To solve this problem, we can use the principle of conservation of momentum and kinetic energy in an elastic collision.
Let's denote the initial velocity of the 2 kg object as v1i, the initial velocity of the 6 kg object as v2i, and the final velocities as v1f and v2f.
According to the principle of conservation of momentum:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
where: m1 = mass of the 2 kg object = 2 kg m2 = mass of the 6 kg object = 6 kg
Using the given information: v1i = 12 m/s (initial velocity of the 2 kg object) v2i = 0 m/s (initial velocity of the stationary 6 kg object)
Substituting these values into the equation, we have:
2 kg * 12 m/s + 6 kg * 0 m/s = 2 kg * v1f + 6 kg * v2f
24 kg·m/s = 2 kg·v1f + 6 kg·v2f
Next, we apply the conservation of kinetic energy in an elastic collision:
(1/2) * m1 * v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
(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
144 J = 1 kg * v1f^2 + 3 kg * v2f^2
Now we have a system of two equations:
- 24 kg·m/s = 2 kg·v1f + 6 kg·v2f
- 144 J = 1 kg·v1f^2 + 3 kg·v2f^2
We can solve this system of equations to find the final velocities v1f and v2f.
Multiplying equation 1 by 3 and subtracting equation 2 from it:
3 * (24 kg·m/s) - 144 J = 6 kg·v1f + 18 kg·v2f - (1 kg·v1f^2 + 3 kg·v2f^2)
72 kg·m/s - 144 J = 6 kg·v1f + 18 kg·v2f - (1 kg·v1f^2 + 3 kg·v2f^2)
Rearranging the terms:
0 = v1f^2 + 6 kg·v1f + 3 kg·v2f^2 + 18 kg·v2f - 144 J + 72 kg·m/s
Simplifying:
v1f^2 + 6 kg·v1f + 3 kg·v2f^2 + 18 kg·v2f - 144 J + 72 kg·m/s = 0
Now we have a quadratic equation in terms of v1f and v2f. By solving this quadratic equation, we can find the final velocities of the two objects.