Question

A particle of mass 5m at rest suddenly breaks on
its own into three fragments. Two fragments of
mass m each move along mutually perpendicular
direction with speed v each. The energy released
during the process is ? [NEET-2019 (Odisha)]​

Answer 1

Given:

Mass of the particle initially = 5m

Mass of two fragments = m

Speed of the two fragments = v

To find:

The energy released during explosion.

Solution:

The velocity of the third particle be v'

Mass of the third particle = 5m - m - m = 3m

3mv' cosθ = mv ......(1)

3mv' sinθ = mv.......(2)

Comparing both the equation we get:

cosθ = sinθ = 1/√2

Putting the value in equation 1 we get:

3mv' *1/√2 = mv

v' = √2v/ 3

Total energy released in the explosion will be the kinetic energies of the three particles:

= 1/2*mv² + 1/2*mv² + 1/2*(3m)* (√2 v/3)²

= 1/2* m * [v² + v² + 2/3v²]

= 1/2*m*8/3*v²

= 4/3*m*v²

Therefore the total energy released will be 4mv²/3.

Answer 2

Answer:

4/3 mv²

Explanation:

plz see the attached image

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