| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Horizontal elastic string on smooth surface |
| Difficulty | Standard +0.3 This is a standard M2 elastic strings question requiring EPE calculation, energy conservation with friction, and application of work-energy principles. Part (a) is direct formula application, part (b) is straightforward energy conservation (answer given), and part (c) adds friction work which is a routine extension. All techniques are standard M2 material with no novel insight required. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Part (a): EPE is \(\frac{\lambda x^2}{2l} = \frac{200(0.5)^2}{2 \times 2} = 12.5\) J | M1, A1 | |
| Part (b): When string becomes slack, using \(\frac{1}{2}mv^2\) = loss in EPE: \(\frac{1}{2} \times 5 \times v^2 = 12.5\); Speed is \(\sqrt{5}\) m s\(^{-1}\) | M1, A1, A1 | NB Using \(\sqrt{5}\) to answer (a) and thus (b) ⟹ no marks; AG |
| Part (c): Resolving vertically, \(R = 5g\); \(F = \mu R\); \(0.4 \times 5g = 2g\); Using change in energy = work done: \(2g \times 0.5 = \frac{1}{2} \times 5 \times (\sqrt{5})^2 - \frac{1}{2} \times 5 \times v^2\); \(9.8 = 12.5 - \frac{5}{2}v^2\); \(v^2 = 1.08\); Speed is 1.04 m s\(^{-1}\) | B1, M1, M1, M1, A1,A1, A1 | M1 for force × distance; A1 first term (or 12.5); A1 second term (inc −); |
| Total: 12 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Part (a):** EPE is $\frac{\lambda x^2}{2l} = \frac{200(0.5)^2}{2 \times 2} = 12.5$ J | M1, A1 | |
| **Part (b):** When string becomes slack, using $\frac{1}{2}mv^2$ = loss in EPE: $\frac{1}{2} \times 5 \times v^2 = 12.5$; Speed is $\sqrt{5}$ m s$^{-1}$ | M1, A1, A1 | NB Using $\sqrt{5}$ to answer (a) and thus (b) ⟹ no marks; AG |
| **Part (c):** Resolving vertically, $R = 5g$; $F = \mu R$; $0.4 \times 5g = 2g$; Using change in energy = work done: $2g \times 0.5 = \frac{1}{2} \times 5 \times (\sqrt{5})^2 - \frac{1}{2} \times 5 \times v^2$; $9.8 = 12.5 - \frac{5}{2}v^2$; $v^2 = 1.08$; Speed is 1.04 m s$^{-1}$ | B1, M1, M1, M1, A1,A1, A1 | M1 for force × distance; A1 first term (or 12.5); A1 second term (inc −); |
| | | **Total: 12** |6 An elastic string has one end attached to a point $O$, fixed on a horizontal table. The other end of the string is attached to a particle of mass 5 kilograms. The elastic string has natural length 2 metres and modulus of elasticity 200 newtons. The particle is pulled so that it is 2.5 metres from the point $O$ and it is then released from rest on the table.
\begin{enumerate}[label=(\alph*)]
\item Calculate the elastic potential energy when the particle is 2.5 m from the point $O$.
\item If the table is smooth, show that the speed of the particle when the string becomes slack is $\sqrt { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item The table is, in fact, rough and the coefficient of friction between the particle and the table is 0.4 .
Find the speed of the particle when the string becomes slack.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q6 [12]}}