A man of mass `m` moves on a plank of mass `M` with a constant veloicty `u` with respect to the plank, as shown in figure.
a.If the plank rest on smooth horizontal surface, determine the velocity of the plank.
b If the man travels a distasnce `L` with respect to the plank, find the distance travelled by the plank with respect to te ground.

Explanation

As no external forces are acting on the system in horizontal direction, so its linear momentum remains constant in that direction. Also there will be no shift in position of centre of mass, i.e `/_\x_(CM)=0`
a. Let `v_(1)` and `v_(2)` be the velocities of the man the plane w.r.t ground. Then, we have
`[vecv_("man")]_("plank")[vecv_("man")]_("ground")-[vecv_("plank")]_("ground")`
Then `[vecv_("man")]_("ground")=[vecv_("man")]_("plank")+[vecv_("plank")]_("ground")`
or `v_(1)=u-v_(2)`

Initially the system is at rest. Therefore,
`0=mv_(1)MV_(2)`
or `0=m(u-v_(2))-Mv_(2)`
or `v_(2)=(mu)/(m+M)`
b. Let `x` be the displacement of plank in backward direction, the displacement of mass is `L-x` in forward direction.
`/_\x_(CM)=(M(-x)+m(L-x))/(M+m)`
`implies0=-Mx+m(L-x)impliesx=(mL)/(M+m)`