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A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Let $m$ be the mass of the rocket flying with a velocity $v$.
So, kinetic energy of the rocket, $K=\frac{1}{2}mv^2$
When the velocity of rocket is tripled suddenly, it becomes $3v$.
Therefore, kinetic energy $K'=\frac{1}{2}m(3v)^2$
$=\frac{9}{2}mv^2$
Now, $\frac{K}{K'}=\frac{\frac{1}{2}mv^2}{\frac{9}{2}mv^2}=\frac{1}{9}$
Or $K:K'=1:9$
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