Using conservation of momentum:
![m1u1+m2u2=m1v1+m2v2](https://img.qammunity.org/2023/formulas/physics/college/guf0ue3jo2xdi2b18ni64usz6lkm6fjanv.png)
Where:
m1 = Mass of the car on the left = 72500kg
m2 = Mass of the car on the right = 90269kg
u1 = Initial speed of the car on the left = 0.59 m/s
u2 = Initial speed of the car on the right = -1.2 m/s
v1 = Final speed of the car on the left
v2 = Final speed of the car on the right
Since the train cars will be linked, we can conclude:
v1 = v2
so:
![\begin{gathered} m1u1+m2u2=v1(m1+m2) \\ so\colon \\ v1=(m1u1+m2u2)/(m1+m2) \\ v1=(72500\cdot0.59+90269(-1.2))/(72500+90269) \\ v1=-(65547.8)/(162769) \\ v1=-0.4(m)/(s) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/u9j4lxynim3191ylpb09y3e35bydp54n4i.png)
Answer:
0.4 m/s