There is no net external force on the two skaters. (Yes, there is a force on each by the spring, but this is internal to the skater-skater-spring system.) Thus the total momentum of the system is conserved:
I'm going to define a coordinate system such that Jo is in the +x direction compared to Al. (In other words, the +x axis is in the direction Jo is going to be moving off in after the spring is released.)
Before the release of the spring the total momentum of the system was 0 kg m/s. So
(Remember this is a vector equation, and Al moves in the -x direction. So his velocity has a negative component in the x direction.)
You can solve this for . For reference I get .