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 `AB=(1)/(2)` (BC). M is much smaller than M and at time t = 0, they are all at rest given in the figure. At subsequent times before any collision takes place. A. m will remain at restB. m will move towards MC. m will move towards 2 MD. m will have oscillatory motion |
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Answer» Correct Answer - C Gravitational force on m due to `2M = F_(BA)` `= (G(2M)m)/((AB)^(2))` along `vec(BA)" "` …. (1) Gravitational force on m due to `M = F_(BC)` `=(G(M)m)/((BC)^(2))` along `vec(BC)` `because BC=2AB " " therefore " " F_(BC)=(GMm)/(4AB^(2))" "` ...... (2) Thus from (1) and (2) we find that `F_(BA)gt F_(BC)` `therefore` m will move towards 2M along `vec(BA)`. |
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