Two masses, initially at rest, attract each other with a constant force. If there is no external force acting on the masses, prove that the centre of mass of the system remains stationary.

Two masses, initially at rest, attract each other with a constant forc

1.	Two masses, initially at rest, attract each other with a constant force. If there is no external force acting on the masses, prove that the centre of mass of the system remains stationary.
Answer» Solution :Let the line joining the two masses be taken as X-axis. Initial POSITION of mass `m_(1) is x_(1)` and that of mass `" "m_(2) ""is" " x_(2) (x_(2) gt x_(1))`. Hence, the initial position of the centre of mass is `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` Suppose a force F acts on the first mass along the positive x direction. Hence a force F will act on the second mass too along the negative direction of the x-axis as per Newton.s third law of motion. Hence, acceleration of the first mass `a_(1)= (F)/(m_(1))` and DISPLACEMENT in time `t=(1)/(2)a_(1)t^(2)`. So, the position of the first one after a time t, `x_(1).=x_(1)+(1)/(2)a_(1)t^(2)=x_(1)+(1)/(2)(F)/(m_(1))t^(2)` Similarly, the position of the second one after a time t, `x_(2).=x_(2)+(1)/(2).((-F))/(m_(2))t^(2)=x_(2)-(1)/(2)(F)/(m_(2))t^(2)` Hence, the position of the centre of mass after time t, `x._("cm")=(m_(1)x_(1).+m_(2)x_(2).)/(m_(1)+m_(2))=(m_(1)x_(1)+(FT^(2))/(2)+m_(2)x_(2)-(Ft^(2))/(2))/(m_(1)+m_(2))` or, `x._("cm")=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=x_("cm")` Hence, the centre of mass remains stationary.