If the velocity of a ball is tripled, what will be the ratio of its initial kinetic energy and final kinetic energy?

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Solution

The correct option is D $1 / 9$
$∵ v_{1} = v; v_{2} = 3 v$
$K E_{1} = \frac{1}{2} m v^{2}$
$K E_{2} = \frac{1}{2} m {(3 v)}^{2} = \frac{9}{2} m v^{2}$
$K E_{1} : K E_{2} = \frac{1}{2} m v^{2} : \frac{9}{2} m v^{2} = 1 : 9$

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If the velocity of a ball is tripled, what will be the ratio of its initial kinetic energy and final kinetic energy?

The correct option is D 1/9∵v1=v; v2=3vKE1=12mv2KE2=12m(3v)2=92mv2KE1:KE2=12mv2:92mv2=1:9

The correct option is D $1 / 9$
$∵ v_{1} = v; v_{2} = 3 v$
$K E_{1} = \frac{1}{2} m v^{2}$
$K E_{2} = \frac{1}{2} m {(3 v)}^{2} = \frac{9}{2} m v^{2}$
$K E_{1} : K E_{2} = \frac{1}{2} m v^{2} : \frac{9}{2} m v^{2} = 1 : 9$