Momentum is conserved if no external force

Similar documents
Physics 111 Lecture 5 Circular Motion

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

Rotational Motion: Statics and Dynamics

Physics 107 TUTORIAL ASSIGNMENT #8

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

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

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

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

Momentum and Collisions

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

Physics 101 Lecture 6 Circular Motion

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

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Potential Energy and Conservation of Energy

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

PHYS Summer Professor Caillault Homework Solutions. Chapter 9

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.

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

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?

Chapters 5-8. Dynamics: Applying Newton s Laws

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


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

Conflict Exam Issue. Sorry, Can t do it. Please see Kevin Pitts if you have any additional questions or concerns about this. Office is 231 Loomis

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

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

Dynamics of Rotational Motion

Chapter 4: The laws of motion. Newton s first law

Exam 3: Equation Summary

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

Chapter 8. Accelerated Circular Motion

Sections and Chapter 10

Department of Physics, Korea University Page 1 of 5

Recap. Centripetal acceleration: v r. a = m/s 2 (towards center of curvature)

Chapter 7-8 Rotational Motion

Chapter 5. really hard to start the object moving and then, once it starts moving, you don t have to push as hard to keep it moving.

Chapter 1: Mathematical Concepts and Vectors

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

Exam 3: Equation Summary

21 MAGNETIC FORCES AND MAGNETIC FIELDS

OSCILLATIONS AND GRAVITATION

Spring 2001 Physics 2048 Test 3 solutions

PHYS 1114, Lecture 21, March 6 Contents:

Chapter 13 Gravitation

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

Force can be exerted by direct contact between bodies: Contact Force.

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

Ch 13 Universal Gravitation

SAMPLE QUIZ 3 - PHYSICS For a right triangle: sin θ = a c, cos θ = b c, tan θ = a b,

Chapter 5. Uniform Circular Motion. a c =v 2 /r

Centripetal Force. Lecture 11. Chapter 8. Course website:

Physics 120 Homework Solutions April 25 through April 30, 2007

Central Force Motion

Gravitation. AP/Honors Physics 1 Mr. Velazquez

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

Physics 11 Chapter 3: Vectors and Motion in Two Dimensions. Problem Solving

A car of mass m, traveling at constant speed, rides over the top of a circularly shaped hill as shown.

( ) ( ) Review of Force. Review of Force. r = =... Example 1. What is the dot product for F r. Solution: Example 2 ( )

Physics 2001 Problem Set 5 Solutions

Chapter. s r. check whether your calculator is in all other parts of the body. When a rigid body rotates through a given angle, all

Uniform Circular Motion

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

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.

Uniform Circular Motion

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.

Kinematics in 2-D (II)

From Newton to Einstein. Mid-Term Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.

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

Chapter 12. Kinetics of Particles: Newton s Second Law

Experiment 09: Angular momentum

Chapter 5 Force and Motion

Chapter 5 Force and Motion

Physics 201 Lecture 18

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)

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

Principles of Physics I

PS113 Chapter 5 Dynamics of Uniform Circular Motion

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

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

Chapter 4. Newton s Laws of Motion. Newton s Law of Motion. Sir Isaac Newton ( ) published in 1687

Objective Notes Summary

Uniform Circular Motion

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

Chapter 5. Applying Newton s Laws. Newton s Laws. r r. 1 st Law: An object at rest or traveling in uniform. 2 nd Law:

Circular-Rotational Motion Mock Exam. Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.

Describing Circular motion

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

Physics 111 Lecture 10. SJ 8th Ed.: Chap Torque, Energy, Rolling. Copyright R. Janow Spring basics, energy methods, 2nd law problems)

Translation and Rotation Kinematics

Physics 1C Fall 2011: Quiz 1 Version A 1

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

Physics 207 Lecture 5. Lecture 5

Lab #9: The Kinematics & Dynamics of. Circular Motion & Rotational Motion

ω = θ θ o = θ θ = s r v = rω

PROJECTILE MOTION. At any given point in the motion, the velocity vector is always a tangent to the path.

Rectilinea Motion. A foce P is applied to the initially stationay cat. Detemine the velocity and displacement at time t=5 s fo each of the foce histoi

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

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

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

Transcription:

Goals: Lectue 13 Chapte 9 v Employ consevation of momentum in 1 D & 2D v Examine foces ove time (aka Impulse) Chapte 10 v Undestand the elationship between motion and enegy Assignments: l HW5, due tomoow l o Wednesday, Read all of Chapte 10 Physics 207: Lectue 13, Pg 1 Momentum is conseved if no extenal foce p befoe = p afte Physics 207: Lectue 13, Pg 2 Page 1

Inelastic collision in 1-D: Example l A block of mass M is initially at est on a fictionless hoizontal suface. A bullet of mass m is fied at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a final speed V. l In tems of m, M, and V : What is the momentum of the bullet with speed v? v V x befoe afte Physics 207: Lectue 13, Pg 3 Inelastic collision in 1-D: Example mv What is the momentum of the bullet with speed v? v Key question: Is thee a net extenal x-di foce? v If not, then momentum in the x-diection is conseved! P Befoe P Afte v mv + M 0 aaaa = ( m + M )V V befoe afte v= (1+M / m)v x Physics 207: Lectue 13, Pg 4 Page 2

p = p p p x y z = = mv = mv Momentum is a vecto mv mv = p x y z x iˆ + p y ˆj + p z kˆ Physics 207: Lectue 13, Pg 5 A pefectly inelastic collision in 2-D l Conside a collision in 2-D (cas cashing at a slippey intesection...no fiction). v 1 m 1 θ V m 1 + m 2 m 2 v 2 befoe afte l If no extenal foce momentum is conseved. l Momentum is a vecto so p x, p y and p z Physics 207: Lectue 13, Pg 6 Page 3

A pefectly inelastic collision in 2-D p 1 p V v 1 θ m 1 + m 2 m 1 m 2 v 2 p 2 befoe afte l x-di p x : m 1 v 1 = (m 1 + m 2 ) V cos θ l y-di p y : m 2 v 2 = (m 1 + m 2 ) V sin θ Physics 207: Lectue 13, Pg 7 Execise Momentum is a Vecto (!) quantity l A block slides down a fictionless amp and then falls and lands in a cat which then olls hoizontally without fiction l In egads to the block landing in the cat is momentum conseved? A. Yes B. No C. Yes & No D. Too little infomation given Physics 207: Lectue 13, Pg 8 Page 4

Execise Momentum is a Vecto (!) quantity l x-diection: No net foce so P x is conseved. l y-diection: Net foce, inteaction with the gound so depending on the system (i.e., do you include the Eath?) P y is not conseved (system is block and cat only) 2 kg 5.0 m Let a 2 kg block stat at est on a 30 incline and slide vetically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cat 30 7.5 m 10 kg What is the final velocity of the cat? Physics 207: Lectue 13, Pg 9 Execise Momentum is a Vecto (!) quantity l x-diection: No net foce so P x is conseved l y-diection: v y of the cat + block will be zeo and we can ignoe v y of the block when it lands in the cat. j N i 5.0 m 30 mg 30 7.5 m 1) a i = g sin 30 = 5 m/s 2 2) d = 5 m / sin 30 = ½ a i t 2 10 m = 2.5 m/s 2 t 2 2s = t v = a i t = 10 m/s v x = v cos 30 = 8.7 m/s y x Physics 207: Lectue 13, Pg 10 Page 5

Execise Momentum is a Vecto (!) quantity l x-diection: No net foce so P x is conseved l y-diection: v y of the cat + block will be zeo and we can ignoe v y of the block when it lands in the cat. j N Initial inal i 5.0 m 30 mg 30 P x : MV x + mv x = (M+m) V x M 0 + mv x = (M+m) V x V x = m v x / (M + m) = 2 (8.7)/ 12 m/s V x = 1.4 m/s y 7.5 m x Physics 207: Lectue 13, Pg 11 Elastic Collisions l Elastic means that the objects do not stick. l Thee ae many moe possible outcomes but, if no extenal foce, then momentum will always be conseved l Stat with a 1-D poblem. Befoe Afte Physics 207: Lectue 13, Pg 12 Page 6

Billiads l Conside the case whee one ball is initially at est. befoe afte p a θ p b P a φ The final diection of the ed ball will depend on whee the balls hit. Physics 207: Lectue 13, Pg 13 Billiads: Without extenal foces, consevation of momentum (and enegy Ch. 10 & 11) l Consevation of Momentum l x-di P x : m v befoe = m v afte cos θ + m V afte cos φ l y-di P y : 0 = m v afte sin θ + m V afte sin φ befoe afte p afte θ p b P afte φ Physics 207: Lectue 13, Pg 14 Page 7

Home Execise Inelastic Collision in 1-D with numbes Do not ty this at home! ice (no fiction) Befoe: A 4000 kg bus, twice the mass of the ca, moving at 30 m/s impacts the ca at est. What is the final speed afte impact if they move togethe? Physics 207: Lectue 13, Pg 15 M = 2m Home execise Inelastic Collision in 1-D m initially MV V 0 M m + M ice v = 0 (no fiction) 2m 2m + m 0 = ( m + M)V o V = V0 = V0 finally 2V o /3 = 20 m/s v f =? Physics 207: Lectue 13, Pg 16 Page 8

Explosions: A collision in evese l A two piece assembly is hanging vetically at est at the end of a 2.0 m long massless sting. The mass of the two pieces ae 60 and 20 kg espectively. Suddenly you obseve that the 20 kg is ejected hoizontally at 30 m/s. The time of the explosion is shot compaed to the swing of the sting. l Does the tension in the sting incease o decease afte the explosion? Befoe Afte Physics 207: Lectue 13, Pg 17 Explosions: A collision in evese l A two piece assembly is hanging vetically at est at the end of a 2.0 m long massless sting. The mass of the two pieces ae 60 and 20 kg espectively. Suddenly you obseve that the 10 kg is ejected hoizontally at 30 m/s. l Deciphe the physics: 1. The geen ball ecoils in the x diection (3 d Law) and, because thee is no net foce in the x-diection the x- momentum is conseved. 2. The motion of the geen ball is constained to a cicula path thee must be centipetal (i.e., adial acceleation) Befoe Afte Physics 207: Lectue 13, Pg 18 Page 9

Explosions: A collision in evese l A two piece assembly is hanging vetically at est at the end of a 20.0 m long massless sting. The mass of the two pieces ae 60 & 20 kg espectively. Suddenly you obseve that the 20 kg mass is suddenly ejected hoizontally at 30 m/s. l Cons. of x-momentum P x befoe = P x afte = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s T befoe = Weight = (60+20) x 10 N = 800 N Befoe Afte Σ y = m a y (adial) = M V 2 / = T Mg T = Mg + MV 2 / = 600 N + 60x(10) 2 /20 N = 900 N Physics 207: Lectue 13, Pg 19 Impulse (A vaiable foce applied fo a given time) l Gavity: At small displacements a constant foce l Spings often povide a linea foce (-k x) towads its equilibium position (Chapte 10) l Collisions often involve a vaying foce (t): 0 maximum 0 l We can plot foce vs time fo a typical collision. The impulse, J, of the foce is a vecto defined as the integal of the foce duing the time of the collision. Physics 207: Lectue 13, Pg 20 Page 10

oce and Impulse (A vaiable foce applied fo a given time) l J a vecto that eflects momentum tansfe J t t dt = ( dp / dt) dt = p dp Impulse J = aea unde this cuve! (Tansfe of momentum!) Impulse has units of Newton-seconds t i t t f t Physics 207: Lectue 13, Pg 21 oce and Impulse l Two diffeent collisions can have the same impulse since J depends only on the momentum tansfe, NOT the natue of the collision. same aea t t big, small t t t small, big t Physics 207: Lectue 13, Pg 22 Page 11

Aveage oce and Impulse av av t t big, av small t t t small, av big t Physics 207: Lectue 13, Pg 23 Execise oce & Impulse l Two boxes, one heavie than the othe, ae initially at est on a hoizontal fictionless suface. The same constant foce acts on each one fo exactly 1 second. Which box has the most momentum afte the foce acts? light heavy A. heavie B. lighte C. same D. can t tell Physics 207: Lectue 13, Pg 24 Page 12

Boxing: Use Momentum and Impulse to estimate g foce Physics 207: Lectue 13, Pg 25 Back of the envelope calculation J (1) m am ~ 7 kg (2) v am ~7 m/s (3) Impact time t ~ 0.01 s Question: Ae these easonable? Impulse J = p ~ m am v am ~ 49 kg m/s avg ~ J/ t ~ 4900 N (1) m head ~ 7 kg = dt a head = / m head ~ 700 m/s 2 ~ 70 g! t = t l Enough to cause unconsciousness ~ 40% of a fatal blow l Only a ough estimate! avg Physics 207: Lectue 13, Pg 26 Page 13

Discussion Execise l The only foce acting on a 2.0 kg object moving along the x- axis. Notice that the plot is foce vs time. l If the velocity v x is +2.0 m/s at 0 sec, what is v x at 4.0 s? l p = m v = Impulse l m v = J 0,1 + J 1,2 + J 2,4 l m v = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s m v = 4 N s v = 2 m/s v x = 2 + 2 m/s = 4 m/s Physics 207: Lectue 13, Pg 27 Ch. 10 : Kinetic & Potential enegies l Kinetic enegy, K = ½ mv 2, is defined to be the lage scale collective motion of one o a set of masses l Potential enegy, U, is defined to be the hidden enegy in an object which, in pinciple, can be conveted back to kinetic enegy l Mechanical enegy, E Mech, is defined to be the sum of U and K. Physics 207: Lectue 13, Pg 28 Page 14

Lectue 13 Assignment: l HW6 up soon l o Wednesday: Read all of chapte 10 Physics 207: Lectue 13, Pg 29 Page 15