question_answer

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 [JEE MAIN Held on 09-01-2020 Evening]

A) \[\frac{5}{8}\frac{{{u}^{2}}}{g}\]

B) \[\frac{3\sqrt{2}}{4}\frac{{{u}^{2}}}{g}\]

C) \[\frac{3\sqrt{3}}{8}\,\,\frac{{{u}^{2}}}{g}\]

D) \[2\sqrt{2}\frac{{{u}^{2}}}{g}\]          

Correct Answer: C

Solution :

[c]  \[2m{{v}_{x}}=mu+mu\,\,\cos 60{}^\circ \]

\[{{v}_{x}}=\frac{3u}{4}\]

Horizontal range after collision

\[=\frac{3u}{4}\times \sqrt{\frac{2H}{g}}\]

\[=\frac{3\sqrt{3}{{u}^{2}}}{8g}\]