Two light elastic strings, each of length \(l\) m and modulus of elasticity \(\lambda\) N, are attached to a particle \(P\) of mass \(m\) kg. The other ends of the strings are attached to fixed points \(A\) and \(B\) on the same horizontal level, where \(AB = 2l\) m. \(P\) is held vertically below the mid-point of \(AB\), with each string taut and inclined at \(30°\) to the horizontal, and released from rest. Given that \(P\) comes to instantaneous rest when each string makes an angle of \(60°\) with the horizontal, show that \(\lambda = \frac{3mg}{6 - 2\sqrt{3}}\). \includegraphics{figure_1} [10 marks]

E.P.E. changes from $2\frac{\lambda}{2l}l(\sec 30° - 1)^2$ to $2\frac{\lambda}{2l}l(\sec 60° - 1)^2$ | M1 A1 A1 |

Gain in E.P.E. $= \lambda l[(\sec 60° - 1)^2 - (\sec 30° - 1)^2] = \lambda l[\frac{4}{3} - \frac{4}{3}]$ | M1 A1 |

Loss in grav. P.E. $= mgl(\tan 60° - \tan 30°) = \frac{2mgl}{\sqrt{3}}$ | M1 A1 |

Hence $\lambda l(\frac{12-4\sqrt{3}}{3\sqrt{3}}) = \frac{2mgl}{\sqrt{3}}$ | M1 A1 A1 |

$\lambda = \frac{3mg}{6-2\sqrt{3}}$ | |

**Total: 10 marks**

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Two light elastic strings, each of length $l$ m and modulus of elasticity $\lambda$ N, are attached to a particle $P$ of mass $m$ kg. The other ends of the strings are attached to fixed points $A$ and $B$ on the same horizontal level, where $AB = 2l$ m. $P$ is held vertically below the mid-point of $AB$, with each string taut and inclined at $30°$ to the horizontal, and released from rest. Given that $P$ comes to instantaneous rest when each string makes an angle of $60°$ with the horizontal, show that $\lambda = \frac{3mg}{6 - 2\sqrt{3}}$.

\includegraphics{figure_1}

[10 marks]

\hfill \mbox{\textit{Edexcel M3  Q4 [10]}}