= Solution
Conservation of momentum in the parent's rest frame gives $p_A+p_B=0$. Equal daughter masses and the <relativistic energy-momentum relation> then give equal daughter energies. Energy conservation fixes
$$
\boxed{E_A=E_B=\frac{Mc^2}{2}.}
$$
Using $E^2=p^2c^2+m^2c^4$ and the specified direction of A,
$$
\boxed{p_A=-\frac c2\sqrt{M^2-4m^2},\qquad
p_B=+\frac c2\sqrt{M^2-4m^2}.}
$$
The velocities follow from $v=pc^2/E$:
$$
\boxed{v_A=-V,\quad v_B=V,\qquad V=c\sqrt{1-4(m/M)^2}.}
$$
These are spatial momenta in the original <inertial frame>. For positive daughter rest mass, $m<M/2$ gives $0<V<c$.
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