A stationary body explodes in to four identical fragments such that three of them fly mutually perpendicular to each other, each with same KE(E0). The energy of explosion will be

Let three of the fragments move along X, Y and Z axes. Therefore their velocities can be given asV→1=Vi^,V→2=Vj^ and V→3=Vk^where V = speed of each of the three fragments. Since, in explosion no net external force is involved, the net momentum of the system remains conserved just before and after the explosion.⇒ Pf=Pi⇒ mV→1+mV→2+mV→3+mV→4=0Pi=0 because the body was stationary), putting the values of V→1,V→2 and V→3, we obtain, V→4=V(i^+j^+k^)Therefore, V4=3VThe energy of explosion (∆KE) system⇒ E=KEf−KEi             =12mV12+12mV22+12mV32+12mV42−(0)Putting v1=v2=v3=V43=V and putting 12mV2=E0, we obtain E=6E0.