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 v each in mutually perpendicular directions. The total energy released in the process of explosion is ............

Options

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

Explanation

Initial linear momentum ` = P_(i) = 0`
Final momentum `P_(f) = 0 = m upsilon hati + m upsilon hatj + vecP_(3)`
`implies P_(3) = m upsilon sqrt2`
Tortl `KE = (P_(3)^(2))/(2 xx 2m) + (1)/(2) m upsilon^(2) + (1)/(2) m upsilon^(2)`
`= (2 m^(2) upsilon^(2))/(4m) + m upsilon^(2) = (3 m upsilon^(2))/(2)`