| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.8 This is a challenging M4 question requiring multiple techniques: deriving potential energy expressions involving geometry (string length via Pythagoras), calculus (differentiation and stability analysis), algebraic manipulation to prove given results, and interpreting equilibrium conditions. The multi-part structure with proof elements and the need to analyze stability using second derivatives and Taylor approximations places it well above average difficulty, though the question provides helpful scaffolding and given results. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02g Hooke's law: T = k*x or T = lambda*x/l6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BC^2 = (6a)^2 + (2a)^2 - 2(6a)(2a)\cos\theta\) | B1 | \(BC = 2a\sqrt{10-6\cos\theta}\) |
| \(V = \lambda mg\left(BC - \sqrt{40}a\right) + \ldots\) | M1 | GPE for \(\lambda mg\) particle – accept \(\lambda mg(6a-(l-BC))\) where BC is a function of \(\theta\) |
| \(\ldots + 2mga\cos\theta\) | B1 | GPE for rod |
| M1 | Differentiate | |
| \(\frac{dV}{d\theta} = \lambda mga(10-6\cos\theta)^{-\frac{1}{2}}(6\sin\theta) + 2mga(-\sin\theta)\) | A1 | AEF |
| \(= 2mga\sin\theta\left(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1\right)\) | E1 | www |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2mga\sin\theta\left(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1\right) = 0 \Rightarrow \sin\theta = 0\) or \(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0\) | M1 | |
| \(\sin\theta = 0 \Rightarrow \theta = 0\) or \(\pi\) while any other positions are dependent on \(\lambda\) | A1 | Must include consideration of \(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0\) | M1 | |
| \(\Rightarrow \cos\theta = \frac{10-9\lambda^2}{6}\) | A1 | |
| Since \(0 < \theta < \pi\), \(-1 < \frac{10-9\lambda^2}{6} < 1\) | M1 | Must state \(0 < \theta < \pi\) or \(-1 < \cos\theta < 1\); condone \(\leq\) for the M mark only |
| \(\Rightarrow \frac{2}{3} < \lambda < \frac{4}{3}\) | E1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2V}{d\theta^2} = 2mga\cos\theta\left(3\lambda(10-6\cos\theta)^{-\frac{1}{2}} - 1\right)\) | M1* | Differentiate using product/quotient rule |
| \(+ 2mga\sin\theta\left(-\frac{3}{2}\lambda(10-6\cos\theta)^{-\frac{3}{2}}(6\sin\theta)\right)\) | A1 | |
| A1 | ||
| \(\frac{d^2V}{d\theta^2} = -2mga\left(\frac{3\lambda}{4} - 1\right) > 0 \Rightarrow\) stable | M1 dep* | Substituting \(\theta = \pi\) into their \(V''\) and attempt to simplify |
| A1 | Correct \(V''\) and conclusion | |
| \(\frac{d^2V}{d\theta^2} = -18mga\lambda\sin^2\theta(10-6\cos\theta)^{-3/2} < 0 \Rightarrow\) unstable | M1 dep* | Setting \(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1\) equal to zero in their \(V''\) |
| A1 | Correct \(V''\) and conclusion | |
| \(\frac{d^2V}{d\theta^2} = 2mga\left(\frac{3\lambda}{2} - 1\right) > 0 \Rightarrow\) stable | M1 dep* | Substituting \(\theta = 0\) into their \(V''\) and attempt to simplify |
| A1 | Correct \(V''\) and conclusion | |
| A1 | Clear justification for all three cases with particular reference to values of \(\lambda\) in the interval \(\frac{2}{3} < \lambda < \frac{4}{3}\) (dependent on all previous marks earned in this part) | |
| [10] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dV}{d\theta} = -\frac{3}{4}mga\theta^3 + \cdots\) | M1 | Substituting \(\lambda = \frac{2}{3}\) and using first derivative test only |
| When \(\theta = 0^-\), \(\frac{dV}{d\theta} > 0\) and when \(\theta = 0^+\), \(\frac{dV}{d\theta} < 0\) \(\therefore\) unstable at \(\lambda = \frac{2}{3}\) | A1 | CAO – allow accurate sketch of the form \(y = -x^3\) |
| [2] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BC^2 = (6a)^2 + (2a)^2 - 2(6a)(2a)\cos\theta$ | B1 | $BC = 2a\sqrt{10-6\cos\theta}$ |
| $V = \lambda mg\left(BC - \sqrt{40}a\right) + \ldots$ | M1 | GPE for $\lambda mg$ particle – accept $\lambda mg(6a-(l-BC))$ where BC is a function of $\theta$ |
| $\ldots + 2mga\cos\theta$ | B1 | GPE for rod |
| | M1 | Differentiate |
| $\frac{dV}{d\theta} = \lambda mga(10-6\cos\theta)^{-\frac{1}{2}}(6\sin\theta) + 2mga(-\sin\theta)$ | A1 | AEF |
| $= 2mga\sin\theta\left(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1\right)$ | E1 | www |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2mga\sin\theta\left(\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1\right) = 0 \Rightarrow \sin\theta = 0$ or $\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0$ | M1 | |
| $\sin\theta = 0 \Rightarrow \theta = 0$ or $\pi$ while any other positions are dependent on $\lambda$ | A1 | Must include consideration of $\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0$ |
| **[2]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1 = 0$ | M1 | |
| $\Rightarrow \cos\theta = \frac{10-9\lambda^2}{6}$ | A1 | |
| Since $0 < \theta < \pi$, $-1 < \frac{10-9\lambda^2}{6} < 1$ | M1 | Must state $0 < \theta < \pi$ or $-1 < \cos\theta < 1$; condone $\leq$ for the M mark only |
| $\Rightarrow \frac{2}{3} < \lambda < \frac{4}{3}$ | E1 | |
| **[4]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2V}{d\theta^2} = 2mga\cos\theta\left(3\lambda(10-6\cos\theta)^{-\frac{1}{2}} - 1\right)$ | M1* | Differentiate using product/quotient rule |
| $+ 2mga\sin\theta\left(-\frac{3}{2}\lambda(10-6\cos\theta)^{-\frac{3}{2}}(6\sin\theta)\right)$ | A1 | |
| | A1 | |
| $\frac{d^2V}{d\theta^2} = -2mga\left(\frac{3\lambda}{4} - 1\right) > 0 \Rightarrow$ stable | M1 dep* | Substituting $\theta = \pi$ into their $V''$ and attempt to simplify |
| | A1 | Correct $V''$ and conclusion |
| $\frac{d^2V}{d\theta^2} = -18mga\lambda\sin^2\theta(10-6\cos\theta)^{-3/2} < 0 \Rightarrow$ unstable | M1 dep* | Setting $\frac{3\lambda}{\sqrt{10-6\cos\theta}} - 1$ equal to zero in their $V''$ |
| | A1 | Correct $V''$ and conclusion |
| $\frac{d^2V}{d\theta^2} = 2mga\left(\frac{3\lambda}{2} - 1\right) > 0 \Rightarrow$ stable | M1 dep* | Substituting $\theta = 0$ into their $V''$ and attempt to simplify |
| | A1 | Correct $V''$ and conclusion |
| | A1 | Clear justification for all three cases with particular reference to values of $\lambda$ in the interval $\frac{2}{3} < \lambda < \frac{4}{3}$ (dependent on all previous marks earned in this part) |
| **[10]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dV}{d\theta} = -\frac{3}{4}mga\theta^3 + \cdots$ | M1 | Substituting $\lambda = \frac{2}{3}$ and using first derivative test **only** |
| When $\theta = 0^-$, $\frac{dV}{d\theta} > 0$ and when $\theta = 0^+$, $\frac{dV}{d\theta} < 0$ $\therefore$ unstable at $\lambda = \frac{2}{3}$ | A1 | CAO – allow accurate sketch of the form $y = -x^3$ |
| **[2]** | | |3 Fig. 3 shows a uniform rigid rod AB of length $2 a$ and mass $2 m$. The rod is freely hinged at A so that it can rotate in a vertical plane. One end of a light inextensible string of length $l$ is attached to B . The string passes over a small smooth fixed pulley at C , where C is vertically above A and $\mathrm { AC } = 6 a$. A particle of mass $\lambda m$, where $\lambda$ is a positive constant, is attached to the other end of the string and hangs freely, vertically below C . The rod makes an angle $\theta$ with the upward vertical, where $0 \leqslant \theta \leqslant \pi$. You may assume that the particle does not interfere with the rod AB or the section of the string BC .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3fdb2cff-0f74-4c88-b25a-759bfab1675a-3_878_615_667_717}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Find the potential energy, $V$, of the system relative to a situation in which the rod AB is horizontal, and hence show that
$$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \sin \theta \left( \frac { 3 \lambda } { \sqrt { 10 - 6 \cos \theta } } - 1 \right) .$$
(ii) Show that $\theta = 0$ and $\theta = \pi$ are the only values of $\theta$ for which the system is in equilibrium whatever the value of $\lambda$.\\
(iii) Show that, if there is a third value of $\theta$ for which the system is in equilibrium, then $\frac { 2 } { 3 } < \lambda < \frac { 4 } { 3 }$.\\
(iv) Given that there are three positions of equilibrium, establish whether each of these positions is stable or unstable.
It is given that, for small values of $\theta$,
$$\frac { \mathrm { d } V } { \mathrm {~d} \theta } \approx 2 m g a \left[ \left( \frac { 3 } { 2 } \lambda - 1 \right) \theta - \left( \frac { 13 } { 16 } \lambda - \frac { 1 } { 6 } \right) \theta ^ { 3 } \right] .$$
(v) Investigate the stability of the equilibrium position given by $\theta = 0$ in the case when $\lambda = \frac { 2 } { 3 }$.
\hfill \mbox{\textit{OCR MEI M4 2016 Q3 [24]}}