Figure shows an overhead view of two particles moving at constant momentum along horizontal paths.
Particle 1, with momentum magnitude `p_(1)=5.0kg*m//s`, has position vector `vecr_(1)` and will pass 2.0 m from point O.
Particle 2, with momentum magnitude `p_(2)=2.0kg*m//s`, has position vector `vecr_(2)` and will pass 4.0 m from point O. What are the magnitude and direction of the net angular momentum `vecL` about point O of the two particle system?

Explanation

To find `vecL`, we can first find the individual angular momenta `vecl_(1)andvecl_(2)` and then add them. To evaluate their magnitudes, we can use any one of Eqs. Through. However, Eq. is easiest, because we are given the perpendicular distances `r_(bot1)(=2.0m)andr_(bot2)(=4.0m)` and the momentum magnitudes `p_(1)andp_(2)`.
Calculations: For particle 1, Eq. yields
`l_(1)=r_(bot1)p_(1)=(2.0)(5.0kg*m//s)`
= `10kg*m^(2)//s`.
To find the direction of vector `vecl_(1)`, we use Eq. and the right-hand rule for vector products. For `vecr_(1)xxvecp_(1)`, the vector product is out of the page, perpendicular to the plane of Fig. This is the positive direction, consistent with the counterclockwise rotation of the particle.s position vector `vecr_(1)` around O as particle 1 moves. Thus, the angular momentum vector for particle 1 is
`l_(1)=+10kg*m^(2)//s`.

Similarly, the magnitude of `vecl_(2)` is
`l=r_(bot2)p_(2)=(4.0m)(2.0kg*m//s)`
= `8.0kg*m^(2)//s`,
and the vector product `vecr_(2)xxvecp_(2)` is into the page, which is the negative direction, consistent with the clockwise rotation momentum vector for particle 2 is
`l_(2)=-8.0kg*m^(2)//s`
The net angular momentum for the two particle system is
`L=l_(1)+l_(2)=+10kg*m^(2)//s+(-8.0kg*m^(2)//s)`
= `+2.0kg*m^(2)//s`.