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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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4 m

7 m

8 m

6 m

answer is A.

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Detailed Solution

If V be the velocity of the boy,

after linear momentum conservation

$30 \times V + 5 \times 2 = 0$

$h e n c e, V = - \frac{1}{3} [i . e, t h e y m o v e i n o p p o s i t e d i r e c t i o n s]$

$t h e i r s e p e r a t i o n a f t e r 3 s e c :$

$d = (2 + \frac{1}{3}) 3 = 7 m$

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