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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 in two mutually perpendicular directions, one with a velocity of `3hati" m s"^(-1)`. And the other with a velocity of `4hatj" m s"^(-1)`. If the explosion occurs in `10^(-4)" s"`, the average force acting on the third piece in newton is

A

`(3hati + 4haj)xx 10^-4`

B

`(3hati - 4haj)xx 10^-4`

C

`(3hai + 4hatj)xx 10^-4`

D

`(3hati + 4hatj ) xx 10^4)`

Text Solution

Verified by Experts

The correct Answer is:
`-(3hati + 4 hatj) xx 10^4`
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