InterviewSolution
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InterviewSolution
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particles of masses 2M m and M are resectively at points A , B and C with ` AB = (1)/(2) (BC)` m is much - much smaller than M and at time ` t=0` they are all at rest as given in figure . As subsequent times before any collision takes palce . A. `m` will remain at restB. `m` will move towards MC. `m` will move towards 2MD. `m` will have oscillatory motion |
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Answer» Correct Answer - C Force on B due to `A = F_(BA) = (G(2Mm))/((AB)^(2))` towards BA Force on B due to `C = F_(BC) = (GMm)/((BC)^(2))` towards BC As, `(BC) = 2AB` `rArr F_(BC) = (GMm)/((2AB)^(2)) = (GMm)/(4(AB)^(2)) le F_(BA)` Hence, `m` will move towards BA (i.e., 2M) |
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