A ring of radius R = 4 m made of a highly dense material has a mass $m_{1} = 5 .4 \times 10^{9} kg$ distributed uniformly over its circumference. A highly dense particle of mass $m_{2} = 6 \times 10^{8} kg$ is placed on the axis of the ring at a distance 3 m from the center of the ring. Find the speed of the particle (in cm/s), when the particle is at the center of the ring. Except mutual gravitational interaction of the two, neglect all other forces.

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answer is 18.

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Detailed Solution

When the particle is at the center, let their respective speeds
are v₁ and v₂.
From conservation of linear momentum,

$\begin{array}{l} 0 = 5 .4 \times 10^{9} v_{1} - 6 \times 10^{8} v_{2} \\ ∴ v_{2} = 9 v_{1} \end{array}$

From conservation of energy, $(U + K)_{i} = (U + K)_{f}$

$\begin{array}{l} - \frac{{Gm}_{1} m_{2}}{\sqrt{4^{2} + 3^{2}}} + 0 = - \frac{{Gm}_{1} m_{2}}{4} + \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} \\ {Gm}_{1} m_{2} (\frac{1}{4} - \frac{1}{5}) = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} (81 v_{1}^{2}) \end{array}$

$Now m_{1} = 9 m_{2} (given)$

$\Rightarrow G \frac{(9 m_{2}^{2})}{20} = \frac{1}{2} (90 m_{2}) v_{1}^{2}$

$\begin{array}{l} \Rightarrow v_{1} = \sqrt{\frac{{Gm}_{2}}{100}} = \sqrt{\frac{6 .67 \times 10^{- 11} \times 6 \times 10^{8}}{100}} = 0 .02 {ms}^{- 1} \\ ∴ v_{2} = 9 v_{1} = 0 .18 {ms}^{- 1} = 18 {cms}^{- 1} \end{array}$