According to the law of conservation of momentum, we have
[tex]p_{i1}+p_{i2}=p_{f1}+p_{f2}_{}[/tex]Where p = mv. So, using the definition of momentum, we have
[tex]m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}[/tex]Let's use the given magnitudes and replace them with their letters.
[tex]1,391\operatorname{kg}\cdot17.4(\frac{m}{s})+1,280\operatorname{kg}\cdot v_{i2}=1,391\operatorname{kg}\cdot0+1,280\operatorname{kg}\cdot0[/tex]Observe that both final velocities are null because they come to stop. Let's solve for v.
[tex]\begin{gathered} 24,203.4\operatorname{kg}\cdot\frac{m}{s}+1,280\operatorname{kg}\cdot v_{i2}=0 \\ 1,280\operatorname{kg}\cdot v_{i2}=-24,203.4\operatorname{kg}\cdot\frac{m}{s} \\ v_{i2}=\frac{-24,203.4\operatorname{kg}\cdot\frac{m}{s}}{1,280\operatorname{kg}} \\ v_{i2}\approx-18.91(\frac{m}{s}) \end{gathered}[/tex]The negative sign shows that the second car is headed in the Western direction.
