/*! 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 25 Two freight trains, each of mass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 38: Problem 25

Two freight trains, each of mass \(6 \times 10^{6} \mathrm{~kg}\) (6 000 metric tons) travel in opposite directions on the same track with equal speeds of \(150 \mathrm{~km} / \mathrm{hr}\). They collide head-on and come to rest. (a) Calculate in joules the kinetic energy \((1 / 2) m v^{2}\) for each train before the collision. (Newtonian expression OK for everyday speeds!) (b) After the collision, the mass of the trains plus the mass of the track plus the mass of the roadbed plus the mass of the surrounding air plus the mass of emitted sound and light has increased by what number of milligrams?

Short Answer

Expert verified
Each train's kinetic energy is \(5.2 \times 10^9\) J. The total energy is \(10.4 \times 10^9\) J, converting into \(0.116\) mg mass increase.

Step by step solution

01

- Convert Speed to Meters per Second

First, convert the speed of the trains from km/hr to m/s. The speed of each train is 150 km/hr. We know that 1 km/hr = 1,000 m/3,600 s. Therefore, \[150 \text{ km/hr} = 150 \times \frac{1,000}{3,600} \text{ m/s} = 41.67 \text{ m/s}\]
02

- Calculate the Kinetic Energy of One Train

Use the kinetic energy formula \( KE = \frac{1}{2} m v^2 \) where \( m = 6 \times 10^6 \) kg and \( v = 41.67 \) m/s.\[KE = \frac{1}{2} \times 6 \times 10^6 \text{ kg} \times (41.67 \text{ m/s})^2\]First, calculate the square of the velocity:\[v^2 = (41.67 \text{ m/s})^2 = 1,736.11 \text{ (m/s)}^2\]Now calculate the kinetic energy:\[KE = \frac{1}{2} \times 6 \times 10^6 \text{ kg} \times 1,736.11 \text{ (m/s)}^2 = 5,208.33 \times 10^6 \text{ J} = 5.2 \times 10^9 \text{ J}\]
03

- Calculate Total Kinetic Energy Before Collision

Since there are two trains with the same kinetic energy, multiply the kinetic energy of one train by 2:\[Total\ KE = 2 \times 5.2 \times 10^9 \text{ J} = 10.4 \times 10^9 \text{ J}\]
04

- Compute the Increase in Mass

According to the principle of energy conservation, the total kinetic energy is converted into mass after the trains come to rest. Use the mass-energy equivalence \( E = mc^2 \) where \( c = 3 \times 10^8 \text{ m/s} \).\[m = \frac{E}{c^2} = \frac{10.4 \times 10^9 \text{ J}}{(3 \times 10^8 \text{ m/s})^2} = \frac{10.4 \times 10^9 \text{ J}}{9 \times 10^{16} \text{ m}^2/\text{s}^2} = 1.156 \times 10^{-7} \text{ kg}\]Convert this mass into milligrams (mg):\[m = 1.156 \times 10^{-7} \text{ kg} \times 10^6 \text{ mg/kg} = 0.116 \text{ mg}\]

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 91影视!

Key Concepts

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

Mass-Energy Equivalence
Mass-energy equivalence is a fundamental concept in physics. It states that mass can be converted into energy and vice versa. This relationship is given by Einstein's famous equation: \( E = mc^2 \). Here, \( E \) is energy, \( m \) is mass, and \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \text{ m/s} \)).
This principle means that when the kinetic energy of the trains in the exercise is converted to other forms of energy, there is a corresponding increase in the system's mass.
The calculated kinetic energy of both trains before collision was found to be \( 10.4 \times 10^9 \text{ J} \). Using the mass-energy equivalence formula, this energy converts into a mass of \( 1.156 \times 10^{-7} \text{ kg} \), or \( 0.116 \text{ mg} \).
This increase is tiny compared to the initial mass but it's measurable, showing the direct impact of energy on mass in moderatly large systems like the one in the exercise.
Collision Physics
In physics, collisions are instances where two or more bodies exert forces on each other for a relatively short time.
They can be elastic, inelastic, or fully inelastic. In an elastic collision, both momentum and kinetic energy are conserved.
In the exercise, the trains undergo a fully inelastic collision, which means they stick together after the impact and come to a complete rest.
This kind of collision does not conserve kinetic energy but conserves momentum.
Before the collision, both trains have equal but opposite momentum, resulting in zero total momentum for the system.
After the collision, since they come to rest, the total momentum remains zero, satisfying the conservation of momentum.
The kinetic energy, however, is not conserved and instead is transformed into other energy forms, such as sound, heat, and the structural deformation of the trains.
Energy Conservation
The law of conservation of energy states that the total energy in an isolated system remains constant over time.
In the exercise, the initial kinetic energy of the two moving trains is a form of mechanical energy.
When the trains collide head-on and come to a stop, this kinetic energy doesn't vanish.
Instead,it transforms into other energy forms such as heat, sound, light, and the energy required to deform the trains.
Additionally, some of the energy transforms into an increase in mass in the system, as shown by mass-energy equivalence.
The key takeaway is that the total energy at the beginning (kinetic) and the end (deformed trains, heat, etc.) of the collision remains the same.
This transformation showcases the principle of energy conservation, reaffirming that energy cannot be destroyed but only changed from one form into another.

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

A flash of light is emitted at an angle \(\phi^{\prime}\) with respect to the \(x^{\prime}\) axis of the rocket frame. (a) Show that the angle \(\phi\) the direction of motion of this flash makes with respect to the \(x\) axis of the laboratory frame is given by the equation $$ \cos \phi=\frac{\cos \phi^{\prime}+v^{\mathrm{rel}} / c}{1+\left(v^{\mathrm{rel}} / c\right) \cos \phi^{\prime}} . $$ Optional: Show that your answer to Problem 34 gives the same result when the velocity \(v^{\prime}\) is given the value \(c .\) (b) A light source at rest in the rocket frame emits light uniformly in all directions. In the rocket frame \(50 \%\) of this light goes into the forward hemisphere of a sphere surrounding the source. Show that in the laboratory frame this \(50 \%\) of the light is concentrated in a narrow forward cone of half-angle \(\phi_{0}\) whose axis lies along the direction of motion of the particle. Derive the following expression for the halfangle \(\phi_{0}\) : $$ \cos \phi_{0}=v^{\mathrm{rel}} / c $$ This result is called the headlight effect. (c) What is the half-angle \(\phi_{0}\) in degrees for a light source moving at \(99 \%\) of the speed of light?

A gamma ray (an energetic photon) falls on a nucleus of initial mass \(m\), initially at rest. The energy \(E_{\mathrm{p}}\) of the incoming gamma ray matches the energy separation between the lowest energy of the nucleus and its first excited state, so the incident photon is absorbed. We want to know the mass \(m^{*}\) of the excited nucleus. (see Fig. \(38-13 .\) ) (a) Show that the conservation of energy and momentum equations are, in an obvious notation: and $$ \begin{array}{c} E_{\mathrm{p}}+m c^{2}=E_{m^{*}} \\ \frac{E_{\mathrm{p}}}{c}=p_{m^{*}}=\frac{\left(E_{m^{*}}^{2}-m^{* 2} c^{4}\right)^{1 / 2}}{c} . \end{array} $$ (b) Combine the two conservation equations to find an expression for \(m^{*}\) as a function of \(E_{\mathrm{p}}, m\), and \(c\). (c) Show that for very small values of \(E_{\mathrm{p}}\) the limiting result is \(m^{*}=m .\) Explain why this limiting result is reasonable.

The famous twin paradox is often introduced as follows: Two identical twins grow up together on Earth. When they reach adulthood, one twin zooms to a distant star and returns to find her stay-at-home sister much older than she is. Thus far no paradox. But Alexis Allen formulates the Twin Paradox for us: "The theory of special relativity tells us that all motion is relative. With respect to the traveling twin, the Earth-bound twin moves away and then returns. Therefore it is the Earthbound twin who should be younger than the 'traveling' twin. But when they meet again at the same place, it cannot possibly be that each twin is younger than the other twin. This Twin Paradox disproves relativity." The paradox is usually resolved by realizing that the traveling twin turns around. Everyone agrees which twin turns around, since the reversal of direction slams the poor traveler against the bulkhead of the decelerating starship, breaking her collarbone. The turnaround, evidenced by the broken collarbone, destroys the symmetry required for the paradox to hold. Good-bye Twin Paradox! Still, Alexis's father Cyril Allen has his doubts about this resolution of the paradox. "Your solution is extremely unsatisfying. It forces me to ask: What if the retro-rockets malfunction and will not fire at all to slow me down as I approach a distant star a thousand light-years from Earth? Then I cannot even stop at that star, much less turn around and head back to Earth. Instead, I continue moving away from Earth forever at the original constant speed. Does this mean that as I pass the distant star, one thousand light-years from Earth, it is no longer possible to say that I have aged less than my Earth-bound twin? But if not, then I would never have even gotten to the distant star at all during my hundred-year lifetime! Your resolution of the Twin Paradox is insufficient and unsatisfying." Write a half-page response to Cyril Allen, answering his objections politely but decisively.

An unpowered rocket moves past you in the positive \(x\) direction at speed \(v^{\text {rel }}=0.9 c\). This rocket fires a bullet out the back that you measure to be moving at speed \(v_{\text {bullet }}=0.3 c\) in the positive \(x\) direction. With what speed relative to the rocket did the rocket observer fire the bullet out the back of her ship?

A meter stick lies at rest in the rocket frame and makes an angle \(\phi^{\prime}\) with the \(x^{\prime}\) axis as measured by the rocket observer. The laboratory observer measures the \(x\) - and \(y\) -components of the meter stick as it streaks past. From these components the laboratory observer computes the angle \(\phi\) that the stick makes with his \(x\) axis. (a) Find an expression for the angle \(\phi\) in terms of the angle \(\phi^{\prime}\) and the relative speed \(v^{\text {rel }}\) between rocket and laboratory frames. (b) What is the length of the "meter" stick measured by the laboratory observer? (c) Optional: Why is your expression in part (a) different from equations derived in Problems 34 and \(35 ?\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Astrophysics

Read Explanation

Electromagnetism

Read Explanation

Torque and Rotational Motion

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

Read Explanation
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.