Question

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 1

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`