/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A proton has a mass of \(1.6726 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A proton has a mass of \(1.6726 \times 10^{-27} \mathrm{~kg}\) and is traveling at a speed where \(y=2.294\) 157; that's \(90.000 \%\) the speed of light. Determine the magnitude of its momentum. How does that compare with its classical momentum? Four significant figures will do. Discuss your answer. [Hint: Relativistically \(p=y m v ;\) classically \(=m v\)

Short Answer

Expert verified
Relativistic momentum is approximately \(1.0327 \times 10^{-18} \mathrm{~kg \cdot m/s}\), much higher than the classical momentum \(4.515 \times 10^{-19} \mathrm{~kg \cdot m/s}\).

Step by step solution

01

Setting Up the Known Values

Identify the known values given in the problem. The mass of the proton is \(m = 1.6726 \times 10^{-27} \mathrm{~kg}\). The relativistic Lorentz factor, \(\gamma\), is 2.294 (note the mistake in text as it should be \(\gamma\), not \(y\)), corresponding to the proton traveling at 90% the speed of light, so \(v = 0.9c\), where \(c = 3 \times 10^8 \mathrm{~m/s}\).
02

Calculate Relativistic Momentum

Use the formula for relativistic momentum: \(p = \gamma m v\). Substitute the known values: \[p = (2.294)(1.6726 \times 10^{-27} \mathrm{~kg})(0.9 \times 3 \times 10^8 \mathrm{~m/s})\]. Compute to find the magnitude of the relativistic momentum.
03

Compute the Expression

Calculate the values step by step:\[p = 2.294 \times 1.6726 \times 10^{-27} \times 0.9 \times 3 \times 10^8\].The expression simplifies to:\[p \approx 1.0327 \times 10^{-18} \mathrm{~kg \cdot m/s}\].
04

Calculate Classical Momentum

Use the formula for classical momentum: \(p = mv\). Substitute the known values: \[p_{\text{classical}} = (1.6726 \times 10^{-27} \mathrm{~kg})(0.9 \times 3 \times 10^8 \mathrm{~m/s})\]. Compute to find classical momentum.
05

Compute Classical Expression

Calculate the values step by step:\[p_{\text{classical}} = 1.6726 \times 10^{-27} \times 0.9 \times 3 \times 10^8\].The expression simplifies to:\[p_{\text{classical}} \approx 4.515 \times 10^{-19} \mathrm{~kg \cdot m/s}\].
06

Compare Relativistic and Classical Momentum

Compare the magnitude of the relativistic momentum to the classical momentum. The relativistic momentum, \(1.0327 \times 10^{-18} \mathrm{~kg \cdot m/s}\), is significantly higher than the classical momentum, \(4.515 \times 10^{-19} \mathrm{~kg \cdot m/s}\), due to the relativistic effects at high speeds.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relativistic Physics
Relativistic physics is a theory in physics that explains how the laws of motion and forms of energy behave at high speeds, particularly speeds close to the speed of light. This approach was developed because classical mechanics, which works well at lower speeds, does not accurately predict behaviors as things go fast. It accounts for the peculiar changes in mass, time, and energy when objects move rapidly. One key idea in relativistic physics is that as you move faster and faster, time and space go through transformations. For instance, time appears to slow down, and lengths contract for objects in motion relative to an observer. Because of these adjustments, calculations like momentum need corrections, and this is where the concept of relativistic momentum comes in. At high speeds, momentum increases beyond the predictions of classical mechanics, making the relativistic corrections crucial to understanding phenomena in the universe properly. This approach helps explain many cosmic phenomena and is essential in particle physics.
Relativistic effects become significant only as speeds approach the speed of light, denoted by the constant c. Traditional calculations start differing, requiring physicists to involve factors that account for relativity.
Lorentz Factor
The Lorentz factor, denoted by \gamma (\gamma), is a crucial concept in the theory of relativity. It adjusts time, length, and relativistic momentum calculations to account for the effects of traveling at significant fractions of the speed of light. It essentially tells us how much these relativistic effects are amplifying our traditional perceptions.
The Lorentz factor is given by the formula: \[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\] where \(v\) is the velocity of the object and \(c\) is the speed of light. As \(v\) approaches \(c\), \(\gamma\) increases dramatically, effectively making time slow down and distances contract from the perspective of an external observer.
This factor also plays a critical role in adjusting the calculation of momentum at high speeds. In the outlined exercise, the Lorentz factor for a proton moving at 90% of the speed of light is given as approximately 2.294. This means the proton's relativistic momentum is over twice the value predicted by classical mechanics. The Lorentz factor helps bridge the gap between classical and relativistic calculations, ensuring that our equations hold true in the realm of high velocities.
Classical Mechanics
Classical mechanics is the branch of physics that deals with the motion of bodies under the influence of forces, particularly at speeds much lower than that of light. It is defined by Newton’s laws of motion and works extremely well for the everyday human-scale and the slow speeds we typically encounter.
In classical mechanics, momentum \(p\) is calculated as the product of an object's mass \(m\) and its velocity \(v\), so \(p = mv\). This formula is sufficient for most low-speed scenarios, such as car collisions or the trajectory of a baseball. However, as speeds increase toward the speed of light, classical mechanics starts to falter, unable to accurately predict the behavior of objects.
This limitation occurs because classical physics does not incorporate the time and space transformations that occur at high velocities, as predicted by Einstein's theory of relativity. As demonstrated in the exercise, classical momentum doesn't account for the significant increase in momentum observed in relativistic physics. Despite its limitations at high speeds, classical mechanics remains fundamental for understanding basic physics, calibrating everyday experiences, and teaching the foundational principles of force and motion.

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

As a spacecraft moving at \(0.92 \mathrm{c}\) travels past an observer on Earth, the Earthbound observer and the occupants of the craft each start identical alarm clocks that are set to ring after \(6.0 \mathrm{~h}\) have passed. According to the Earthling, what does the Earth clock read when the spacecraft clock rings?

A proton has a mass of \(1.6726 \times 10^{-27} \mathrm{~kg}\) and is traveling at \(\mathrm{a}\) speed where \(y=2.294157\); that's \(90 \%\) the speed of light. Determine its total energy. Four significant figures will do. Discuss your answer as it compares with classical energy.

A spacecraft moving at \(0.95 \mathrm{c}\) travels from the Earth to the star Alpha Centauri, which is \(4.5\) light years away. How long will the trip take according to \((a)\) Earth clocks and \((b)\) spacecraft clocks? (c) How far is it from Earth to the star according to spacecraft occupants? ( \(d\) ) What do they compute their speed to be? A light year is the distance light travels in 1 year, namely 1 light year \(=\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\left(3.16 \times 10^{7} \mathrm{~s}\right)=9.47 \times 10^{15} \mathrm{~m}\) Hence the distance to the star (according to earthlings) is $$ \begin{array}{l} d_{e}=(4.5)\left(9.47 \times 10^{15} \mathrm{~m}\right)=4.3 \times 10^{16} \mathrm{~m} \\ \text { (a) } \Delta t_{e}=\frac{d_{e}}{v}=\frac{4.3 \times 10^{16} \mathrm{~m}}{(0.95)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=1.5 \times 10^{8} \mathrm{~s} \end{array} $$ (b) Because clocks on the moving spacecraft run slower, $$ \Delta t_{\text {cnth }}=\Delta t_{\varepsilon} \sqrt{1-(v / \mathrm{c})^{2}}=\left(1.51 \times 10^{8} \mathrm{~s}\right)(0.312)=4.7 \times 10^{7} \mathrm{~s} $$ (c) For the spacecraft occupants, the Earth-star distance is moving past them with speed 0.95c. Therefore, that distance is shortened for them; they find it to be $$ d_{\text {cath }}=\left(4.3 \times 10^{16} \mathrm{~m}\right) \sqrt{1-(0.95)^{2}}=1.3 \times 10^{16} \mathrm{~m} $$ (d) For the spacecraft occupants, their relative speed is $$ v=\frac{d_{\text {cnft }}}{\Delta t_{\text {caff }}}=\frac{1.34 \times 10^{16} \mathrm{~m}}{4.71 \times 10^{7} \mathrm{~s}}=2.8 \times 10^{8} \mathrm{~m} / \mathrm{s} $$ which is 0.95c. Both Earth and spacecraft observers measure the same relative speed.

Two cells that subdivide on Earth every \(10.0 \mathrm{~s}\) start from the Earth on a journey to the Sun \(\left(1.50 \times 10^{11} \mathrm{~m}\right.\) away) in \(\mathrm{a}\) spacecraft moving at \(0.850 \mathrm{c}\). How many cells will exist when the spacecraft crashes into the Sun? According to Earth observers, with respect to whom the cells are moving, the time taken for the trip to the Sun is the distance traveled \((x)\) over the speed \((v)\) $$ \Delta t_{\mathrm{M}}=\frac{x}{v}=\frac{1.50 \times 10^{11} \mathrm{~m}}{(0.850)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=588 \mathrm{~s} $$ Because spacecraft clocks are moving with respect to the planet, they appear from Earth to run more slowly. The time these clocks read is $$ \Delta t_{\mathrm{s}}=\Delta f_{\mathrm{M}} / \gamma=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}} $$ and so $$ \Delta l_{\mathrm{s}}=310 \mathrm{~s} $$ The cells divide according to the spacecraft clock, a clock that is at rest relative to them. They therefore undergo 31 divisions in this time, since they divide each \(10.0 \mathrm{~s}\). Therefore, the total number of cells present on crashing is $$ (2)^{31}=2.1 \times 10^{9} \text { cells } $$

A proton \(\left(\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}\right)\) is accelerated to a kinetic energy of \(200 \mathrm{MeV}\). What is its speed at this energy?

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.