| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Simple harmonic motion with elastic string |
| Difficulty | Challenging +1.2 This is a multi-part M3 question combining elastic strings, equilibrium, SHM, and energy methods. Part (a) is routine equilibrium using Hooke's law. Parts (b)-(c) require standard SHM identification and period calculation. Part (d) tests conceptual understanding of when the string becomes slack. Part (e) requires energy conservation with careful handling of the elastic potential energy formula. While lengthy (5 parts), each component uses standard M3 techniques without requiring novel insight—slightly above average difficulty due to the extended nature and need for careful bookkeeping across multiple scenarios. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
**Part (a)**
M1 for using Hooke's Law, resolving vertically.
$$T = \frac{9mg}{6a}pa - mg$$
A1 cso for $p = \frac{2}{3}$
**Part (b)**
M1 for an equation of motion vertically. Must have a tension, a weight and a mass $\times$ acceleration. Allow with $a$ for acceleration. Must be dimensionally correct, but allow for misuse of brackets.
$$T - mg = m\ddot{x}$$
A1 for a correct equation, can still have $a$
M1 dep for rearranging to the form $\ddot{x} = -\omega^2 x$. Acceleration as $a$ scores M0.
$$\ddot{x} = -\frac{3g}{2a}x$$
A1 for a correct equation and a conclusion eg $\equiv$ SHM. Accept "shown".
**Part (c)**
M1 for using period $T = \frac{2\pi}{\omega}$ with their $\omega$ to obtain the period.
$$T = 2\pi\sqrt{\frac{2a}{3g}}$$
A1 ft for $2\pi\sqrt{\frac{2a}{3g}}$
**Part (d)**
B1 for any statement equivalent to those shown. The string never becomes slack or the SHM is complete.
**Part (e)**
B1 for the EPE lost or initial EPE. Need not be simplified.
$$\text{Loss of EPE} = 3mga - \frac{9mg(6. A particle of mass $m$ is attached to one end of a light elastic string, of natural length $6 a$ and modulus of elasticity 9 mg . The other end of the string is attached to a fixed point $A$ on a ceiling. The particle hangs in equilibrium at the point $B$, where $B$ is vertically below $A$ and $A B = ( 6 + p ) a$.
\begin{enumerate}[label=(\alph*)]
\item Show that $p = \frac { 2 } { 3 }$
The particle is now released from rest at a point $C$ vertically below $B$, where $A C < \frac { 22 } { 3 } a$.
\item Show that the particle moves with simple harmonic motion.
\item Find the period of this motion.
\item Explain briefly the significance of the condition $A C < \frac { 22 } { 3 } a$.
The point $D$ is vertically below $A$ and $A D = 8 a$. The particle is now released from rest at $D$. The particle first comes to instantaneous rest at the point $E$.
\item Find, in terms of $a$, the distance $A E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q6 [13]}}