Edexcel M4 2003 June — Question 4 15 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2003
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic potential energy calculations
DifficultyChallenging +1.2 This is a standard M4 elastic string energy problem requiring systematic application of PE formulas (gravitational + elastic) and differentiation for equilibrium. While it involves multiple components and careful geometry (finding extension from coordinates), the method is routine for this module: express total PE, differentiate and set to zero, then use second derivative test. The algebra is moderately involved but follows a predictable template for this topic.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.04e Rigid body equilibrium: coplanar forces
4. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{47e1d96b-4582-4324-a946-66989a2c66fc-3_581_826_801_648}
\end{figure} A uniform rod \(A B\), of length \(2 a\) and mass \(8 m\), is free to rotate in a vertical plane about a fixed smooth horizontal axis through \(A\). One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(\frac { 4 } { 5 } \mathrm { mg }\), is fixed to \(B\). The other end of the string is attached to a small ring which is free to slide on a smooth straight horizontal wire which is fixed in the same vertical plane as \(A B\) at a height 7a vertically above \(A\). The rod \(A B\) makes an angle \(\theta\) with the upward vertical at \(A\), as shown in Fig. 2.
  1. Show that the potential energy \(V\) of the system is given by $$V = \frac { 8 } { 5 } m g a \left( \cos ^ { 2 } \theta - \cos \theta \right) + \text { constant. }$$
  2. Hence find the values of \(\theta , 0 \leq \theta \leq \pi\), for which the system is in equilibrium.
  3. Determine the nature of these positions of equilibrium.
Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Extension of string \(= 7a - 2a\cos\theta - a = 2a(3-\cos\theta)\)B1
\(\text{PE} = 8mga\cos\theta + \frac{4mg}{5}\times\frac{4a^2}{2a}(3-\cos\theta)^2\)B1, M1 A1
\(= 8mga\cos\theta + \frac{8mga}{5}(9 - 6\cos\theta - \cos^2\theta)\)M1
\(= \frac{8mga}{5}(\cos^2\theta - \cos\theta) + C\)A1 (6) (*)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dV}{d\theta} = \frac{8mga}{5}(-2\cos\theta\sin\theta + \sin\theta)\)M1 A1
\(= 0\)M1
\(\Rightarrow \sin\theta = 0\) or \(\cos\theta = \frac{1}{2}\)
\(\Rightarrow \theta = 0\) or \(\pi\), or \(\theta = \frac{\pi}{3}\)A1, A1 (5)
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d^2V}{d\theta^2} = \frac{8mga}{5}(\cos\theta + 2\sin^2\theta - 2\cos^2\theta)\)M1 A1
\(\theta = 0\): \(V'' < 0\ \left(= -\frac{8mga}{5}\right)\) unstable
\(\theta = \pi\): \(V'' < 0\ \left(= -3\times\frac{8mga}{5}\right)\) unstableA1
\(\theta = \frac{\pi}{3}\): \(V'' > 0\ \left(= 3\times\frac{8mga}{5}\right)\) stableA1 (4) 
# Question 4:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Extension of string $= 7a - 2a\cos\theta - a = 2a(3-\cos\theta)$ | B1 | |
| $\text{PE} = 8mga\cos\theta + \frac{4mg}{5}\times\frac{4a^2}{2a}(3-\cos\theta)^2$ | B1, M1 A1 | |
| $= 8mga\cos\theta + \frac{8mga}{5}(9 - 6\cos\theta - \cos^2\theta)$ | M1 | |
| $= \frac{8mga}{5}(\cos^2\theta - \cos\theta) + C$ | A1 | (6) (*) |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{d\theta} = \frac{8mga}{5}(-2\cos\theta\sin\theta + \sin\theta)$ | M1 A1 | |
| $= 0$ | M1 | |
| $\Rightarrow \sin\theta = 0$ or $\cos\theta = \frac{1}{2}$ | | |
| $\Rightarrow \theta = 0$ or $\pi$, or $\theta = \frac{\pi}{3}$ | A1, A1 | (5) |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2V}{d\theta^2} = \frac{8mga}{5}(\cos\theta + 2\sin^2\theta - 2\cos^2\theta)$ | M1 A1 | |
| $\theta = 0$: $V'' < 0\ \left(= -\frac{8mga}{5}\right)$ unstable | | |
| $\theta = \pi$: $V'' < 0\ \left(= -3\times\frac{8mga}{5}\right)$ unstable | A1 | |
| $\theta = \frac{\pi}{3}$: $V'' > 0\ \left(= 3\times\frac{8mga}{5}\right)$ stable | A1 | (4) |

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4.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
  \includegraphics[alt={},max width=\textwidth]{47e1d96b-4582-4324-a946-66989a2c66fc-3_581_826_801_648}
\end{center}
\end{figure}

A uniform rod $A B$, of length $2 a$ and mass $8 m$, is free to rotate in a vertical plane about a fixed smooth horizontal axis through $A$. One end of a light elastic string, of natural length $a$ and modulus of elasticity $\frac { 4 } { 5 } \mathrm { mg }$, is fixed to $B$. The other end of the string is attached to a small ring which is free to slide on a smooth straight horizontal wire which is fixed in the same vertical plane as $A B$ at a height 7a vertically above $A$. The rod $A B$ makes an angle $\theta$ with the upward vertical at $A$, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy $V$ of the system is given by

$$V = \frac { 8 } { 5 } m g a \left( \cos ^ { 2 } \theta - \cos \theta \right) + \text { constant. }$$
\item Hence find the values of $\theta , 0 \leq \theta \leq \pi$, for which the system is in equilibrium.
\item Determine the nature of these positions of equilibrium.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M4 2003 Q4 [15]}}
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