AP Physics 1 - Circular Motion and Gravitation Practice Test (Multiple Choice Section) Answer Section

Similar documents
DYNAMICS OF UNIFORM CIRCULAR MOTION

Unit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.

Unit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.

Circular Motion. x-y coordinate systems. Other coordinates... PHY circular-motion - J. Hedberg

Answers to test yourself questions

Chapter 5: Uniform Circular Motion

r cos, and y r sin with the origin of coordinate system located at

Chap13. Universal Gravitation

Motion in a Plane Uniform Circular Motion

AP * PHYSICS B. Circular Motion, Gravity, & Orbits. Teacher Packet

PHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1

Physics 231 Lecture 17

Chap 5. Circular Motion: Gravitation

ΣF = r r v. Question 213. Checkpoints Chapter 6 CIRCULAR MOTION

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

3.3 Centripetal Force

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

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

Uniform Circular Motion

Extra notes for circular motion: Circular motion : v keeps changing, maybe both speed and

1) Consider a particle moving with constant speed that experiences no net force. What path must this particle be taking?

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

2013 Checkpoints Chapter 6 CIRCULAR MOTION

Objective Notes Summary

Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.

Physics 101 Lecture 6 Circular Motion

10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.

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

Sections and Chapter 10

Physics 111. Lecture 14 (Walker: Ch. 6.5) Circular Motion Centripetal Acceleration Centripetal Force February 27, 2009

ω = θ θ o = θ θ = s r v = rω

PS113 Chapter 5 Dynamics of Uniform Circular Motion

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

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

Chapter 13 Gravitation

Uniform Circular Motion

Physics 111 Lecture 5 Circular Motion

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?

OSCILLATIONS AND GRAVITATION

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

Midterm Exam #2, Part A

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

CHAPTER 5: Circular Motion; Gravitation


Gravitation. AP/Honors Physics 1 Mr. Velazquez

Chapter 8. Accelerated Circular Motion

Circular Orbits. and g =

rt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)

Motion in a Circle. Content 1. Kinematics of uniform circular motion 2. Centripetal acceleration 3. Centripetal force.

Universal Gravitation

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

Spring 2001 Physics 2048 Test 3 solutions

Physics 201 Homework 4

The study of the motion of a body along a general curve. the unit vector normal to the curve. Clearly, these unit vectors change with time, u ˆ

Uniform Circular Motion

Chapter 13: Gravitation

NEETIIT.COM. Angular Displacement. Page - 1

Unit 6 Test Review Gravitation & Oscillation Chapters 13 & 15

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Shree Datta Coaching Classes, Contact No Circular Motion

PROBLEM (page 126, 12 th edition)

Discover the answer to this question in this chapter.

Chapters 5-8. Dynamics: Applying Newton s Laws

Discover the answer to this question in this chapter.

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

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

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

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

3.2 Centripetal Acceleration

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

Physics 181. Assignment 4

Introduction to Mechanics Centripetal Force

m1 m2 M 2 = M -1 L 3 T -2

Rotational Motion. Lecture 6. Chapter 4. Physics I. Course website:

Force of gravity and its potential function

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

Motion along curved path *

Chapter 7 Rotational Motion and the Law of Gravity

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

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Recitation PHYS 131. must be one-half of T 2

Name. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?

Understanding the Concepts

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

UCSD Phys 4A Intro Mechanics Winter 2016 Ch 5 Solutions

PHYSICS NOTES GRAVITATION

Chapter 12. Kinetics of Particles: Newton s Second Law

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.

4. Two and Three Dimensional Motion

Physics 2001 Problem Set 5 Solutions

Physics 312 Introduction to Astrophysics Lecture 7

MAGNETIC FIELD INTRODUCTION

Ch 13 Universal Gravitation

kg 2 ) 1.9!10 27 kg = Gm 1

Lecture 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:

Circular Motion. Mr. Velazquez AP/Honors Physics

1131 T Question 1

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

Section 6.2: Orbits. Gm r. v = v 2 = Gm r. m = rv2 G. Solution: m = rv2 G ( )( 7.5!10 5 m/s ) 2. = 5.34!1017 m m kg # # m2. kg 2

Physics: Work & Energy Beyond Earth Guided Inquiry

Transcription:

AP Physics 1 - Cicula Motion and Gaitation Pactice est (Multiple Choice Section) Answe Section MULIPLE CHOICE 1. B he centipetal foce must be fiction since, lacking any fiction, the coin would slip off. he fiction is static because the coin is not slipping.. D he net foce and the acceleation on any object in unifom cicula motion ae both always towad the cente of the cicle. 3. D When the block is on the plane, the acceleation is 4 because the block is speeding up while going down and to the ight. When the block is at the bottom, it is actually going along (pat of) a cicula path. his means acceleation must be towad the cente of the cicle, which is 1. 4. B If both cues ae banked at the same angle, then they both poide the same centipetal acceleation: ( }) a = { c { ( same) ( 4) 5. A When an object moes in a cicle, its elocity is continuously changing diection, so since the elocity changes, the object must be acceleating. Velocity is nee the same as speed because elocity has diection and speed does not. o moe in a cicle, thee must be a net foce (called the centipetal foce) that points towad the cente of the cicle (in this case, the tension in the sting). If the sting beaks, the object will moe in a staight line tangent to the cicle. 6. C Objects feel weightless because they ae in fee-fall (the only foce on them in obit is gaity), theefoe the astonauts and the space ship all fall togethe at the same ate. 1 014 John R. Fensley, he College Entance Examination Boad, et.al. All ights eseed.

7. B GmM Foce of gaity is both F g = mg and F G =. Set these equal and cancel out m to get the gaity on a planet of mass M: GM g = Half the diamete and one-tenth the mass means that mass and adius ae multiplied by: ( 1/10 } ) G M { g = { (?) ( 1/ ) 1 1 Note that the denominato is squaed, so g is multiplied by 10 4 = 4/10. Multiplying Eath s gaity by 4/10 yields 4 m/s. 8. E Inese-squae law: As distance inceases, the foce deceases as the inese-squae of the distance. If the distance is multiplied by 4, the foce is multiplied by the inese-squae of 4, which is 1/16. 9. C ( }) m F c = { ( 4) he speed (which is squaed) must be doubled in the numeato to balance the multiplication of (in the denominato) by fou. 10. D he acceleation must be pointed towad the cente of the cicle, so if the acceleation points west then the ca is at the east end of the tack. If the ca is at the east end of the tack and elocity points noth, then the ca must be going counteclockwise aound the tack. Now sole fo acceleation: a = / () = /(800) = 1600 = 40 m/s 11. D When any ca goes aound a cue on a oadway, the fictional foce is what poides the centipetal foce to make the ca go in a cicle. he lage ca has twice the mass, but half the elocity on the same-adius oad: (}) ( ) 1/ } m { F = ( 1/ ) { ( 1) 014 John R. Fensley, he College Entance Examination Boad, et.al. All ights eseed.

1. D If 10 kg is just baely enough to beak the sting, then the sting s beaking stength is 100 N (the weight of the 10 kg). Using the sting to otate an object on a fictionless table will mean that the sting s tension is the centipetal foce causing the cicula motion (the object s weight and nomal foce cancel). Find the maximum speed: m F = ( ) ( 0.5) 100 = ( ) 400 = = 0 m/s 13. D When the ship is the same distance between the Eath and the Moon, the gaitational foce fom the Eath is geate than the Moon s foce because Eath has moe mass. Because these two gaitational foces ae unequal (Eath s is geate), thee is a net foce on the spaceship. 14. A Conside setting up a cicula obit equation by setting gaity foce equal to centipetal foce. how in the fact that speed = cicumfeence/peiod. Let M = sta s mass, and m = planet s mass. GmM m π = and = GM = (note that the planet s mass doesn t matte!) GM π = 3 GM = 4π Note the left-hand side the sta s mass is inesely popotional to the peiod (squaed). hat means that a longe obital peiod belongs to a sta with less mass. 15. C Apply Keple s hid Law, which states that two objects obiting the same massie object hae obital adii and peiods that obey the atio 1 /R 3 1 = /R 3. Let Eath s peiod and adius both be epesented by 1, so the asteoid s adius is : (1) /(1) 3 = /() 3 = 8 =.8 yeas 3 014 John R. Fensley, he College Entance Examination Boad, et.al. All ights eseed.

16. D Because the ties ae not sliding on the oadway, the fiction is static and we use µ = 0.8. he fictional foce is F f = µf N = µmg (since F N = mg on a flat suface). he fiction is what poides the centipetal foce, so set fictional foce equal to centipetal foce: µ mg = m = µ g = ( 0.8)( 10)( 50) = 400 = 0 m/s 17. B At the bottom of the swing, thee is an upwad tension (and upwad acceleation) and downwad weight foce: F F = ma (If weight is 4 N, mass is 0.4 kg) g ( 6 ) ( 4) = ( 0.4) a c c a c = 5 m/s = ½g 18. D hink in tems of foces acting on the ca; the only foces acting on the ca ae weight (down) and the nomal foce (to the ight). his makes the net foce down and to the ight. Acceleation has the same diection as net foce, so the acceleation is also down and to the ight. OHER 19. C&D wo objects in the same obit hae the same speed and magnitude of acceleation, since speed and acceleation of an object in obit depends on the planet s mass, not the obiting object s. But because F = ma and K = ½m, foce and kinetic enegy double if the mass of the object is doubled. 0. A&D he net foce is the centipetal foce because the obit is cicula. he ectos and F ae NO constant because thei diections ae continuously changing (een though the speed and stength of the net foce ae both constant). In unifom cicula motion (including cicula obits), the net amount of wok done on the object is always zeo. 4 014 John R. Fensley, he College Entance Examination Boad, et.al. All ights eseed.

AP Physics 1 Cicula Motion and Gaitation Pactice est (Fee Response Section SOLUIONS) Poblem 1 (7 Points) (eache Made) (a) 1 point Vecto points tangent to the cicle (eithe diection) and ectos a and F point towad the cente. No points awaded fo unlabeled ectos o if thee ae any extaneous ectos. (b) (c) points he student maks peiod of otation and fequency of otation only. he student indicates that both coins ae on the same otating suface, so they complete cicles in the same time. his point can be awaded fo this easoning een if point was not awaded. points he student maks B = 3 A π he student shows = and explains that thee times the adius esults in thee times the speed. (d) points he student maks µ B = 3µ A m he student indicates that the fictional foce is equal to the centipetal foce, so µ mg =. Masses cancel, so mass does not affect this answe ( µ g = ). If speed is multiplied by thee, then the numeato is multiplied by 9, and if adius is multiplied by 3, then the denominato is multiplied by 3. his means that the coefficient of fiction must be thee ties as much. 015 John Fensley. All ights eseed.

Poblem (13 Points) (eache Made) (a-i) points GmM he student shows equations such as F = and he student uses the equations to deie GM = o m F = GM = o an equialent equation. (a-ii) points (b) π he student shows thei esult fom pat (a) as well as the equation =. 3 4π he student uses the equations to deie GM = 4π R, =, o an equialent equation 3 R GM 1 point Radius R (m) Peiod (sec) R 3 (m 3 ) (sec ) 1.0 10 10 8.31 10 5 1.73 10 30 6.91 10 11 1.98 10 10 1.77 10 6 7.76 10 30 3.13 10 1.83 10 10.99 10 6.7 10 31 8.94 10 1 3.71 10 10 4.47 10 6 5.11 10 31.00 10 13 he student fills in at least thee coect numbes in each column. 015 John Fensley. All ights eseed.

(c) 3 points (d) he scale on each axis is linea and causes at least half of the gaph space to be used etically and hoizontally. Coect indications of scientific notation exponents ae shown. Fou points ae plotted coectly based on the data table. he best-fit does not pass fathe than two small gid-squae lengths fom any point on the gaph. points Fo cubing the gien adius and identifying the point (R 3 = 3.38 10 30 m 3, = 1.38 10 1 sec ) on the best-fit line. could be between 1.34 10 1 sec and 1.4 10 1 sec. Fo taking the squae oot of the alue of, giing 1.17 10 6 seconds (between 1.15 10 6 s and 1.0 10 6 s accepted). (e) 3 points Fo indicating that the slope of the line is 3.9 10 19 (between 3.7 10 19 and 4.1 10 19 is acceptable). 4π Fo setting the slope equal to. GM Fo the mass of the sta between 1.44 10 30 kg and 1.60 10 30 kg 015 John Fensley. All ights eseed.