# Question #86795

$K = \frac{1}{2} m {v}^{2}$
as the velocity becomes ${v}_{1} = \frac{1}{3} v$ we get a new kinetic energy:
${K}_{1} = \frac{1}{2} m {v}_{1}^{2} = \frac{1}{2} m {\left(\frac{1}{3} v\right)}^{2} = \frac{1}{2} m \frac{1}{9} {v}^{2} = \frac{1}{9} \left(\frac{1}{2} m {v}^{2}\right) = \frac{1}{9} K$