Three solid spheres each of mass m and radius R are placed at three corners of an equilateral triangle of side ‘d’ released. The speed of any one sphere at the time of collision would be d>2R [Assume there is no external gravitational force acting on the system of three spheres].
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a
Gm1d−3R
b
Gm3d−1R
c
Gm2R−1d
d
Gm1R−2d
answer is D.
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Detailed Solution
From conservation of mechanical energy Energy of system at the instance of release = Energy of system at the instance of Collision∴ 3−Gm2d=3−Gm22R+312mv2 and v=Gm1R−2dTherefore, the correct answer is (D).