The fastest vehicle flown by humans on a regular basis is the Space Shuttle which has a mass of about \(1 \times 10^{5} \mathrm{kg}\) and travels at a speed of about \(8 \times 10^{3} \mathrm{m} / \mathrm{s} .\) Find the momentum of the Space Shuttle using relativistic formulae and then again by using nonrelativistic formulae. By what percent do the relativistic and nonrelativistic momenta of the Space Shuttle differ? [Hint: You might want to use one of the approximations in Appendix A.5.]

Short Answer

Expert verified
Question: Calculate the relativistic and nonrelativistic momenta of a Space Shuttle with mass \(1 \times 10^5 kg\) and speed \(8 \times 10^3 m/s\), and find the percentage difference between the two momenta. To calculate the relativistic and nonrelativistic momenta of the Space Shuttle, follow the three steps: 1. Find the relativistic momentum using the formula \(p = \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}\), where m is the mass, v is the speed, and c is the speed of light. 2. Find the nonrelativistic momentum using the formula \(p = mv\), where m is the mass and v is the speed. 3. Compute the percentage difference between the momenta using the formula \(\dfrac{|p_{rel} - p_{nr}|}{p_{nr}} \times 100\%\). Upon following this procedure, you will obtain the relativistic momentum \(p_{rel}\), the nonrelativistic momentum \(p_{nr}\), and the percentage difference between the two momenta.

Step by step solution

01

Find relativistic momentum

To find the relativistic momentum, we'll use the formula: \(p = \dfrac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}}\) Plug in the values: - Mass (m) = \(1 \times 10^5 kg\) - Speed (v) = \(8 \times 10^3 m/s\) - Speed of light (c) = \(3 \times 10^8 m/s\) \(p_{rel} = \dfrac{(1 \times 10^5)(8 \times 10^3)}{\sqrt{1 - \dfrac{(8 \times 10^3)^2}{(3 \times 10^8)^2}}}\) We can now calculate the relativistic momentum \(p_{rel}\).
02

Find nonrelativistic momentum

Now let's find the nonrelativistic momentum using the formula \(p = mv\). Plug in the values: - Mass (m) = \(1 \times 10^5 kg\) - Speed (v) = \(8 \times 10^3 m/s\) \(p_{nr} = (1 \times 10^5)(8 \times 10^3)\) We can compute the nonrelativistic momentum \(p_{nr}\).
03

Compute the percentage difference between the momenta

Finally, we'll find the percentage difference between the relativistic and nonrelativistic momenta using the formula: \(\dfrac{|p_{rel} - p_{nr}|}{p_{nr}} \times 100\%\) Plug in the values of \(p_{rel}\) and \(p_{nr}\) which we have computed in earlier steps. We can now find the percentage difference between the relativistic and nonrelativistic momenta. By following these steps, you will be able to find both the relativistic and nonrelativistic momenta and the percentage difference between them.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Kurt is measuring the speed of light in an evacuated chamber aboard a spaceship traveling with a constant velocity of \(0.60 c\) with respect to Earth. The light is moving in the direction of motion of the spaceship. Siu- Ling is on Earth watching the experiment. With what speed does the light in the vacuum chamber travel, according to Siu-Ling's observations?
Radon decays as follows: $^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .$ The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium- 218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. How much energy is released in the decay? \(\left(1 \mathrm{u}=931.494 \mathrm{MeV} / \mathrm{c}^{2} .\right)\)
When an electron travels at \(0.60 c,\) what is its total energy in mega- electron-volts?
Find the conversion between the momentum unit MeV/c and the SI unit of momentum.
Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free