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

Initially rear buggy $+$ man were moving with velocity $v_0$ After the man jump into the front buggy, let the velocity of the rear buggy becomes $v_{rear}$. Since the velocity of the man relative to rear buggy is $u$, it follows that

                      $\overrightarrow{u} = {\overrightarrow{V}}_{man}-{\overrightarrow{V}}_{rear}$

where ${\overrightarrow{V}}_{man}$ is the velocity of the man with respect to the ground.

                          $V_{man }= u+ v_{rear}$

Applying momentum conservation for the rear buggy, we have

$$\begin{gathered} \left(M +m\right)v_0 = M{v}_{rear}0+m\left(u+v_{rear}\right) \\ v_{rear}= v_0-\frac{mu}{\left(M+m\right)} \left(i\right) \end{gathered}$$

When the man jumps into the front buggy, let the velocity of this buggy becomes $v_{front}$. For the two buggies system from momentum conservation, we have

     $M{v}_0+\left(M+m\right)v_0 = M{v}_{rear}+(M+m)v_{front} \left(ii\right)$

after solving above equations we get

$v_{front}= v_0+\frac{mMu}{{\left(M+m\right)}^2}$