Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Q19PE (page 289)

Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum.

Short Answer

Expert verified

The equation for the kinetic energy of a particle expressed as a function of its momentum is, \(K.E. = \dfrac{1}{2}pv\)

Step by step solution

01

Explaining the equation for kinetic energy.

The equation for the kinetic energy of a particle having mass, m and having a velocity, vis given by,

\(K.E. = \dfrac{1}{2}m{v^2}\) …………………. (1)

02

Explaining the equation for momentum

The momentum of the same particle (as described above) having a mass mand having velocity v is,

\(p = mv\) ………………….2

On dividing both sides of the equation 2 by v, we get

\(\therefore \dfrac{p}{v} = \dfrac{{mv}}{v}\)

\(\therefore \dfrac{p}{v} = m\)…………………. 3

03

Derivation of the equation of kinetic energy of a particle expressed as a function of its momentum.

Substitute equation 3 in equation1, we get

\(K.E. = \dfrac{1}{2}\dfrac{p}{v}{v^2} = \dfrac{1}{2}pv\)

One vin the numerator cancels with the other v in the denominator.

Thus,

\(K.E. = \dfrac{1}{2}pv\)

The above equation is the equation for the kinetic energy of a particle expressed as a function of its momentum

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!

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\(90.0\;{\rm{kg}}\)ice hockey player hits a\(0.150\;{\rm{kg}}\)puck, giving the puck a velocity of\(45.0\;{\rm{m/s}}\). If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal\(15.0\;{\rm{m}}\)away?

What is the acceleration of a \[5000\;kg\]rocket taking off from the Moon, where the acceleration due to gravity is only \[1.6\;m/{s^2}\], if the rocket expels \[8.00\;kg\] of gas per second at an exhaust velocity of \[{\rm{2}}{\rm{.20 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}\]?

A\(5.50\;kg\)bowling ball moving at\(9.00\;{\rm{m/s}}\)collides with a\(0.850\;{\rm{kg}}\)bowling pin, which is scattered at an angle of\(85.{0^ \circ }\)to the initial direction of the bowling ball and with a speed of\(15.0\;{\rm{m/s}}\). (a) Calculate the final velocity (magnitude and direction) of the owling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.

A battleship that is6.00×107kg and is originally at rest fires a1100-kg artillery shell horizontally with a velocity of575m/s . (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs.

What is an inelastic collision? What is a perfectly inelastic collision?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Electromagnetism

Read Explanation

Electricity

Read Explanation

Solid State Physics

Read Explanation

Conservation of Energy and Momentum

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