In a perfectly elastic collision, both momentum and kinetic energy are conserved. We can use these conservation principles to determine the velocities of the two balls after the collision.
Let's denote the initial velocity of the 4 kg ball as "v1" and the final velocities of the 4 kg ball and the 2 kg ball as "v1f" and "v2f" respectively.
Since the 2 kg ball is initially at rest, its initial velocity is 0 m/s (v2 = 0).
Using the conservation of momentum: Initial momentum = Final momentum
m1 * v1 + m2 * v2 = m1 * v1f + m2 * v2f
where m1 = mass of the 4 kg ball = 4 kg v1 = initial velocity of the 4 kg ball = 5 m/s m2 = mass of the 2 kg ball = 2 kg v2 = initial velocity of the 2 kg ball = 0 m/s (at rest)
Plugging in the values: (4 kg) * (5 m/s) + (2 kg) * (0 m/s) = (4 kg) * v1f + (2 kg) * v2f
20 kg m/s = 4 kg * v1f + 0 kg m/s + 2 kg * v2f
20 kg m/s = 4 kg * v1f + 2 kg * v2f
Since the collision is perfectly elastic, kinetic energy is also conserved:
Initial kinetic energy = Final kinetic energy
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
Plugging in the values: (1/2) * (4 kg) * (5 m/s)^2 + (1/2) * (2 kg) * (0 m/s)^2 = (1/2) * (4 kg) * v1f^2 + (1/2) * (2 kg) * v2f^2
(1/2) * (4 kg) * (25 m^2/s^2) = (1/2) * (4 kg) * v1f^2 + (1/2) * (2 kg) * v2f^2
50 kg m^2/s^2 = 2 kg * v1f^2 + 0 kg m^2/s^2 + 1 kg * v2f^2
50 kg m^2/s^2 = 2 kg * v1f^2 + 1 kg * v2f^2
Now we have a system of equations:
20 kg m/s = 4 kg * v1f + 2 kg * v2f
50 kg m^2/s^2 = 2 kg * v1f^2 + 1 kg * v2f^2
We can solve these equations simultaneously to find the velocities.
From the first equation, we can solve for v1f in terms of v2f:
v1f = (20 kg m/s - 2 kg * v2f) / 4 kg
Substituting this into the second equation:
50 kg m^2/s^2 = 2 kg * [(20 kg m/s - 2 kg * v2f) / 4 kg]^2 + 1 kg * v2f^2
Simplifying and rearranging:
50 kg m^2/s^2 = (20 kg m/s - 2 kg * v2f)^2 / 2 kg + 1 kg * v2f^2
Multiplying through by 2 kg:
100 kg m^2/s^2 = (20 kg m/s - 2 kg * v2f)^2 + 2 kg^2 * v2f^2
Expanding and rearranging:
100 kg m^2/s^2 = 400 kg^2 m^2/s^2 - 80 kg m/s * v2f + 4 kg^2 * v2f^2 + 2 kg^2 * v2f^2
100 kg m^2/s^2 - 400 kg^2 m^2/s^2 = 6 kg^2 * v2f^2 - 80 kg m/s * v2f
-300 kg m^2/s^2 = 6 kg^2 * v2f^2 - 80 kg m/s * v2f
Rearranging and dividing through by 10 kg m/s:
6 kg * v2f^2 - 80 m/s * v2f - 30 kg m/s^2 = 0
Now we have a quadratic equation in terms of v2f. We can solve this equation to find the values of v2f.
Solving this quadratic equation, we get two possible solutions for v2f:
v2f ≈ -2.19 m/s or v2f ≈ 5.19 m/s
Substituting these values back into the first equation:
v1f = (20 kg m/s - 2 kg * -2.19 m/s) / 4 kg ≈ 10.59 m/s
or
v1f = (20 kg m/s - 2 kg * 5.19 m/s) / 4 kg ≈ 6.93 m/s
Therefore, assuming a perfectly elastic collision, the velocities of the two balls after the collision are approximately:
The 4 kg ball: v1f ≈ 10.59 m/s (moving east) The 2 kg ball: v2f ≈ -2.19 m/s (moving west) or v2f ≈ 5.19 m/s (moving east)