| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle in circular motion with string/rod |
| Difficulty | Challenging +1.8 This is a challenging M4 question requiring elastic potential energy formulation with trigonometry, energy minimization for equilibrium, and second derivative test for stability. The geometric setup is complex (semicircular wire with string to external point), requiring careful coordinate work and algebraic manipulation to reach the given expression. However, it's a standard energy methods question with clear structure and guided parts, making it accessible to well-prepared M4 students despite the technical demands. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a): Take O as zero p.e. | B1 | Reference point |
| Mechanical potential energy: \(mgh = -mga\cos\theta\) | B1 | Gravitational PE |
| Elastic potential energy: \(\frac{\lambda x^2}{2l} = \frac{1}{2} \times \frac{4mg}{5a}(2a\cos 2\theta - \frac{5}{4}a)^2\) | M1 | Elastic PE formula with extension |
| Total p.e.: \(= -2mga\cos^2\theta + mga + \frac{8mg}{5a}\left(\frac{8a\cos\theta - 5a}{4}\right)^2\) | M1 | Combination of PE terms |
| \(= -2mga\cos^2\theta + mga + \frac{mga}{10}(8\cos\theta - 5)^2\) | A1 | Simplified form |
| \(= \frac{mga}{10}(8\cos\theta - 5)^2 - 2mga\cos^2\theta + c\) | A1 | Final expression with constant |
| Part (b): Equilibrium when \(\frac{dE}{d\theta} = 0\) | B1 | Condition for equilibrium |
| \(\frac{dE}{d\theta} = \frac{mga}{10} \times 16 \times (8\cos\theta - 5)(-\sin\theta) + 4mga\cos\theta\sin\theta = 0\) | M1 | Differentiation |
| \(mga\sin\theta\left(8 - \frac{64}{5}\cos\theta\right) = 0\) | M1 | Factorization |
| \(\sin\theta = 0, \theta = 0°\) or \(\cos\theta = \frac{10}{11}, \theta = 24.6°\) | A1 A1 | Two equilibrium positions |
| Part (c): \(\frac{d^2E}{d\theta^2} = mga\sin\theta\left(\frac{64}{5}\sin\theta\right) + mga\cos\theta\left(8 - \frac{44}{5}\cos\theta\right)\) | M1 | Second derivative |
| When \(\theta = 0°\): \(\frac{d^2E}{d\theta^2} = mga\left(8 - \frac{44}{5}\right) = -\frac{4}{5}mga < 0\) so max E, unstable | M1 A1 | Check stability at \(\theta = 0°\) |
| When \(\theta = 24.6°\): \(\frac{d^2E}{d\theta^2} = mga\left(\frac{64}{5}\left(1 - \left(\frac{10}{11}\right)^2\right) + \frac{10}{11}\left(8 - \frac{44}{5} \times \frac{10}{11}\right)\right) = \frac{83}{55}mga > 0\) so min E, stable | M1 A1 | Check stability at \(\theta = 24.6°\) |
**Part (a):** Take O as zero p.e. | B1 | Reference point |
Mechanical potential energy: $mgh = -mga\cos\theta$ | B1 | Gravitational PE |
Elastic potential energy: $\frac{\lambda x^2}{2l} = \frac{1}{2} \times \frac{4mg}{5a}(2a\cos 2\theta - \frac{5}{4}a)^2$ | M1 | Elastic PE formula with extension |
Total p.e.: $= -2mga\cos^2\theta + mga + \frac{8mg}{5a}\left(\frac{8a\cos\theta - 5a}{4}\right)^2$ | M1 | Combination of PE terms |
$= -2mga\cos^2\theta + mga + \frac{mga}{10}(8\cos\theta - 5)^2$ | A1 | Simplified form |
$= \frac{mga}{10}(8\cos\theta - 5)^2 - 2mga\cos^2\theta + c$ | A1 | Final expression with constant |
**Part (b):** Equilibrium when $\frac{dE}{d\theta} = 0$ | B1 | Condition for equilibrium |
$\frac{dE}{d\theta} = \frac{mga}{10} \times 16 \times (8\cos\theta - 5)(-\sin\theta) + 4mga\cos\theta\sin\theta = 0$ | M1 | Differentiation |
$mga\sin\theta\left(8 - \frac{64}{5}\cos\theta\right) = 0$ | M1 | Factorization |
$\sin\theta = 0, \theta = 0°$ or $\cos\theta = \frac{10}{11}, \theta = 24.6°$ | A1 A1 | Two equilibrium positions |
**Part (c):** $\frac{d^2E}{d\theta^2} = mga\sin\theta\left(\frac{64}{5}\sin\theta\right) + mga\cos\theta\left(8 - \frac{44}{5}\cos\theta\right)$ | M1 | Second derivative |
When $\theta = 0°$: $\frac{d^2E}{d\theta^2} = mga\left(8 - \frac{44}{5}\right) = -\frac{4}{5}mga < 0$ so max E, unstable | M1 A1 | Check stability at $\theta = 0°$ |
When $\theta = 24.6°$: $\frac{d^2E}{d\theta^2} = mga\left(\frac{64}{5}\left(1 - \left(\frac{10}{11}\right)^2\right) + \frac{10}{11}\left(8 - \frac{44}{5} \times \frac{10}{11}\right)\right) = \frac{83}{55}mga > 0$ so min E, stable | M1 A1 | Check stability at $\theta = 24.6°$ |
---\includegraphics{figure_1}
A smooth wire $PMQ$ is in the shape of a semicircle with centre $O$ and radius $a$. The wire is fixed in a vertical plane with $PQ$ horizontal and the mid-point $M$ of the wire vertically below $O$. A smooth bead $B$ of mass $m$ is threaded on the wire and is attached to one end of a light elastic string. The string has modulus of elasticity $4mg$ and natural length $\frac{3}{4}a$. The other end of the string is attached to a fixed point $P$ which is a distance $a$ vertically above $O$, as shown in Fig. 1.
\begin{enumerate}[label=(\alph*)]
\item Show that, when $\angle BFO = \theta$, the potential energy of the system is
$$\frac{1}{16}mga(8 \cos \theta - 5)^2 - 2mga \cos^2\theta + \text{constant}.$$ [6]
\item Hence find the values of $\theta$ for which the system is in equilibrium. [6]
\item Determine the nature of the equilibrium at each of these positions. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q6 [17]}}