Question

Two identical buggies move one after the other due to inertia (without friction) with the same velocity ${\overrightarrow{v}}_0$. 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"
$(M+m)\overrightarrow{v_0}=m(\overrightarrow{u}+\overrightarrow{v_R})+M\overrightarrow{v_R}$ (1)
From momentum conservation, for the system (front buggy $+$ man coming from rear buggy)
$M\overrightarrow{v_0}+m(\overrightarrow{u}+\overrightarrow{v_R})=(M+m)\overrightarrow{v_F}$
So, $\overrightarrow{v_F}=\frac{M\overrightarrow{v_0}}{M+m}=\frac{m}{M+m}(\overrightarrow{u}+\overrightarrow{v_R})$
Putting the value of $\overrightarrow{v_R}$ from (1), we get
$\overrightarrow{v_F}=\overrightarrow{v_0}+\frac{mM}{{(M+m)}^2}\overrightarrow{u}$