I consider two classic cases: first, the sideways momentum is conserved in the form m*vs + (M-m)*Vs = 0. Which is the relation which would hold if you stroked to the side while neither skate rested on the ground; i.e. no momentum transfer to the ground as the body and leg/skate have equal and opposite momenta. The second case will assume vs=0, i.e. no sideways motion of the body at all. To do this the skate under the body has to react with the ground so as to cancel the momentum the powering skate would otherwise deliver.
2a. Skater Sideways Velocity from Momentum ConservationM*vf2 + (M-m)*Vs2 +m*vs2= M*v2 .
(M-m)*Vs + m*vs = 0
M*vf2 + [M*m/(M-m)]*vs2 = M*v2 .
The solution is: vf=v*sin(t), with, vs=[(M-m)/m]1/2 *v*cos(t),
along with: Vf=v*sin(t), and, Vs=-[m/(M-m)]1/2 *v*cos(t).
2b. Skater Sideways Velocity Is ZeroThe results are plotted in the next section.M*vf2 + m*vs2 = M*v2 .
The solution is: vf=v*sin(t), with, vs=(M/m)1/2 *v*cos(t),
along with: Vf=v*sin(t), and, Vs=0.