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

Under what conditions is the momentum of a system constant? Can the x component of momentum be constant even if they 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 the ycomponent 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 xzplane. In this condition, gravitational force and centripetal force will act in ydirection only that will result in the change in the ycomponent of the momentum. There will not be any force that will act in xdirection , so there is no change in the xcomponent 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.

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

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 donโ€™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.

Consider a head-on collision between two objects. Object 1, which has mass m1, is initially in motion, and collides head-on with object 2, which has massm2and is initially at rest. Which of the following statements about the collision are true?

(1)pโ†’1,initial=pโ†’1,final+pโ†’2,final.

(2)|pโ†’1,final|<|pโ†’1, initial|.

(3) Ifm2โ‰ซm1, then|ฮ”pโ†’1|>|ฮ”pโ†’2|.

(4) Ifm1โ‰ซm2, then the final speed of object 2 is less than the initial speed of object 1.

(5) Ifm2โ‰ซm1, then the final speed of object 1 is greater than the final speed of object 2.

Redo the analysis of the Rutherford experiment, this time using the concept of the centre-of-momentum reference frame. Let m = the mass of the alpha particle and M = the mass of the gold nucleus. Consider the specific case of the alpha particle rebounding straight back. The incoming alpha particle has a momentum p1 , the outgoing alpha particle has a momentum p3 , and the gold nucleus picks up a momentum p4 . (a) Determine the velocity of the centre of momentum of the system. (b) Transform the initial momenta to that frame (by subtracting the centre-of-momentum velocity from the original velocities). (c) Show that if the momenta in the centre-of-momentum frame simply turn around (180โ—ฆ), with no change in their magnitudes, both momentum and energy conservation are satisfied, whereas no other possibilitysatisfies both conservation principles. (Try drawing some other
momentum diagrams.) (d) After the collision, transform back to
the original reference frame (by adding the center-of-momentum
velocity to the velocities of the particles in the center-of-mass
frame). Although using the center-of-momentum frame may be
conceptually more difficult, the algebra for solving for the final
speeds is much simpler

An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10 MeV (10 ร— 106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Assuming that all speeds are small compared to the speed of light, answer the following questions about the collision. (a) What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? (b) What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? (c) What is the final kinetic energy of the alpha particle? (d) What is the final kinetic energy of the gold nucleus? (e) Assuming that the movement of the gold nucleus is negligible, calculate how close the alpha particle will get to the gold nucleus in this head-on collision.

A hydrogen atom is at rest, in the first excited state, when it emits a photon of energy 10.2 eV. (a) What is the speed of the ground-state hydrogen atom when it recoils due to the photon emission? Remember that the magnitude of the momentum of a photon of energy E is p = E/c. Make the initial assumption that the kinetic energy of the recoiling atom is negligible compared to the photon energy. (b) Calculate the kinetic energy of the recoiling atom. Is this kinetic energy indeed negligible compared to the photon energy?

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