The answer is $\frac{1}{m_1}$ and $\frac{1}{m_2}$.
I used the conservation of momentum (taking into consideration initial momentum is zero) , coefficient of restitution , and conservation of kinetic energy (since it is elastic collision) and did not approach to a solution.
With the usual notation, by conservation of momentum, $$0=m_1u_1-m_2u_2=-m_1v_1+m_2v_2$$ And by Newton's Law of Restitution, $$u_1+u_2=v_1+v_2$$
Eliminating $v_1$ gives $$v_2=\frac{m_{1} u_{1}+m_{1} u_{2}}{m_{1}+m_{2}}$$ $$=\frac{I+m_1\frac{I}{m_2}}{m_1+m_2}$$ $$=\frac{I}{m_2}$$
Similarly for $v_1$.