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GRE Subject Test: Physics : Energy and Momentum

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Example Questions

Example Question #1 : Energy And Momentum

A relativistic particle of mass m has a total energy 37 times its rest energy. What is the momentum of the particle, in units of mc?

Possible Answers:

144

Correct answer:

Explanation:

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

$E=\gamma E_{o}$

Where gamma is related to the momentum by:

$p=\gamma mc$

Combining the equations and solving for p, we get:

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

Which, in the units specified, is 37.

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Example Question #1 : Special Relativity

A particle of mass m traveling at a relativistic speed has a momentum of 50 mc. What is the total energy E of the particle, expressed in units of the rest mass energy E_o?

Possible Answers:

5 E_o

75 E_o

$5\sqrt{2}E_o$

50 E_o

2500 E_o

Correct answer:

50 E_o

Explanation:

For a relativistic particle, momentum is given by:

$p=\gamma mc$

from which we can solve for gamma:

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

Total energy of a relativistic particle is given by:

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

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Example Question #1 : Special Relativity

The rest mass energy E_o of a particle with mass is one quarter of its total energy . What is the of the particle's momentum, in units of ?

Possible Answers:

$\sqrt{3}$

$\sqrt{15}$

Correct answer:

$\sqrt{15}$

Explanation:

The question tells us:

$E=4E_{o}$

Where the rest mass energy of a particle is:

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

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

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

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

Solving this for , we find:

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

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

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Example Question #1 : Energy And Momentum

The relativistic momentum of a particle with mass is 3mc . What is the total energy of the particle, given in units of the rest mass energy E_o ?

Possible Answers:

$2E_{o}$

$4E_{o}$

$10E_{o}$

$\sqrt{10}E_{o}$

$\sqrt{2}E_{o}$

Correct answer:

$\sqrt{10}E_{o}$

Explanation:

The total energy of a relativistic particle is given by:

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

Substituting the momentum, we get:

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

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

$E_{o}=mc^2$

The total energy is:

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

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Example Question #1 : Energy And Momentum

A scientist measures the spectrum of relativistic jet emitted from a black hole. He finds that the a particular spectral line, which has a stationary wavelength of 212.5 nm, has a Doppler shifted wavelength of 643.7 nm. What is the radial velocity of the relativistic jet?

Possible Answers:

$2.2*10^8 \frac{m}{s}\textup{ away from the Earth}$

$2.2*10^8 \frac{m}{s}\textup{ towards the Earth}$

$2.4*10^8 \frac{m}{s} \textup{ towards the Earth}$

$3.2*10^8 \frac{m}{s} \textup{ towards the Earth}$

$2.4*10^8 \frac{m}{s} \textup{ away from the Earth}$

Correct answer:

$2.4*10^8 \frac{m}{s} \textup{ away from the Earth}$

Explanation:

The relativistic Doppler shift equation is given by:

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

Where $\beta$ is defined as:

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

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 $3*10^8 \frac{m}{s}$ , so $3.2*10^8 \frac{m}{s}$ is not a possible answer.

Making the approximation that

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

Combining this with the first equation:

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

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

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

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

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

From beta, we can find the velocity:

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

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Example Question #1 : Energy And Momentum

A particle in an accelerator has a total energy E=5 GeV and a relativistic momentum $p=4 \frac{GeV}{c}$ . What is the rest mass of the particle, in units of $\frac{1}{c^2}$ ?

Possible Answers:

$\sqrt{2}$

Correct answer:

Explanation:

The total energy of a relativistic particle is given by:

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

Solving for the rest mass :

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

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

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GRE Subject Test: Physics : Energy and Momentum

Study concepts, example questions & explanations for GRE Subject Test: Physics

All GRE Subject Test: Physics Resources

Example Questions

Example Question #1 : Energy And Momentum

Example Question #1 : Special Relativity

Example Question #1 : Special Relativity

Example Question #1 : Energy And Momentum

Example Question #1 : Energy And Momentum

Example Question #1 : Energy And Momentum

All GRE Subject Test: Physics Resources

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