A stationary body of mass 3kg explodes into three equal pieces. Two of the pieces fly off at at right angles to each other, one with a velocity2i^ m/s and the other with a velocity 3j^ m/s. If the explosion takes place in 10-5sec, the average force acting on the third piece in newton is
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a
(2i^+ 3j^) 10-5
b
-2i^ +3j^ 105
c
3j^+2i^ 105
d
2i^ -3j^ 105
answer is B.
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Detailed Solution
according to conservation of linear momentum 0 =m32i^+m32j^+m3v m3v =-m32i^+m32j^ force on third piece is m3vt =-2i^+2j^ 105