A `20 kg` kid is sitting in a `60 kg` boat in a lake. The distance of kid from the bank of lake is `20 m`. If the kid moves `8m` on the boat towards the bank, then find the distance of kid from the bank. The system is initially at rest and there is no friction between boat and water.

Explanation

Let `c.m.` of boat is at distance `x_(0)` from bank
`c.m. of ( boat + kid) ("before")`
`x_(c.m.) = ( 20 xx 20 + 60 x_(0))/(20 + 60)` (i) when the kid moves towards the bank , by `8m`, the boat moves away from the bank by `x` , because here `vec(F)_(ext) = 0`. Initially `vec(v)_(c.m.) = 0` and hence location of `c.m.` is fixed.
`x'_(c.m.) = (20 ( 20 - 8 + x) + 60(x_(0) + x))/(20 + 60)`
As the location of `c.m.` is fixed,
`x_(c.m.) = x'_(c.m.)`
`x = 2 m`
Distance of kid from the bank `= 20 - 8 + x = 14 m`
Short - cut
When the location of center of mass is fixed , i.e. `vec(v)_(c.m.) = 0`
`vec(v)_(c.m.) = (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))/(m_(1) + m_(2)) = 0`
`m_(1) vec(v)_(1) + m_(2) vec(v)_(2) = 0`
`m_(1) (Delta vec(r)_(1))/(Delta t) + m_(2) (Delta vec(r)_(2))/(Delta t) = 0`
`m_(1) Delta vec( r)_(1) + m_(2) Delta vec(r)_(2) = 0`
`m_(1) d_(1) = m_(2) d_(2)`
where `d_(1) , d_(2)` : distance travelled by `m_(1)` and `m_(2)` with respect to the ground.
Alternative solution for example 18 : Since the kid moves `8m` towards the bank and (boat + kid) move by `x` away from the bank.
Distance travelled by the boat with respect to the ground `= x`
Distance travelled by the kid with respect to the ground `= 8 -x`
`m_(1) d_(1) = m_(2) d_(2)`
`20(8 - x) = 60s rArr x = 2m`
Distance of the kid from the bank of lake is
`20 - 8 + x = 12 + 2 = 14 m`
OR
Since the kid moves towards the bank , c.m. of the system will be shifted towards the bank , but the location of c.m. is fixed. For this , the boat has to move away from the bank, (on the boat , kid is there) , so (boat + kid) have to move away from the bank.
`20 xx 8 = (20 + 60)x rArr x = 2 m`