Answer
Since the kinetic energy after the collision is less than the kinetic energy before the collision, kinetic energy is not conserved.
Work Step by Step
In part (b), we found that the original speed of the cue ball is $~~4.75~m/s$
We can find an expression for the kinetic energy before the collision:
$K_i = \frac{1}{2}Mv_i^2$
$K_i = \frac{1}{2}M(4.75)^2$
$K_i = (11.3~M)~J$
We can find an expression for the total kinetic energy after the collision:
$K_f = \frac{1}{2}M(3.50)^2+\frac{1}{2}M(2.00)^2$
$K_f = (6.125~M)~J+(2.00~M)~J$
$K_f = (8.125~M)~J$
Since the kinetic energy after the collision is less than the kinetic energy before the collision, kinetic energy is not conserved.
