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 `2hati m//s` and the other withl velocity of `3hati m//s`. If the explosion takes place in `10^(-5)` the average force acting on the third piece in newtons

Options

A. `(2hati+3hatj)xx10^(-5)`
B. `-(2hati+3hatj)xx10^(5)`
C. `(3hatj+2hatj)xx10^(5)`
D. `-(2hati+3hatj)xx10^(5)`

Answer

B

Explanation

Since the body explodes into three equal parts, therefore
`m_(1)=m_(2)=m_(3)=m/3=1kg`
Let the velocity of the third part be `vecv`. According to the principle of conservation of the linear momentum.
Momentum of sytem before explosion = momentum of system after explosion
or `mv=m_(1)v_(1)+m_(2)v_(2)+m_(3)v_(3)`
or `3xx0=1xx2hati+3hatj+1xxvecv`
or `v=-(2hati+3hatj)m/s`
Average force acting on the third particle is
`vecF=(vec(mv))/t=(-1xx(2hati+3hatj))/10^(-5)=(2hati+3hatj)x10^(5)N`