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Two point masses, each of mass M are fixed at point A and B respectively. A third point m is
released from infinity, so that it can move along y-axis under the influence of mutual gravitational attraction on it due to point masses kept at A and B respectively as shown in the figure (i). Figure (ii) represents the potential energy of system (includes m, M at A and M at B) with position of m at y-axis. (Neglect any forces other than gravity on particle C). (Given : $\frac{{Gm}^{2}}{d} = 12 joule, m = 6 k g)$ Given AC=d.
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point mass m will perform periodic motion

maximum speed of particle is 24 m/s

maximum speed of particle is 4 m/s.

value of $U_{1}$ must be equal to –24 joule if gravitational potential energy of two point mass system
is taken to be zero when separation between them is infinite.

answer is A, C.

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Detailed Solution

$U_{1} = - \frac{GMm}{(2 d)} = - \frac{4 {Gm}^{2}}{(2 d)}$

$U_{1} = - 24 J$

From energy conservation

$0 + 0 - 24 = \frac{1}{2} m v^{2} + (- \frac{GMm}{d}) \times 2 + (- 24)$

$\frac{2 GMm}{d} = \frac{1}{2} {mv}^{2}$

$\frac{4 {Gm}^{2}}{d} = \frac{1}{2} {mv}^{2} \Rightarrow v = 4 m s^{- 2}$

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