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Two identical buggies 1 and 2 with one man in each move along parallel rails. When the buggies are opposite to each other, the men jump in a direction perpendicular to the direction of motion of buggies, so as to exchange their places. As a consequence, buggy 1 stops and buggy 2 keeps moving in the same direction with its final velocity v. Find the initial velocities v_(1) and v_(2) of buggies. Mass of each buggy (without man) equals M mass of each man is ignore frictional effects anywhere and the buggies are constrained to move along the rails only. |
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Answer» <P> Solution :In the PROBLEMS involving mass EJECTION and mass addtions, the ejected and added mass both are part of a system. When the men jump they push on the floor of the bugy (action), the same force acts on them as well (reaction).This action reaction pair is intenal force for the system of man and buggy. similarly, when men jump in, there is an action reaction force between them and the buggy. Let buggy 1 moves to the right and buggy 2 to the left, when the men jump. Man in 1 brings momentum `mv_(1)` directed towards right in buggy 2. similarly, the main 2 brings momentum `mv_(2)` in `2` directed towards left. for system 1: `P_(i)(M+m)v_(1)-mv_(1)-v_(2)=Mv_(1)-mv_(2)` `P_(f)=0` Form law of conservation of momentum.  `P_(1)=P_(f)` `Mv_(1)=mv_(2)=0.............i` For system 2: `P_(i)=(M+m)v_(2)-mv_(1)-mv_(2)` `P_(f)=(M+m)v` `P_(i)=P_(f)`.............ii `(M_(m))v_(2)-mv_(1)-mv_(2)=(M+m)v`..........ii  On solving eqn i and ii simultaneouysly we obtain `v_(1)=(mv)/((M-m)), v_(2)=(Mv)/((M-m))` |
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