Physics, 05.05.2020 04:01 ghari112345

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.

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Answer from: carolinamleal04

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

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

Answer from: Quest

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Answer from: Quest

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A particle of rest energy mc2 is moving with speed v in the positive x direction. The particle decay...