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Question

# A stationary body of mass 3 kg, explodes into three equal parts. Two of the pieces fly off at right angles to each other, with velocities of 2^i m/s and 3^j m/s. If the explosion takes place in 10−3 s, find out the average force exerted on the third piece.

A
(2^i3^j)×103 N
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B
(2^i+3^j)×103 N
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C
(2^i3^j)×103 N
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D
None of these
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Solution

## The correct option is A (−2^i−3^j)×103 NAccording to conservation of momentum principle, →pi=→p1f+→p2f+→p3f 0=2^i+3^j+→p3f ∴→p3f=−2^i−3^j For the third piece, average force, →F=Δ→pΔt=(−2^i−3^j)−010−3 ⇒→F=(−2^i−3^j)×103 N

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