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Q.

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 2i^ m/s and the other with a velocity 3j^ m/s. If the explosion takes place in 10-5 sec, 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^)×10−5

answer is B.

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Detailed Solution

Mass of each piece (m) = 1 kg. Initial momentum = 0. Final momentum =p1+p2+p3. From the principle of conservation of momentum, we have p1+p2+p3=0or p3=−p1+p2=−mv1+mv2=−mv1+v2=−1kg×(2i^+3j^) ms−1 =−(2i^+3j^) kgms−1ForceF=p3t=−( 2i^+3j^ ) kgms−110−5 s=−(2i^+3j^)×105 newton

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