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. `(4E_(0))/3`
C. `4E_(0)`
D. `8E_(0)`

Answer

A

Explanation

Let the three mutually perpendicular directions be along `xy` and `z`- axis respectively.
`vecp_(1)=mv_(0)hati,vecp_(2)=mv_(0)hatj`
where
`1/2mv_(0)^(2)=E_(0)`
`vecp_(3)=mv_(0)hatk` and `vecp_(4)=mvecv`
By linear momentum conservation
`0=vecp_(1)+vecp_(2)+vecp_(3)+vecp_(4)`
or `vecv=-v_(0)(hati+hatj+hatk)`
or `v=v_(0)sqrt(1^(2)+1^(2)+1^(2))=v_(0)sqrt(3)`
Total energy `=3(1/2mv_(0)^(2))+1/2mv^(2)=3E_(0)+3E_(0)=6E_(0)`