| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic potential energy calculations |
| Difficulty | Standard +0.3 This is a standard M2 elastic strings question with routine application of Hooke's law, energy conservation, and equilibrium. Part (a) is a bookwork proof requiring simple integration. Parts (b)(i)-(iv) follow a predictable structure: find equilibrium extension, calculate EPE using the given formula, apply energy conservation to derive the given equation, then solve a quadratic. All steps are mechanical applications of standard techniques with no novel problem-solving required. Slightly easier than average due to the scaffolded nature and 'show that' parts. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Work done \(= \int_0^e \frac{\lambda x}{l}dx = \left[\frac{\lambda x^2}{2l}\right]_0^e = \frac{\lambda e^2}{2l}\) | M1 A1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(T = \frac{\lambda x}{l}\): \(5g = \frac{150 \times x}{0.6}\), Extension is 0.196 m | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| EPE \(= \frac{\lambda x^2}{2l} = \frac{150 \times (0.3)^2}{2 \times 0.6} = 11.25\) J | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x\) above P: EPE \(= \frac{150 \times (0.3 - x)^2}{2 \times 0.6}\), PE[relative to P] \(= (-5) \times g \times x\). KE + EPE [at new point] = EPE [at P] - gain in PE: \(\frac{1}{2}mv^2 + \frac{150 \times (0.3-x)^2}{2 \times 0.6} = \frac{150 \times (0.3)^2}{2 \times 0.6} - 5gx\). Equation: \(\frac{1}{2}mv^2 + \frac{150(x^2 - 0.6x)}{2 \times 0.6} = -5gx\), \(\frac{1}{2}(5.3)v^2 + 125x^2 - 75x = -49x\), \(v^2 = 10.4x - 50x^2\) | M1 A1 M1 M1 m1 A1 | 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Particle is at rest when \(v = 0\): \(10.4x - 50x^2 = 0\), \(x = 0\) [not required] or \(x = \frac{10.4}{50} = 0.208\) m above P | M1 A1 | 2 marks |
**Part (a)**
Work done $= \int_0^e \frac{\lambda x}{l}dx = \left[\frac{\lambda x^2}{2l}\right]_0^e = \frac{\lambda e^2}{2l}$ | M1 A1 A1 | 3 marks | Needs limit of 0. AG
**Part (b)(i)**
Using $T = \frac{\lambda x}{l}$: $5g = \frac{150 \times x}{0.6}$, Extension is 0.196 m | M1 A1 | 2 marks
**Part (b)(ii)**
EPE $= \frac{\lambda x^2}{2l} = \frac{150 \times (0.3)^2}{2 \times 0.6} = 11.25$ J | M1 A1 | 2 marks
**Part (b)(iii)**
When $x$ above P: EPE $= \frac{150 \times (0.3 - x)^2}{2 \times 0.6}$, PE[relative to P] $= (-5) \times g \times x$. KE + EPE [at new point] = EPE [at P] - gain in PE: $\frac{1}{2}mv^2 + \frac{150 \times (0.3-x)^2}{2 \times 0.6} = \frac{150 \times (0.3)^2}{2 \times 0.6} - 5gx$. Equation: $\frac{1}{2}mv^2 + \frac{150(x^2 - 0.6x)}{2 \times 0.6} = -5gx$, $\frac{1}{2}(5.3)v^2 + 125x^2 - 75x = -49x$, $v^2 = 10.4x - 50x^2$ | M1 A1 M1 M1 m1 A1 | 7 marks | For $\frac{150 \times (...-x)^2}{2 \times 0.6}$. For $5 \times g \times$ distance. 4 terms all signs correct, 2 terms correct. Equation involving terms in $v^2, x^2$ and $x$ only
**Part (b)(iv)**
Particle is at rest when $v = 0$: $10.4x - 50x^2 = 0$, $x = 0$ [not required] or $x = \frac{10.4}{50} = 0.208$ m above P | M1 A1 | 2 marks
**Total: 16 marks**
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## GRAND TOTAL: 75 marks8
\begin{enumerate}[label=(\alph*)]
\item Hooke's law states that the tension in a stretched string of natural length $l$ and modulus of elasticity $\lambda$ is $\frac { \lambda x } { l }$ when its extension is $x$.
Using this formula, prove that the work done in stretching a string from an unstretched position to a position in which its extension is $e$ is $\frac { \lambda e ^ { 2 } } { 2 l }$.\\
(3 marks)
\item A particle, of mass 5 kg , is attached to one end of a light elastic string of natural length 0.6 metres and modulus of elasticity 150 N . The other end of the string is fixed to a point $O$.
\begin{enumerate}[label=(\roman*)]
\item Find the extension of the elastic string when the particle hangs in equilibrium directly below $O$.
\item The particle is pulled down and held at the point $P$, which is 0.9 metres vertically below $O$.
Show that the elastic potential energy of the string when the particle is in this position is 11.25 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 above $\boldsymbol { P }$.
Show that, while the string is taut,
$$v ^ { 2 } = 10.4 x - 50 x ^ { 2 }$$
\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 2008 Q8 [16]}}