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A particle of mass m is projected from the ground with an initial speed u at an angle $α$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial velocity u. The angle that the composite system makes with the horizontal immediately after the collision is

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$\frac{π}{4} + α$

$\frac{π}{4}$

$\frac{π}{2} - α$

$\frac{π}{2}$

answer is A.

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Detailed Solution

Velocity at highest point $u \cos α$ along horizontal velocity at the same point for the other body

$V = \sqrt{u^{2} - 2 g H} = \sqrt{u^{2} - \frac{u^{2} \sin^{2} α}{2 g} 2 g} = u \cos α$
Along vertical direction after inelastic collision

$\begin{array}{l} m u \cos α \hat{i} + m u \sin α \hat{j} = 2 m \vec{V} \\ d i r e c t i o n, \tan θ = \frac{V_{y}}{V_{x}} = 1 \\ θ = 45^{o} = \frac{π}{4} \end{array}$

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