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Question

# A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other. One with a velocity of 2 ˆim/s and the other with a velocity of 3 ˆj m/s. If the explosion takes place in 10−5s, the average force acting on the third piece in newtons is :

A
(2ˆi+3ˆj)×105
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B
(2ˆi+3ˆj)×105
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C
(3ˆi+2ˆj)×105
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D
(2ˆi+3ˆj)×105
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Solution

## The correct option is A −(2ˆi+3ˆj)×105Mass of each fragment m=M3=3kg3=1kgLet the momentum of third piece be →P3 and that of other two be →P1 and →P2∴ →P1=mv1=2^i and →P2=mv2=3^jInitial momentum of the bomb is zero i.e →Pi=0Applying conservation of momentum:0=→P1+→P2+→P30=2^i+3^j+→P3 ⟹→P3=−(2^i+3^j) kgms−1Now average force acting on third piece →F=→P3t=−(2^i+3^j)10−5⟹→F=−(2^i+3^j)×105 N

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