Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 104

Many nuclear collisions are truly elastic. If a proton with kinetic energy \(E_{0}\) collides elastically with another proton at rest and travels at an angle of \(25^{\circ}\) with respect to its initial path, what is its energy after the collision with respect to its original energy? What is the final energy of the proton that was originally at rest?

Short Answer

Expert verified
#tag_title# Step 6: Calculate the final energies #tag_content# We know the relationship between the final velocities and initial velocity (from Step 5). To find the final energies, we substitute the expressions for \(v_{1f}\) and \(v_{2f}\) into the equation for the proportion of energies: \(1 = \left(\frac{v_{1f}^2 + v_{2f}^2}{v_0^2}\right)\) Now, substitute the expressions from Step 5 for \(v_{1f}\) and \(v_{2f}\): \(1 = \frac{(\frac{v_0}{\cos^2{25^\circ} + \sin^2{25^\circ}})^2 + (\frac{v_0}{\cos^2{25^\circ} + \sin^2{25^\circ}})^2}{v_0^2}\) Simplify the equation, and the proportionality constant for both final energies is found to be: \(\frac{1}{2}\) Thus, the final energies of both protons are: \(E_{1f} = \frac{1}{2}E_0\) \(E_{2f} = \frac{1}{2}E_0\) So, both protons have equal final energies, which are half of the initial kinetic energy (\(E_0\)) of the initially moving proton.

Step by step solution

01

Identify the conservation laws in the problem

Since the collision is elastic, both linear momentum and kinetic energy are conserved. The conservation of linear momentum can be represented by the equation: \(\vec{p}_{i} = \vec{p}_{f}\) where \(\vec{p}_{i}\) is the initial momentum and \(\vec{p}_{f}\) is the final momentum. For kinetic energy, the conservation is given by: \(E_{i} = E_{f}\) where \(E_{i}\) is the initial kinetic energy and \(E_{f}\) is the final kinetic energy.
02

Establish the initial conditions

Initially, one proton has kinetic energy \(E_0\) and the other proton is at rest, i.e., its kinetic energy is zero. Let the mass of each proton be denoted as \(m\). Then the initial momentum can be written as: \(\vec{p}_{i} = m\vec{v}_{0}\) where \(\vec{v}_{0}\) is the initial velocity of the proton with kinetic energy \(E_0\).
03

Set up the conservation of linear momentum

Let \(\vec{v}_{1f}\) and \(\vec{v}_{2f}\) be the final velocities of the initially moving proton and initially stationary proton, respectively. The conservation equations, separated by components, will then be: \(m v_{0} = m v_{1f} \cos{25^\circ} + m v_{2f} \sin{25^\circ}\) (for the x-axis component) \(0 = -m v_{1f} \sin{25^\circ} + m v_{2f} \cos{25^\circ}\) (for the y-axis component)
04

Set up the conservation of kinetic energy

We will now set up the conservation of kinetic energy equation: \(E_0 = \frac{1}{2}m v_{1f}^2 + \frac{1}{2}m v_{2f}^2\)
05

Solve the equations

We have three equations and two unknowns, \(v_{1f}\) and \(v_{2f}\). Firstly, divide both x and y components of the momentum conservation equation by \(m\): \(v_0 = v_{1f} \cos{25^\circ} + v_{2f} \sin{25^\circ}\) \(0 = -v_{1f} \sin{25^\circ} + v_{2f} \cos{25^\circ}\) We can now solve these equations for \(v_{1f}\) and\(v_{2f}\). We get: \(v_{1f} = \frac{v_0}{\cos^2{25^\circ} + \sin^2{25^\circ}}\) \(v_{2f} = \frac{v_0}{\cos^2{25^\circ} + \sin^2{25^\circ}}\) Since the energy is proportional to the square of the velocity, the proportion between the final energies and the initial energy can be found, using the kinetic energy conservation equation: \(E_0 = \frac{1}{2}m \left( v_{1f}^2 + v_{2f}^2 \right)\) Divide both sides by \(E_0\): \(1 = \frac{v_{1f}^2}{v_0^2} + \frac{v_{2f}^2}{v_0^2}\)

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.

Start your free trial

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

Key Concepts

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

Conservation of Linear Momentum
When we tackle problems involving collisions in nuclear physics, one of the fundamental principles we use is the conservation of linear momentum. This law states that if no external forces are acting on a closed system, the total linear momentum of that system remains constant. In the context of an elastic collision between protons, as described in the exercise, this conservation can be expressed mathematically by the equation
\[\begin{equation}\vec{p}_{i} = \vec{p}_{f}\end{equation}\]
where \(\vec{p}_{i}\) refers to the initial total momentum of the system and \(\vec{p}_{f}\) refers to the final total momentum after the collision. Since one of the protons is at rest initially, the entire system's momentum is carried by the moving proton. After the collision, the momentum is shared between the two protons, but the total remains the same.
Conservation of Kinetic Energy
In an elastic collision, alongside momentum, kinetic energy is also conserved. This means that the total kinetic energy of the colliding particles before the collision is equal to the total kinetic energy after. This conservation is crucial for understanding nuclear collisions, as it dictates the distribution of energy between particles post-collision. For the case in our exercise:
\[\begin{equation}E_{i} = E_{f}\end{equation}\]
This equation implies that the kinetic energy of the moving proton before collision \(E_0\), is equal to the sum of the kinetic energies of both protons after the collision. This principle guides us to solve for the final velocities and consequently the energies of the protons post-collision.
Elastic Collision Equations
To solve elastic collision problems, we apply specific equations that embody the conservation laws. These equations allow us to calculate unknown variables such as the final velocities or energies of particles after a collision. The process outlined in the exercise first involves splitting the conservation of momentum into component forms – one for each direction, x and y. Additionally, we set an equation for the conservation of kinetic energy:
  • For x-axis momentum: \(m v_{0} = m v_{1f} \cos{25^\circ} + m v_{2f} \sin{25^\circ}\)
  • For y-axis momentum: \(0 = -m v_{1f} \sin{25^\circ} + m v_{2f} \cos{25^\circ}\)
  • For kinetic energy: \(E_0 = \frac{1}{2}m v_{1f}^2 + \frac{1}{2}m v_{2f}^2\)

By solving these equations simultaneously, we find the final velocities and can derive the sharing of kinetic energy between the protons after they've collided.
Momentum and Energy Conservation in Collisions
In nuclear physics, understanding how momentum and energy are conserved during collisions is indispensable. These conservations are not just separate concepts but are intertwined. The beauty of physics lies in how the conservation of momentum can sometimes be used to infer energy distribution, as in the case of elastic collisions. When two particles collide elastically, as we've seen in this proton collision exercise, the total system kinetic energy remains unaltered, despite individual particle energies changing. The process of solving the collision problems entails using both momentum and energy conservation equations in tandem. This dual consideration leads us to solve for the final states of the colliding protons by establishing a relationship between their velocities before and after the impact and hence their energies.

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

Using momentum and force principles, explain why an air bag reduces injury in an automobile collision.

Rank the following objects from highest to lowest in terms of momentum and from highest to lowest in terms of energy. a) an asteroid with mass \(10^{6} \mathrm{~kg}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) b) a high-speed train with a mass of \(180,000 \mathrm{~kg}\) and a speed of \(300 \mathrm{~km} / \mathrm{h}\) c) a 120 -kg linebacker with a speed of \(10 \mathrm{~m} / \mathrm{s}\) d) a 10 -kg cannonball with a speed of \(120 \mathrm{~m} / \mathrm{s}\) e) a proton with a mass of \(6 \cdot 10^{-27} \mathrm{~kg}\) and a speed of \(2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\).

Three birds are flying in a compact formation. The first bird, with a mass of \(100 . \mathrm{g}\) is flying \(35.0^{\circ}\) east of north at a speed of \(8.00 \mathrm{~m} / \mathrm{s}\). The second bird, with a mass of \(123 \mathrm{~g}\), is flying \(2.00^{\circ}\) east of north at a speed of \(11.0 \mathrm{~m} / \mathrm{s}\). The third bird, with a mass of \(112 \mathrm{~g}\), is flying \(22.0^{\circ}\) west of north at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). What is the momentum vector of the formation? What would be the speed and direction of a \(115-\mathrm{g}\) bird with the same momentum?

Moessbauer spectroscopy is a technique for studying molecules by looking at a particular atom within them. For example, Moessbauer measurements of iron (Fe) inside hemoglobin, the molecule responsible for transporting oxygen in the blood, can be used to determine the hemoglobin's flexibility. The technique starts with X-rays emitted from the nuclei of \({ }^{57}\) Co atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14 \mathrm{keV}\) and \(14 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(\mathrm{A}^{57}\) Co nucleus recoils as an \(\mathrm{X}\) -ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26} \mathrm{~kg} .\) What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

A small car of mass 1000 . kg traveling at a speed of \(33.0 \mathrm{~m} / \mathrm{s}\) collides head on with a large car of mass \(3000 \mathrm{~kg}\) traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The two cars stick together. The duration of the collision is \(100 . \mathrm{ms}\). What acceleration (in \(g\) ) do the occupants of the small car experience? What acceleration (in \(g\) ) do the occupants of the large car experience?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Torque and Rotational Motion

Read Explanation

Wave Optics

Read Explanation

Electrostatics

Read Explanation

Medical Physics

Read Explanation

Astrophysics

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.

Sign-up for free