MCQ
A particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ 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 speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is :
  • $\frac{\pi}{4}$
  • B
    $\frac{\pi}{4}+\alpha$
  • C
    $\frac{\pi}{4}-\alpha$
  • D
    $\frac{\pi}{2}$

Answer

Correct option: A.
$\frac{\pi}{4}$
a
At the highest point

$Image$

$v _1=\frac{ u _0 \cos \alpha}{2} \quad \text { (by applying momentum conservation in horizontal direction) } $

$v _2=\frac{ u _0 \cos \alpha}{2} \quad \text { (by applying momentum conservation in vertical direction) } $

$\left( H =\frac{ u _0^2 \sin ^2 \alpha}{2 g }\right. \text { ) } $

$\theta=45^{\circ}$

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