Two particles \(A\) and \(B\) of equal mass \(M\) are moving with the same speed \(v\) as shown in the figure . They collide completely inelastically and move as a single particle \(C\). The angle \(\theta \) that the path of \(C\) makes with the \(X-\) axis is given by 

According to law of conservation of linear momentum, verticle component,

\(2\,mv'\,\sin \,\theta  = mv\sin {60^ \circ } + mv\,\sin {45^ \circ }\)

\(2mv'\sin \theta  = \frac{{mv}}{{\sqrt 2 }} + \frac{{mv\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)

Horizonatal component,

\(2\,mv'\,\sin \,\theta  = mv\sin {60^ \circ } - mv\,\sin {45^ \circ }\)

\(2mv'\cos \theta  = \frac{{mv}}{2} + \frac{{mv}}{{\sqrt 2 }}\,\,\,\,\,\,...\left( {ii} \right)\)

Dividing \(e{q^n}\) \((i)\) by \(e{q^n}\) \((ii),\)

\(\tan \theta  = \frac{{\frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2}}}{{\frac{1}{2} - \frac{1}{{\sqrt 2 }}}} = \frac{{\sqrt 2  + \sqrt 3 }}{{1 - \sqrt 2 }}\)