Essentially in this problem, the momentum of each buggy is conserved in their direction of motion, since there are no external forces in this direction acting on the buggies. Initially (before the people do the exchange jump) the momentum of the buggies are (m + M) v1 and (m+M)v2 respectively. When the people jump across their buggies, they do not loose their respective momenta in the air and so when they land in the opposite buggy, they add their respective momenta to the opposite buggy.

After the jump, for buggy 1 , the total momentum is the sum of the momentum of the buggy and the momentum of the man in buggy 2 (since he jumped into buggy 1), i.e. Mv1 + mv2. The momentum for buggy 2 after the jump is, the sum of the momentum of buggy 2 and the man in buggy 1 (since now he has jumped into buggy 2), i.e. Mv2 + mv1. The problem tells us that finally the buggy 1 stopped while buggy 2 now moves at a velocity v. This means that,

Solving these two equations we have,

## Monday, May 12, 2008

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