A boy of mass m =50 kg stands at the end A of a flat plank AB of wood of mass M=100 kg and length  $l$ =10 m floating in the still water in a lake. The end B of the plank is at a distance of 30 m from the shore of the lake as shown in fig. the boy walks a distance of 6 m on the plank towards the shore. How far is the boy from the same shore now? Neglect viscosity of water.
 

Initially the centre of mass of the system (plank +boy) is at rest. To walk, the boy exerts a force in the backward direction. The plank in turns exerts a reaction force on the boy in the forward direction. These forces are internal to the system. Sine no external force acts, the centre of mass of the system remains at rest even when the boy walks on the plank.
Let the shore be at the origin $o(x=0).$  initially let $x$ be the distance of the centre of mass of the plank from $o$ .  Then the distance of the centre of mass of the system (plank +boy) from $o$  will be

                $x_{CM}=\frac{M\times x+m\times (30+10)}{M+m}=\frac{100x+50\times 40}{100+50}$
                                                                       $=\frac{100x+2000}{150}$

Since the boy moves towards the shore and the centre of mass of the system has to remain at rest, the plank will move away from the shore. If  is the distance moved by the plank, the distance of the centre of mass from the shore when the boy walks 6 m on the plank is given by

                       $x_{CM}=\frac{100(x+x)+50(40-6+x)}{100+50}$
                                $=\frac{100x+150\times 1700}{150}$
Since  $x_{CM}=x_{CM}$
         $\frac{100x+2000}{150}=\frac{100x+150\times 1700}{150}$
$\Rightarrow x=\frac{300}{150}$ = 2m
$\therefore$Distance of the boy from the shore $=40-6+2=36$  m