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(D) Let the mass of each body be $m$. The kinetic energy $(KE)$ of the first body is given by $K_{1} = \frac{1}{2}mv^{2} \quad \dots(i)$. The kinetic

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$9:1$

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(D) Let the mass of each body be $m$. The kinetic energy $(KE)$ of the first body is given by $K_{1} = \frac{1}{2}mv^{2} \quad \dots(i)$. The kinetic energy $(KE)$ of the second body is given by $K_{2} = \frac{1}{2}m(3v)^{2} = \frac{1}{2}m(9v^{2}) = \frac{9}{2}mv^{2} \quad \dots(ii)$. To find the ratio of their kinetic energies,we divide equation $(ii)$ by equation $(i)$: $\frac{K_{2}}{K_{1}} = \frac{\frac{9}{2}mv^{2}}{\frac{1}{2}mv^{2}} = \frac{9}{1}$. Thus,the ratio of their kinetic energies is $9:1$.

Two bodies of equal masses move with the uniform velocities $v$ and $3v$ respectively. Find the ratio of their kinetic energies.

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