A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with speed v each in mutually perpendicular directions. The total energy released in the process of explosion is _____________.

Options

A. `(2//3)mv^(2)`
B. `(3//2) mv^(2)`
C. `(4//3) mv^(2)`
D. `(3//4) mv^(2)`

Answer

B

Explanation

`mv_(1)+mvj+2mv_(3)=0`
`vec(v)_(3)=-((vi+vj))/2=-v/2(i+j)=-v/(sqrt(2))`
`k_(f)=1/2 mv^(2)+1/2 mv^(2)+1/2 2m(v^(2))/2`
`k_(f)=(3mv^(2))/2`