Two billiard balls of identical mass move straight toward each other (head-on, i

Question

Two billiard balls of identical mass move straight toward each other (head-on, in a straight line towards each other). Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are +28.5 cm/s and -20.2 cm/s, what is the velocity of each ball after the collision? Assume friction and rotation are unimportant.
v1f = -20.2 cm/s
v2f = 28.5 cm/s

Find the final velocity of the two balls if the ball with velocity v2i = -20.2 cm/s has a mass equal to half that of the ball with initial velocity v1i = +28.5 cm/s.

Try as I might, I cannot figure this out - could be a conceptual or simple mathematical mistake I am making, I am not sure. Hence, can you please provide a step-by-step explanation along with correct answers? Thank you!!!

Explanation / Answer

The first case for which both momentum and kinetic energy are conserved i.e. for elastic collision, as the masses of the balls are same, the velocities are exchanged. If m2 = 0.5m1 Applying the law of conservation of liear momentum and the conservation of kinetic energy we get the final velocities of the masses v1f = (m1-m2)v1i/(m1+m2) + (2m2)v2i/(m1+m2) v1f = 0.5m1v1i/1.5m1 + m1v2i/1.5m1 v1f = v1i/3 + 2v2i/3 v1f = 28.5/3 + 2*-20.2/3 v1f = -3.967 cm/s v2f = (m1-m2)v2i/(m1+m2) + (2m2)v1i/(m1+m2) v2f = 0.5m1v2i/1.5m1 + m1v1i/1.5m1 v2f = v2i/3 + 2v1i/3 v2f = -20.2/3 + 2*28.5/3 v2f = 12.267 cm/s v2f = (m1-m2)v2i/(m1+m2) + (2m2)v1i/(m1+m2) v2f = 0.5m1v2i/1.5m1 + m1v1i/1.5m1 v2f = v2i/3 + 2v1i/3 v2f = -20.2/3 + 2*28.5/3 v2f = 12.267 cm/s v2f = 0.5m1v2i/1.5m1 + m1v1i/1.5m1 v2f = v2i/3 + 2v1i/3 v2f = -20.2/3 + 2*28.5/3 v2f = 12.267 cm/s v2f = v2i/3 + 2v1i/3 v2f = -20.2/3 + 2*28.5/3 v2f = 12.267 cm/s v2f = -20.2/3 + 2*28.5/3 v2f = 12.267 cm/s

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