/*! 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 46 An alpha particle (mass \(=4.00 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 46

An alpha particle (mass \(=4.00 \mathrm{u}\) ) has a head-on, elastic collision with a nucleus (mass \(=166 \mathrm{u}\) ) that is initially at rest. What percentage of the kinetic energy of the alpha particle is transferred to the nucleus in the collision?

Short Answer

Expert verified
Answer: To calculate the percentage of kinetic energy transferred, we need the initial or final velocities of the alpha particle, which were not provided in this exercise. If such values were provided or could be calculated from given information, the remaining calculations could be completed using the conservation of momentum and conservation of energy equations to find the percentage of kinetic energy transferred.

Step by step solution

01

Conservation of Momentum

According to the law of conservation of momentum, the total momentum of the system before the collision equals the total momentum of the system after the collision. Let's denote the masses of alpha particle and nucleus as m1 and m2 respectively, and their initial and final velocities as u1, u2, v1, and v2 respectively. Since the nucleus is initially at rest, u2 = 0. Initially, Momentum of alpha particle = m1u1 Initially, Momentum of nucleus = m2u2 (which is 0) Finally, Momentum of alpha particle = m1v1 Finally, Momentum of nucleus = m2v2 Conservation of momentum equation can be written as: m1u1 + m2u2 = m1v1 + m2v2 Plugging in the known values, we get: m1u1 = m1v1 + m2v2
02

Conservation of Energy

For an elastic collision, the total mechanical energy is conserved. Hence, the initial kinetic energy (KE_initial) equals the final kinetic energy (KE_final). KE_initial = KE_final Initial kinetic energy of alpha particle = \(\frac{1}{2}m1u1^2\) Initial kinetic energy of nucleus = 0 (since it's initially at rest) Final kinetic energy of alpha particle = \(\frac{1}{2}m1v1^2\) Final kinetic energy of nucleus = \(\frac{1}{2}m2v2^2\) Now, equating the initial and final kinetic energies, we get: \(\frac{1}{2}m1u1^2 = \frac{1}{2}m1v1^2 + \frac{1}{2}m2v2^2\)
03

Solve for v2

To find the percentage of kinetic energy transferred, we need to find the final velocity of the nucleus (v2) after the collision. Using the conservation of momentum equation from Step 1, we can solve for v2: v2 = \(\frac{m1(u1-v1)}{m2}\)
04

Substitute values and find the percentage of transferred kinetic energy

The mass of alpha particle m1 = 4.00 u and mass of nucleus m2 = 166 u. Calculate the final kinetic energy of the nucleus by substituting the values: KE_final_nucleus = \(\frac{1}{2}m2v2^2\) Percentage of kinetic energy transferred = \(\frac{KE_{final\_nucleus}}{KE_{initial}}\) 脳 100 Plug in the values and compute the final answer. Note that we have not been provided the initial or final velocities of the alpha particle. If such values were provided or could be calculated from given information, the remaining calculations could be completed to find the percentage of kinetic energy transferred.

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!

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 bungee jumper is concerned that his elastic cord might break if it is overstretched and is considering replacing the cord with a high-tensile- strength steel cable. Is this a good idea?

A soccer ball with mass \(0.265 \mathrm{~kg}\) is initially at rest and is kicked at an angle of \(20.8^{\circ}\) with respect to the horizontal. The soccer ball travels a horizontal distance of \(52.8 \mathrm{~m}\) after it is kicked. What is the impulse received by the soccer ball during the kick? Assume there is no air resistance.

A baseball pitcher delivers a fastball that crosses the plate at an angle of \(7.25^{\circ}\) relative to the horizontal and a speed of \(88.5 \mathrm{mph}\). The ball (of mass \(0.149 \mathrm{~kg}\) ) is hit back over the head of the pitcher at an angle of \(35.53^{\circ}\) with respect to the horizontal and a speed of \(102.7 \mathrm{mph}\). What is the magnitude of the impulse received by the ball?

Stuck in the middle of a frozen pond with only your physics book, you decide to put physics in action and throw the 5.00 -kg book. If your mass is \(62.0 \mathrm{~kg}\) and you throw the book at \(13.0 \mathrm{~m} / \mathrm{s}\), how fast do you then slide across the ice? (Assume the absence of friction.)

A 170.-g hockey puck moving in the positive \(x\) direction at \(30.0 \mathrm{~m} / \mathrm{s}\) is struck by a stick at time \(t=2.00 \mathrm{~s}\) and moves in the opposite direction at \(25.0 \mathrm{~m} / \mathrm{s}\). If the puck is in contact with the stick for \(0.200 \mathrm{~s}\), plot the momentum and the position of the puck, and the force acting on it as a function of time, from 0 to \(5.00 \mathrm{~s}\). Be sure to label the coordinate axes with reasonable numbers.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Mechanics and Materials

Read Explanation

Particle Model of Matter

Read Explanation

Force

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.