Answer
$-1.6m/s$
Work Step by Step
We know that
$v_2=\pm\frac{m_1v_1}{\sqrt{m_2^2-(m_2^2-m_1^2)\frac{v_1^2}{c^2}}}$
We plug in the known values to obtain:
$v_2=\pm\frac{(88Kg)(2.0m/s)}{\sqrt{(120kg)^2-(120Kg)^2-(88Kg)^2\frac{(2.0m/s)^2}{(3.0m/s)^2}}}$
$v_2=-1.6m/s$