Answer
$0.376m/s$
Work Step by Step
We can determine the required speed as follows:
$p_1=\frac{m_1v_1}{\sqrt{1-\frac{v_1^2}{c^2}}}$
We plug in the known values to obtain:
$p_1=\frac{(88Kg)(2.0m/s)}{\sqrt{1-(\frac{2.0m/s}{3.0m/s})^2}}$
$\implies p_1=236.13Kgm/s$
and $p_2=\frac{(120Kg)(-1.2m/s)}{\sqrt{1-(\frac{-1.2m/s}{3.0m/s})^2}}$
$\implies p_2=-157.117Kgm/s$
According to the law of conservation of momentum
$p_1+p_2=p$
$\implies 236.13-157.117=\frac{(88Kg+120Kg)v}{\sqrt{1-\frac{v^2}{c^2}}}$
This simplifies to:
$v=0.376m/s$
