A particle of mass $m$ is projected from the ground with an initial speed $u$ 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…

Question

A particle of mass $m$ is projected from the ground with an initial speed $u$ 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 $u$. The angle that the composite system makes with the horizontal immediately after the collision is

Aicharya · Accepted Answer

At the highest point $v_{1}=\frac{u_{0} \cos \alpha}{2} \quad$ (by applying momentum conservation in horizontal direction) $\mathrm{v}_{2}=\frac{\mathrm{u}_{0} \cos \alpha}{2} \quad$ (by applying momentum conservation in vertical direction) $\theta=45^{\circ}\left(\mathrm{H}=\frac{\mathrm{u}_{0}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}\right)$

A particle of mass $m$ is projected from the ground with an initial speed $u$ at an angle $\alpha$ with the…

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