Answer to Question #91612 in Classical Mechanics for Neha

Question

Two identical buggies 1 and 2 with one man in each move without friction due to
inertia along the parallel rails toward each other. When the buggies get opposite each other, the
men exchange their places by jumping in the direction perpendicular to the motion direction. As
a consequence, buggy 1 stops and buggy 1 keeps moving in the same direction, with its velocity
becoming equal to v . Find the initial velocities of the buggies 1v and 2v if the mass of each buggy
(without a man) equals M and the mass of each man m.

Answer 1

Initially, two buggies have velocities $v_1$ and $v_2$ respectively.

In this situation, we only care about parallel velocity/momentum changes.

At the moment of jumping, parallel momentum of the buggies are $p_1=Mv_1,\;\;\; p_2=Mv_2$, and momentum of men: $P_1=mv_1,\;\;\; P_2=mv_2$

According to momentum conservation principle, and taking into the account that first buggy stopped after the jump:

$Mv_1-mv_2=0$

Second buggy, continued the movement with the velocity $v$:

$Mv_2-mv_1=(m+M)v$

Using these two equations, we find:

$v_2=\frac{M}{M-m}v$

$v_1=\frac{m}{M-m}v$