Two identical railroad cars sit on a horizontal track, with a distance $D$ between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. $(a)$ Describe their relative motion. (b) Repeat the analysis if the mass of one car is now three times that of the other.
Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass,
$$
x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}
$$
where $x_{1}$ and $x_{2}$ are the positions of the centers of mass of the two
cars.
(a) If $m_{1}=m_{2}$, this equation becomes
$$
0=\frac{x_{1}+x_{2}}{2} \text { or } x_{1}=-x_{2}
$$
The two cars approach the center of mass, which is originally midway between the two cars (that is, $D / 2$ from each), in such a way that their centers of mass are always equidistant from it.
(b) If $m_{1}=3 m_{2}$, then we have
$$
0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4}
$$
from which $x_{1}=-x_{2} / 3$. Since $m_{1}>m_{2}$, it must be that $x_{1}<x_{2}$ proportionately. The two cars approach each other in such a way that the center of mass of the system remains motionless
and the heavier car is always one-third as far away from it as the lighter car.
Originally, because $\left|x_{1}\right|+\left|x_{2}\right|=D$, we had $x_{2} / 3+x_{2}=D$. So $m_{2}$ was originally a distance $x_{2}=3 D / 4$ from the center of mass, and $m_{1}$ was a distance $D / 4$ from it.