Question 110 of 130

A particle of mass m is projected with a speed u from the ground at an angle $\theta =\dfrac{\pi }{3}$ 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:

A$\dfrac{3\sqrt{3}}{8}\dfrac{{u}^{2}}{g}$B$\dfrac{3\sqrt{2}}{4}\dfrac{{u}^{2}}{g}$C$\dfrac{5}{8}\dfrac{{u}^{2}}{g}$D$2\sqrt{2}\dfrac{{u}^{2}}{g}$

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Answer: (A) $\dfrac{3\sqrt{3}}{8}\dfrac{{u}^{2}}{g}$

${p}_{i}={p}_{f}$
$mu+mucos\theta =2mv$
$\Rightarrow v=\dfrac{u\left(1+\cos ⁡{60}^{0}\right)}{2}=\dfrac{3}{4}u$
So horizontal range after collision $=vt$
$=v\sqrt{\dfrac{{2H}_{max}}{g}}$
$=\dfrac{3}{4}u\sqrt{\dfrac{2{u}^{2}{\sin }^{2}⁡\left({60}^{0}\right)}{2{g}^{2}}}$
$=\dfrac{3}{4}{u}^{2}\dfrac{\sqrt{\dfrac{3}{4}}}{g}=\dfrac{3\sqrt{3}{u}^{2}}{8g}$