question_answer

Consider a uniform rod of mass M = 4 m and length l pivoted about its centre. A mass moving with velocity v making angle\[\theta =\frac{\pi }{4}\]to the rod’s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is                [JEE MAIN Held On 08-01-2020 Morning]

A)   \[\frac{4}{7}\frac{V}{l}\]                    

B)        \[\frac{3}{7}\frac{V}{l}\]

C)   \[\frac{3}{7\sqrt{2}}\frac{V}{l}\]         

D)        \[\frac{3\sqrt{2}}{7}\frac{V}{l}\]

Correct Answer: D

Solution :

[d]

About point P, angular momentum (L) of the

System

\[{{L}_{\operatorname{initial}}}=\frac{mv}{\sqrt{2}}\times \frac{1}{2}\]

\[{{L}_{final}}=\left[ \frac{(4m){{l}^{2}}}{12}+\frac{m{{l}^{2}}}{4} \right]\omega \]

As \[{{L}_{initial}}={{L}_{final}}\]

\[\omega =\frac{3\sqrt{2}v}{7l}\]