| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with elastic strings/springs |
| Difficulty | Challenging +1.8 This M4 question requires setting up potential energy expressions for both gravitational PE (involving centers of mass of four rods in a geometric configuration) and elastic PE (using Hooke's law with extension), then applying calculus to find equilibrium and test stability. While systematic, it demands careful geometric reasoning, energy formulation for multiple components, and second derivative analysis—significantly beyond routine mechanics but follows established M4 patterns. |
| Spec | 6.02d 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(AC = 4a\cos\theta \Rightarrow\) extension \(= 4a\cos\theta - 2a\) | B1 | |
| \(V = -2mga\cos\theta - 2mg\times 3a\cos\theta + \frac{4mg}{4a}(4a\cos\theta-2a)(+C)\) | M1 A1 A1 | |
| \(= -8mga\cos\theta + 4mg(2\cos\theta-1)^2(+C)\) | ||
| \(= 4mga[(2\cos\theta-1)^2 - 2\cos\theta](+C)\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{dV}{d\theta} = 4mga[2(2\cos\theta-1)(-2\sin\theta)+2\sin\theta]\) | M1 A1 | |
| \(= 8mga\sin\theta(3-4\cos\theta)\) | ||
| \(= 0 \Rightarrow \cos\theta = \tfrac{3}{4}\;(\theta\neq 0,\pi)\) | M1 A1 | |
| \(\Rightarrow \theta = 0.723\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{d^2V}{d\theta^2} = 8mga\cos\theta(3-4\cos\theta)+32mga\sin^2\theta\) | M1 A1 | |
| As \(\theta=\tfrac{3}{4} \Rightarrow \frac{d^2V}{d\theta^2} = 0 + 32mga\times\tfrac{7}{16}\) | M1 | |
| \(= 14mga\) | ||
| \(> 0 \Rightarrow\) stable | A1 | (4) |
# Question 4:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $AC = 4a\cos\theta \Rightarrow$ extension $= 4a\cos\theta - 2a$ | B1 | |
| $V = -2mga\cos\theta - 2mg\times 3a\cos\theta + \frac{4mg}{4a}(4a\cos\theta-2a)(+C)$ | M1 A1 A1 | |
| $= -8mga\cos\theta + 4mg(2\cos\theta-1)^2(+C)$ | | |
| $= 4mga[(2\cos\theta-1)^2 - 2\cos\theta](+C)$ | A1 | (5) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{dV}{d\theta} = 4mga[2(2\cos\theta-1)(-2\sin\theta)+2\sin\theta]$ | M1 A1 | |
| $= 8mga\sin\theta(3-4\cos\theta)$ | | |
| $= 0 \Rightarrow \cos\theta = \tfrac{3}{4}\;(\theta\neq 0,\pi)$ | M1 A1 | |
| $\Rightarrow \theta = 0.723$ | A1 | (4) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{d^2V}{d\theta^2} = 8mga\cos\theta(3-4\cos\theta)+32mga\sin^2\theta$ | M1 A1 | |
| As $\theta=\tfrac{3}{4} \Rightarrow \frac{d^2V}{d\theta^2} = 0 + 32mga\times\tfrac{7}{16}$ | M1 | |
| $= 14mga$ | | |
| $> 0 \Rightarrow$ stable | A1 | (4) |
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{c68c85a1-9d80-4ced-bfb6-c7b5347e9bb8-3_424_422_1181_844}
\end{center}
\end{figure}
Four identical uniform rods, each of mass $m$ and length $2 a$, are freely jointed to form a rhombus $A B C D$. The rhombus is suspended from $A$ and is prevented from collapsing by an elastic string which joins $A$ to $C$, with $\angle B A D = 2 \theta , 0 \leq \theta \leq \frac { 1 } { 3 } \pi$, as shown in Fig. 2. The natural length of the elastic string is $2 a$ and its modulus of elasticity is $4 m g$.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy, $V$, of the system is given by
$$V = 4 m g a \left[ ( 2 \cos \theta - 1 ) ^ { 2 } - 2 \cos \theta \right] + \text { constant } .$$
\item Hence find the non-zero value of $\theta$ for which the system is in equilibrium.
\item Determine whether this position of equilibrium is stable or unstable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2002 Q4 [13]}}