A body of mass 4m at rest explodes into three fragments. Two of the fragments each of mass m move with speed v  in mutually perpendicular directions. 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 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` .