A particle of rest energy mc2 is moving with speed v in the positive x direction. The particle decays into two particles, each of rest energy 140 MeV. One particle, with kinetic energy 282 MeV, moves in the positive x direction, and the other particle, with kinetic energy 25 MeV, moves in the negative x direction. Find the rest energy of the original particle and its speed.

The rest energy of the particle is  

The velocity of the particle is  

Explanation:

From the question we are told that

         The rest energy of the original particle is

         The speed of the particle is  

         The  rest energy the first  particle is  

          The rest energy of the second particle is  

         The first particle has a kinetic energy of positive x

           The second particle has a kinetic energy of negative x

The total energy of the first  particle is  

Substituting values

The total energy of the first  particle is  

The momentum of the first particle can be mathematically evaluated using this expression

The final energy of both particle is mathematically evaluated as

The momentum of the first particle can be mathematically evaluated using this expression

The negative sign shows that he direction is  negative x-axis

The final momentum of both particle is mathematically evaluated as

    Substituting value  

According to the law of conservation of momentum and the conservation of energy the final energy and momentum of both particle must be equal to the initial energy of the original  particle

Then  the rest energy is mathematically represented as

       Substituting values

The velocity of the original particle can be obtained from the mathematical expression as follows

Where c is the speed of light with value  

Substituting value

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