/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q4CP Car 1 headed north and car 2 hea... [FREE SOLUTION] | 91影视

91影视

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掳(so car 2 was deflected 60掳). What can you conclude about the cars before the collision?

Short Answer

Expert verified

Car 1鈥檚 momentum was 3times greater than the momentum of car 2.

Step by step solution

01

 Step 1: Identification of the given data 

The given data can be listed below as,

  • Momentum of car 1, p1
  • Momentum of car 2, p2
  • Initial momentum, pi
  • Final momentum, pf
  • Car 1鈥檚 deflection,30掳
  • Car 2鈥檚 deflection,60掳

The space diagram is as follows:

02

Understanding Impact parameter 

The distance between centres perpendicular to the incoming velocity is calledthe 鈥渋mpact parameter鈥 and is often denoted byb.

A head-on collision has an impact parameter of zero.The smaller the impact parameter, the more severe is the collision, and thelarger the deflection angle of the incoming particle (larger 鈥渟cattering鈥), exceptfor a head-on collision, where if the masses are equal the incoming ball stopsdead and the target ball gets the entire momentum.

03

Determination of the momentum of the cars.

The net force on the system is zero, then the momentum of the system must be conserved:

pi=pf

Applying the momentum conservation inx direction:

p2,i=pfcos(60)鈥︹︹. first equation

Applying the momentum conservation iny direction:

p1,i=pfsin(60)鈥︹︹...second equation

By dividing the second equation by the first equation:

p1,ip2,i=pfsin(60)pfcos(60)=tan(60)

tan (60) = 鈭3

So,

p1,i=3p2,i.

Hence, car 1鈥檚 momentum was 3times greater than the momentum of car 2.

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 91影视!

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

Two asteroids in outer space collide and stick together. The mass of each asteroid, and the velocity of each asteroid before the impact, are known. To find the momentum of the stuck-together asteroids after the impact, what approach would be useful? (1) Use the Energy Principle. (2) Use the Momentum Principle. (3) It depends on whether or not the speed of the asteroids was near the speed of light. (4) Use the relationship among velocity, displacement, and time. (5) It depends on whether the collision was elastic or inelastic.

A 6kg mass traveling at speed 10m/s strikes a stationarymass head-on, and the two masses stick together.

(a) What was the initial total kinetic energy?

(b) What is the final speed?

(c) What is the final total kinetic energy?

(d) What was the increase in internal energy of the two masses?

A beam of high-energy 蟺 鈭 (negative pions) is shot at a flask of liquid hydrogen, and sometimes a pion interacts through the strong interaction with a proton in the hydrogen, in the reaction 蟿蟿-+p+蟿蟿-+X+, where X + is a positively charged particle of unknown mass. The incoming pion momentum is 3 GeV/c (1GeV = 1000 MeV = 1 脳 109 electron-volts). The pion is scattered through , and its momentum is measured to be 1510 MeV/c (this is done by observing the radius of curvature of its circular trajectory in a magnetic field). A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. What is the rest mass of the unknown X+ particle, in MeVc2? Explain your work carefully. It is advantageous to write the equations not in terms of v but rather in terms of E and p; remember that E2-(pc)2=(mc2)2.

A Fe-57 nucleus is at rest and in its first excited state, 14.4 keV above the ground state (14.4 脳 103 eV, where 1 eV = 1.6脳10鈭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 鈥淢枚ssbauer effect鈥 is the name given to a related phenomenon discovered by Rudolf M枚ssbauer 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 outer space, rock 1 whose mass is5Kg and whose velocity was(3300,-3100,3400)m/s struck rock 2, which was at rest. After the collision, rock 1鈥檚 velocity is(2800,-2400,3700)m/s . (a) What is the 铿乶al momentum of rock 2? (b) Before the collision, what was the kinetic energy of rock 1? (c) Before the collision, what was the kinetic energy of rock 2? (d) After the collision, what is the kinetic energy of rock 1? (e) Suppose that the collision was elastic (that is, there was no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2? (f) On the other hand, suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, whereEthermal,1+Ethermal,2=7.16106J . What is the 铿乶al kinetic energy of rock 2? (g) In this case (some of the kinetic energy being converted to thermal energy), what was the transfer of energy Q (microscopic work) from the surroundings into the two-rock system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

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.