Question

Two identical buggies each of mass $M$ moves one after due to inertia (without friction) with the some velocity. A man of mass $m$ rides the rear buggy. At a certain moment, the man jumps into the front buggy with velocity relative to this buggy. Knowing that the mass of each buggy is equal to $M$. Find the velocity with which the buggies will move after that.

Answer 1

If we take rear 'buggy + man' as system. There is no external force acting on system in horizontal direction. Hence linear momentum of the system should be conserved. Let velocity of rear buggy just after jumping of man just after jumping will be $({\overrightarrow{V}}_r+\overrightarrow{u})$.
${\overrightarrow{p}}_i=(M+m){\overrightarrow{V}}_0$
${\overrightarrow{p}}_j=M{\overrightarrow{V}}_r+m({\overrightarrow{V}}_r+\overrightarrow{u})$ and ${\overrightarrow{p}}_i={\overrightarrow{p}}_j$
$⟹(M+m){\overrightarrow{V}}_0=M{\overrightarrow{V}}_r+m({\overrightarrow{V}}_r+\overrightarrow{u})$
${\overrightarrow{V}}_r={\overrightarrow{V}}_0-\frac{m\overrightarrow{u}}{(M+m)}$ (i)