InterviewSolution
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InterviewSolution
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Particles of masses 2M, m and M are respectively at points A, B and C with `sqrt((GM)/(R))(sqrt(2)+1)` is much-much smaller than M and at time `t=0`, they are all at rest as given in figure At subsequent time before any collision takes place.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 Fore 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))lt F_(BA)` Hence, `m` will move towards BA (i.e., 2M) |
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