InterviewSolution
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InterviewSolution
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Two points mass `m` and `2m` are kept at a distance `a`. Find the speed of particles and their relative velocity of approach when separation becomes `a//2`. |
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Answer» Reduced mass `mu=(m.2m)/(m+2m)=(2m)/(3)` Changining in potential energy is `U_(1)-U_(2)=-(Gm.2m)/(a)-(-(Gm.2m)/(a//2))` `=(2Gm^(2))/(a)` `(1)/(2)mu v_(1//2)^(2)=(2Gm^(2))/(a)implies(1)/(2)=sqrt((6Gm)/(a))` `v_(1//2)=v_(1)+v_(2)=sqrt((6Gm)/(a))` ......(i) `mv_(1)=2mv_(2)` `v_(1)=2v_(2)` `v_(1)+v_(2)=3v_(2)=sqrt((6Gm)/(a))implies v_(2)=(1)/(3)sqrt((6Gm)/(a))=sqrt((2Gm)/(3a))` `v_(1)=2v_(2)=2aqrt((2Gm)/(3a))` |
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