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

The mass distribution of the two pieces does not matter, so we will assume they are equal. (What you assume them to be does not matter.) Thus, the initial kinetic energy is: $=\frac{1}{2}(2m)v^2=mv^2$ The final kinetic energy is: $=\frac{1}{2}m(2v)^2+\frac{1}{2}m(2v)^2=4mv^2$ Thus, the kinetic energy after the explosion is four times as great.