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

Similar documents
Momentum is conserved if no external force

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

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

PHYS Summer Professor Caillault Homework Solutions. Chapter 9

FZX: Personal Lecture Notes from Daniel W. Koon St. Lawrence University Physics Department CHAPTER 7

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

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

Chapter 13 Gravitation

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

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 6: motion in two and three dimensions III. Slide 6-1

Chapter 7-8 Rotational Motion

Physics 107 TUTORIAL ASSIGNMENT #8

PHYS 172: Modern Mechanics. Summer Lecture 4 The Momentum Principle & Predicting Motion Read

Explain to each other your movie project and the variable you are calculating. Discuss how will you know if it could occur in real life.

Physics 181. Assignment 4

Physics 121 Hour Exam #5 Solution

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

OSCILLATIONS AND GRAVITATION

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.

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

CHAPTER 25 ELECTRIC POTENTIAL

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.

Motion in Two Dimensions

Chapter 5 Force and Motion

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

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

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

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

EN40: Dynamics and Vibrations. Midterm Examination Thursday March

Chapter 5 Force and Motion

MAGNETIC FIELD INTRODUCTION

Section 26 The Laws of Rotational Motion

Central Force Motion

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

System of Particles: Center-of-Mass

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

Center of Mass and Linear

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

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.

KEPLER S LAWS OF PLANETARY MOTION

Flux. Area Vector. Flux of Electric Field. Gauss s Law

PHYSICS NOTES GRAVITATION

Momentum and Collisions

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

20-9 ELECTRIC FIELD LINES 20-9 ELECTRIC POTENTIAL. Answers to the Conceptual Questions. Chapter 20 Electricity 241

7.2. Coulomb s Law. The Electric Force

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

Easy. P4.2 Since the car is moving with constant speed and in a straight line, the. resultant force on it must be regardless of whether it is moving

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?

PHYSICS 272 Electric & Magnetic Interactions

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

Chapters 5-8. Dynamics: Applying Newton s Laws

Department of Physics, Korea University Page 1 of 5

Physics 11 Chapter 4: Forces and Newton s Laws of Motion. Problem Solving

ATMO 551a Fall 08. Diffusion

Homework 7 Solutions

Diffusion and Transport. 10. Friction and the Langevin Equation. Langevin Equation. f d. f ext. f () t f () t. Then Newton s second law is ma f f f t.

Physics 2212 GH Quiz #2 Solutions Spring 2016

Potential Energy and Conservation of Energy

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

Ch 13 Universal Gravitation

Centripetal Force OBJECTIVE INTRODUCTION APPARATUS THEORY

16.1 Permanent magnets

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

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

1 2 U CV. K dq I dt J nqv d J V IR P VI

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism

Physics 235 Chapter 5. Chapter 5 Gravitation

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

ASTR415: Problem Set #6

Chapter 5 Page 5.1 CHAPTER 5. r Force times distance has units of energy. Therefore, fxr=u, or f / is dimensionless.

Physics 201 Lecture 18

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Between any two masses, there exists a mutual attractive force.

Gravitation. Chapter 12. PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman. Lectures by James Pazun

Gauss Law. Physics 231 Lecture 2-1

Chapter 2: Basic Physics and Math Supplements

Escape Velocity. GMm ] B

3.2 Centripetal Acceleration

Electric Charge and Field

3. Electromagnetic Waves II

PreClass Notes: Chapter 9, Sections

Lecture 19 Angular momentum. Chapter

An Exact Solution of Navier Stokes Equation

4. Two and Three Dimensional Motion

Lecture 3.7 ELECTRICITY. Electric charge Coulomb s law Electric field

12.1 Introduction 12-1

Physics NYB problem set 5 solution

Central Force Problem. Central Force Motion. Two Body Problem: Center of Mass Coordinates. Reduction of Two Body Problem 8.01 W14D1. + m 2. m 2.

Physics: Work & Energy Beyond Earth Guided Inquiry

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

1121 T Question 1

q r 1 4πε Review: Two ways to find V at any point in space: Integrate E dl: Sum or Integrate over charges: q 1 r 1 q 2 r 2 r 3 q 3

Sections and Chapter 10

Lab 10: Newton s Second Law in Rotation

Your Comments. Do we still get the 80% back on homework? It doesn't seem to be showing that. Also, this is really starting to make sense to me!

MONTE CARLO SIMULATION OF FLUID FLOW

Solutions to Problems : Chapter 19 Problems appeared on the end of chapter 19 of the Textbook

Physics 207 Lecture 5. Lecture 5

Transcription:

Physics 1501 Fall 2008 Mechanics, Themodynamics, Waves, Fluids Lectue 18: System of Paticles II Slide 18-1

Recap: cente of mass The cente of mass of a composite object o system of paticles is the point whee, fom the standpoint of Newton s second law, the mass acts as though it wee concentated. The position of the cente of mass is a weighted aveage of the positions of the individual paticles: Fo a system of discete paticles, cm = m i i M Fo a continuous distibution of matte, cm = dm M In both cases, M is the system s total mass. Slide 18-2

Recap: motion of the cente of mass The cente of mass obeys Newton s second law: F net extenal = Ma Hee most pats of the skie s body undego complex motions, but his cente of mass descibes the paabolic tajectoy of a pojectile: cm Slide 18-3

Finding the cente of mass A system of individual paticles x = mx 1 + mx 3 cm 4m y cm = my 1 + my 3 4m = m(x 1 x 1 ) = 0 4m = 2my 1 4m = 1 2 y 1 = 3 4 L = 0.43L A system of continuous matte Expess the mass element dm in tems of the geometical vaiable x: dm M = (w/l)x dx 1 2 wl Evaluate the integal: x cm = 1 M x dm = 1 M 0 = 2x dx L 2 L 2Mx x dx = 2 L L 2 0 L 2 x2 dx so x cm = 2 L 2 0 L x 2 dx = 2 x 3 L 2 3 0 L= 2L3 3L = 2 2 3 L Slide 18-4

Moe on cente of mass The cente of mass of a composite object can be found fom the CMs of its individual pats. An object s cente of mass need not lie within the object! Which point is the CM? The high jumpe cleas the ba, but his CM doesn t. Slide 18-5

question A thick wie is bent into a semicicle, as shown in the figue. Which of the points shown is the cente of mass of the wie? A. Point A B. Point B C. Point C Slide 18-6

Motion of the cente of mass Absent any extenal foces on a system, the cente of mass motion emains unchanged; if it s at est, it emains in the same place no matte what intenal foces may act. c. Hee Jumbo walks, but the CM of the ail ca plus elephant doesn t move. This allows us to find the ca s final position: x cm = m Jx Jf + m c x cf M = m J(x Ji +19 m+ x cf )+ m c x cf M (19 m)m J (19 m)(4.8 t) x cm = 4 6 (m J + m c ) = (15 t + 4.8 t) = 4.6 m Slide 18-7

Momentum and the cente of mass The cente of mass obeys Newton s law, which can be witten F = Ma o, equivalently, F net extenal = Ma cm = d P F net extenal dt P P = m vi i whee is the total momentum of the system: = M v cm with the velocity of the cente of mass. v cm Slide 18-8

question A 500-g fiewoks ocket is moving with velocity v =60 ˆj m s at the instant it explodes. If you wee to add the momentum vectos of all its fagments just afte the explosion, what would be the esult? A. v= 60 ˆj kg m s B. v = 30 ˆ j kg ms C. v = 60000 ˆ j kg ms D. v = 30000 ˆj kg m s Slide 18-9

Consevation of momentum When the net extenal foce is zeo, dpdt= 0. Theefoe the total momentum of the system is unchanged: P = constant This is the consevation of linea momentum. A system of thee billiad balls: Initially two ae at est; all the momentum is in the lefthand ball: Now they e all moving, but the total momentum emains the same: Slide 18-10

Collisions A collision is a bief, intense inteaction between objects. The collision time is shot compaed with the timescale of the objects oveall motion. Intenal foces of the collision ae so lage that we can neglect any extenal foces acting on the system duing the bief collision time. Theefoe linea momentum is essentially conseved duing collisions. Slide 18-11

Elastic and inelastic collisions In an elastic collision, the intenal foces of the collision ae consevative. Theefoe an elastic collision conseves kinetic enegy as well as linea momentum. In an inelastic collision, the foces ae not consevative and mechanical enegy is lost. In a totally inelastic collision, the colliding objects stick togethe th to fom a single composite object. But if a collision is totally inelastic, that doesn t necessaily mean that all kinetic enegy is lost. Slide 18-12

Totally inelastic collisions Totally inelastic collisions ae govened entiely by consevation of momentum. Since the colliding objects join to fom a single composite object, thee s only one final velocity: Befoe collision Afte collision Theefoe consevation of momentum eads m ) v 1 v 1 + m 2 v 2 = ( m 1 + m 2 )v f Slide 18-13

Elastic collisions Elastic collisions conseve both momentum and kinetic enegy: Befoe collision Afte collision Theefoe the consevation laws ead m v1i 1 + m v2i 2 = m v1f 1 + m v2f 2 1 m v 2 + 1 m v 2 = 1 m v 2 + 1 m v 2 2 1 1i 2 2 2i 2 1 1f 2 2 2f Slide 18-14

question Two skates toss a basketball back and foth on fictionless ice. Which one of the following does not change? A. The momentum of an individual skate B. The momentum of the system consisting of one skate and dthe basketball C. The momentum of the basketball D. The momentum of the system consisting of both skates and the basketball Slide 18-15

Elastic collisions in one dimension In geneal, the consevation laws don t detemine the outcome of an elastic collision. Othe infomation is needed, such as the diection of one of the outgoing paticles. But fo one-dimensional collisions, when paticles collide head-on, then the initial velocities detemine the outcome: Solving both consevation laws in this case gives Slide 18-16

question Which one of the following qualifies as an inelastic collision? A. Two magnets appoach, thei noth poles facing; they epel and evese diection without touching. B. A tuck stikes a paked ca and the two slide off togethe, cumpled metal hopelessly entwined. C. A basketball flies though the ai on a paabolic tajectoy. D. A basketball ebounds off the backboad. Slide 18-17

Special cases: 1-D elastic collisions; m 2 initially iti at est 1) m 1 << m 2 Incident object ebounds with essentially its incident velocity 2) m 1 = m 2 Incident object stops; stuck object moves away with initial speed of incident object 3) m 1 >> m 2 Incident object continues with essentially its initial velocity; stuck object moves away with twice that velocity Slide 18-18

Clicke question Ball A is at est on a level floo. Ball B collides elastically with Ball A, and the two move off sepaately, but in the same diection. What can you conclude about the masses of the two balls? A. Ball A and Ball B have the same mass. B. BllBh Ball B has a geate mass than Ball BllA. C. Ball A has a geate mass than Ball B. D. You cannot conclude anything without moe infomation. Slide 18-19

Summay A composite system behaves as though its mass is concentated at the cente of mass: cm = m i i M (discete paticles) cm = The cente of mass obeys Newton s laws, so dm M F net extenal = Ma cm o, equivalently, F net extenal = d P dt (continuous matte) In the absence of a net extenal foce, a system s linea momentum is conseved, egadless of what happens intenally to the system. Collisions ae bief, intense inteactions that conseve momentum. Elastic collisions also conseve kinetic enegy. Totally inelastic collisions occu when colliding objects join to make a single composite object. Slide 18-20