 Solution

The familiar right-angle result requires an 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 ${\mathbf{v}}_1,{\mathbf{v}}_2$,Squaring the first equality and using the second givesThus the two nonzero outgoing directions are 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 $7m{v}^2/10$, 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.