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:

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 $$2i^$$ $$m/s$$ 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:

Infinity Learn · Accepted Answer

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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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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