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.

Short Answer

Expert verified

1) the total momentum of the system does not change during the collision, 2) the sum of the final kinetic energies must equal to the sum of the initial kinetic energies for a two-object system, 4) The forces the object exert on each other are internal to the system and don’t change the total momentum of the system and 5) during the time interval from just before to just after the collision, external forces are negligible.

Step by step solution

01

Significance of the law of conservation of momentum of a system

This law states that the momentum of a particular system before and after the collision is constant if no external force acts on the system.

The law of the conservation of momentum gives the effect of the collision between the electron and the hydrogen atom.

02

Determination of the collision between the hydrogen and electron

From the law of conservation of momentum, momentum mainly describes the product of velocity and mass. Moreover, the total momentum of a body before and after the collision remains constant. Furthermore, as both the electron and the hydrogen atom have been selected as a system, then their total momentum does not change during the collision.

Also, if the collision is considered elastic, then, the net change in the kinetic energy is also zero.

Thus the following conditions will be applicable, 1) the total momentum of the system does not change during the collision, 2) the sum of the final kinetic energies must equal to the sum of the initial kinetic energies for a two-object system, 4) The forces the object exerts on each other are internal to the system and don’t change the total momentum of the system and 5) during the time interval from just before to just after the collision, external forces are negligible.

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

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)p1,initial=p1,final+p2,final.

(2)|p1,final|<|p1, initial|.

(3) Ifm2m1, then|Δp1|>|Δp2|.

(4) Ifm1m2, then the final speed of object 2 is less than the initial speed of object 1.

(5) Ifm2m1, then the final speed of object 1 is greater than the final speed of object 2.

A car moving east at 30m/s runs head-on into a 3000 kg truck moving west at 20m/s. The vehicles stick together. Use the concept of the centre of momentum frame to determine how much kinetic energy is lost.

Car 1 headed north and car 2 headed west collide. They stick together and leave skid marks on the pavement, which show that car 1 was deflected 30°(so car 2 was deflected 60°). What can you conclude about the cars before the collision?

We can use our results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall embedded in the Earth. Suppose that a professional baseball pitcher hurts a baseball(m=155g)with a speed ofrole="math" localid="1668602437434" 100mih(v1=44m/s)at a wall that is securely anchored to the Earth, and the ball bounces back with little loss of kinetic energy.

(a) What is the approximate recoil speed of the Earth(M=6×1024kg)?

(b) Calculate the approximate recoil kinetic energy of the Earth and compare it to the kinetic energy of the baseball.

The Earth gets lots of momentum (twice the momentum of the baseball) but very little kinetic energy.

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

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