question_answer

A particle of mass m moves in the AY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin 0 is \[{{L}_{A}}\]when it is at A and \[{{L}_{B}}\]when it is at B, then

A) \[{{L}_{A}}>{{L}_{B}}\]

B) \[{{L}_{A}}={{L}_{B}}\] 

C)  the relationship between\[{{L}_{A}}\]and\[{{L}_{B}}\]depends upon the slope of the line AB

D) \[{{L}_{A}}<{{L}_{B}}\]

Correct Answer: B

Solution :

 From the definition of angular momentum, \[\vec{L}=\vec{r}\times \vec{p}=rmv\,\sin \phi (-\vec{k})\] Therefore, the magnitude of L is \[L=mvr\sin \phi =mvd\] where\[d=r\sin \phi \]is the distance of closest approach of the particle to the origin. As d is the same for both the cases, hence\[{{L}_{A}}={{L}_{B}}.\]