A particle of mass `4m` which is at rest explodes into four equal fragments. All four fragments scattered in the same horizontal plane. Three fragments are found to move with velocity `v` as shown in the figure. The total energy released in the process is

Options

A. (a) `mv^2(3-sqrt2)`
B. (b) `1/2mv^2(3-sqrt2)`
C. (c) `2mv^2`
D. (d) `1/2mv^2(1+sqrt2)`

Answer

A

Explanation

Net momentum of three fragements

`p_i=p_f=0`
`:.` Momentum of fourth part is `(sqrt2-1)mv` in opposite direction.
So, velocity of fourth part is `(sqrt2-1)v`
`:. KE=3(1/2mv^2)+1/2m[(sqrt2-1)v]^2`
`=(3-sqrt2)mv^2`