Answer
17 percent
Work Step by Step
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:
$=7+10=\fbox {17 percent}$
