Answer
a) See solution below
b) $\arctan(1.07)=-46.9^o$
$v_B=1.23\frac{m}{s}$
Work Step by Step
a) $m_Av_A+m_Bv_B=m_Av_A'+m_Bv_B'$
For the x direction
$(0.120kg)(2.8\frac{m}{s})+(0.140kg)(0\frac{m}{s})=(0.120kg)(2.8\frac{m}{s})\cos(30^o)+(0.140kg)v_{Bx}$
For the y direction
$0=(0.120kg)(\sin(30.0^o))+(0.140kg)(\sin(\theta))$
b) For the x direction
$(0.336kg\frac{m}{s})+0=(0.218kg\frac{m}{s})+(0.140kg)v_{Bx}$
$v_B\cos(\theta)=0.841\frac{m}{s}$
For the y direction
$-0.9=v_B(\sin(\theta))$
Combining the two equations, we get the following result
$\arctan(1.07)=-46.9^o$
$v_B=1.23\frac{m}{s}$