= Solution
The familiar right-angle result requires an \b[elastic collision] with conserved horizontal <momentum> and the appropriate treatment of <rotational kinetic energy>. In the ideal smooth-sphere model the spins do not change during the collision, so the translational <kinetic energy> is separately conserved. For equal masses, with initial <velocity> $\mathbf u$ and outgoing <velocities> $\mathbf v_1,\mathbf v_2$,
$$
\mathbf u=\mathbf v_1+\mathbf v_2,\qquad u^2=v_1^2+v_2^2.
$$
Squaring the first equality and using the second gives
$$
\boxed{\mathbf v_1\cdot\mathbf v_2=0.}
$$
Thus the two nonzero outgoing directions are \b[perpendicular]. In the head-on limiting case one sphere stops, so an angle for its zero velocity is undefined. An ideal elastic model that enforces <rolling without slipping> at both ends gives the same energy equation: each solid sphere has total <kinetic energy> $7mv^2/10$, and the common factor cancels.
Elasticity is not stated explicitly in the question. It cannot be omitted from the argument: for a smooth equal-mass collision with normal incoming component $u_n$ and <coefficient of restitution> $e<1$, the outgoing scalar product is $(1-e^2)u_n^2/4>0$. Moreover, after a smooth impact the spheres need not still be rolling; subsequent frictional relaxation can change their directions. The right-angle statement therefore concerns the specified ideal elastic collision, not arbitrary later rolling trajectories.
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