## Essential University Physics: Volume 1 (4th Edition)

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.