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A particle of mass 4m which is at rest expodes into 3 fragments.2 of the fragments move with speed v in mutually perpendicular directions.What is the energy released in the process?

A particle of mass 4m which is at rest expodes into 3 fragments.2 of the fragments move with speed v in mutually perpendicular directions.What is the energy released in the process?

Grade:12

1 Answers

Shobhit Varshney IIT Roorkee
askIITians Faculty 33 Points
6 years ago
Dear Student,

The question is a bit unclear. Since the mass of the fragments is not mentioned, it is not possible to predice the energy released in the process.
But if we consider the mass of the 2 fragments as ‘m’. Then, mass of third fragment = 4m-2m= 2m.
Equation momentum in x-direction,
m*v = (2m)*(vx), vx= v/2
Similarly in y-direction,
m*v = (2m)*(vy), vy= v/2
Therefore, speed of fragment ‘2m’ =v/√2
Engery released = 0.5mv2 + 0.5mv2 + 0.5*2m* (v/√2)2 = 1.5 mv2

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