Suppose that the final velocities of the rear and front buggies after jump became v1 and v2 respectively.

For the rear buggy, momentum before the jump is (M+m)v0. After the jump the rear buggy moves at v1 and so with a momentum Mv1. The person in the air moves with a velocity u relative to the rear buggy and so his velocity relative to a stationary observer is v1 + u. The momentum of the person in the air is thus m(v1+u). Since, there was no external force acting on the (rear buggy + man) system their total momentum must be conserved immediately before and after jump. This means that,

For the front buggy, the momentum of the (man + buggy) system just before and after he lands is conserved. Before he lands into the front buggy his momentum in the air is m(v1 + u) and the momentum of the buggy in front is Mv0 . After her lands, both the buggy and man move at a speed v2 together and so their combined momentum is (M+m)v2. In other words,

## Monday, May 12, 2008

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## 1 comment:

wow thnx a lot

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