Answer
The final speed is 0.20 km/s in the original direction of asteroid A.
Work Step by Step
We can use conservation of momentum to solve this question:
$m_A~v_A + m_B~v_B = (m_A+m_B)~v$
$(7.5\times 10^{12}~kg)(3.3~km/s)+(1.45\times 10^{13}~kg)(-1.4~km/s) = (7.5\times 10^{12}~kg +1.45\times 10^{13}~kg)~v$
$v = \frac{4.45\times 10^{12}~kg\cdot m/s}{2.2\times 10^{13}~kg}$
$v = 0.20~km/s$
The final speed is 0.20 km/s in the original direction of asteroid A.