Determine the velocities of A abd B after collision, if the coefficient of restitution between the blocks is
e = 0.1 if a block of mass 5 kg is sliding forward on the smooth surface with a velocity (vA)1 = 10 m/s when it strikes the 2.5 kg block B, which is originally at rest.

1) 4.2 m/s and 3.6 m/s
2) 5.5 m/s and 4.2 m/s
3) 8.7 m/s and 9.5 m/s
4) 4.8 m/s and 10 m/s

The equation of momentum can be written as
mB (vB)1 + mA (vA)1 = mB (vB)2 + mA (vA)2
0 + 5 (6) = 2.5 (vB)2 + 5 (vA)2   --------(1)

For the restution we have:
e = (vA)2 - (vB)2 / (vB)1 - (vA)1
0.1 = (vA)2 - (vB)2 / 6 - 0
(vA)2 - (vB)2 = 0.6  --------------------(2)

Solving equations (1) and (2) we have :
(vA)2 = 4.2 m/s and also
(vB)2 = 3.6 m/s in the direction of the block A

So the final velocities of blocks A and B are
(vA)2 = 4.2 m/s and (vB)2 = 3.6 m/s