Question

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.

Answer 1

Answer:

The rest energy of the particle is  $mc^2 = 498 MeV$

The velocity of the particle is  $v = 1.587 *10^8 m/s$

Explanation:

From the question we are told that

         The rest energy of the original particle is $E = mc^2$

         The speed of the particle is  $v$

         The  rest energy the first  particle is  $(mc^2)_1 = 140 MeV1$

          The rest energy of the second particle is  $(mc^2)_2= 140MeV$

         The first particle has a kinetic energy of $KE = 282 MeV$ positive x

           The second particle has a kinetic energy of $KE_2 = 25 MeV$ negative x

           

The total energy of the first  particle is  

             $E_1 = (mc^2)_1 + KE_1$

Substituting values

           $E_1 = 140 + 282$

            $E_1 = 422 MeV$

The total energy of the first  particle is  

             $E_1 = (mc^2)_2 + KE_2$

             $E_1 = 140 + 25$

              $E_1 =165MeV$

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

           $p__{1 }}= \sqrt{E^2_1 - ((mc^2)_1) ^2}$

           $p__{1 }}= \sqrt{422^2 - 140^2}$

            $p__{1 }}= 398MeV$

The final energy of both particle is mathematically evaluated as

             $E = E_1 + E_2$

             $E = 422 + 165$

              $E = 587 MeV$  

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

           $p__{2 }}= - \sqrt{E^2_2 - ((mc^2)_2) ^2}$

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

           $p__{2 }}= -\sqrt{165^2 - 140^2}$

            $p__{2 }}= -87MeV$

The final momentum of both particle is mathematically evaluated as

                $P = p_1 + p_2$

    Substituting value  

                $P = 398 - 87$

                $P = 311 MeV$

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

                 $mc^2 = \sqrt{E^2 - P^2}$

       Substituting values

                 $mc^2 = \sqrt{587 ^2 - 311^2}$

                  $mc^2 = 498 MeV$

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

                  $v = c\sqrt{1 - [\frac{(mc^2)}{E}]^2 }$

Where c is the speed of light with value   $c = 3.0 *10^8 m/s$

Substituting value

                  $v = 3*10^8 *\sqrt{1 - [ \frac{498}{587}] ^2}$

                 $v = 1.587 *10^8 m/s$