A stationary body explodes in to four identical fragments such that three of them fly mutually perpendicular to each other, each with same `KE(E_(0))`. The energy of explosion will be

Options

A. `6E_(0)`
B. `3E_(0)`
C. `4E_(0)`
D. `2E_(0)`

Answer

A

Explanation

Let three of the fragments move along `X,Y` and `Z` axes. Therefore their velocities can be given as
`vecV_(1)=Vhati, vecV_(2)=VhatJ` and `vecV_(3)=Vhatk`
where `V=`speed of each of the three fragments. Since, in explosion no net external force in involved, the net of momentum of the system remains conserved just before and after the explosion.
`implies (P)_(f)=(P_(i))`
`=mvecV_(1)+mvecV_(2)=mvecV_(3)+mvecV_(4)=0`
`(P_(i)=0` because the body was stationary), putting the value of `vecV_(1),vecV_(2)` and `vecV_(3)` we obtain `vecV_(4)=V(hati+hatj+hatk)`
Therefore `V_(4)=sqrt(3)V`
The energy of explosion `(/_\KE)` system
`implies E=KE_(f)-KE_(f)`
`=(1/2mV_(1)^(2)+1/2mV_(2)^(2)+1/2MV_(3)^(2)+1/2mV_(4)^(2))-(0)`
Putting `v_(1)=v_(2)=v_(3)=(V_(4))/(sqrt(3))=V` and putting `1/2mV^(2)=E_(0)`.
We obtain `E=6E_(0)`.