Final answer:
After applying the conservation of momentum principle to the two colliding balls, the velocity of the 0.32 kg ball after the collision is found to be -0.47875 m/s, indicating it has reversed direction post-collision.
Step-by-step explanation:
The question asks us to determine the velocity of a 0.32 kg ball after it collides head-on with a 0.75 kg ball in a scenario implying a conservation of momentum because no external forces are acting on the balls. In an equation form, this principle can be written as m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the two balls, v1 and v2 are their initial velocities, and v1' and v2' are their velocities after the collision. Given that the 0.75 kg ball was initially at rest, v2 = 0 m/s, and we know m1 = 0.32 kg, v1 = 5.1 m/s, m2 = 0.75 kg, and v2' = 2.38 m/s after the collision. We can solve for v1' (the velocity of the 0.32 kg ball after the collision) by plugging in the known values and rearranging the equation as follows:
m1v1 + m2v2 = m1v1' + m2v2'
0.32 kg * 5.1 m/s + 0.75 kg * 0 m/s = 0.32 kg * v1' + 0.75 kg * 2.38 m/s
1.632 kg*m/s + 0 = 0.32 kg * v1' + 1.785 kg*m/s
v1' = (1.632 kg*m/s - 1.785 kg*m/s) / 0.32 kg
v1' = -0.47875 m/s
Therefore, the velocity of the 0.32 kg ball after the collision is -0.47875 m/s, which indicates that it has reversed direction due to the collision.