Answer
$195.12\space km/s$
$Direction = 25.9^{\circ} $ from the spacecraft initial direction
Work Step by Step
Please see the attached image first.
Here we use the principle of conservation of momentum.
$\vec P= Constant$
We can write,
$m_{SC}\vec V_{SC}= m_{O}\vec V_{O}+m_{L}\vec V_{L}$
Let's choose the x-axis along the direction of $\vec V_{SC}.$Then the two components of the momentum conservation equation become,
x component : $m_{SC}V_{SC}= m_{O}V_{Ox}+m_{L}V_{L x}$
y component : $0= m_{O}V_{Oy}+m_{L}V_{L y}$
$\rightarrow V_{Lx}= \frac{m_{SC}V_{SC}-m_{O}V_{O x}}{m_{L}}= \frac{m_{SC}V_{SC}-m_{O}V_{O}cos\theta}{m_{L}} $
Let's plug known values into this equation.
$V_{Lx}=\frac{784\space kg\times81.6\space km/s-549\space kg\times55.2cos(41.4^{\circ})\space km/s}{235\space kg}$
$V_{lx}=175.5\space km/s$
$\downarrow V_{Ly}= -\frac{m_{O}V_{Oy}}{m_{L}}= -\frac{m_{O}V_{O }sin\theta}{m_{L}}$
$V_{Ly}=\frac{-549\space kg\times[-55.2\space km/s\times sin(41.4^{\circ})]}{235\space kg}=85.28\space km/s$
$V_{L}=\sqrt {V_{Lx}^{2}+V_{Ly}^{2}}= \sqrt {175.5^{2}+85.28^{2}}km/s$
$V_{L}=195.12\space km/s $
$tan\alpha=\frac{V_{Ly}}{V_{Lx}}=\frac{85.28\space km/s}{175.5\space km/s}$
$\alpha=25.9^{\circ}$
