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

Options

A. `1/2 mv^(2)`
B. `mv^(2)`
C. `5/3 mv^(2)`
D. `3/2 mv^(2)`

Answer

d

Explanation

According to law of conservation of momentumn,
` M_(3) =sqrt(M_(1)^(2)+M_(2)^(2))`
`(4m-m-m)v_(1)=sqrt((mv)^(2)+(mv)^(2))=mvsqrt(2)`
`v_(1) =vsqrt(2)/2`
`Energy released, E =(1)/(2)mv^(2)+(1)/(2)mv^(2)+1/2(2m)v^(2)`
`mv^(2)+m[vsqrt/2]^(2)`
` mv^(2)+mv^(2)/2`
`3/2mv^(2)`