Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Q1Q (page 410)

Under what conditions is the momentum of a system constant? Can the x component of momentum be constant even if the y component is changing? In what circumstances? Give an example of such behavior.

Short Answer

Expert verified

The momentum of a system is constant if there is no net vector force on the system.

Yes,the xcomponent of momentum be constant even if they component is changing if the object is moving in xz plane.

Step by step solution

01

Significance of the momentum of a system

Momentum is defined as the product of the mass and velocity of the object and it can be changed by applying the force to the object. So, it can be constant if no external force acts on the system.

02

Identification of the condition of the constant momentum of a system

In the case when net vector forces are not acting on the system, then the momentum is said to be constant.

Assume that a ball is moving in xz plane. In this condition, gravitational force and centripetal force will act in y direction only that will result in the change in the y component of the momentum. There will not be any force that will act in x direction , so there is no change in the x component of the 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 Fe-57 nucleus is at rest and in its first excited state, 14.4 keV above the ground state (14.4 × 103 eV, where 1 eV = 1.6×10−19 J). The nucleus then decays to the ground state with the emission of a gamma ray (a high-energy photon). (a) Wthe recoil speed of the nucleus? (b) Calculate the slight difference in eV between the gamma-ray energy and the 14.4 keV difference between the initial and final nuclear states. (c) The “Mössbauer effect” is the name given to a related phenomenon discovered by Rudolf Mössbauer in 1957, for which he received the 1961 Nobel Prize for physics. If the Fe-57 nucleus is in a solid block of iron, occasionally when the nucleus emits a gamma ray the entire solid recoils as one object. This can happen due to the fact that neighbouring atoms and nuclei are connected by the electric interatomic force. In this case, repeat the calculation of part (a) and compare with your previous result. Explain briefly

In a collision between an electron and a hydrogen atom, why is it useful to select both objects as the system? Pick all that apply: (1) The total momentum of the system does not change during the collision. (2) The sum of the final kinetic energies must equal the sum of the initial kinetic energies for a two-object system. (3) The kinetic energy of a two-object system is nearly zero. (4) The forces the objects exert on each other are internal to the system and don’t change the total momentum of the system. (5) During the time interval from just before to just after the collision, external forces are negligible.

A bullet of mass 0.102 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? (b) Calculate the kinetic energy of the bullet plus the block before the collision. (c) Calculate the kinetic energy of the bullet plus the block after the collision. (d) Was this collision elastic or inelastic? (e) Calculate the rise in internal energy of the bullet plus block as a result of the collision.

In outer space a rock whose mass is 3kg and whose velocity was(3900,-2900,3000)m/sstruck a rock with mass 13kg and velocity(220,-260,300)m/s. After the collision, the 3kg rock’s velocity is(3500,-2300,3500)m/s. (a) What is the final velocity of the 13kg rock? (b) What is the change in the internal energy of the rocks? (c) Which of the following statements about Q (transfer of energy into the system because of a temperature difference between system and surroundings) are correct? (1)Q≈0 because the duration of the collision was very short. (2)Q=Ethermal of the rocks. (3)Q≈0 because there are no significant objects in the surroundings. (4)Q=∆k of the rocks.

Redo Problem P21, this time using the concept of the center-of-momentum reference frame.

A car of mass 2300 kg collides with a truck of mass 4300 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car’s velocity just before the collision was⟨38, 0, 0⟩m/s, and the truck’s velocity just before the collision was⟨−16, 0, 27⟩m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. What system and principle should you use? (1) Energy Principle (2) Car plus truck (3) Momentum Principle (4) Car alone (5) Truck alone (b) What is the velocity of the stuck-together car and truck just after the collision? (c) In your analysis in part (b), why can you neglect the effect of the force of the road on the car and truck? (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)? (e) Is this collision elastic or inelastic?
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Linear Momentum

Read Explanation

Measurements

Read Explanation

Fluids

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

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