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 the momentum conservation, for the system "near buggy with man"
$(M+m){\overrightarrow{\nu }}_0=m(\overrightarrow{u}+{\overrightarrow{\nu }}_R)+M{\overrightarrow{\nu }}_R\left(1\right)$
From momentum conservation, fro the system (front buggy $+$ man coming from nearbuggy)
$M{\overrightarrow{\nu }}_0+m(\overrightarrow{u}+{\overrightarrow{\nu }}_R)$$=(M+m){\overrightarrow{\nu }}_R$
So, ${\overrightarrow{\nu }}_f=\frac{M{\overrightarrow{\nu }}_0}{M+m}+\frac{m}{M+m}(\overrightarrow{u}+{\overrightarrow{\nu }}_r)$
Putting the value of ${\overrightarrow{\nu }}_R$ from$\left(1\right)$ we get
${\overrightarrow{\nu }}_F={\overrightarrow{\nu }}_{)}+\frac{mM}{(M+m{)}^2}\overrightarrow{u}$