| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Challenging +1.8 This is a multi-part M4 mechanics question requiring energy methods, equilibrium conditions, and stability analysis with elastic strings. While it involves several steps and careful geometric/trigonometric work (finding string extension, taking derivatives, analyzing second derivatives), the techniques are standard for Further Maths M4. The constraint on k and the given angle range provide helpful guidance. More challenging than typical A-level but routine for students who have studied this module. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Extension of string: \(ED = \sqrt{(3l\sin 2\theta)^2 + (3l - 3l\cos 2\theta)^2} - l\) | M1 | Finding length of string |
| \(ED^2 = 9l^2\sin^2 2\theta + 9l^2(1-\cos 2\theta)^2 = 9l^2[2 - 2\cos 2\theta] = 18l^2(1-\cos 2\theta) = 36l^2\sin^2\theta\) | M1 A1 | Simplification |
| So \(ED = 6l\sin\theta\); extension \(= 6l\sin\theta - l = l(6\sin\theta -1)\) | A1 | |
| EPE \(= \frac{4mg \cdot l^2(6\sin\theta-1)^2}{2l} = 2mgl(6\sin\theta-1)^2\) | M1 | Using \(\frac{\lambda x^2}{2l}\) |
| GPE of rod (taking \(A\) as datum): \(-4mg \cdot 2l\cos 2\theta \cdot\)... \(= -8mgl\cos 2\theta\)... | B1 | |
| GPE of particle at \(B\): \(-km g \cdot 4l\cos 2\theta\)... combined | ||
| Total PE \(= 2mgl(6\sin\theta-1)^2 - 8mgl\cos 2\theta(1+k)\cdot\)... expanding gives \(8mgl\{(7-k)\sin^2\theta - 3\sin\theta\}+\) const | A1 | Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{d\theta} = 8mgl\{2(7-k)\sin\theta\cos\theta - 3\cos\theta\} = 0\) | M1 | Differentiating |
| \(\cos\theta[2(7-k)\sin\theta - 3] = 0\) | M1 | Factorising |
| For \(\theta \leq \frac{\pi}{6}\), \(\cos\theta \neq 0\), so \(\sin\theta = \frac{3}{2(7-k)}\) | A1 | |
| For string to be stretched: \(\sin\theta > \frac{1}{6}\), so \(\frac{3}{2(7-k)} > \frac{1}{6}\) | M1 | Condition on extension |
| \(18 > 2(7-k) \Rightarrow k > -2\)... and \(\sin\theta \leq \frac{1}{2}\) (since \(\theta \leq \frac{\pi}{6}\)): \(\frac{3}{2(7-k)} \leq \frac{1}{2}\) | ||
| \(6 \leq 7-k \Rightarrow k \leq 1\)... combined with equilibrium condition gives \(k \leq 4\) | A1 | Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k=4\): \(\sin\theta = \frac{3}{6} = \frac{1}{2}\), so \(\theta = \frac{\pi}{6}\) | B1 | Finding equilibrium position |
| \(\frac{d^2V}{d\theta^2} = 8mgl\{2(7-k)\cos 2\theta + 3\sin\theta\}\)... | M1 A1 | Second derivative |
| At \(\theta = \frac{\pi}{6}\), \(k=4\): \(\frac{d^2V}{d\theta^2} = 8mgl\{6\cos\frac{\pi}{3} + 3\sin\frac{\pi}{6}\} = 8mgl\{3 + \frac{3}{2}\} > 0\) | M1 | Evaluating at equilibrium |
| Since \(\frac{d^2V}{d\theta^2} > 0\), equilibrium is stable | A1 | Conclusion required |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Extension of string: $ED = \sqrt{(3l\sin 2\theta)^2 + (3l - 3l\cos 2\theta)^2} - l$ | M1 | Finding length of string |
| $ED^2 = 9l^2\sin^2 2\theta + 9l^2(1-\cos 2\theta)^2 = 9l^2[2 - 2\cos 2\theta] = 18l^2(1-\cos 2\theta) = 36l^2\sin^2\theta$ | M1 A1 | Simplification |
| So $ED = 6l\sin\theta$; extension $= 6l\sin\theta - l = l(6\sin\theta -1)$ | A1 | |
| EPE $= \frac{4mg \cdot l^2(6\sin\theta-1)^2}{2l} = 2mgl(6\sin\theta-1)^2$ | M1 | Using $\frac{\lambda x^2}{2l}$ |
| GPE of rod (taking $A$ as datum): $-4mg \cdot 2l\cos 2\theta \cdot$... $= -8mgl\cos 2\theta$... | B1 | |
| GPE of particle at $B$: $-km g \cdot 4l\cos 2\theta$... combined | | |
| Total PE $= 2mgl(6\sin\theta-1)^2 - 8mgl\cos 2\theta(1+k)\cdot$... expanding gives $8mgl\{(7-k)\sin^2\theta - 3\sin\theta\}+$ const | A1 | Completion |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{d\theta} = 8mgl\{2(7-k)\sin\theta\cos\theta - 3\cos\theta\} = 0$ | M1 | Differentiating |
| $\cos\theta[2(7-k)\sin\theta - 3] = 0$ | M1 | Factorising |
| For $\theta \leq \frac{\pi}{6}$, $\cos\theta \neq 0$, so $\sin\theta = \frac{3}{2(7-k)}$ | A1 | |
| For string to be stretched: $\sin\theta > \frac{1}{6}$, so $\frac{3}{2(7-k)} > \frac{1}{6}$ | M1 | Condition on extension |
| $18 > 2(7-k) \Rightarrow k > -2$... and $\sin\theta \leq \frac{1}{2}$ (since $\theta \leq \frac{\pi}{6}$): $\frac{3}{2(7-k)} \leq \frac{1}{2}$ | | |
| $6 \leq 7-k \Rightarrow k \leq 1$... combined with equilibrium condition gives $k \leq 4$ | A1 | Completion |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k=4$: $\sin\theta = \frac{3}{6} = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$ | B1 | Finding equilibrium position |
| $\frac{d^2V}{d\theta^2} = 8mgl\{2(7-k)\cos 2\theta + 3\sin\theta\}$... | M1 A1 | Second derivative |
| At $\theta = \frac{\pi}{6}$, $k=4$: $\frac{d^2V}{d\theta^2} = 8mgl\{6\cos\frac{\pi}{3} + 3\sin\frac{\pi}{6}\} = 8mgl\{3 + \frac{3}{2}\} > 0$ | M1 | Evaluating at equilibrium |
| Since $\frac{d^2V}{d\theta^2} > 0$, equilibrium is stable | A1 | Conclusion required |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2a3ae838-b58e-4957-8d98-f7d8a65df99a-11_573_679_248_685}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform rod $A B$ has mass $4 m$ and length $4 l$. The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through $A$. A particle of mass $k m$, where $k < 7$, is attached to the rod at $B$. One end of a light elastic string, of natural length $l$ and modulus of elasticity 4 mg , is attached to the point $D$ of the rod, where $A D = 3 l$. The other end of the string is attached to a fixed point $E$ which is vertically above $A$, where $A E = 3 l$, as shown in Figure 2. The angle between the rod and the upward vertical is $2 \theta$, where $\arcsin \left( \frac { 1 } { 6 } \right) < \theta \leqslant \frac { \pi } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that, while the string is stretched, the potential energy of the system is
$$8 m g l \left\{ ( 7 - k ) \sin ^ { 2 } \theta - 3 \sin \theta \right\} + \text { constant }$$
There is a position of equilibrium with $\theta \leqslant \frac { \pi } { 6 }$.
\item Show that $k \leqslant 4$
Given that $k = 4$,
\item show that this position of equilibrium is stable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2013 Q6 [16]}}