The particle of mass 5 m breaks in three fragments of mass m,m and 3 m respectively. Two fragments of mass meach, move in perpendicular direction with velocity v and the left fragment will move in a direction with velocity v′ such that the total momentum of the system must remain conserved. 
By law of conservation of momentum, 5 m×0⇒v′∴∣v′∣=mvi^+mvj^+3mv′=−3vi^−3vj^=(−3v)2+(−3v)2=3v2 ∴ Energy released E=21mv2+21mv2+21×3 m(3v2)2=mv2+3mv2=34mv2