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 i m/s and the other with a velocity of 3 j m/s. If the explosion takes place in 10^(5)s, the average force acting on the third piece in newton is:

A stationary body of mass 3 kg explodes into three equal pieces. Two o

1.	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 i m/s and the other with a velocity of 3 j m/s. If the explosion takes place in 10^(5)s, the average force acting on the third piece in newton is:
Answer» `(3overset(WEDGE)(i)+3overset(wedge)(j))xx10^(-5)` `(2overset(wedge)(i)+3overset(wedge)(j))xx10^(5)` `(2overset(wedge)(i)-3overset(wedge)(j))xx10^(5)` `(2overset(wedge)(i)-3overset(wedge)(j))xx10^(-5)` SOLUTION :According to law of conservation of momentum `mvecupsilon=mxx2hati+mxx3hatj` `:.vecupsilon=2hati+3hatj` Now `veca=(vecupsilon)/(t)=(2hati+3hatj)/(10^(-5))` and`vecF=mveca=1xx(2hati+3hatj)xx10^(5)N` `=-(2hati+3hatj)xx10^(5)N`-ve SIGN because it is in OPPOSITE direction (b) is the choice

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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 of 2 i m/s and the other with a velocity of 3 j m/s. If the explosion takes place in 10^(5)s, the average force acting on the third piece in newton is:

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