| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.3 This is a standard M2 elastic strings problem with clear structure: (a) direct EPE formula application, (b) energy conservation with given answer to verify, (c) checking if particle reaches point A. All parts use routine techniques (Hooke's law, energy methods) with straightforward setup on an inclined plane. The 'show that' format reduces difficulty further by confirming the working direction. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{300 \times (1.5)^2}{2 \times 4} = 84.375 = 84.4\) J | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 3.66\) | M1 A1 m1 A1 A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore\) particle will not reach A | B1 B1 E1 | 3 marks |
**(a)** EPE = $\frac{\lambda x^2}{2l}$
$= \frac{300 \times (1.5)^2}{2 \times 4} = 84.375 = 84.4$ J | M1 A1 | 2 marks
**(b)** When string is slack, gain in PE is $mgh = 6 \times g \times 1.5\sin 30 = 44.1$ J
KE = EPE − gain in PE = $84.375 - 44.1 = 40.275$
$\frac{1}{2} \cdot 6 \cdot v^2 = 40.275$
$v = 3.66$ | M1 A1 m1 A1 A1 | 5 marks | AG
**(c)** At A, PE gained above initial position is $6 \times g \times 5.5 \sin 30 = 161.7$ J
This is more than initial elastic potential energy
$\therefore$ particle will not reach A | B1 B1 E1 | 3 marks | Or PE above position string slack is 117.6; KE at A is −77.3; Or Using $v^2 = u^2 + 2as$ with $a = 0.5g$ and $s = 1.37$ or 1.366; Hence stops before A; Vertical height above sling slack is 0.683; Vertical height above starting point is 1.435
**Total: 10 marks**
---6 A light elastic string has one end attached to a point $A$ fixed on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The other end of the string is attached to a particle of mass 6 kg . The elastic string has natural length 4 metres and modulus of elasticity 300 newtons.
The particle is pulled down the plane in the direction of the line of greatest slope through $A$. The particle is released from rest when it is 5.5 metres from $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{1bc18163-b20e-4dc6-bd35-496efec8dc73-4_314_713_1900_660}
\begin{enumerate}[label=(\alph*)]
\item Calculate the elastic potential energy of the string when the particle is 5.5 metres from the point $A$.
\item Show that the speed of the particle when the string becomes slack is $3.66 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to three significant figures.
\item Show that the particle will not reach point $A$ in the subsequent motion.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q6 [10]}}