= Solution
For equal <masses>, let the incoming <laboratory frame> <velocity> be $\mathbf v$ and the outgoing <velocities> be $\mathbf a,\mathbf b$. <Conservation of momentum> and <kinetic energy> give
$$
\mathbf v=\mathbf a+\mathbf b,\qquad v^2=a^2+b^2.
$$
Squaring the first equation and subtracting the second yields $2\mathbf a\cdot\mathbf b=0$. Hence
$$
\boxed{\mathbf a\cdot\mathbf b=0,\qquad \text{outgoing directions differ by }90^\circ.}
$$
The stated oblique <collision> ensures both outgoing <speeds> are nonzero. In the degenerate head-on case one particle can stop, and a direction cannot be assigned to its zero <velocity>.
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