A dog of mass 10 kg is standing on a flat 10 m long boat so that it is 20 m meters from the shore. It walks 8 m on the boat towards the shore and then stops. The mass of the boat is 40 kg and friction between the boat and the water surface is negligible. How far is the dog from the shore now ?

Explanation

Given that initially the system is at rest so initial momentum of the system (Dog + boat) is zero. Now as in motion of dog no external force is applied to the system final momentum of the system zero.
So, `m vec(v)_(1)+M vec(v)_(2)=0` [as (m+M)= Finite]
or `m(Delta vec(r )_(1))/(dt)+M(Delta vec(r )_(2))/(dt)=0 " " [as vec(v)=(d vec(r ))/(dt)]`
or `m Delta vec(r )_(1)+M Delta vec(r )_(2)=0`
[as `Delta vec(r )=vec(d)=` displacement]
`md_(1)-Md_(2)=0` [as `vec(d)_(2)` is opposite to `vec(d)_(1)`]
i.e., `md_(1)=Md_(2) " "` ......(1)

Now when dog moves 4 m towards shore relative to boat, the boat will shift a distance `d_(2)` relative to shore opposite to the displacement of dog so, the displacement of dog relative to shore (towards shore) will be
`d_(2)=4-d_(1)(because d_(1)+d_(2)=d_(rel)=4)` .....(2)
substituting the value of `d_(2)` from Eqn. (2) in (1)
`md_(1)=M(4-d_(1))` or `d_(1)=(M xx 4)/((m+M))=(20xx4)/(5+20)=3.2 m`
As initially the dog was 10 m from the shore, so now he will be 10 - 3.2 = 6.8 m away from the shore.