You have two equations of momentum conservation to work with: One for the original direction of motion and another for the second, perpendicular, direction.
The two pucks must leave in opposite sides of the original line of motion.
The M's are the same and cancel out of the momentum equations. Thus
v1 sin 30 - v2 sin 60 = 0
v1 cos 30 + v2 cos 60 = Vo = 5.0 m/s
v1 - v2*sqrt3 = 0
v1*sqrt3 + v2 = 10
3*v1+ v2*sqrt3 = 10 sqrt3
4*v1 = 10*sqrt3
v1 = (5/2)*sqrt3 = 4.33 m/s
v2 = v1/sqrt3 = 2.50 m/s
Kinetic energy happens to be conserved in this case, but I did not have to use that fact to solve the problem. You will note that
(M/2)v1^2 + (M/2)v2^2 = (M/2)Vo^2.
A puck of mass 0.60 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the moving puck is 5.0 m/s. After the collision, one puck leaves with a speed v1 at 30° to the original line of motion. The second puck leaves with speed v2 at 60°.
Calculate v1 and v2.
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