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

A body of mass 4m at rest explodes into three fragments. Two of the fr

1.	A body of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed vin mutually perpendicular directions. Total energy released in the process is
Answer» `mv^2` `3/2 mv^2` `2mv^2` `3mv^2` Solution :Here initial momentum `VECP = 0`. Since no external force EXISTS, hence momentum must remian 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, so `\|vec(p_1) + vec(p_2)\| = msqrt(v^2 + v^2) = sqrt(2) mv` `:. \|vecp_3\| = \|vecp_1 + vecp_2\| = sqrt(2) mv` The mass of third fragment is 2m. `:. ` 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` .