(i) Find the acceleration of the centre of mass of two particle approaching towards each other under their own grabitational field.
(ii) A boy of mass 30 kg is standing on a flat boat so that he is 20 meter from the shore. He walks 8 m on the boat towards the shore and then stops. The mass of the boat is 90 kg and friction between the boat and the water surface is negligible. How far is the boy 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, `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.