We can use the conservation of momentum and kinetic energy to solve for the final velocities of the two objects.
Conservation of momentum:
m1v1i + m2v2i = m1v1f + m2v2f
where m1 and v1 are the mass and velocity of object 1 before the collision, and m2 and v2 are the mass and velocity of object 2 before the collision.
Plugging in the values:
(2 kg)(12 m/s) + (6 kg)(0 m/s) = (2 kg)(v1f) + (6 kg)(v2f)
Simplifying:
24 kg m/s = 2 kg v1f + 6 kg v2f
Conservation of kinetic energy:
(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2
Plugging in the values:
(1/2)(2 kg)(12 m/s)^2 + (1/2)(6 kg)(0 m/s)^2 = (1/2)(2 kg)(v1f)^2 + (1/2)(6 kg)(v2f)^2
Simplifying:
144 J = 1 kg v1f^2 + 3 kg v2f^2
Now we have two equations with two unknowns (v1f and v2f). Solving for v1f in terms of v2f in the first equation:
v1f = (24 kg m/s - 6 kg v2f)/2 kg = 12 m/s - 3v2f
Plugging this into the second equation:
144 J = 1 kg (12 m/s - 3v2f)^2 + 3 kg v2f^2
Simplifying and solving for v2f:
144 J = 1 kg (144 m^2/s^2 - 72 v2f + 9 v2f^2) + 3 kg v2f^2
144 J = 144 J - 72 kg m/s v2f + 9 kg m^2/s^2 v2f^2 + 3 kg v2f^2
6 kg v2f^2 - 72 kg m/s v2f + 144 J = 0
Dividing by 6 kg:
v2f^2 - 12 kg m/s v2f + 24 J/kg = 0
Using the quadratic formula:
v2f = [12 kg m/s ± sqrt((12 kg m/s)^2 - 4(1)(24 J/kg))]/(2)
v2f = [12 kg m/s ± sqrt(96) m/s]/2
v2f = 6 kg m/s ± 2sqrt(6) m/s
v2f ≈ 9.90 m/s or v2f ≈ 2.10 m/s
Plugging these values into the equation we found for v1f:
v1f = 12 m/s - 3v2f
v1f ≈ -16.70 m/s or v1f ≈ 38.70 m/s
Since the negative velocity doesn't make physical sense, the final velocities of the two objects are:
v1f ≈ 38.70 m/s and v2f ≈ 2.10 m/s