A dog weighting 5 kg is standing on a flat boat so that he is 10 metre from the shore. He walks 4 metre on the boat toward shore and then halts. The boat weight 20 kg and one can assume that there is no friction between it and the water. How fat is the dog from the shore at the end of this time ?

A dog weighting 5 kg is standing on a flat boat so that he is 10 metre

1.	A dog weighting 5 kg is standing on a flat boat so that he is 10 metre from the shore. He walks 4 metre on the boat toward shore and then halts. The boat weight 20 kg and one can assume that there is no friction between it and the water. How fat is the dog from the shore at the end of this time ?
Answer» Solution :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 log 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)=3.2m` 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.