chemistry
posted by john .
The kinetic energy of a rolling billiard ball is given by {\rm{KE}} = 1/2\,mv^2 . Suppose a 0.17{\rm kg} billiard ball is rolling down a pool table with an initial speed of 4.5 m/s. As it travels, it loses some of its energy as heat. The ball slows down to 3.5 m/s and then collides straighton with a second billiard ball of equal mass. The first billiard ball completely stops and the second one rolls away with a velocity of 3.5 m/s. Assume the first billiard ball is the system

Not a chemistry question. Someone in physics, please.

There is no question here. The statement that the KE of the ball is (1/2) M V^2 is incorrect if it is rolling, as it says it is. The KE of a rolling solid sphere is (7/10) M V^2.
I have never seen a rolling billard ball lose 22% of its speed while crossing a pool table.
Under these circumstqances, there is no properly posed question for us to answer.
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