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)`