Take WEST to be the negative direction, and EAST to be positive.
The total initial momentum of the system is
(4.00 kg) (-5.00 m/s) + (6.00 kg) (-2.00 m/s) = -32.0 kg•m/s
Let v₄ and v₆ denote the final speeds of the 4.00-kg and 6.00-kg objects, respectively. The total final momentum is then
(4.00 kg) v₄ + (6.00 kg) v₆
Momentum is conserved, so
(4.00 kg) v₄ + (6.00 kg) v₆ = -32.0 kg•m/s
The total kinetic energy of the system at the start is
1/2 (4.00 kg) (-5.00 m/s)² + 1/2 (6.00 kg) (-2.00 m/s)² = 62 J
and this energy is also conserved, so that the total kinetic energy after the collision is
1/2 (4.00 kg) v₄² + 1/2 (6.00 kg) v₆² = 62 J
(2.00 kg) v₄² + (3.00 kg) v₆² = 62 J
Solve these two equations for v₄ and v₆. You would end up with two solutions, one of which corresponds to the initial conditions, which we throw out. Then
v₄ = -1.40 m/s
v₆ = -4.40 m/s
which is to say, the 6.00-kg object has a velocity of (D) 4.40 m/s WEST after the collision.
