| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| 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 setting up elastic potential energy expressions for two strings with different moduli, combining with gravitational PE, then differentiating to find equilibrium. The geometry is moderately complex (requiring careful tracking of string extensions and angles), and the stability analysis adds another layer. However, it follows a standard energy method framework for elastic string problems, making it challenging but within expected M4 scope rather than requiring exceptional insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| GPE is \(-mg(\frac{1}{2}a\sin 2\theta)\) | B1 | Negative sign is essential, but may be implied later |
| EPE is \(\frac{3mg}{2a}AD^2 + \frac{4mg}{2a}BD^2\) | M1 | |
| \(= \frac{3mg}{2a}(2a\cos\theta)^2 + \frac{4mg}{2a}(2a\sin\theta)^2\) | A1 | Any correct form |
| \(= mga(6\cos^2\theta + 8\sin^2\theta)\) | ||
| \(= mga(3 + 3\cos 2\theta + 4 - 4\cos 2\theta)\) | M1 | Expressing EPE in terms of \(\cos 2\theta\) |
| \(= mga(7 - \cos 2\theta)\) | ||
| Total PE is \(V = mga(7-\cos 2\theta) - \frac{1}{2}mga\sin 2\theta\) | ||
| \(= \frac{1}{2}mga(14 - 2\cos 2\theta - \sin 2\theta)\) | A1 ag | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\mathrm{d}V}{\mathrm{d}\theta} = \frac{1}{2}mga(4\sin 2\theta - 2\cos 2\theta)\) | B1 | |
| \(\frac{\mathrm{d}V}{\mathrm{d}\theta} = 0\) when \(4\sin 2\theta = 2\cos 2\theta\) | M1 | Equating to zero and solving |
| \(\tan 2\theta = 0.5\) | ||
| \(\theta = 0.232\) (3 sf) | A1 | Accept 13.3° |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = \frac{1}{2}mga(8\cos 2\theta + 4\sin 2\theta)\) | ||
| When \(\theta = 0.232\), \(\frac{\mathrm{d}^2V}{\mathrm{d}\theta^2} > 0\) | M1 | |
| So the equilibrium is stable | A1 | Fully correct working only |
| [2] |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| GPE is $-mg(\frac{1}{2}a\sin 2\theta)$ | B1 | Negative sign is essential, but may be implied later |
| EPE is $\frac{3mg}{2a}AD^2 + \frac{4mg}{2a}BD^2$ | M1 | |
| $= \frac{3mg}{2a}(2a\cos\theta)^2 + \frac{4mg}{2a}(2a\sin\theta)^2$ | A1 | Any correct form |
| $= mga(6\cos^2\theta + 8\sin^2\theta)$ | | |
| $= mga(3 + 3\cos 2\theta + 4 - 4\cos 2\theta)$ | M1 | Expressing EPE in terms of $\cos 2\theta$ |
| $= mga(7 - \cos 2\theta)$ | | |
| Total PE is $V = mga(7-\cos 2\theta) - \frac{1}{2}mga\sin 2\theta$ | | |
| $= \frac{1}{2}mga(14 - 2\cos 2\theta - \sin 2\theta)$ | A1 ag | |
| | [5] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\mathrm{d}V}{\mathrm{d}\theta} = \frac{1}{2}mga(4\sin 2\theta - 2\cos 2\theta)$ | B1 | |
| $\frac{\mathrm{d}V}{\mathrm{d}\theta} = 0$ when $4\sin 2\theta = 2\cos 2\theta$ | M1 | Equating to zero and solving |
| $\tan 2\theta = 0.5$ | | |
| $\theta = 0.232$ (3 sf) | A1 | Accept 13.3° |
| | [3] | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = \frac{1}{2}mga(8\cos 2\theta + 4\sin 2\theta)$ | | |
| When $\theta = 0.232$, $\frac{\mathrm{d}^2V}{\mathrm{d}\theta^2} > 0$ | M1 | |
| So the equilibrium is stable | A1 | Fully correct working only |
| | [2] | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{337dd1f9-a691-4e99-9aa7-7a93d8bb13be-2_439_1045_1512_550}
Two small smooth pegs $A$ and $B$ are fixed at a distance $2 a$ apart on the same horizontal level, and $C$ is the mid-point of $A B$. A uniform rod $C D$, of mass $m$ and length $a$, is freely pivoted at $C$ and can rotate in the vertical plane containing $A B$, with $D$ below the level of $A B$. A light elastic string, of natural length $a$ and modulus of elasticity $3 m g$, passes round the peg $A$ and its ends are attached to $C$ and $D$. Another light elastic string, of natural length $a$ and modulus of elasticity $4 m g$, passes round the peg $B$ and its ends are also attached to $C$ and $D$. The angle $C A D$ is $\theta$, where $0 < \theta < \frac { 1 } { 2 } \pi$, so that the angle $B C D$ is $2 \theta$ (see diagram).\\
(i) Taking $A B$ as the reference level for gravitational potential energy, show that the total potential energy of the system is
$$\frac { 1 } { 2 } m g a ( 14 - 2 \cos 2 \theta - \sin 2 \theta )$$
(ii) Find the value of $\theta$ for which the system is in equilibrium.\\
(iii) Determine whether this position of equilibrium is stable or unstable.
\hfill \mbox{\textit{OCR M4 2011 Q4 [10]}}