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Question and Answer

Two points masses of mass \(4m\) and \(m\) respectively separated by distance \(r\), are revolved under mutual force of attraction. Ratio of their kinetic energies will be
(A) \(1 : 4\)
(B) \(1 : 5\)
(C) \(1 : 1\)
(D) \(1 : 2\)

Answer

Answer: A
Considering the origin of the coordinate system at \(4m\), we evaluate the position of the center of mass as
\(\dfrac{4m\times 0+m\times r}{4m+m}=\dfrac{r}{5}\)
Thus the center of mass is \(\dfrac{r}{5}\) from \(4m\) and \(\dfrac{4r}{5}\) from m.
The ratio of their kinetic energy is given as
\(\dfrac{\dfrac{1}{2}[I\omega^2]_{4m}}{\dfrac{1}{2}[I\omega^2]_m}\)

As the angular velocity of the both the masses would be same we get the ratio of kinetic energy as
\(\dfrac{4m(\dfrac{r}{5})^2}{m(\dfrac{4r}{5})^2}=\dfrac{1}{4}\)
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Solution

Considering the origin of the coordinate system at 4m, we evaluate the position of the center of mass as
\(\frac{{4m \times 0 + m \times r}}{{4m + m}} = \frac{r}{5}\)
Thus the center of mass is \(\frac {r}{5}\) from \(4m\) and \(\frac {4r}{5}\) from m.
The ratio of their kinetic energy is given as
\(\frac{{\frac{1}{2}{{[I{\omega ^2}]}_{4m}}}}{{\frac{1}{2}{{[I{\omega ^2}]}_m}}}\)
As the angular velocity of the both the masses would be same we get the ratio of kinetic energy as
\(\frac{{4m{{(\frac{r}{5})}^2}}}{{m{{(\frac{{4r}}{5})}^2}}} = \frac{1}{4}\)
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