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  • A particle of mass m is projected with a speed u from the ground at an angle  w.r.t. horizontal (x-axis). When

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 :
Option: 1 \frac{3\sqrt{3}}{8}\; \frac{u ^{2}}{g}      
Option: 2  2\sqrt{2}\; \frac{u^{2}}{g}       
Option: 3  \frac{3\sqrt{2}}{4}\; \frac{u ^{2}}{g}      
Option: 4 \frac{5}{8}\; \frac{u ^{2}}{g}
 

Answers (1)

best_answer

Apply momentum conservation 

p_{i}=p_{f}

mu+mu\cos \theta =2mv

\Rightarrow v=\frac{u(1+\cos 60^{\circ})}{2}=\frac{3}{4}u

So horizontal range after collision =vt=v\sqrt{\frac{2H_{max}}{g}}

                                                        =\frac{3}{4}u\sqrt{\frac{2u^{2}\sin ^{2}(60^{\circ})}{2g^{2}}}

                                                         =\frac{3}{4}u^{2}\frac{\sqrt{\frac{3}{4}}}{g}=\frac{3\sqrt{3}u^{2}}{8g}

Hence the correct option is (1).

Posted by

avinash.dongre

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