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 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 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:
Answer» <P>`(2hati+3hatj)10^(-5)` `-(2hati+3hatj)10^(+5)` `(3hatj-2hati)10^(-5)` `(2hatj-2hati)10^(-5)` SOLUTION :`vec(p_(3))=-(vec(p_(1))+vec(p_(2))),F=(p_(3))/(t)`