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Two identical buggies, each of mass M, glide one behind the other along a frictionless track with the same velocity v0 (motion sustained by inertia). A man of mass m sits in the rear buggy. He suddenly leaps into the leading buggy with a velocity v measured relative to the rear buggy. Determine the final velocities of the rear and front buggies after the jump.

1. (Mv0 + mv)/(M + m) and (Mv0 - mv)/(M + m)
2. (2Mv0 + mv)/(2M + m) and (2Mv0 - mv)/(2M + m)
3. v0 + mv/M and v0 - mv/M
4. v0 and v0

Correct answer: (2Mv0 + mv)/(2M + m) and (2Mv0 - mv)/(2M + m)

Solution

When the man jumps backward relative to the rear buggy, that buggy speeds up; when he lands in the front buggy, the combined mass (M + m) gains the man's momentum, giving the standard Irodov result.

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