After 1st collision
mvA = mv'A + 2mv'B
0 = v'B - v'A
0 = v'B - v'A
v'B - v'A = -(vA - vB) = v'B - 0 = 9 - v'A
Also, v'B - v'A = -(0 - 9) = 9
Solving these equations,
v'A = 3 m/s
v'B = 6 m/s
After the 2nd collision
2mv'B = (2m + m)vC
2v'B = 3vC
vC = (2/3)v'B
vC = (2/3)(6 m/s)
vC = 4 m/s