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

We first find the total energy: $E_t=\frac{1}{2}mv^2 + (\frac{1}{2}mv^2 )(.4) \\ E_t=.7MR^2$ We know that the mass declined by 10 percent, so we obtain: $\Delta E_t=.7\times.1=.07$ Adding the 10 percent change in inertia, it follows: $=7+10=\fbox {17 percent}$