| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.2 This is a Further Maths M4 question on potential energy and equilibrium, requiring systematic application of PE formulas, differentiation, and stability analysis. While it involves multiple parts and careful geometry (string length constraint), the techniques are standard for this module: expressing PE in terms of one variable, setting dV/dθ=0 for equilibrium, and using d²V/dθ² for stability. The 'show that' part guides students through the PE derivation, and subsequent parts follow predictable patterns. Harder than typical A-level due to Further Maths content, but routine within M4. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| PE of rod \(= -kMga\sin2\theta\) | B1 | |
| \(BP = 2\times 2a\sin\theta = 4a\sin\theta\) | M1 A1 | |
| PE of mass \(= -Mg(6a-4a\sin\theta)\) | M1 | |
| \(V = -Mg(6a-4a\sin\theta)-kMga\sin2\theta\) | ||
| \(= Mga(4\sin\theta - k\sin2\theta)+\text{constant}\) * | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = Mga(4\cos\theta - 2k\cos2\theta)\) | M1 A1 | |
| \(4\times\frac{3}{4} - 2k\!\left(2\!\left(\frac{3}{4}\right)^2-1\right)=0\) | M1 M1 | |
| \(\Rightarrow k=12\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(4\cos\theta - 24(2\cos^2\theta-1)=0\) | M1 | |
| \(12\cos^2\theta - \cos\theta - 6 = 0\) | DM1 | |
| \((4\cos\theta-3)(3\cos\theta+2)=0\) | ||
| \(\cos\theta = -\dfrac{2}{3}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)(-4\sin\theta+4k\sin2\theta)\) | M1 A1 | |
| When \(\cos\theta=\frac{3}{4}\): \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)\times 44.97.. \Rightarrow\) stable | M1 A1 | |
| When \(\cos\theta=-\frac{2}{3}\): \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)\times -50.68.. \Rightarrow\) unstable | A1 | (5), Total: 18 |
## Question 7(a):
| Working/Answer | Marks | Notes |
|---|---|---|
| PE of rod $= -kMga\sin2\theta$ | B1 | |
| $BP = 2\times 2a\sin\theta = 4a\sin\theta$ | M1 A1 | |
| PE of mass $= -Mg(6a-4a\sin\theta)$ | M1 | |
| $V = -Mg(6a-4a\sin\theta)-kMga\sin2\theta$ | | |
| $= Mga(4\sin\theta - k\sin2\theta)+\text{constant}$ * | A1 | **(5)** |
## Question 7(b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = Mga(4\cos\theta - 2k\cos2\theta)$ | M1 A1 | |
| $4\times\frac{3}{4} - 2k\!\left(2\!\left(\frac{3}{4}\right)^2-1\right)=0$ | M1 M1 | |
| $\Rightarrow k=12$ | A1 | **(5)** |
## Question 7(c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $4\cos\theta - 24(2\cos^2\theta-1)=0$ | M1 | |
| $12\cos^2\theta - \cos\theta - 6 = 0$ | DM1 | |
| $(4\cos\theta-3)(3\cos\theta+2)=0$ | | |
| $\cos\theta = -\dfrac{2}{3}$ | A1 | **(3)** |
## Question 7(d):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)(-4\sin\theta+4k\sin2\theta)$ | M1 A1 | |
| When $\cos\theta=\frac{3}{4}$: $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)\times 44.97.. \Rightarrow$ stable | M1 A1 | |
| When $\cos\theta=-\frac{2}{3}$: $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = (Mga)\times -50.68.. \Rightarrow$ unstable | A1 | **(5), Total: 18** |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{376d12ab-022c-4070-a1e0-88eacc2fe48e-5_917_814_303_587}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform rod $A B$, of length $2 a$ and mass $k M$ where $k$ is a constant, is free to rotate in a vertical plane about the fixed point $A$. One end of a light inextensible string of length $6 a$ is attached to the end $B$ of the rod and passes over a small smooth pulley which is fixed at the point $P$. The line $A P$ is horizontal and of length $2 a$. The other end of the string is attached to a particle of mass $M$ which hangs vertically below the point $P$, as shown in Figure 3. The angle $P A B$ is $2 \theta$, where $0 ^ { \circ } \leq \theta \leq 180 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system is
$$M g a ( 4 \sin \theta - k \sin 2 \theta ) + \text { constant. }$$
The system has a position of equilibrium when $\cos \theta = \frac { 3 } { 4 }$.
\item Find the value of $k$.
\item Hence find the value of $\cos \theta$ at the other position of equilibrium.
\item Determine the stability of each of the two positions of equilibrium.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2008 Q7 [18]}}