| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Standard +0.3 This is a standard M2 elastic string problem requiring routine application of Hooke's law, EPE formula, and energy conservation. Part (a) uses equilibrium conditions, (b) is direct substitution into EPE = λx²/2l, and (c) applies conservation of energy with given answer to show. All techniques are textbook exercises with clear structure and no novel insight required. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
8 A particle, of mass 10 kg , is attached to one end of a light elastic string of natural length 0.4 metres and modulus of elasticity 100 N . The other end of the string is fixed to the point $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the elastic string when the particle hangs in equilibrium directly below $O$.
\item The particle is pulled down and held at a point $P$, which is 1 metre vertically below $O$.
Show that the elastic potential energy of the string when the particle is in this position is 45 J .
\item The particle is released from rest at the point $P$. In the subsequent motion, the particle has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it is $x$ metres below $\boldsymbol { O }$.
\begin{enumerate}[label=(\roman*)]
\item Show that, while the string is taut,
$$v ^ { 2 } = 39.6 x - 25 x ^ { 2 } - 14.6$$
\item Find the value of $x$ when the particle comes to rest for the first time after being released, given that the string is still taut.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q8 [14]}}