A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same kinetic energy.,E0. The energy of explosion is given bynE0. Find the value ofn.Correct answer is '6'. Can you explain this answer?

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Answer

Answer:

Given:
- A stationary body explodes into four identical fragments.
- Three of the fragments fly off mutually perpendicular to each other.
- Each fragment has the same kinetic energy, denoted as E0.
- The energy of the explosion is given by nE0.

To find the value of n, we need to analyze the situation and apply the principles of conservation of momentum and conservation of energy.

Conservation of Momentum:
- Before the explosion, the body is stationary, so its momentum is zero.
- After the explosion, the momentum of the system should still be zero, as there are no external forces acting on it.
- Since the fragments fly off mutually perpendicular to each other, their momentum vectors will also be mutually perpendicular.
- Let's assume the magnitude of the momentum of each fragment is p, then the total momentum of the system after the explosion is given by:
p + p + p = 3p

Conservation of Energy:
- The energy of the explosion is given by the sum of the kinetic energies of the fragments.
- Since each fragment has the same kinetic energy E0, the total energy of the explosion is given by:
E0 + E0 + E0 + E0 = 4E0

Now, using the given relationship between the energy of the explosion and E0:
nE0 = 4E0

Simplifying the equation:
n = 4/1

Therefore, the value of n is 4.

The correct answer is '4', not '6'. It seems there is a mistake in the given solution.

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