Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For a nonrelativistic particle of mass \(m,\) show that \(K=p^{2} /(2 m) .\) [Hint: Start with the nonrelativistic expressions for kinetic energy \(K\) and momentum \(p\).]

Short Answer

Expert verified
Answer: For a nonrelativistic particle of mass m, the relationship between kinetic energy (K) and momentum (p) is given by the equation: \(K=p^{2} /(2 m)\).

Step by step solution

01

Write down the expression for kinetic energy

Kinetic energy (K) for a nonrelativistic particle of mass (m) and velocity (v) is given by the formula: K = \(\frac{1}{2} mv^2\)
02

Write down the expression for momentum

Momentum (p) is defined as the product of an object's mass (m) and its velocity (v): p = mv
03

Solve for velocity in the momentum equation

In order to relate the two expressions, we'll first solve for velocity (v) in the momentum equation above by dividing both sides by mass (m): v = \(\frac{p}{m}\)
04

Substitute the expression for velocity in the kinetic energy equation

Now that we have an expression for velocity (v) in terms of momentum (p) and mass (m), we can substitute it into the kinetic energy equation: K = \(\frac{1}{2}m\left(\frac{p}{m}\right)^2\)
05

Simplify the equation to show K = \(p^2 / (2m)\)

Simplify the above equation to show that the kinetic energy (K) is equal to the square of the momentum (p) divided by twice the mass (2m): K = \(\frac{1}{2}m\left(\frac{p^2}{m^2}\right)\) K = \(\frac{p^2}{2m}\) Thus, we have shown that for a nonrelativistic particle of mass m, \(K=p^{2} /(2 m)\).

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

Radon decays as \(^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .\) 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. \((1 u=931.5 \mathrm{MeV} / c^{2} .\) ) If the radon nucleus is initially at rest in the lab frame, at what speeds (in the lab frame) do the (a) polonium-218 nucleus and (b) alpha particle move?

The Tevatron is a particle accelerator at Fermilab that accelerates protons and antiprotons to high energies in an underground ring. Scientists observe the results of collisions between the particles. The protons are accelerated until they have speeds only \(100 \mathrm{m} / \mathrm{s}\) slower than the speed of light. The circumference of the ring is \(6.3 \mathrm{km} .\) What is the circumference according to an observer moving with the protons? [Hint: Let \(v=c-u\) where $v \text { is the proton speed and } u=100 \mathrm{m} / \mathrm{s} .$]
An engineer in a train moving toward the station with a velocity \(v=0.60 c\) lights a signal flare as he reaches a marker \(1.0 \mathrm{km}\) from the station (according to a scale laid out on the ground). By how much time, on the stationmaster's clock, does the arrival of the optical signal precede the arrival of the train?
Particle \(A\) is moving with a constant velocity \(v_{\mathrm{AE}}=+0.90 c\) relative to an Earth observer. Particle \(B\) moves in the opposite direction with a constant velocity \(v_{\mathrm{BE}}=-0.90 c\) relative to the same Earth observer. What is the velocity of particle \(B\) as seen by particle \(A ?\)
An astronaut in a rocket moving at \(0.50 c\) toward the Sun finds himself halfway between Earth and the Sun. According to the astronaut, how far is he from Earth? In the frame of the Sun, the distance from Earth to the Sun is \(1.50 \times 10^{11} \mathrm{m}\).
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