This is a case of conservation of momentum in which;
mv1 + 2(mv2) = mfvf
Where m = mass of first railroad cars = 2.30e+4 kg = 23000 kg; v2 = velocity of first railroad car = 3.00 m/s; v2 = velocity of the two other railroad cars = 1.10 m/s; mf = total mass of the three cars after coupling = 23000*3 = 69000 kg; vf = velocity of the coupled cars
Substituting;
23000*3 + 2(23000*1.1) = 69000*vf
69000vf = 119600
vf = 119600/69000 = 1.73 m/s
The velocity the three cars couple will be 1.73 m/s.
