A particle of mass 5 m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along a mutually perpendicular direction with speed v each. The energy released during the process is,

Explanation:

From Newton's 2nd law of motion we have;

F_ext = 

where Fext = external force acting on the object.

& P = momentum = mv

When there is no external force exists then, Fext = 0.

⇒ = 0 ⇒P = const.     -----(1)

Calculation:

Given:

The total mass of the object = 5m,

Mass of two smaller fragments = m, 

Mass of larger fragment = 5m - 2m = 3m,

The velocity of smaller fragments = v which are perpendicular.

Let us consider the velocity of 3rd fragment as V.

As there are no external forces acting on the object we can apply momentum conservation law as:

5m×u = mv î + mv ĵ + 3mV  

⇒3mV = -mv(î + ĵ)

⇒V = (î + ĵ) 

⇒IVI = 

Energy will be released in the form of kinetic energy.

∴ The total energy released = mv² + ​​mv2 + (3m)V2

=mv2

Hence option 2) is the correct choice.