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A stationary body of mass 3 kg explodes ...

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 `2hati m//s` and the other withl velocity of `3hati m//s`. If the explosion takes place in `10^(-5)` the average force acting on the third piece in newtons

A

`(2hati+3hatj)xx10^(-5)`

B

`-(2hati+3hatj)xx10^(5)`

C

`(3hatj+2hatj)xx10^(5)`

D

`-(2hati+3hatj)xx10^(5)`

Text Solution

Verified by Experts

The correct Answer is:
B
Since the body explodes into three equal parts, therefore
`m_(1)=m_(2)=m_(3)=m/3=1kg`
Let the velocity of the third part be `vecv`. According to the principle of conservation of the linear momentum.
Momentum of sytem before explosion = momentum of system after explosion
or `mv=m_(1)v_(1)+m_(2)v_(2)+m_(3)v_(3)`
or `3xx0=1xx2hati+3hatj+1xxvecv`
or `v=-(2hati+3hatj)m/s`
Average force acting on the third particle is
`vecF=(vec(mv))/t=(-1xx(2hati+3hatj))/10^(-5)=(2hati+3hatj)x10^(5)N`
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Knowledge Check

  • 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 2 hatjm//s and the other with a velocity 3 hatjm//s . If the explosion takes place in 10^(-5) sec, the average force acting on the third piece in Newtons is:

    A
    `(2hati+3hatj)10^(-5)`
    B
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    C
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    D
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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 2hati m/s and the other with a velocity 3hatj m/s. If the explosion takes place in 10^(-5) s, the average force acting on the third piece in newton is

    A
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    B
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    D
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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 2hati m//s and the other with a velocity 3hatj m//s .If the explosion takes place in 10 sec ,the average force acting on the third place in Newtons is:

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    B
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    C
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