Initially, the whole system is at rest, so `v_(CM)=0`. As there is no external force acting on the system`vecv_(CM)=`constant `= 0`. So position of centre of mass of system remains fixed.
`x_(CM)=(mx_(1)+Mx_(2))/(m+M)`…………i
where in is mass of the cannon balls and `M` that of the (car `+` cannon) system.
`As /_x_(CM)=0` therefore
`m/_x_(1)+m/_x_(2)=0`.....ii
As cannon balls cannot leave the car, so maximum displacement of the balls relative to the car is `L` and in doing so the car will shift a distance `/_x_(2)=D`(say) relative to the ground, opposite to the displacement of the balls, then the displacement of balls relative to ground will be
`/_x_(1)=L-D`.........iii
Substituting the value of `/_x_(1)` from eqn iii in eqn ii we get
`m(L-D)-MD=0`
`impliesD=(mL)/(M+m)=L/(1+M/m)`
`implies DltL`
i.e. rail road car cannot travel more than `L`,