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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Updated on: 10-Oct-2022

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