Motion in Two Dimensions
|
|
- Cecilia Harper
- 5 years ago
- Views:
Transcription
1 SOLUTIONS TO PROBLEMS Motion in Two Dimensions Section 3.1 The Position, Velocity, and Acceleation Vectos P3.1 x( m) 0!3 000!1 70!4 70 m y( m)! ! 330 m (a) Net displacement x + y 4.87 km at 8.6 S of W FIG. P3.1 (b) Aveage speed ( 0.0 m s) ( 180 s) + ( 5.0 m s) ( 10 s) + ( 30.0 m s) ( 60.0 s) Aveage velocity 4.87! 103 m 360 s 180 s + 10 s s 13.5 m s along R 3.3 m s 1
2 Motion in Two Dimensions Section 3. P3.3 Two-Dimensional Motion with Constant Acceleation ( 4.00î ĵ ) m s and v ( 0.0) ( 0.0î! 5.00ĵ ) m s (a) a x! v x! t 0.0 " m s m s a y! v y! t "5.00 " m s "0.300 m s (b)! tan "1 " ( " fom + x axis At t 5.0 s x f x i + v xi t + 1 a x t ( 5.0) + 1 ( 0.800) ( 5.0) 360 m y i + v yi t + 1 a y t! ( 5.0) + 1 (!0.300)( 5.0)!7.7 m v xf v xi + a x t ( ) 4 m s v yf v yi + a y t 1! 0.3( 5)!6.5 m s v " tan!1 y (!6.50 tan!1 4.0 (!15. v x P3.5 a 3.00ĵ m s ; 5.00î m s ; i 0î + 0ĵ (a) f i + v i t + 1 a t! 5.00tî t ĵ " m + a t ( 5.00î tĵ ) m s (b) t.00 s, f 5.00(.00) î + 1 ( 3.00) (.00) ĵ ( 10.0î ĵ ) m so x f 10.0 m, 6.00 m 5.00î v f v xf + v yf ( ) m s ( ) ĵ 5.00î ĵ ( 5.00) + ( 6.00) 7.81 m s
3 Chapte 3 3 P3.6 (a) Fo the x-component of the motion we have x f x i + v xi t + 1 a x t m 0 + ( 1.80! 10 7 m s)t + 1 ( 8! 1014 m s )t ( 4! m s )t + ( 1.80! 10 7 m s)t " 10 " m 0 t "1.80! 107 m s ± ( )("10 " m) ( ) ( 1.8! 10 7 m s) " 4 4! m s 4! m s "1.8! 107 ± 1.84! 10 7 m s 8! m s We choose the + sign to epesent the physical situation Hee 4.39! 105 m s t 8! m s 5.49! 10"10 s. y i + v yi t + 1 a y t ( 1.6! 1015 m s )( 5.49! 10 "10 s).41! 10 "4 m. So, f ( 10.0 î ĵ ) mm. (b) ( )( 5.49! 10 "10 s) ( ) î + ( 4.39! 105 m s) î + ( 8.78! 105 m s) ĵ ( ) î + ( 8.78! 105 m s) ĵ + at 1.80! 10 7 m s î + 8! 1014 m s î + 1.6! m s ĵ 1.80! 10 7 m s 1.84! 10 7 m s ( 1.84! 10 7 m s) + ( 8.78! 10 5 m s) 1.85! 10 7 m s v (d)! tan "1 y ( 8.78 ) 10 5 tan" ) 10 7 (.73 v x
4 4 Motion in Two Dimensions Section 3.3 Pojectile Motion *P3.9 At the maximum height v y 0, and the time to each this height is found fom v yf y + a y t as t v yf! y a y 0! y!g y g. The vetical displacement that has occued duing this time is Thus, if!y ( ) max 1 ft " (!y) max v y,avg t v yf + y t " 0 + y " " 1 m 3.81 ft 3.66 m, then y g y g. y g (!y) max ( 9.80 m s )( 3.66 m) 8.47 m s, and if the angle of pojection is! 45, the launch speed is P3.11 Take the oigin at the mouth of the cannon. y sin! 8.47 m s 1.0 m s. sin 45 x f v xi t Theefoe, v yi t + 1 a y t : 000 m ( m s) t t.00 s 800 m ( m s)sin! i t + 1 ("9.80 m s )t ( ) ".00 s 800 m ( m s)sin! i ( m s 800 m( cos! i ) 000 m( sin! i ) " 19.6 m 19.6 m m( cos! i ) 000 m 1 " cos! i ( ) ".00 s cos! i (31360)cos! i + (640000)cos 4! i ( )cos! i " ( )cos 4! i cos 4! i " cos! i cos! i ± ( ) " 4( )(384) o ! i.4 o 89.4 (Both solutions ae valid.)
5 Chapte 3 5 P3.1 (a) x f v xi t 8.00cos 0.0 ( 3.00).6 m (b) Taking y positive downwads, and the final point just befoe gound impact, v yi t + 1 g t 8.00sin 0.0 ( 3.00) + 1 ( 9.80) ( 3.00) 5.3 m. Now take the final point 10 m below the window ( sin 0.0 )t + 1 ( 9.80)t 4.90t +.74t! t!.74 ± (.74 ) s 9.80 P3.13 Conside the motion fom oiginal zeo height to maximum height h: v yf v yi + a y (! y i ) gives 0 v yi! g h! 0 v yi gh ( ) o Now conside the motion fom the oiginal point to half the maximum height: v yf v yi + a y (! y i ) gives v yh gh +!g v yh gh ( ) " 1 h! 0 so At maximum height, the speed is v x 1 v x + v yh 1 v x + gh Solving, v x gh 3 Now the pojection angle is! i tan "1 v yi v x tan "1 gh gh/3 tan"
6 6 Motion in Two Dimensions P3.15 (a) We use the tajectoy equation: x f tan! i " v i cos. With x f 36.0 m,! i 0.0 m s, and! 53.0 we find ( 9.80 m s ) 36.0 m ( 36.0 m)tan 53.0! 0.0 m s ( 3.94! 3.05) m m. ( ) ( ) cos ( 53.0 ) gx f 3.94 m. The ball cleas the ba by (b) The time the ball takes to each the maximum height is t 1 sin! i g ( )( sin53.0 ) 0.0 m s 1.63 s m s The time to tavel 36.0 m hoizontally is t x f x t 36.0 m (0.0 m s )( cos 53.0 ).99 s. Since t > t 1 the ball cleas the goal on its way down. P3.16 The hoizontal component of displacement is x f v xi t ( )t. Theefoe, the time equied to each the building a distance d away is t d. At this time, the altitude of the wate is v yi t + 1 a " d y t sin! i ( g " d Theefoe the wate stikes the building at a height h above gound level of h d tan! i " gd cos! i.. Section 3.4 The Paticle in Unifom Cicula Motion ( ) P3.3 a c v 0.0 m s 1.06 m 377 m s The mass is unnecessay infomation.
7 Chapte 3 7 P m ; v t! T a v R ( )! m ( ) 60.0 s 00 ev m s 10.5 m s 19 m s inwad P3.7 The satellite is in fee fall. Its acceleation is due to gavity and is by effect a centipetal acceleation. a c g so v g. Solving fo the velocity, v g ( 6, ) ( 103 m) ( 8.1 m s ) 7.58! 10 3 m s. v! T and T! v! ( 7,000 " 10 3 m) 5.80 " 10 3 s 7.58 " 10 3 m s T 5.80 " 10 3 s 1 min 60 s ( 96.7 min. Section 3.5 Tangential and Radial Acceleation P3.8 (b) We do pat (b) fist. The tangential speed is descibed by + a t t 0.7 m s 0 + a t ( 1.75 s) so a t m s fowad (a) Now at t 1.5 s, + a t t 0 + ( 0.4 m s )1.5 s 0.5 m s so ( ) a c v 0.5 m s 0. m 1.5 m s towad the cente a a + a t 0.4 m s fowad m s inwad a fowad and inwad at! tan " ( a 1.31 m s fowad and 7.3 inwad
8 8 Motion in Two Dimensions P3.30 (a) See figue to the ight. (b) The components of the 0. and the.5 m s along the ope togethe constitute the centipetal acceleation: a c (.5 m s )cos( 90.0! 36.9 ) + ( 0. m s )cos m s Section 3.6 a c v so v a c 9.7 m s ( 1.50 m) 6.67 m s tangent to cicle v 6.67 m s at 36.9 above the hoizontal Relative Velocity FIG. P3.30 P3.33 Total time in still wate t d v ! 103 s. Total time time upsteam plus time downsteam: t up (1.0! 0.500) 1.43 " 103 s t down 588 s Theefoe, t total 1.43! ! 10 3 s, 1.0 moe than if the wate wee still.. *P3.34 v ce the velocity of the ca elative to the eath. v wc the velocity of the wate elative to the ca. v we the velocity of the wate elative to the eath. These velocities ae elated as shown in the diagam at the ight. (a) (b) Since v we is vetical, v wc sin 60.0 v ce 50.0 km h o v wc 57.7 km h at 60.0 west of vetical. Since v ce has zeo vetical component, FIG. P3.34 v we v wc cos 60.0 ( 57.7 km h)cos km h downwad.
9 Additional Poblems Chapte 3 9 P3.47 x f x t t cos 40.0 Thus, when x f 10.0 m, t 10.0 m cos At this time, should be 3.05 m!.00 m 1.05 m. Thus, 1.05 m ( sin 40.0 )10.0 m cos 40.0 Fom this, 10.7 m s. + 1 " 10.0 m (!9.80 m s ) cos 40.0.
10 10 Motion in Two Dimensions P3.50 Measue heights above the level gound. The elevation y b of the ball follows y b R + 0! 1 gt with x t so y b R! gx. (a) The elevation y of points on the ock is descibed by y + x R. We will have y b y at x 0, but fo all othe x we equie the ball to be above the ock suface as in y b > y. Then y b + x > R R! gx R " R! gx + x > R + g x x > R g x x > gx R. If this inequality is satisfied fo x appoaching zeo, it will be tue fo all x. If the ball s paabolic tajectoy has lage enough adius of cuvatue at the stat, the ball will clea the whole ock: 1 > gr > gr. (b) With gr and y b 0, we have 0 R! gx gr o x R. The distance fom the ock s base is x! R (! 1)R.
Phys 201A. Homework 5 Solutions
Phys 201A Homewok 5 Solutions 3. In each of the thee cases, you can find the changes in the velocity vectos by adding the second vecto to the additive invese of the fist and dawing the esultant, and by
More informationDescribing Circular motion
Unifom Cicula Motion Descibing Cicula motion In ode to undestand cicula motion, we fist need to discuss how to subtact vectos. The easiest way to explain subtacting vectos is to descibe it as adding a
More informationPhysics 4A Chapter 8: Dynamics II Motion in a Plane
Physics 4A Chapte 8: Dynamics II Motion in a Plane Conceptual Questions and Example Poblems fom Chapte 8 Conceptual Question 8.5 The figue below shows two balls of equal mass moving in vetical cicles.
More informationPS113 Chapter 5 Dynamics of Uniform Circular Motion
PS113 Chapte 5 Dynamics of Unifom Cicula Motion 1 Unifom cicula motion Unifom cicula motion is the motion of an object taveling at a constant (unifom) speed on a cicula path. The peiod T is the time equied
More informationPhysics 111 Lecture 5 Circular Motion
Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight
More information4. Two and Three Dimensional Motion
4. Two and Thee Dimensional Motion 1 Descibe motion using position, displacement, elocity, and acceleation ectos Position ecto: ecto fom oigin to location of the object. = x i ˆ + y ˆ j + z k ˆ Displacement:
More informationPhysics 11 Chapter 3: Vectors and Motion in Two Dimensions. Problem Solving
Physics 11 Chapte 3: Vectos and Motion in Two Dimensions The only thing in life that is achieved without effot is failue. Souce unknown "We ae what we epeatedly do. Excellence, theefoe, is not an act,
More informationCentripetal Force. Lecture 11. Chapter 8. Course website:
Lectue 11 Chapte 8 Centipetal Foce Couse website: http://faculty.uml.edu/andiy_danylov/teaching/physicsi PHYS.1410 Lectue 11 Danylov Depatment of Physics and Applied Physics Today we ae going to discuss:
More information6.4 Period and Frequency for Uniform Circular Motion
6.4 Peiod and Fequency fo Unifom Cicula Motion If the object is constained to move in a cicle and the total tangential foce acting on the total object is zeo, F θ = 0, then (Newton s Second Law), the tangential
More information= 4 3 π( m) 3 (5480 kg m 3 ) = kg.
CHAPTER 11 THE GRAVITATIONAL FIELD Newton s Law of Gavitation m 1 m A foce of attaction occus between two masses given by Newton s Law of Gavitation Inetial mass and gavitational mass Gavitational potential
More informationUniform Circular Motion
Unifom Cicula Motion Intoduction Ealie we defined acceleation as being the change in velocity with time: a = v t Until now we have only talked about changes in the magnitude of the acceleation: the speeding
More informationPHYS Summer Professor Caillault Homework Solutions
PHYS 1111 - Summe 007 - Pofesso Caillault Homewok Solutions Chapte 4 3. Pictue the Poblem: The ca moves up the 5.5 incline with constant acceleation, changing both its hoizontal and vetical displacement
More informationDYNAMICS OF UNIFORM CIRCULAR MOTION
Chapte 5 Dynamics of Unifom Cicula Motion Chapte 5 DYNAMICS OF UNIFOM CICULA MOTION PEVIEW An object which is moing in a cicula path with a constant speed is said to be in unifom cicula motion. Fo an object
More informationb) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?
Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 13, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with
More informationPhysics 101 Lecture 6 Circular Motion
Physics 101 Lectue 6 Cicula Motion Assist. Pof. D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Equilibium, Example 1 q What is the smallest value of the foce F such that the.0-kg block will not slide
More informationc) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?
Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 10, 2012 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with
More information- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.
- 5 - TEST 1R This is the epeat vesion of TEST 1, which was held duing Session. This epeat test should be attempted by those students who missed Test 1, o who wish to impove thei mak in Test 1. IF YOU
More informationEasy. 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
Chapte 4 Homewok Solutions Easy P4. Since the ca is moving with constant speed and in a staight line, the zeo esultant foce on it must be egadless of whethe it is moving (a) towad the ight o the left.
More informationThe 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
SOLUTIONS TO PROBLEMS The Laws of Motion Section 4.3 Mass P4. Since the ca is moving with constant speed and in a staight line, the esultant foce on it must be zeo egadless of whethe it is moving (a) towad
More informationRelative motion (Translating axes)
Relative motion (Tanslating axes) Paticle to be studied This topic Moving obseve (Refeence) Fome study Obseve (no motion) bsolute motion Relative motion If motion of the efeence is known, absolute motion
More informationMotion in a Plane Uniform Circular Motion
Lectue 11 Chapte 8 Physics I Motion in a Plane Unifom Cicula Motion Couse website: http://faculty.uml.edu/andiy_danylo/teaching/physicsi PHYS.1410 Lectue 11 Danylo Depatment of Physics and Applied Physics
More informationEasy. 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
Chapte 8 Homewok Solutions Easy P8. Assume the velocity of the blood is constant ove the 0.60 s. Then the patient s body and pallet will have a constant velocity of 6 0 5 m 3.75 0 4 m/ s 0.60 s in the
More informationCircular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.
Honos Physics Fall, 2016 Cicula Motion & Toque Test Review Name: M. Leonad Instuctions: Complete the following woksheet. SHOW ALL OF YOUR WORK ON A SEPARATE SHEET OF PAPER. 1. Detemine whethe each statement
More informationPhysics C Rotational Motion Name: ANSWER KEY_ AP Review Packet
Linea and angula analogs Linea Rotation x position x displacement v velocity a T tangential acceleation Vectos in otational motion Use the ight hand ule to detemine diection of the vecto! Don t foget centipetal
More informationCh04: Motion in two and three dimensions (2D and 3D)
Ch4: Motion in two and thee dimensions (D and 3D) Displacement, elocity and acceleation ectos Pojectile motion Cicula motion Relatie motion 4.: Position and displacement Position of an object in D o 3D
More informationChapter 5: Uniform Circular Motion
Chapte 5: Unifom Cicula Motion Motion at constant speed in a cicle Centipetal acceleation Banked cuves Obital motion Weightlessness, atificial gavity Vetical cicula motion Centipetal Foce Acceleation towad
More informationr cos, and y r sin with the origin of coordinate system located at
Lectue 3-3 Kinematics of Rotation Duing ou peious lectues we hae consideed diffeent examples of motion in one and seeal dimensions. But in each case the moing object was consideed as a paticle-like object,
More informationPhysics 2001 Problem Set 5 Solutions
Physics 2001 Poblem Set 5 Solutions Jeff Kissel Octobe 16, 2006 1. A puck attached to a sting undegoes cicula motion on an ai table. If the sting beaks at the point indicated in the figue, which path (A,
More informationPhysics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 6: motion in two and three dimensions III. Slide 6-1
Physics 1501 Fall 2008 Mechanics, Themodynamics, Waves, Fluids Lectue 6: motion in two and thee dimensions III Slide 6-1 Recap: elative motion An object moves with velocity v elative to one fame of efeence.
More informationSections and Chapter 10
Cicula and Rotational Motion Sections 5.-5.5 and Chapte 10 Basic Definitions Unifom Cicula Motion Unifom cicula motion efes to the motion of a paticle in a cicula path at constant speed. The instantaneous
More informationPhysics 107 TUTORIAL ASSIGNMENT #8
Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type
More informationKinematics in 2-D (II)
Kinematics in 2-D (II) Unifom cicula motion Tangential and adial components of Relative velocity and acceleation a Seway and Jewett 4.4 to 4.6 Pactice Poblems: Chapte 4, Objective Questions 5, 11 Chapte
More informationChapter 5. Uniform Circular Motion. a c =v 2 /r
Chapte 5 Unifom Cicula Motion a c =v 2 / Unifom cicula motion: Motion in a cicula path with constant speed s v 1) Speed and peiod Peiod, T: time fo one evolution Speed is elated to peiod: Path fo one evolution:
More informationCircular Motion. Mr. Velazquez AP/Honors Physics
Cicula Motion M. Velazquez AP/Honos Physics Objects in Cicula Motion Accoding to Newton s Laws, if no foce acts on an object, it will move with constant speed in a constant diection. Theefoe, if an object
More informationHoizontal Cicula Motion 1. A paticle of mass m is tied to a light sting and otated with a speed v along a cicula path of adius. If T is tension in the sting and mg is gavitational foce on the paticle then,
More informationModeling Ballistics and Planetary Motion
Discipline Couses-I Semeste-I Pape: Calculus-I Lesson: Lesson Develope: Chaitanya Kuma College/Depatment: Depatment of Mathematics, Delhi College of Ats and Commece, Univesity of Delhi Institute of Lifelong
More informationObjective Notes Summary
Objective Notes Summay An object moving in unifom cicula motion has constant speed but not constant velocity because the diection is changing. The velocity vecto in tangent to the cicle, the acceleation
More informationPROJECTILE MOTION. At any given point in the motion, the velocity vector is always a tangent to the path.
PROJECTILE MOTION A pojectile is any object that has been thown though the ai. A foce must necessaily set the object in motion initially but, while it is moing though the ai, no foce othe than gaity acts
More informationName. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?
AP Physics 1 Lesson 9.a Unifom Cicula Motion Outcomes 1. Define unifom cicula motion. 2. Detemine the tangential velocity of an object moving with unifom cicula motion. 3. Detemine the centipetal acceleation
More informationChapters 5-8. Dynamics: Applying Newton s Laws
Chaptes 5-8 Dynamics: Applying Newton s Laws Systems of Inteacting Objects The Fee Body Diagam Technique Examples: Masses Inteacting ia Nomal Foces Masses Inteacting ia Tensions in Ropes. Ideal Pulleys
More informationChapter 4. Newton s Laws of Motion
Chapte 4 Newton s Laws of Motion 4.1 Foces and Inteactions A foce is a push o a pull. It is that which causes an object to acceleate. The unit of foce in the metic system is the Newton. Foce is a vecto
More informationChapter 8. Accelerated Circular Motion
Chapte 8 Acceleated Cicula Motion 8.1 Rotational Motion and Angula Displacement A new unit, adians, is eally useful fo angles. Radian measue θ(adians) = s = θ s (ac length) (adius) (s in same units as
More informationAP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet
AP * PHYSICS B Cicula Motion, Gavity, & Obits Teache Packet AP* is a tademak of the College Entance Examination Boad. The College Entance Examination Boad was not involved in the poduction of this mateial.
More informationChapter 7-8 Rotational Motion
Chapte 7-8 Rotational Motion What is a Rigid Body? Rotational Kinematics Angula Velocity ω and Acceleation α Unifom Rotational Motion: Kinematics Unifom Cicula Motion: Kinematics and Dynamics The Toque,
More informationOSCILLATIONS AND GRAVITATION
1. SIMPLE HARMONIC MOTION Simple hamonic motion is any motion that is equivalent to a single component of unifom cicula motion. In this situation the velocity is always geatest in the middle of the motion,
More informationChapter. s r. check whether your calculator is in all other parts of the body. When a rigid body rotates through a given angle, all
conveted to adians. Also, be sue to vanced to a new position (Fig. 7.2b). In this inteval, the line OP has moved check whethe you calculato is in all othe pats of the body. When a igid body otates though
More informationRotational Motion: Statics and Dynamics
Physics 07 Lectue 17 Goals: Lectue 17 Chapte 1 Define cente of mass Analyze olling motion Intoduce and analyze toque Undestand the equilibium dynamics of an extended object in esponse to foces Employ consevation
More informationUniform Circular Motion
Unifom Cicula Motion constant speed Pick a point in the objects motion... What diection is the velocity? HINT Think about what diection the object would tavel if the sting wee cut Unifom Cicula Motion
More informationCentripetal Force OBJECTIVE INTRODUCTION APPARATUS THEORY
Centipetal Foce OBJECTIVE To veify that a mass moving in cicula motion expeiences a foce diected towad the cente of its cicula path. To detemine how the mass, velocity, and adius affect a paticle's centipetal
More informationBetween any two masses, there exists a mutual attractive force.
YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONS Name: QUESTION 1 (1995 EXAM) (a) State Newton s Univesal Law of Gavitation in wods Between any two masses, thee exists a mutual attactive foce. This foce
More informationPhysics 201 Homework 4
Physics 201 Homewok 4 Jan 30, 2013 1. Thee is a cleve kitchen gadget fo dying lettuce leaves afte you wash them. 19 m/s 2 It consists of a cylindical containe mounted so that it can be otated about its
More informationb) (5) What average force magnitude was applied by the students working together?
Geneal Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibium Nov. 3, 2010 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults
More informationPHYS 1114, Lecture 21, March 6 Contents:
PHYS 1114, Lectue 21, Mach 6 Contents: 1 This class is o cially cancelled, being eplaced by the common exam Tuesday, Mach 7, 5:30 PM. A eview and Q&A session is scheduled instead duing class time. 2 Exam
More informationPHYS Summer Professor Caillault Homework Solutions. Chapter 5
PHYS 1111 - Summe 2007 - Pofesso Caillault Homewok Solutions Chapte 5 7. Pictue the Poblem: The ball is acceleated hoizontally fom est to 98 mi/h ove a distance of 1.7 m. Stategy: Use equation 2-12 to
More informationLab #9: The Kinematics & Dynamics of. Circular Motion & Rotational Motion
Reading Assignment: Lab #9: The Kinematics & Dynamics of Cicula Motion & Rotational Motion Chapte 6 Section 4 Chapte 11 Section 1 though Section 5 Intoduction: When discussing motion, it is impotant to
More informationMotions and Coordinates
Chapte Kinematics of Paticles Motions and Coodinates Motion Constained motion Unconstained motion Coodinates Used to descibe the motion of paticles 1 ectilinea motion (1-D) Motion Plane cuvilinea motion
More informationCIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line
1 CIRCULAR MOTION 1. ANGULAR DISPLACEMENT Intoduction: Angle subtended by position vecto of a paticle moving along any abitay path w..t. some fixed point is called angula displacement. (a) Paticle moving
More informationQUESTION 1 [25 points]
(Fist) QUESTION 1 [5 points] An object moves in 1 dimension It stats at est and unifomly acceleates at 5m/s fo s It then moves with constant velocity fo 4s It then unifomly acceleates at m/s until it comes
More informationRecap. Centripetal acceleration: v r. a = m/s 2 (towards center of curvature)
a = c v 2 Recap Centipetal acceleation: m/s 2 (towads cente of cuvatue) A centipetal foce F c is equied to keep a body in cicula motion: This foce poduces centipetal acceleation that continuously changes
More informationGravitation. AP/Honors Physics 1 Mr. Velazquez
Gavitation AP/Honos Physics 1 M. Velazquez Newton s Law of Gavitation Newton was the fist to make the connection between objects falling on Eath and the motion of the planets To illustate this connection
More informationPhysics 181. Assignment 4
Physics 181 Assignment 4 Solutions 1. A sphee has within it a gavitational field given by g = g, whee g is constant and is the position vecto of the field point elative to the cente of the sphee. This
More informationINTRODUCTION. 2. Vectors in Physics 1
INTRODUCTION Vectos ae used in physics to extend the study of motion fom one dimension to two dimensions Vectos ae indispensable when a physical quantity has a diection associated with it As an example,
More informationMomentum and Collisions
SOLUTIONS TO PROBLES Section 8. P8. m 3.00 kg, (a) omentum and Collisions Linea omentum and Its Consevation v ( 3.00î 4.00ĵ ) m s p mv ( 9.00î.0ĵ ) kg m s Thus, p x 9.00 kg m s and p y.0 kg m s. p p x
More informationChapter 5 Force and Motion
Chapte 5 Foce and Motion In Chaptes 2 and 4 we have studied kinematics, i.e., we descibed the motion of objects using paametes such as the position vecto, velocity, and acceleation without any insights
More informationF g. = G mm. m 1. = 7.0 kg m 2. = 5.5 kg r = 0.60 m G = N m 2 kg 2 = = N
Chapte answes Heinemann Physics 4e Section. Woked example: Ty youself.. GRAVITATIONAL ATTRACTION BETWEEN SMALL OBJECTS Two bowling balls ae sitting next to each othe on a shelf so that the centes of the
More informationRotational Motion. Lecture 6. Chapter 4. Physics I. Course website:
Lectue 6 Chapte 4 Physics I Rotational Motion Couse website: http://faculty.uml.edu/andiy_danylov/teaching/physicsi Today we ae going to discuss: Chapte 4: Unifom Cicula Motion: Section 4.4 Nonunifom Cicula
More informationkg 2 ) 1.9!10 27 kg = Gm 1
Section 6.1: Newtonian Gavitation Tutoial 1 Pactice, page 93 1. Given: 1.0 10 0 kg; m 3.0 10 0 kg;. 10 9 N; G 6.67 10 11 N m /kg Requied: Analysis: G m ; G m G m Solution: G m N m 6.67!10 11 kg ) 1.0!100
More informationExtra notes for circular motion: Circular motion : v keeps changing, maybe both speed and
Exta notes fo cicula motion: Cicula motion : v keeps changing, maybe both speed and diection ae changing. At least v diection is changing. Hence a 0. Acceleation NEEDED to stay on cicula obit: a cp v /,
More informationCircular-Rotational Motion Mock Exam. Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.
AP Physics C Sping, 2017 Cicula-Rotational Motion Mock Exam Name: Answe Key M. Leonad Instuctions: (92 points) Answe the following questions. SHOW ALL OF YOUR WORK. ( ) 1. A stuntman dives a motocycle
More informationUniform Circular Motion
Unifom Cicula Motion Have you eve idden on the amusement pak ide shown below? As it spins you feel as though you ae being pessed tightly against the wall. The ide then begins to tilt but you emain glued
More informationChapter 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.
Chapte 5 Fiction When an object is in motion it is usually in contact with a viscous mateial (wate o ai) o some othe suface. So fa, we have assumed that moving objects don t inteact with thei suoundings
More informationPotential Energy and Conservation of Energy
Potential Enegy and Consevation of Enegy Consevative Foces Definition: Consevative Foce If the wok done by a foce in moving an object fom an initial point to a final point is independent of the path (A
More informationω = θ θ o = θ θ = s r v = rω
Unifom Cicula Motion Unifom cicula motion is the motion of an object taveling at a constant(unifom) speed in a cicula path. Fist we must define the angula displacement and angula velocity The angula displacement
More informationObjects usually are charged up through the transfer of electrons from one object to the other.
1 Pat 1: Electic Foce 1.1: Review of Vectos Review you vectos! You should know how to convet fom pola fom to component fom and vice vesa add and subtact vectos multiply vectos by scalas Find the esultant
More informationChapter 5 Force and Motion
Chapte 5 Foce and Motion In chaptes 2 and 4 we have studied kinematics i.e. descibed the motion of objects using paametes such as the position vecto, velocity and acceleation without any insights as to
More information7.2. Coulomb s Law. The Electric Force
Coulomb s aw Recall that chaged objects attact some objects and epel othes at a distance, without making any contact with those objects Electic foce,, o the foce acting between two chaged objects, is somewhat
More informationm1 m2 M 2 = M -1 L 3 T -2
GAVITATION Newton s Univesal law of gavitation. Evey paticle of matte in this univese attacts evey othe paticle with a foce which vaies diectly as the poduct of thei masses and invesely as the squae of
More informationconstant t [rad.s -1 ] v / r r [m.s -2 ] (direction: towards centre of circle / perpendicular to circle)
VISUAL PHYSICS ONLINE MODULE 5 ADVANCED MECHANICS NON-UNIFORM CIRCULAR MOTION Equation of a cicle x y Angula displacement [ad] Angula speed d constant t [ad.s -1 ] dt Tangential velocity v v [m.s -1 ]
More information06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )
06 - ROTATIONAL MOTION Page ) A body A of mass M while falling vetically downwads unde gavity beaks into two pats, a body B of mass ( / ) M and a body C of mass ( / ) M. The cente of mass of bodies B and
More informationSPH4U Unit 6.3 Gravitational Potential Energy Page 1 of 9
SPH4 nit 6.3 Gavitational Potential negy Page of Notes Physics ool box he gavitational potential enegy of a syste of two (spheical) asses is diectly popotional to the poduct of thei asses, and invesely
More informationTrigonometric Functions of Any Angle 9.3 (, 3. Essential Question How can you use the unit circle to define the trigonometric functions of any angle?
9. Tigonometic Functions of An Angle Essential Question How can ou use the unit cicle to define the tigonometic functions of an angle? Let be an angle in standad position with, ) a point on the teminal
More informationVectors and 2D Motion. Vectors and Scalars
Vectos and 2D Motion Vectos and Scalas Vecto aithmetic Vecto desciption of 2D motion Pojectile Motion Relative Motion -- Refeence Fames Vectos and Scalas Scala quantities: equie magnitude & unit fo complete
More informationNiraj Sir. circular motion;; SOLUTIONS TO CONCEPTS CHAPTER 7
SOLUIONS O CONCEPS CHAPE 7 cicula otion;;. Distance between Eath & Moon.85 0 5 k.85 0 8 7. days 4 600 (7.) sec.6 0 6 sec.4.85 0 v 6.6 0 8 05.4/sec v (05.4) a 0.007/sec.7 0 /sec 8.85 0. Diaete of eath 800k
More information10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.
NSWRS - P Physics Multiple hoice Pactice Gavitation Solution nswe 1. m mv Obital speed is found fom setting which gives v whee M is the object being obited. Notice that satellite mass does not affect obital
More informationCircular Orbits. and g =
using analyse planetay and satellite motion modelled as unifom cicula motion in a univesal gavitation field, a = v = 4π and g = T GM1 GM and F = 1M SATELLITES IN OBIT A satellite is any object that is
More information2013 Checkpoints Chapter 6 CIRCULAR MOTION
013 Checkpoints Chapte 6 CIRCULAR MOTIO Question 09 In unifom cicula motion, thee is a net foce acting adially inwads. This net foce causes the elocity to change (in diection). Since the speed is constant,
More informationMODULE 5 ADVANCED MECHANICS GRAVITATIONAL FIELD: MOTION OF PLANETS AND SATELLITES VISUAL PHYSICS ONLINE
VISUAL PHYSICS ONLIN MODUL 5 ADVANCD MCHANICS GRAVITATIONAL FILD: MOTION OF PLANTS AND SATLLITS SATLLITS: Obital motion of object of mass m about a massive object of mass M (m
More informationME 210 Applied Mathematics for Mechanical Engineers
Tangent and Ac Length of a Cuve The tangent to a cuve C at a point A on it is defined as the limiting position of the staight line L though A and B, as B appoaches A along the cuve as illustated in the
More informationPHYS Summer Professor Caillault Homework Solutions. Chapter 9
PHYS - Summe 007 - Pofesso Caillault Homewok Solutions Chapte 9 3. Pictue the Poblem The owne walks slowly towad the notheast while the cat uns eastwad and the dog uns nothwad. Stategy Sum the momenta
More information3.2 Centripetal Acceleration
unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tack-and-field event in which an athlete thows a hamme
More informationRotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart
Rotational Motion & Angula Momentum Rotational Motion Evey quantity that we have studied with tanslational motion has a otational countepat TRANSLATIONAL ROTATIONAL Displacement x Angula Position Velocity
More information21 MAGNETIC FORCES AND MAGNETIC FIELDS
CHAPTER 1 MAGNETIC ORCES AND MAGNETIC IELDS ANSWERS TO OCUS ON CONCEPTS QUESTIONS 1. (d) Right-Hand Rule No. 1 gives the diection of the magnetic foce as x fo both dawings A and. In dawing C, the velocity
More informationTAMPINES JUNIOR COLLEGE 2009 JC1 H2 PHYSICS GRAVITATIONAL FIELD
TAMPINES JUNIOR COLLEGE 009 JC1 H PHYSICS GRAVITATIONAL FIELD OBJECTIVES Candidates should be able to: (a) show an undestanding of the concept of a gavitational field as an example of field of foce and
More informationrt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)
Cicula Motion Fom ancient times cicula tajectoies hae occupied a special place in ou model of the Uniese. Although these obits hae been eplaced by the moe geneal elliptical geomety, cicula motion is still
More informationto point uphill and to be equal to its maximum value, in which case f s, max = μsfn
Chapte 6 16. (a) In this situation, we take f s to point uphill and to be equal to its maximum value, in which case f s, max = μsf applies, whee μ s = 0.5. pplying ewton s second law to the block of mass
More informationLecture 22. PE = GMm r TE = GMm 2a. T 2 = 4π 2 GM. Main points of today s lecture: Gravitational potential energy: Total energy of orbit:
Lectue Main points of today s lectue: Gavitational potential enegy: Total enegy of obit: PE = GMm TE = GMm a Keple s laws and the elation between the obital peiod and obital adius. T = 4π GM a3 Midtem
More information1121 T Question 1
1121 T1 2008 Question 1 ( aks) You ae cycling, on a long staight path, at a constant speed of 6.0.s 1. Anothe cyclist passes you, tavelling on the sae path in the sae diection as you, at a constant speed
More informationRevision Guide for Chapter 11
Revision Guide fo Chapte 11 Contents Revision Checklist Revision Notes Momentum... 4 Newton's laws of motion... 4 Wok... 5 Gavitational field... 5 Potential enegy... 7 Kinetic enegy... 8 Pojectile... 9
More informationMAGNETIC FIELD INTRODUCTION
MAGNETIC FIELD INTRODUCTION It was found when a magnet suspended fom its cente, it tends to line itself up in a noth-south diection (the compass needle). The noth end is called the Noth Pole (N-pole),
More informationChapter 2: Basic Physics and Math Supplements
Chapte 2: Basic Physics and Math Supplements Decembe 1, 215 1 Supplement 2.1: Centipetal Acceleation This supplement expands on a topic addessed on page 19 of the textbook. Ou task hee is to calculate
More information