| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Work-energy with multiple stages |
| Difficulty | Standard +0.3 This is a standard M2 elastic strings question with routine applications of Hooke's law and energy conservation. Part (a) is a bookwork proof, part (b)(i) is straightforward equilibrium, and parts (ii)-(iii) use standard energy methods with clear setups. The multi-part structure and energy conservation make it slightly above trivial, but all techniques are standard textbook exercises requiring no novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Tension at extension \(x\): \(T = \frac{\lambda x}{l}\) | B1 | |
| Work done \(= \int_0^e \frac{\lambda x}{l}\, dx\) | M1 | Integrating tension |
| \(= \left[\frac{\lambda x^2}{2l}\right]_0^e = \frac{\lambda e^2}{2l}\) | A1 | Completion, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At equilibrium: \(T = (4+3)g = 7g\) | M1 | Resolving for whole system |
| \(\frac{196 \times e}{2} = 7g\) | A1 | Correct equation |
| \(e = \frac{7g}{98} = 0.7\) m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using energy conservation from initial position (extension 0.7m) to rest | M1 | |
| Let new extension be \(x\). Change in EPE + change in GPE = 0 | M1 | Setting up energy equation |
| \(\frac{196x^2}{4} - \frac{196(0.7)^2}{4} = 4g(x - 0.7)\) | A1 A1 | |
| \(49x^2 - 24.01 = 39.2x - 27.44\) | DM1 | Solving |
| \(49x^2 - 39.2x + 3.43 = 0\) | A1 | \(x = 0.1\) m or \(x = 0.7\) m, so extension \(= 0.1\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Maximum speed when acceleration = 0, i.e. when \(T = 4g\) | M1 A1 | |
| \(\frac{196e}{2} = 4g \Rightarrow e = \frac{4g}{98} = 0.4\) m | A1 | cao |
# Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Tension at extension $x$: $T = \frac{\lambda x}{l}$ | B1 | |
| Work done $= \int_0^e \frac{\lambda x}{l}\, dx$ | M1 | Integrating tension |
| $= \left[\frac{\lambda x^2}{2l}\right]_0^e = \frac{\lambda e^2}{2l}$ | A1 | Completion, no errors |
# Question 7(b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At equilibrium: $T = (4+3)g = 7g$ | M1 | Resolving for whole system |
| $\frac{196 \times e}{2} = 7g$ | A1 | Correct equation |
| $e = \frac{7g}{98} = 0.7$ m | A1 | cao |
# Question 7(b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using energy conservation from initial position (extension 0.7m) to rest | M1 | |
| Let new extension be $x$. Change in EPE + change in GPE = 0 | M1 | Setting up energy equation |
| $\frac{196x^2}{4} - \frac{196(0.7)^2}{4} = 4g(x - 0.7)$ | A1 A1 | |
| $49x^2 - 24.01 = 39.2x - 27.44$ | DM1 | Solving |
| $49x^2 - 39.2x + 3.43 = 0$ | A1 | $x = 0.1$ m or $x = 0.7$ m, so extension $= 0.1$ m |
# Question 7(b)(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Maximum speed when acceleration = 0, i.e. when $T = 4g$ | M1 A1 | |
| $\frac{196e}{2} = 4g \Rightarrow e = \frac{4g}{98} = 0.4$ m | A1 | cao |7
\begin{enumerate}[label=(\alph*)]
\item An elastic string has natural length $l$ and modulus of elasticity $\lambda$. The string is stretched from length $l$ to length $l + e$.
Show, by integration, that the work done in stretching the string is $\frac { \lambda e ^ { 2 } } { 2 l }$.
\item A block, of mass 4 kg , is attached to one end of a light elastic string. The string has natural length 2 m and modulus of elasticity 196 N . The other end of the string is attached to a fixed point $O$.
\begin{enumerate}[label=(\roman*)]
\item A second block, of mass 3 kg , is attached to the 4 kg block and the system hangs in equilibrium, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{9d039ec3-fd0a-40ae-9afe-7627439081df-16_374_291_890_877}
Find the extension in the string.
\item The block of mass 3 kg becomes detached from the 4 kg block and falls to the ground. The 4 kg block now begins to move vertically upwards.
Find the extension of the string when the 4 kg block is next at rest.
\item Find the extension of the string when the speed of the 4 kg block is a maximum.\\
(3 marks)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-18_2486_1714_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-19_2486_1714_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q7 [15]}}