A particle \(P\), of mass \(m\) kg, is attached to two light elastic strings, each of natural length \(l\) m and modulus of elasticity \(3mg\) N. The other ends of the strings are attached to the fixed points \(A\) and \(B\), where \(AB\) is horizontal and \(AB = 2l\) m. \includegraphics{figure_4} If \(P\) rests in equilibrium vertically below the mid-point of \(AB\), with each string making an angle \(\theta\) with the vertical, show that $$\cot \theta - \cos \theta = \frac{1}{6}.$$ [8 marks]

Symmetric, so tensions in strings are equal | B1 M1 A1 | 
$2T\cos\theta = mg$ | | 
$AP\sin\theta = l$, so $AP = \frac{l}{\sin\theta}$ | M1 A1 | 
$T = \frac{3mg}{2}\left(\frac{1}{\sin\theta} - l\right)$ | | 
Hence $2 \times 3mg\left(\frac{1}{\sin\theta} - 1\right)\cos\theta = mg$ | M1 A1 A1 | 
$6(\cot\theta - \cos\theta) = 1$, etc | | 

**Total: 8 marks**

A particle $P$, of mass $m$ kg, is attached to two light elastic strings, each of natural length $l$ m and modulus of elasticity $3mg$ N. The other ends of the strings are attached to the fixed points $A$ and $B$, where $AB$ is horizontal and $AB = 2l$ m.

\includegraphics{figure_4}

If $P$ rests in equilibrium vertically below the mid-point of $AB$, with each string making an angle $\theta$ with the vertical, show that
$$\cot \theta - \cos \theta = \frac{1}{6}.$$ [8 marks]

\hfill \mbox{\textit{Edexcel M3  Q4 [8]}}