Momentum and Relativistic Energy

To understand the speed of a particle with a given momentum and total relativistic energy, we need to analyze the relationship between momentum, energy, and speed in the context of special relativity.

Momentum (p)
Momentum is defined as the product of an object's mass (m) and its velocity (v). In the context of relativistic physics, momentum is given by the equation:

p = γ * m * v

where γ is the Lorentz factor, given by:

γ = 1 / sqrt(1 - (v^2 / c^2))

Here, c represents the speed of light in a vacuum, approximately 3 x 10^8 meters per second.

Relativistic Energy (E)
The relativistic energy of a particle is given by the equation:

E = γ * m * c^2

where m is the mass of the particle and c is the speed of light.

Given Information
In this question, we are given that the momentum of the particle is 5 MeV/c and the total relativistic energy is 10 MeV.

Calculating the Speed
To find the speed of the particle, we can equate the equations for momentum and relativistic energy:

γ * m * v = p
γ * m * c^2 = E

Using these equations, we can eliminate γ and m to find the speed (v):

v = p / (γ * m) = p / (E / (γ * c^2))

Substituting the given values into the equation:

v = (5 MeV/c) / (10 MeV / (γ * c^2))

Simplifying further:

v = (5 MeV/c) / (10 MeV / (1 / sqrt(1 - (v^2 / c^2)) * c^2))

v = (5 MeV/c) / (10 MeV * sqrt(1 - (v^2 / c^2)) / c^2)

v = (5 / 10) * sqrt(1 - (v^2 / c^2))

v = 0.5 * sqrt(1 - (v^2 / c^2))

To solve this equation, we can square both sides:

v^2 = 0.25 * (1 - (v^2 / c^2))

v^2 = 0.25 - 0.25 * (v^2 / c^2)

1.25 * (v^2 / c^2) = 0.25

v^2 / c^2 = 0.25 / 1.25

v^2 / c^2 = 0.2

v^2 = 0.2 * c^2

v = sqrt(0.2 * c^2)

v ≈ 0.447 c

Therefore, the speed of the particle is approximately 0.447 times the speed of light (c).