A wooden plank of mass `20kg` is resting on a smooth horizontal floor. A man of mass `60kg` starts moving from one end of the plank to the other end. The length of the plank is `10m`. Find the displacement of the plank over the floor when the man reaches the other end of the plank.

Explanation

Here, the system is man+plank. Net force on this system in horizontal direction is zero and initially the centre of mass of the system is at rest. Therefore, the centre of mass does not move in horizontal direction.
Let x be the displacement of the Plank. Assuming the origin, i.e. `x=0` at the position show in figure

As we said earlier also, the centre of mass will not move in horizontal direction (x-axis). Therefore, for centre of mass to remain stationary,
`x_i=x_f`
`((60)(0)+20(10/2))/(60+20)=((60)(10-x)+20(10/2-x))/(60+20)`
`5/4=(6(10-x)+2(10/2-x))/(8)=(60-6x+10-2x)/(8)`
or `5=30-3x+5-x`
or `4x=30`
or `x=30/4m` or `x=7.5m`
Alternate Method
`x_L="displacement of plank towards left"=x`
`m_L="mass of plank displaced towards left"=20kg`
`x_R="displacement of man relative to ground towards right" =10-x`
and `m_R="mass of man displaced towards right"=60kg`
Applying `x_Rm_R=x_Lm_L`, we get
`(10-x)(60)=20x`
or `x=30-3x`
or `4x=30`
`:. x=30/4=7.5m`