Answer
$m_A=40.0u$
Work Step by Step
In the horizontal direction
$m_Nv_N+m_Av_A=m_Nv'_N+m_Av'_A$
$(20.0u)v_N=(20.0u)v'_{Nx}+m_Av'_{Ax}$
In the vertical direction
$(20.0u)v'_{Ny}=m_Av'_{Ay}$
$\frac{m_Nv_N^2}{2}+\frac{m_Av_A^2}{2}=\frac{m_Nv_N^{'2}}{2}+\frac{m_Av_A^{'2}}{2}$
$m_Nv_N^2=m_Nv_N^{'2}+m_Av_A^{'2}$
$m_N=m_N\bigg(\frac{v'_N}{v_N}\bigg)^{2}+m_A\bigg(\frac{v_A}{v_N}\bigg)^{2}$
$\frac{m_Nv_N}{\sin(105.6^o)}=\frac{m_Nv'_N}{\sin(130^o)}$
$\frac{v'_N}{v_N}=0.795$
$\frac{m_Nv_N}{\sin(105.6^o)}=\frac{m_Av_A}{\sin(124.4^o)}$
$\frac{v_A}{v_N}=0.857\frac{m_N}{m_A}$
$20.0u=(20.0u)(0.795)^{2}+m_A\big(0.857\frac{20.0u}{m_A}\big)^{2}$
$m_A=40.0u$
