To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy.
Let's denote the initial velocity of the 2 kg mass object as v1 and the initial velocity of the 6 kg mass object as v2. We'll also denote the final velocities of the two objects as vf1 and vf2, respectively.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
m1 * v1_initial + m2 * v2_initial = m1 * vf1 + m2 * vf2
where m1 and m2 are the masses of the respective objects.
In this case, the 2 kg mass object is moving with a velocity of 12 m/s, while the 6 kg mass object is stationary (v2_initial = 0 m/s). Substituting the given values into the momentum conservation equation, we have:
2 kg * 12 m/s + 6 kg * 0 m/s = 2 kg * vf1 + 6 kg * vf2
24 kg·m/s = 2 kg·vf1 + 6 kg·vf2
Simplifying, we get:
24 kg·m/s = 2 kg·vf1 + 6 kg·vf2
Dividing through by 2 kg, we have:
12 m/s = vf1 + 3 kg·vf2 ----(1)
Next, we can apply the principle of conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, this can be expressed as:
(1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * vf1^2 + (1/2) * m2 * vf2^2
Plugging in the given values, we have:
(1/2) * 2 kg * (12 m/s)^2 + (1/2) * 6 kg * (0 m/s)^2 = (1/2) * 2 kg * vf1^2 + (1/2) * 6 kg * vf2^2
144 J = vf1^2 + 3 kg·vf2^2 ----(2)
We now have a system of two equations (Equations 1 and 2) with two variables (vf1 and vf2). We can solve these equations simultaneously to find the values of vf1 and vf2.
Solving Equations 1 and 2 simultaneously, we find:
12 = vf1 + 3 vf2 ----(3) (Dividing Equation 1 by 1 kg)
144 = vf1^2 + 3 vf2^2 ----(4)
From Equation 3, we can express vf1 in terms of vf2:
vf1 = 12 - 3 vf2
Substituting this value into Equation 4:
144 = (12 - 3 vf2)^2 + 3 vf2^2
Expanding and simplifying:
144 = 144 - 72 vf2 + 9 vf2^2 + 3 vf2^2
Combining like terms:
0 = 12 vf2^2 - 72 vf2
Dividing through by 12:
0 = vf2^2 - 6 vf2
Factoring:
0 = vf2 (vf2 - 6)
Setting each factor equal to zero:
vf2 = 0 or vf2 - 6 = 0
If vf2 = 0, then from Equation 3, vf1 = 12 - 3(0) = 12 m/s.
If vf2 - 6 = 0, then vf2 = 6. Substituting this value into Equation 3, vf1 = 12 - 3(6) = -6 m/s.
Therefore, the final velocities of the two objects are vf1 = 12 m/s and vf2 = 0 m/s for one case, and vf1 = -6 m/s and vf2 = 6 m/s for the other case.