Answer
The magnitude of the net angular momentum is $~~9.8~kg~m^2/s$
Work Step by Step
We can find the magnitude of the angular momentum of $P_1$ about the point:
$L_1 = r_{\perp}~mv$
$L_1 = (1.5~m)(6.5~kg)(2.2~m/s)$
$L_1 = 21.45~kg~m^2/s$
By the right hand rule, the direction of $L_1$ is in the -z direction.
We can find the magnitude of the angular momentum of $P_2$ about the point:
$L_2 = r_{\perp}~mv$
$L_2 = (2.8~m)(3.1~kg)(3.6~m/s)$
$L_2 = 31.248~kg~m^2/s$
By the right hand rule, the direction of $L_2$ is in the +z direction.
We can find the magnitude of the net angular momentum:
$L_{net} = L_2-L_1$
$L_{net} = (31.248~kg~m^2/s)-(21.45~kg~m^2/s)$
$L_{net} = 9.8~kg~m^2/s$
The magnitude of the net angular momentum is $~~9.8~kg~m^2/s$
