Answer:
[tex]L=m(xv_y-yv_x)k[/tex]
Explanation:
It is given that,
Velocity of a particle, [tex]v=v_xi+v_yj[/tex]
Position vector of a particle, [tex]r=xi+yj[/tex]
We need to find the angular momentum of the particle. It is given by :
[tex]L=r\times p[/tex], p = linear momentum
[tex]L=r\times (mv)[/tex]
[tex]L=m(r\times v)[/tex]
[tex]L=m((xi+yj)\times (v_xi+v_yj))[/tex]
[tex]L=m(xv_y-yv_x)k[/tex]
So, the angular momentum of the particle is [tex](xv_y-yv_x)k[/tex]. Hence, this is the required solution.