Figure shows two identical particles `1` and `2`, each of mass `m`, moving in opposite directions with same speed `vec V` along parallel lines. At a particular instant, `vec r_1` and `vec r_2` are their respective position vectors drawn from point `A` which is in the plane of the parallel lines. Which of the following is the correct statement ?
.

Options

A. Agular momentum `I_(1)` of particle a about A is `I = mv (d_(1))o.`
B. Angular momentum `I_(2)` of particle 2 about A is ` I_(2)=mvr_(2) o.`
C. total angular momentum of the system about A is ` I = mv(r_(1)+r_(2))o.`
D. total angular momentum of the system about A is `I=mv(d_(2)-d_(1))ox`

Answer

A,B

Explanation

the angular momentum L of a particle with respect toorigin is deined to be `L=rxxp` where , r is the podsition vector of the particle and p is the linear momentum , The direction of L is is perpendicular to both d r and p by right hand rule .
For particle 1. `I_(1) =r_(1)xxmv` is not of plane of the paper and perpendicular to `r_(1)` and p(mv) Similarly `I_(2)=r_(2)xxm(-v)` is into the plane of the paper and perpendicular to `r_(2) and -p`
Hence , total angular momentum
`l=l_(1)+l_(2)=r_(1)xxmv+(-r_(2)xxmv)`
`|l|=mvd_(1)- mvd_(2)as d_(2)gt d_(1)` total angular momentum will be inward
Hence `I=mv(d_(2)-d_(1))ox`
note : in the expression of angular momentum `I=rxxp` the direction of l is taken by right hand rule .