Two identical buggies move one after the other due to inertia (without friction) with the same velocity `v_0`. A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy with a velocity u relative to his buggy. Knowing that the mass of each buggy is equal to M, find the velocities with which the buggies will move after that.

Two identical buggies move one after the other due to inertia (without

1.	Two identical buggies move one after the other due to inertia (without friction) with the same velocity `v_0`. A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy with a velocity u relative to his buggy. Knowing that the mass of each buggy is equal to M, find the velocities with which the buggies will move after that.
Answer» From momentum conservation, for the system "rear buggy with man" `(M+m)vecv_0=m(vecu+vecv_R)+Mvecv_R` (1) From the momentum conservation, for the system (front buggy+man coming from near buggy) `Mvecv_0+m(vecu+vecv_R)=(M+m)vecv_F` So, `vecv_F=(Mvecv_0)/(M+m)+(m)/(M+m)(vecu+vecv_R)` Putting the value of `vecv_R` from (1), we get `vecv_F=vecv_0+(mM)/((M+m)^2)vecu`

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