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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 `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:

A

`(2hati+3hatj)10^(-5)`

B

`-(2hati+3hatj)10^(+5)`

C

`(3hatj-2hati)10^(-5)`

D

`(2hatj-2hati)10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
B
`vecP_(3)=-(vecP_(1)+vecP_(2)),F=p_(3)/t`
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