1, M1 A1, A1, M1, M1, A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \theta = 0,\) or \(\theta = \pm\arccos\frac{3}{4}\) (\(\pm 0.723\)) | M1 A1, M1, M1, A1, A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(<0\) ∴ Unstable | M1 A1, M1, A1 | (4 marks) |
## Part (a)
PE of R = $-\sqrt{2}mga\cos 2\theta$ (+c) | (1)
PE of LH mass = $-\frac{3}{2}mg(2a - 2a\sin(45 + \theta))$ (+c) | (2)
PE of RH mass = $-\frac{3}{2}mg(2a - 2a\sin(45 - \theta))$ (+c) | (3)
$V = (1) + (2) + (3)$ (in terms of $\theta$ etc.)
$= -\sqrt{2}mga\cos 2\theta - \frac{3}{2}mg[4a - a\sqrt{2}(\cos\theta + \sin\theta + \cos\theta - \sin\theta)]$
$= -\sqrt{2}mga\cos 2\theta - \frac{3}{2}mga(-2\sqrt{2}\cos\theta + 4)$
$= \sqrt{2}mga(3\cos\theta - \cos 2\theta) + \text{constant}$ (*) | B1, M1 A1, A1, M1, M1, A1 | (7 marks)
## Part (b)
$\frac{dV}{d\theta} = \sqrt{2}mga(-3\sin\theta + 2\sin 2\theta)$
$\frac{dV}{d\theta} = 0 \Rightarrow 2\sin 2\theta - 3\sin\theta = 0$
$\Rightarrow \sin\theta(4\cos\theta - 3) = 0$
$\Rightarrow \theta = 0,$ or $\theta = \pm\arccos\frac{3}{4}$ ($\pm 0.723$) | M1 A1, M1, M1, A1, A1 | (6 marks)
## Part (c)
$\frac{d^2V}{d\theta^2} = \sqrt{2}mga(-3\cos\theta + 4\cos 2\theta)$
$\cos\theta = \frac{3}{4}: \frac{d^2V}{d\theta^2} = \sqrt{2}mga(-3\cdot\frac{3}{4} + 4(2\cdot\frac{9}{16} - 1))$
$= \sqrt{2}mga(-\frac{9}{4} + \frac{1}{2})$
$<0$ ∴ Unstable | M1 A1, M1, A1 | (4 marks)
**Total: 17 marks**\includegraphics{figure_1}
A smooth wire with ends $A$ and $B$ is in the shape of a semi-circle of radius $a$. The mid-point of $AB$ is $O$ and is fixed in a vertical plane and hangs below $AB$ which is horizontal. A small ring $R$, of mass $m\sqrt{2}$, is threaded on the wire and is attached to two light inextensible strings. The other end of each string is attached to a particle of mass $\frac{3m}{2}$. The particles hang vertically under gravity, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that, when the radius $OR$ makes an angle $2\theta$ with the vertical, the potential energy, $V$, of the system is given by
$$V = \sqrt{2}mga(3 \cos \theta - \cos 2\theta) + \text{constant}.$$
[7]
\item Find the values of $\theta$ for which the system is in equilibrium.
[6]
\item Determine the stability of the position of equilibrium for which $\theta > 0$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2006 Q6 [17]}}