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 (v_A)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
m_B (v_B)₁ + m_A (v_A)₁ = m_B (v_B)₂ + m_A (v_A)₂
0 + 5 (6) = 2.5 (v_B)₂ + 5 (v_A)₂   --------(1)

For the restution we have:
e = (v_A)₂ - (v_B)₂ / (v_B)₁ - (v_A)₁
0.1 = (v_A)₂ - (v_B)₂ / 6 - 0
(v_A)₂ - (v_B)_{2 = 0.6  --------------------(2)}

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

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