A body of mass 4 m at rest explodes into three fragment . Two of the fragment each of mass move with speed v in mutually perpendicular direction. Total energy released in the process is

Options

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

Answer

B

Explanation

Here initial momentum `vecp = 0` Since no external force exists, hence momentum must remain conserved i.e., `vecp_(1) + vecp_(2) + vecp_(3) = 0`
as two fragments of mass m each are moving with speed v each at right angles,
`|vecp_(1) + vecp_(2)| = msqrt(v^(2) + v^(2)) = sqrt(2)mv therefore |vecp_(3)| = |vecp_(1) + vecp_(2)| = sqrt(2) mv`
The mass of third fragment is 2m. `therefore` Kinetic energies of three fragments are
`K_(1) = (P_(1)^(2))/(2m) = (1)/(2)mv^(2), K_(2) = (p_(2)^(2))/(2m) = (1)/(2)mv^(2) and K_(3) = (P_(3)^(2))/(2(2m)) = (1)/(2)mv^(2)`
Total energy released during explosion ` = K_(1) + K_(2) + K_(3) = (3)/(2)mv^(2)`