Answer
The proof is below.
Work Step by Step
We use our answers from the last problem. The initial kinetic energy is:
$K_i=\frac{1}{2}(4m)v^2+\frac{1}{2}(m)(2v)^2=4mv^2$
The final kinetic energy is
$K_f=\frac{1}{2}(4m)(1.4v)^2+\frac{1}{2}(m)(.4v)^2=4mv^2$
Thus, kinetic energy is conserved.
