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 2 i m/s and the other with a velocity of 3 j m/s. If the explosion takes place in `10^(5)s`, the average force acting on the third piece in newton is:

Options

A. `(3overset(wedge)(i)+3overset(wedge)(j))xx10^(-5)`
B. `(2overset(wedge)(i)+3overset(wedge)(j))xx10^(5)`
C. `(2overset(wedge)(i)-3overset(wedge)(j))xx10^(5)`
D. `(2overset(wedge)(i)-3overset(wedge)(j))xx10^(-5)`

Answer

b

Explanation

According to law of conservation of momentum
`mvecupsilon=mxx2hati+mxx3hatj`
`:.vecupsilon=2hati+3hatj`
Now `veca=(vecupsilon)/(t)=(2hati+3hatj)/(10^(-5))`
and`vecF=mveca=1xx(2hati+3hatj)xx10^(5)N`
`=-(2hati+3hatj)xx10^(5)N`-ve sign because it is in opposite direction (b) is the choice