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:

Options

A. `(2hati+3hatj)10^(-5)`
B. `-(2hati+3hatj)10^(+5)`
C. `(3hatj-2hati)10^(-5)`
D. `(2hatj-2hati)10^(-5)`

Answer

B

Explanation

`vec(p_(3))=-(vec(p_(1))+vec(p_(2))),F=(p_(3))/(t)`