Answer
$2.51Kg,5.21Kg$
Work Step by Step
We can find $m_1$ and $m_2$ as follows:
$p_1=p_{total,x}=p_{total}cos\theta$
$p_1=(17.6)cos66.5^{\circ}=7.02\frac{Kg.m}{s}$
Now $m_1=\frac{p_1}{v_1}$
$\implies m_1=\frac{7.02}{2.80}=2.51Kg$
We can find $m_2$ as
$p_2=p_{total,y}=p_{total}sin\theta$
$p_2=(17.6)sin66.5^{\circ}$
$m_2=\frac{p_2}{v_2}$
We plug in the known values to obtain:
$m_2=\frac{(17.6)sin66.5^{\circ}}{3.10}=5.21Kg$
