Question #86795

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Question #86795

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Answer 1 · 2017-08-17T14:16:03

The kinetic energy should become one ninth of the initial one.

Explanation:

The initial kinetic energy will be:
$K = \frac{1}{2} m v^{2}$ $K=1/2mv^2$
as the velocity becomes $v_{1} = \frac{1}{3} v$ $v_1=1/3v$ we get a new kinetic energy:
$K_{1} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} m {(\frac{1}{3} v)}^{2} = \frac{1}{2} m \frac{1}{9} v^{2} = \frac{1}{9} (\frac{1}{2} m v^{2}) = \frac{1}{9} K$ $K_1=1/2mv_1^2=1/2m(1/3v)^2=1/2m1/9v^2=1/9(1/2mv^2)=1/9K$

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