Answer
The magnitude of the angular momentum of the system is $~~0$
Work Step by Step
By the right hand rule, the direction of the angular momentum of each particle about the point is in the opposite direction.
We can find the magnitude of the angular momentum of the system:
$L = r_{\perp}~m_1v_1-r_{\perp}~m_2v_2$
$L = (0.021~m)(2.90\times 10^{-4}~kg)(5.46~m/s)- (0.021~m)(2.90\times 10^{-4}~kg)(5.46~m/s)$
$L = 0$
The magnitude of the angular momentum of the system is $~~0$.
