| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Challenging +1.2 This is a standard M4 elastic string energy problem requiring systematic application of formulas for elastic PE and gravitational PE, followed by differentiation to find equilibrium positions. While it involves multiple steps and careful algebra (showing a specific expression, solving a quadratic for equilibrium, testing stability via second derivative), these are well-practiced techniques for M4 students with no novel geometric insight required. Slightly above average due to the algebraic manipulation needed and the three-part structure. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| GPE is \((-) mg(2a\cos\theta)\cos\theta\) | B1 | or \(mg(a+a\cos2\theta)\) |
| M1 | Using \(\frac{\lambda x^2}{2l}\) (allow one error) | |
| EPE is \(\frac{2mg}{2(\frac{1}{2}a)}(2a\cos\theta - \frac{1}{2}a)^2\) | A1 | |
| \(V = 2mga(4\cos^2\theta - 2\cos\theta + \frac{1}{4}) - 2mga\cos^2\theta\) | ||
| \(= mga(6\cos^2\theta - 4\cos\theta + \frac{1}{2})\) | A1 | AG |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dV}{d\theta} = mga(-12\cos\theta\sin\theta + 4\sin\theta)\) | B1 | or \(mga(-6\sin2\theta + 4\sin\theta)\) \ |
| Positions of equilibrium occur when \(\frac{dV}{d\theta} = 0\) | M1 | |
| \(\theta = 0\) and \(\theta = \cos^{-1}\frac{1}{3}\ (=1.23\ \text{or}\ 70.5°)\) | A1A1 | Can be awarded when B1 has not been given |
| (Hence two positions) | ||
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2V}{d\theta^2} = mga(-12\cos^2\theta + 12\sin^2\theta + 4\cos\theta)\) | B1 | FT if comparable \ |
| M1 | Considering the sign of \(\frac{d^2V}{d\theta^2}\) \ | Or M2 (replacing B1M1) for another method to determine stability |
| When \(\theta=0\), \(\frac{d^2V}{d\theta^2} = -8mga < 0\) | ||
| so this position is unstable | A1 | CWO \ |
| When \(\theta = \cos^{-1}\frac{1}{3}\), \(\frac{d^2V}{d\theta^2} = mga\left(-12\times\frac{1}{9}+12\times\frac{8}{9}+4\times\frac{1}{3}\right) = \frac{32}{3}mga > 0\) | ||
| so this position is stable | A1 | CWO \ |
| [4] |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| GPE is $(-) mg(2a\cos\theta)\cos\theta$ | B1 | or $mg(a+a\cos2\theta)$ |
| | M1 | Using $\frac{\lambda x^2}{2l}$ (allow one error) |
| EPE is $\frac{2mg}{2(\frac{1}{2}a)}(2a\cos\theta - \frac{1}{2}a)^2$ | A1 | |
| $V = 2mga(4\cos^2\theta - 2\cos\theta + \frac{1}{4}) - 2mga\cos^2\theta$ | | |
| $= mga(6\cos^2\theta - 4\cos\theta + \frac{1}{2})$ | A1 | AG |
| **[4]** | | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dV}{d\theta} = mga(-12\cos\theta\sin\theta + 4\sin\theta)$ | B1 | or $mga(-6\sin2\theta + 4\sin\theta)$ \| Condone $mga$ omitted, but penalise wrong sign |
| Positions of equilibrium occur when $\frac{dV}{d\theta} = 0$ | M1 | |
| $\theta = 0$ and $\theta = \cos^{-1}\frac{1}{3}\ (=1.23\ \text{or}\ 70.5°)$ | A1A1 | Can be awarded when B1 has not been given |
| (Hence two positions) | | |
| **[4]** | | |
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2V}{d\theta^2} = mga(-12\cos^2\theta + 12\sin^2\theta + 4\cos\theta)$ | B1 | FT if comparable \| or $mga(-12\cos2\theta + 4\cos\theta)$ |
| | M1 | Considering the sign of $\frac{d^2V}{d\theta^2}$ \| Or M2 (replacing B1M1) for another method to determine stability |
| When $\theta=0$, $\frac{d^2V}{d\theta^2} = -8mga < 0$ | | |
| so this position is unstable | A1 | CWO \| *www give BOD, but M1 (or M2) must be explicitly earned* |
| When $\theta = \cos^{-1}\frac{1}{3}$, $\frac{d^2V}{d\theta^2} = mga\left(-12\times\frac{1}{9}+12\times\frac{8}{9}+4\times\frac{1}{3}\right) = \frac{32}{3}mga > 0$ | | |
| so this position is stable | A1 | CWO \| *As above* |
| **[4]** | | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-4_640_608_267_715}
A smooth wire forms a circle with centre $O$ and radius $a$, and is fixed in a vertical plane. The highest point on the wire is $A$. A small ring $R$ of mass $m$ moves along the wire. A light elastic string, with natural length $\frac { 1 } { 2 } a$ and modulus of elasticity $2 m g$, has one end attached to $A$ and the other end attached to $R$. The string $A R$ makes an angle $\theta$ (measured anticlockwise) with the downward vertical (see diagram), and you may assume that the string does not become slack.\\
(i) Taking $A$ as the reference level for gravitational potential energy, show that the total potential energy of the system is $m g a \left( 6 \cos ^ { 2 } \theta - 4 \cos \theta + \frac { 1 } { 2 } \right)$.\\
(ii) Show that there are two positions of equilibrium for which $0 \leqslant \theta < \frac { 1 } { 2 } \pi$.\\
(iii) For each of these positions of equilibrium, determine whether it is stable or unstable.
\hfill \mbox{\textit{OCR M4 2013 Q6 [12]}}