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Q.

A boy of 30 kg mass, standing in a gravity free space throws a stone of mass 5 kg with a velocity of 2 m/s. what will be the separation between the boy and the stone 3 second after the stone is thrown? 

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a

7 m

b

8 m

c

6 m

d

4 m

answer is A.

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Detailed Solution

If V be the velocity of the boy,

 after linear momentum conservation

30×V+5×2 =0

hence, V=-13     [i.e, they move in opposite directions]

their seperation after 3 sec : 

d=2+133 =7m

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