# A particle of mass m is projected with a speed u from the ground at an angle $\theta =\frac{\pi }{3}$ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $u\hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is

Answers (1)

\begin{aligned} &\text { By momentum conservation, }\\ &\frac{\mathrm{mu}}{2}+\mathrm{mu}=2 \mathrm{mv}^{\prime}\\ &v^{\prime}=\frac{3 v}{4} \end{aligned}

\begin{aligned} &\text { Range after collision }=\frac{3 \mathrm{v}}{4} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\\ &=\frac{3 \mathrm{v}}{4} \sqrt{\frac{2 \cdot \mathrm{u}^{2} \sin ^{2} 60^{\circ}}{\mathrm{g} 2 \mathrm{g}}}\\ &=\frac{3}{4} \frac{\sqrt{3}}{2} \cdot \frac{u^{2}}{g}=\frac{3 \sqrt{3} u^{2}}{8 g} \end{aligned}

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