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A dog weighing 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 weighs 20 kg and one can assume that there is no friction between it and the water. How far is the dog from the shore at the end of this time ? |
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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, `mvecv_(1) + Mvecv_(2) =0` [as (m+M) = Finite] or `m(Deltavecr_(1))/(dt) + M(Deltavecr_(2))/(dt)=0` [as `vecv =(dvecr)/(dt)`] or `mDeltavecr_(1) + MDeltar_(2)=0` `md_(1) -Md_(2) =0` [as `vecd_(2)` is OPPOSITE to `vecd_(1)`] i.e. `md_(1) = Md_(2)`......(1)  Now when dog moves 4 m towards shoie lelative 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) (therefore d_(1) + d_(2) =d_("ret") =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) = (20 xx 4)/(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. |
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