Question:
A rocket is moving up with a velocity $u$. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Solution:
$\mathrm{K}_{1}=\frac{1}{2} m v^{2}$
$\mathrm{K}_{2}=\frac{1}{2} m(3 v)^{2}=\frac{9}{2} m v^{2}$
$\therefore \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{1}{9}$ or $\mathrm{K}_{2}=9 \mathrm{~K}_{1}$
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