LINEAR MOMENTUM Physical quantities that we have been using to characterize the motion of a particle

Similar documents
1121 T Question 1

Lecture 23: Central Force Motion

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Chapter 13 Gravitation

Center of Mass and Linear

r dt dt Momentum (specifically Linear Momentum) defined r r so r r note: momentum is a vector p x , p y = mv x = mv y , p z = mv z

ATMO 551a Fall 08. Diffusion

Physics 107 TUTORIAL ASSIGNMENT #8

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 18: System of Particles II. Slide 18-1

Momentum is conserved if no external force

Momentum and Collisions

Chapter 7-8 Rotational Motion

Easy. r p 2 f : r p 2i. r p 1i. r p 1 f. m blood g kg. P8.2 (a) The momentum is p = mv, so v = p/m and the kinetic energy is

Rotational Motion: Statics and Dynamics

Department of Physics, Korea University Page 1 of 5

SPH4U Unit 6.3 Gravitational Potential Energy Page 1 of 9

EN40: Dynamics and Vibrations. Midterm Examination Tuesday March

30 The Electric Field Due to a Continuous Distribution of Charge on a Line

C3 Interactions transfer momentum. C4 - Particles and Systems. General Physics 1

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

CHAPTER 25 ELECTRIC POTENTIAL

Orbital Angular Momentum Eigenfunctions

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

Ch. 4: FOC 9, 13, 16, 18. Problems 20, 24, 38, 48, 77, 83 & 115;

Phys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.

Chapter 5 Force and Motion

Chapter 5 Force and Motion

Physics 181. Assignment 4

Phys 331: Ch 4. Conservative Forces & Curvi-linear 1-D Systems 1

Class 6 - Circular Motion and Gravitation

Dynamics of Rotational Motion

EN40: Dynamics and Vibrations. Midterm Examination Thursday March

Physics Tutorial V1 2D Vectors

arxiv: v1 [physics.pop-ph] 3 Jun 2013

AP-C WEP. h. Students should be able to recognize and solve problems that call for application both of conservation of energy and Newton s Laws.

to point uphill and to be equal to its maximum value, in which case f s, max = μsfn

PHYSICS NOTES GRAVITATION

PHYSICS 1210 Exam 2 University of Wyoming 14 March ( Day!) points

Centripetal Force OBJECTIVE INTRODUCTION APPARATUS THEORY

Lab #0. Tutorial Exercises on Work and Fields

b) (5) What average force magnitude was applied by the students working together?

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

Chapters 5-8. Dynamics: Applying Newton s Laws

Physics 1114: Unit 5 Hand-out Homework (Answers)

HW Solutions # MIT - Prof. Please study example 12.5 "from the earth to the moon". 2GmA v esc

Objects usually are charged up through the transfer of electrons from one object to the other.

Physics 1A (a) Fall 2010: FINAL Version A 1. Comments:

06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )

CHAPTER 5: Circular Motion; Gravitation

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II September 15, 2012 Prof. Alan Guth PROBLEM SET 2

When a mass moves because of a force, we can define several types of problem.

Physics 121 Hour Exam #5 Solution

Rotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart

b) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?

Study on Application of New Theorem of Kinetic Energy

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1

Physics 120 Homework Solutions April 25 through April 30, 2007

On the Correct Formulation of the Starting Point of Classical Mechanics

τ TOT = r F Tang = r F sin φ Chapter 13 notes: Key issues for exam: The explicit formulas

To Feel a Force Chapter 7 Static equilibrium - torque and friction

Astronomy 421 Concepts of Astrophysics I. Astrophysics Talks at UNM. Course Logistics. Backgrounds. Other Opportunities

PHYS Summer Professor Caillault Homework Solutions. Chapter 9

Physics 4A Chapter 8: Dynamics II Motion in a Plane

1131 T Question 1

Electrostatics. 3) positive object: lack of electrons negative object: excess of electrons

Physics 111 Lecture 5 Circular Motion

On the velocity autocorrelation function of a Brownian particle

Physics 2212 GH Quiz #2 Solutions Spring 2016

2/24/2014. The point mass. Impulse for a single collision The impulse of a force is a vector. The Center of Mass. System of particles


PHYS 1410, 11 Nov 2015, 12:30pm.

The Millikan Experiment: Determining the Elementary Charge

Movie Review Part One due Tuesday (in class) please print

Principles of Physics I

Rigid Body Dynamics 2. CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2018

Potential Energy and Conservation of Energy

Tidal forces. m r. m 1 m 2. x r 2. r 1

EELE 3331 Electromagnetic I Chapter 4. Electrostatic fields. Islamic University of Gaza Electrical Engineering Department Dr.

Pendulum in Orbit. Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ (December 1, 2017)

CHAPTER 10 ELECTRIC POTENTIAL AND CAPACITANCE

Homework 7 Solutions

r ˆr F = Section 2: Newton s Law of Gravitation m 2 m 1 Consider two masses and, separated by distance Gravitational force on due to is

Chapter 7. Impulse and Momentum

17.1 Electric Potential Energy. Equipotential Lines. PE = energy associated with an arrangement of objects that exert forces on each other

Chapter 8. Linear Momentum, Impulse, and Collisions

CIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line

DYNAMICS OF UNIFORM CIRCULAR MOTION

You are to turn in this final and the scantron to the front when finished.

Algebra-based Physics II

The Laws of Motion ( ) N SOLUTIONS TO PROBLEMS ! F = ( 6.00) 2 + ( 15.0) 2 N = 16.2 N. Section 4.4. Newton s Second Law The Particle Under a Net Force

Newton s Laws, Kepler s Laws, and Planetary Orbits

F g. = G mm. m 1. = 7.0 kg m 2. = 5.5 kg r = 0.60 m G = N m 2 kg 2 = = N

An Exact Solution of Navier Stokes Equation

Multiple choice questions [100 points] As shown in the figure, a mass M is hanging by three massless strings from the ceiling of a room.

( ) rad ( 2.0 s) = 168 rad


OSCILLATIONS AND GRAVITATION

10. Universal Gravitation

LECTURE 15. Phase-amplitude variables. Non-linear transverse motion

Transcription:

LINEAR MOMENTUM Physical quantities that we have been using to chaacteize the otion of a paticle v Mass Velocity v Kinetic enegy v F Mechanical enegy + U Linea oentu of a paticle (1) is a vecto!

Siple exaple about collisions Wall 0.1 Kg Wall 0.1 Kg B E F O R E A F T E R What is the change in LINEAR MOMENTUM?

Execise B E F O R E A F T E R

What is the LINEAR MOMENTUM of this acoscopic object? Answe: Add up the LINEAR MOMENTUM of evey individual icoscopic constituents of the acoscopic object

LINEAR MOMENTUM of a syste coposed of N paticles Next task: To descibe the otion of a syste of paticles

d dt (3) CASE 1: No extenal foces acting on the syste (only intenal ones) Notice: 0 (4)

No fiction when no net extenal foces ae acting on the acoscopic object (5) Exaple 10 k/h 6 k/h 5 k/h 11 k/h M 1 150 kg 2 100 kg

Question M 1 100 Kg 2 167 Kg

CASE 2: Extenal foces (as well as intenal foces, of couse) act on the syste

Afte the kick

The Cente of Mass

Exaple

Exaple 0 c î + 0 c

Exaple Exaple 0.5 c î - 0.5 c Exaple

These types of pobles can be solved by syety aguents and soe judicious ticks. To that effect, let s show fist a geneal esult that states that the cente of ass of a coplicated geoety object can obtained by a) fist finding the cente of ass of sub-egions pats, and b) then finding the final cente of ass based on the knowledge of the cente of ass of those sub-egions. Conside an abitay distibution of N paticles, as shown in the figue below. The paticles have been nubeed, accodingly. In geneal, the position of the cente of ass (with espect to soe efeence) is given by,... R 1 1 2 2 N N CM... 1 2 N N 1 1 2 3 4 5 6 7 8 N Notice, we can beak down the nueato into an abitay nube of subsets goups of paticles (see also the othe figue below): R CM 1 1 2 2 3 3 1 2 4 4... We also egoup in the denoinato 5 5 N 6 6 7 7...

...... 3 3 7 6 5 4 3 2 1 7 7 6 6 5 5 4 4 2 2 1 1 CM R 1 2 3 4 5 6 7 8 1 A B Goup-A has a ass equal to ( 1 + 2 + 3 ), let s call it M A Goup-B has a ass equal to ( 4 + 5 + 6 + 7 ), let s call it M B etc Notice the nueato can be e-witten as follows B B B )...... 3 3 2 2 1 1 ( A 7 7 6 6 5 5 4 4 A A M M M M M M CM R B B B )...... 3 3 2 2 1 1 ( A 7 7 6 6 5 5 4 4 A A M M M M M M CM R CM of goup A CM of goup B That is, when tying to find the CM of an object whose ass is distibuted in a soewhat coplicated geoetical distibution, we can sub-divide the object into salle siple objects. Each division can be epesented by its coesponding CM.

Exaple

Motion of the Cente of Mass So it is like all the ass of the syste wee concentated at. This syste of paticles can be eplaced by 2 3 M 1

We want to pove the following: Fo otion analysis, the acoscopic ball can be eplaced by a point object located at the cente of ass.

1

Expession 1 also iplies d

Exaple: The ballistic pendulu The objective is to find the speed v of the bullet A block of wood is hanging fo two long cods. A bullet is fied into the block, coing quickly to est. The block+bullet syste then swing upwads aising a vetical distance h6.3 c v M g h M 5.4 kg 9.5 g h 6.3 c Pocedue-1 Initial kinetic enegy: (1/2) v 2 Initial potential enegy: 0 Final kinetic enegy: 0 Final potential enegy: (+M)gh The consevation of the echanical enegy iplies (1/2) v 2 (+M)gh Fo which we can solve fo v. What is wong with the pocedue above? Answe: The consevation of echanical-enegy is not valid in this case. Pat of the bullet s initial kinetic enegy is dissipated as heat

Is the wok / kinetic-enegy theoe K W valid in this case? Answe: Yes Pocedue-2 Step-1 Consevation of linea oentu We assue that the collision is vey bief, such that a) Duing the collision the foces on the block (gavitation and tension fo the cod) ae balanced. That is, no net extenal foce is acting on the bullet-block syste. Hence, the syste can be consideed isolated and, theefoe, its total linea oentu is conseved b) The collision is in one diension (just afte the collision the syste oves hoizontally.) Befoe the collision P i v Just afte the collision, the syste oves with velocity V (unknown yet) P f ( + M)V The consevation of the linea oentu iplies v ( + M)V V [ / ( + M) ] v (1) Step-2 Consevation of the echanical enegy Afte the collision thee ae not dissipative foces. Initial kinetic enegy: (1/2) ( +M) V 2 (1/2) ( +M) [ 2 / ( + M) 2 ] v 2 (1/2) [ 2 / ( + M) ] v 2 Initial potential enegy: 0

Final kinetic enegy: 0 Final potential enegy: (+M)gh The consevation of the echanical enegy iplies (1/2) [ 2 / ( + M) ] v 2 (+M)gh v 2 2[ (+M) 2 / 2 ]gh Substituting values v 630 /s