question_answer

A particle is moving along a line \[y=x+a\] with a constant velocity \[v\]. Find the angular momentum of the particle about the origin.

A) \[mva\]

B) \[mva\sqrt{2}\]

C) \[\frac{mva}{\sqrt{2}}\]

D) \[\frac{mvy}{x\sqrt{2}}\]

Correct Answer: C

Solution :

[c] \[\angle POQ=45{}^\circ \,\,So,\text{ }OQ=a\cos 45{}^\circ =a/\sqrt{2}\] \[L=mv(OQ)=mva/\sqrt{2}\]