# A 1.55 kg particle moves in the xy plane with a velocity of v = ( 3.51 , -3.39 ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r = ( 1.22 , 1.26 ) m. ?

Let,the velocity vector is $\vec{v} = 3.51 \hat{i} - 3.39 \hat{j}$
So,$m \vec{v} = \left(5.43 \hat{i} - 5.24 \hat{j}\right)$
And,position vector is $\vec{r} = 1.22 \hat{i} + 1.26 \hat{j}$
So,angular momentum about origin is vec r ×m vec v=(1.22hati +1.26hatj)×(5.43hati-5.24 hat j)=-6.4hatk-6.83hatk=-13.23hatk
So,the magnitude is $13.23 K g {m}^{2} {s}^{-} 1$