The total energy E of a relativistic particle is related to its rest mass energy E_o by:

$\displaystyle E=\gamma E_{o}$

Where gamma is related to the momentum by:

$\displaystyle p=\gamma mc$

Combining the equations and solving for p, we get:

$\displaystyle p=\frac{E}{E_{o}}mc=37 mc$

Which, in the units specified, is 37.

For a relativistic particle, momentum is given by:

$\displaystyle p=\gamma mc$

from which we can solve for gamma:

$\displaystyle \gamma=\frac{p}{mc}=\frac{50 mc}{mc}=50$

Total energy of a relativistic particle is given by:

$\displaystyle E=\gamma E_{o}=50E_{0}$

The question tells us:

$\displaystyle E=4E_{o}$

Where the rest mass energy of a particle is:

$\displaystyle E_{o}=mc^{2}$

Using the equation for the total energy of a particle, we can substitute:

$\displaystyle E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}$

$\displaystyle 4mc^2=\sqrt{p^2c^2+m^2c^4}$

Solving this for $\displaystyle p$, we find:

$\displaystyle p=\sqrt{15m^2c^2}=\sqrt{15}\,\,mc$

Which, in units of mc, gives us the correct answer.

The total energy of a relativistic particle is given by:

$\displaystyle E=\sqrt{p^2c^2+m^2c^4}$

Substituting the momentum, we get:

$\displaystyle E=\sqrt{9m^2c^4+m^2c^4}=\sqrt{10}mc^2$

Because the rest mass energy of a particle is given by:

$\displaystyle E_{o}=mc^2$

The total energy is:

$\displaystyle E=\sqrt{10}E_{o}$

The relativistic Doppler shift equation is given by:

$\displaystyle \lambda_{moving}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\lambda_{stationary}$

Where $\displaystyle \beta$ is defined as:

$\displaystyle \beta=\frac{v_{source}}{c}$

Because the stationary wavelength is shorter than the moving wavelength, the object must be receding from the Earth, eliminating two answers.

The speed of light is approximately $\displaystyle 3*10^8 \frac{m}{s}$, so $\displaystyle 3.2*10^8 \frac{m}{s}$ is not a possible answer.

Making the approximation that

$\displaystyle \frac{\lambda_{m}}{\lambda_{s}}=\frac{643.7}{212.5}\approx3$

Combining this with the first equation:

$\displaystyle \frac{\lambda_m}{\lambda_s}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}=3$

$\displaystyle 3\sqrt{1-\beta}=\sqrt{1+\beta}$

$\displaystyle 9(1-\beta)=1+\beta$

$\displaystyle 9-9\beta=1+\beta$

$\displaystyle 10\beta=8,\,\,\,\beta=0.8$

From beta, we can find the velocity:

$\displaystyle \beta=\frac{v_{source}}{c}\rightarrow v_{source}=\beta c=(0.8)(3\times10^8 \frac{m}{s})=2.4\times10^8 \frac{m}{s}$

The total energy of a relativistic particle is given by:

$\displaystyle E=\sqrt{p^2c^2+m^2c^4}$

Solving for the rest mass $\displaystyle m$:

$\displaystyle m=\frac{\sqrt{E^2-p^2c^2}}{c^2}=\frac{\sqrt{25-16}}{c^2}=\frac{3}{c^2}$

Which, in the units specified, gives the answer 3.