FULL MECHANICS SOLUTION
|
|
- Mark Hill
- 5 years ago
- Views:
Transcription
1 FULL MECHANICS SOLUION. m f For long the tngentil direction m 3g cos 3 sin 3 f N m 3g sin 3 cos3 from soling 3. ( N 4) ( N 8) N gsin 3. = ut + t = ut g sin cos t u t = gsin cos = = s] 3 4 o 4. cos 6 o u i ( u sin 6 gt) j o o u cos 6 j ( u sin 6 gt) k o. ( u sin 6 gt) u cos 6 cos o. ( u cos 6 ) ( u sin 6 gt) 5. In fig. () = 5 In fig.(b) 8 = = 5 =, 3 8 = 8 = 6, 3 6. By constrint motion cos = u = u sec u= m/s h= 9 m or = d = usectn d sin N cos CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #
2 When contct lees, N= i.e. sin = nd cos = m tn = g or = g cot lso = tn hcot = So nd d h cosec d u sec tn d cos =m nd sin + N = or h cosec u sec tn or tn = or = 6 gh u u sec h cosec or g cot u sec tn or tn= 3 7. (b) Friction force = cos sin 8. (b) l s u = mu l cos 6 o m m o cos 6 l ension will be mimum t the bottom (lower most point) Mg m l m f(s) m 8 g For m N 8g o ensure no slipping, f N 8g s fs CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #
3 m 4 min 9. (d) u= m R Q M R m gr ill then the bll reches the lowest point, liner momentum will not be consered. Post tht, liner momentum shll lso be consered. m gr M m m gr M m. (c) N =constnt f k F= N fk F N N.5g 5N otl contct force 5 3 N. () ˆ ˆ ˆ ˆ 4 4i 4 3j 8V perfectly inelstic collision 4 34i 3j ˆ 3 i ˆ j 8 9 Loss of mechnicl energy KE KE J 4 CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 3
4 . (c) m/5 4 kg 4 kg m/sec 4t the string will be tut gin fter tencet. he hnging mss will be under free fll. 4t t gt t t t 4 / t 4sec At this instnt the elocities of ech mss. 4m/sec 4m/sec =+g(.4)=6m/sec After the jirk N V V N sec m / sec 3. (c) 3 Gien 3 k where k is constnt of proportionlity d 3 k d Integrting d k d k C [C is constnt of integrtion] When, C k k () Also when, k k CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 4
5 From (), we get When () In ccelertion motion slope ds increses wheres for retrded motion it decreses. 5. (c) Using ector form of kinemtic eqution gh gh g g h u g 6. () Since the mss is sme therefore the length of the plnk should be twice the rnge u sin R 4 3m g 7. () 5 t t t 4 t 5 t So, t sec. i.e., horizontl distnce m () AC sec. So, u sin u sin g m / s g Now y u t t h u sin 3 g 3 y y h 3 9 h 3 45 CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 5
6 h m 9. (b) m mss of the block Applying Newton s Lw on the block in erticl direction N m.. () 3 N (gien reding of spring blnce) N 5 N m s (gien reding of W.M.) N 5 3 m m 4kg g. (b) F.B.D. of mn nd plteform N 5g N 5g N. () he F.B.D. of block A nd B re From constrint, the ccelertion of A & B re nd respectiely. Applying Newton s second lw to blocks A nd B, we we get 4g 4. (i) 5g 5.. (ii) Soling we get ccelertion of A nd B s g 7 Downwrd, g upwrd respectiely 7. (c) CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 6
7 he F.B.D. of pulley is s shown Let, nd FS be the forces Eerted by the horizontl string, erticl String by the support on the mssless pulley Respectiely. hen FS or FS ension in ech string is 3. () 4. (d) m ilblet mn 'sfeet fr fr. N fr fr. 4 4 N ilblet ground 5. (A,C) So net friction on mn is 5 5 N. 6. ime depends only on y-direction ( perpendiculr to rier flow) elocity which is sme for ll the four 7. Suppose only block (A) nd (C) moe = m = m = 9 m = 7 8 g & = 7 9 g CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 7
8 = m. 7g 9 = 4 9 < ] 8. (A,C,D) 9. (C,D) direction of friction does not chnge friction is constnt. 3. (, c, d) As there is no friction between the bll nd the grooe so there will be no centripetl force i.e., No net force on the bll towrds the center hence the bll moes in rdilly outwrd direction w.r.t disc 3. (b, c, d) After Impct e gh cos Before Impct gh cos e gh cos A H gh sin O gh cos N c toa gh cos gc gh sin gh cos c tao e gh cos Due to the impct only horizontl component of elocity gets ffected so totl time of flight remins the sme t t OA AO c c gh sin gh cos e gh cos g c gh sin gh cos e g h sin e c h sin c e c c e h sin c CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 8
9 3. (, b, c) h u g / mu cos m () As N = t ngle of rottion mv So cos () Using eq. () g m cos cos cos cos cos 4 5 cos 6 cos 5/ 6 5 h cos (, c, d) k m F k m k k F d rel d F k k d m m d F k d m m m d F k d d m m m At mimum compression or mimum etension reltie elocity should become zero. F k m m m F/ m or K m m mf K m m CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 9
10 When m m m then m F / K Compression the spring will be zero. 34. (, d) 3 t 33t 8t d 6t 66t 8 d t 66 Direction chnges when t 5, t 6 Minimum speed is when it is zero i.e. t 5 & t 6 Retrded mens elocity is opposite of ccelertion i.e. t 5 & 5.5 t (, c, d) time t t / t / distnce left t / t n n t / s Distnce trelled in time t t t d t d t. t t 36. (, c, d) Slope of displcement-time cure gies elocity. () During OA slope is + e but decresing hence elocity is positie nd ccelertion is negtie. (c) During BC slope is e nd going to zero hence elocity is e but ccelertion is +e. (d) During DE slope is +e nd incresing hence elocity is +e nd incresing +e ccelertion 37. (b, c) Initil elocity of prticle in ector from cn be written s cos ˆi sin ˆ j p.() Velocity of prticle t ny time t will be: ˆ Q cosi sin gtˆj..() Gien tht p Q CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #
11 . p Q cos sin sin gt Or Or sin gt or t cos ec g Substituting this lue of t in Eq. () we get: ˆ ˆ Q cosi sin j sin Or Q cos sin sin cot 38. (, c) At highest point ngle between nd is ccelertion. n R 9. Hence, totl ccelertion is only norml or rdil g u cos R 39. (b,c) 4. (,c) Sol: u cos Or R g At point of projection component of ccelertion (= g) long elocity ector is g cos 9 or g sin. g tn m g CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #
12 4. (b,d) Since the pprent weight is incresing, hence ccelertion of the lift should be upwrds. his is possible in cse of (b) nd (d) 4. (,b,c,d) here is no horizontl force on block A, therefore it does not moe in -directing, wheres there is net downwrd fore N is cting on it, mking its ccelertion long negtie y-direction. Block B moes downwrd s well s in negtie -direction. Downwrd ccelertion of A nd B will be equl due to constrin, thus w.r.t. B, A moes in positie -direction. Due to the component of norml eerted by C on B, it moes in negtie -direction 43. (,d) Mimum lue of friction force between 4m nd inclined plne 4 cos45 4 Here pulling force F 4 cos 45 P Block will not moe Accelertion of 4m block, frictionl force on 4m block 44. (b,c) fm.4 8N For equilibrium Minimum lue of m m 8 m kg Mimum lue of m m 8 m 8kg So, (B) nd (c) re correct 45. (CD) 46. (BD) 47. (D) 48. (BC) 49. (AB) 5. (), 5. (b), 5. () Slope of V ersus t grph is g y g t As displcement long y-is is zero k Vy CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA #
13 u y tn u 53. (D) 54. (D) 55. = 5 = 5 = 4g m ' = = r = = 9g m rp = 9g ( g) = g r 5m 56. Let = A Bt d At t, tn A 4 Agin, t = 3 sec, = 3 B3 B So t 4 4 t 3 t d 4 t = 57. [6] f sin + mr µ cos sin + mr r or µ tn + gcos 58. () sin + mrw K = 5 N/m Initilly the spring is reled M If the block moes down by, then the spring is etended by s the string connecting the spring is in etensible. Using energy consertion principle, m K he etension in the spring will be mimum when the block comes to rest momentrily. i.e., = CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 3
14 k k4 k / k Mimum etension in spring cm k 59. (5) If M remins sttionry = Mg Mg cos cos sin m sin m m sin sin m sin Mg m. rd / sec re / sec 6. (5) A V A y B V B C V CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 4 C V Using energy consertion, ma mcg maa mbb mcc g B
15 VA cos V A y VA sin VB sin V B V sin V cos A B VA VB l 6 VA 38 8 VB cos CENERS: MUMBAI / DELHI / AKOLA / KOLKAA / LUCKNOW / NASHIK / GOA # 5
SOLUTIONS TO CONCEPTS CHAPTER
1. m = kg S = 10m Let, ccelertion =, Initil velocity u = 0. S= ut + 1/ t 10 = ½ ( ) 10 = = 5 m/s orce: = = 5 = 10N (ns) SOLUIONS O CONCEPS CHPE 5 40000. u = 40 km/hr = = 11.11 m/s. 3600 m = 000 kg ; v
More informationForces from Strings Under Tension A string under tension medites force: the mgnitude of the force from section of string is the tension T nd the direc
Physics 170 Summry of Results from Lecture Kinemticl Vribles The position vector ~r(t) cn be resolved into its Crtesin components: ~r(t) =x(t)^i + y(t)^j + z(t)^k. Rtes of Chnge Velocity ~v(t) = d~r(t)=
More information1/31/ :33 PM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
1/31/18 1:33 PM Chpter 11 Kinemtics of Prticles 1 1/31/18 1:33 PM First Em Sturdy 1//18 3 1/31/18 1:33 PM Introduction Mechnics Mechnics = science which describes nd predicts conditions of rest or motion
More information2/20/ :21 AM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
//15 11:1 M Chpter 11 Kinemtics of Prticles 1 //15 11:1 M Introduction Mechnics Mechnics = science which describes nd predicts the conditions of rest or motion of bodies under the ction of forces It is
More information2/2/ :36 AM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.
//16 1:36 AM Chpter 11 Kinemtics of Prticles 1 //16 1:36 AM First Em Wednesdy 4//16 3 //16 1:36 AM Introduction Mechnics Mechnics = science which describes nd predicts the conditions of rest or motion
More informationPHYSICS 211 MIDTERM I 21 April 2004
PHYSICS MIDERM I April 004 Exm is closed book, closed notes. Use only your formul sheet. Write ll work nd nswers in exm booklets. he bcks of pges will not be grded unless you so request on the front of
More informationNarayana IIT Academy
INDIA Sec: Sr. IIT_IZ Jee-Advnced Dte: --7 Time: 09:00 AM to :00 Noon 0_P Model M.Mrks: 0 KEY SHEET CHEMISTRY C D 3 D B 5 A 6 D 7 B 8 AC 9 BC 0 ABD ABD A 3 C D 5 B 6 B 7 9 8 9 0 7 8 3 3 6 PHYSICS B 5 D
More informationJURONG JUNIOR COLLEGE
JURONG JUNIOR COLLEGE 2010 JC1 H1 8866 hysics utoril : Dynmics Lerning Outcomes Sub-topic utoril Questions Newton's lws of motion 1 1 st Lw, b, e f 2 nd Lw, including drwing FBDs nd solving problems by
More informationSOLUTIONS TO CONCEPTS CHAPTER 6
SOLUIONS O CONCEPS CHAPE 6 1. Let ss of the block ro the freebody digr, 0...(1) velocity Agin 0 (fro (1)) g 4 g 4/g 4/10 0.4 he co-efficient of kinetic friction between the block nd the plne is 0.4. Due
More informationChapter 5 Exercise 5A
Chpter Exercise Q. 1. (i) 00 N,00 N F =,00 00 =,000 F = m,000 = 1,000 = m/s (ii) =, u = 0, t = 0, s =? s = ut + 1 t = 0(0) + 1 ()(00) = 00 m Q.. 0 N 100 N F = 100 0 = 60 F = m 60 = 10 = 1 m/s F = m 60
More informationPhysicsAndMathsTutor.com
1. A uniform circulr disc hs mss m, centre O nd rdius. It is free to rotte bout fixed smooth horizontl xis L which lies in the sme plne s the disc nd which is tngentil to the disc t the point A. The disc
More informationMotion. Acceleration. Part 2: Constant Acceleration. October Lab Phyiscs. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Motion ccelertion Prt : Constnt ccelertion ccelertion ccelertion ccelertion is the rte of chnge of elocity. = - o t = Δ Δt ccelertion = = - o t chnge of elocity elpsed time ccelertion is ector, lthough
More informationPhysics 105 Exam 2 10/31/2008 Name A
Physics 105 Exm 2 10/31/2008 Nme_ A As student t NJIT I will conduct myself in professionl mnner nd will comply with the proisions of the NJIT Acdemic Honor Code. I lso understnd tht I must subscribe to
More informationSECTION B Circular Motion
SECTION B Circulr Motion 1. When person stnds on rotting merry-go-round, the frictionl force exerted on the person by the merry-go-round is (A) greter in mgnitude thn the frictionl force exerted on the
More informationCoimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 2010 MARKING SCHEME APPLIED MATHEMATICS HIGHER LEVEL
Coimisiún n Scrúduithe Stáit Stte Emintions Commission LEAVING CERTIFICATE 00 MARKING SCHEME APPLIED MATHEMATICS HIGHER LEVEL Generl Guidelines Penlties of three types re pplied to cndidtes' work s follows:
More informationPhysics Honors. Final Exam Review Free Response Problems
Physics Honors inl Exm Review ree Response Problems m t m h 1. A 40 kg mss is pulled cross frictionless tble by string which goes over the pulley nd is connected to 20 kg mss.. Drw free body digrm, indicting
More informationSOLUTIONS TO CONCEPTS CHAPTER 10
SOLUTIONS TO CONCEPTS CHPTE 0. 0 0 ; 00 rev/s ; ; 00 rd/s 0 t t (00 )/4 50 rd /s or 5 rev/s 0 t + / t 8 50 400 rd 50 rd/s or 5 rev/s s 400 rd.. 00 ; t 5 sec / t 00 / 5 8 5 40 rd/s 0 rev/s 8 rd/s 4 rev/s
More informationPhys 7221, Fall 2006: Homework # 6
Phys 7221, Fll 2006: Homework # 6 Gbriel González October 29, 2006 Problem 3-7 In the lbortory system, the scttering ngle of the incident prticle is ϑ, nd tht of the initilly sttionry trget prticle, which
More informationMEE 214 (Dynamics) Tuesday Dr. Soratos Tantideeravit (สรทศ ต นต ธ รว ทย )
MEE 14 (Dynmics) Tuesdy 8.30-11.0 Dr. Sortos Tntideerit (สรทศ ต นต ธ รว ทย ) sortos@oep.go.th Lecture Notes, Course updtes, Extr problems, etc No Homework Finl Exm (Dte & Time TBD) 1/03/58 MEE14 Dynmics
More information13.4 Work done by Constant Forces
13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push
More informationMASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS
MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES SECTION MULTIPLE CHOICE QUESTIONS QUESTION QUESTION
More informationPlane curvilinear motion is the motion of a particle along a curved path which lies in a single plane.
Plne curiliner motion is the motion of prticle long cured pth which lies in single plne. Before the description of plne curiliner motion in n specific set of coordintes, we will use ector nlsis to describe
More informationLinear Motion. Kinematics Quantities
Liner Motion Physics 101 Eyres Kinemtics Quntities Time Instnt t Fundmentl Time Interl Defined Position x Fundmentl Displcement Defined Aerge Velocity g Defined Aerge Accelertion g Defined 1 Kinemtics
More informationThe momentum of a body of constant mass m moving with velocity u is, by definition, equal to the product of mass and velocity, that is
Newtons Lws 1 Newton s Lws There re three lws which ber Newton s nme nd they re the fundmentls lws upon which the study of dynmics is bsed. The lws re set of sttements tht we believe to be true in most
More information4-6 ROTATIONAL MOTION
Chpter 4 Motions in Spce 51 Reinforce the ide tht net force is needed for orbitl motion Content We discuss the trnsition from projectile motion to orbitl motion when bll is thrown horizontlly with eer
More informationCorrect answer: 0 m/s 2. Explanation: 8 N
Version 001 HW#3 - orces rts (00223) 1 his print-out should hve 15 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Angled orce on Block 01 001
More informationHomework: 5, 9, 19, 25, 31, 34, 39 (p )
Hoework: 5, 9, 19, 5, 31, 34, 39 (p 130-134) 5. A 3.0 kg block is initilly t rest on horizontl surfce. A force of gnitude 6.0 nd erticl force P re then pplied to the block. The coefficients of friction
More informationLecture 8. Newton s Laws. Applications of the Newton s Laws Problem-Solving Tactics. Physics 105; Fall Inertial Frames: T = mg
Lecture 8 Applictions of the ewton s Lws Problem-Solving ctics http://web.njit.edu/~sireno/ ewton s Lws I. If no net force ocects on body, then the body s velocity cnnot chnge. II. he net force on body
More informationFORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81
FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first
More informationE S dition event Vector Mechanics for Engineers: Dynamics h Due, next Wednesday, 07/19/2006! 1-30
Vector Mechnics for Engineers: Dynmics nnouncement Reminders Wednesdy s clss will strt t 1:00PM. Summry of the chpter 11 ws posted on website nd ws sent you by emil. For the students, who needs hrdcopy,
More informationNumerical Problems With Solutions(STD:-XI)
Numericl Problems With Solutions(STD:-XI) Topic:-Uniform Circulr Motion. An irplne executes horizontl loop of rdius 000m with stedy speed of 900kmh -. Wht is its centripetl ccelertion? Ans:- Centripetl
More informationModel Solutions to Assignment 4
Oberlin College Physics 110, Fll 2011 Model Solutions to Assignment 4 Additionl problem 56: A girl, sled, nd n ice-covered lke geometry digrm: girl shore rope sled ice free body digrms: force on girl by
More information(3.2.3) r x x x y y y. 2. Average Velocity and Instantaneous Velocity 2 1, (3.2.2)
Lecture 3- Kinemtics in Two Dimensions Durin our preious discussions we he been tlkin bout objects moin lon the striht line. In relity, howeer, it rrely hppens when somethin moes lon the striht pth. For
More informationPlane curvilinear motion is the motion of a particle along a curved path which lies in a single plane.
Plne curiliner motion is the motion of prticle long cured pth which lies in single plne. Before the description of plne curiliner motion in n specific set of coordintes, we will use ector nlsis to describe
More informationDynamics: Newton s Laws of Motion
Lecture 7 Chpter 4 Physics I 09.25.2013 Dynmics: Newton s Lws of Motion Solving Problems using Newton s lws Course website: http://fculty.uml.edu/andriy_dnylov/teching/physicsi Lecture Cpture: http://echo360.uml.edu/dnylov2013/physics1fll.html
More informationANSWERS, HINTS & SOLUTIONS PART TEST I
AITS-PT-I-PCM (Sol.)-JEE(Min)/9 IITJEE JEE(Min)-09 ANSWERS, HINTS & SLUTINS PART TEST I (Min) ALL INDIA TEST SERIES Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS. A. B 6. B. A. C 6. C. A. B 6. A.
More informationDO NOT OPEN THIS EXAM BOOKLET UNTIL INSTRUCTED TO DO SO.
PHYSICS 1 Fll 017 EXAM 1: October 3rd, 017 8:15pm 10:15pm Nme (printed): Recittion Instructor: Section #: DO NOT OPEN THIS EXAM BOOKLET UNTIL INSTRUCTED TO DO SO. This exm contins 5 multiple-choice questions,
More informationPhysics 207 Lecture 5
Phsics 07 Lecture 5 Agend Phsics 07, Lecture 5, Sept. 0 Chpter 4 Kinemtics in or 3 dimensions Independence of, nd/or z components Circulr motion Cured pths nd projectile motion Frmes of reference dil nd
More informationME 141. Lecture 10: Kinetics of particles: Newton s 2 nd Law
ME 141 Engineering Mechnics Lecture 10: Kinetics of prticles: Newton s nd Lw Ahmd Shhedi Shkil Lecturer, Dept. of Mechnicl Engg, BUET E-mil: sshkil@me.buet.c.bd, shkil6791@gmil.com Website: techer.buet.c.bd/sshkil
More informationA wire. 100 kg. Fig. 1.1
1 Fig. 1.1 shows circulr cylinder of mss 100 kg being rised by light, inextensible verticl wire. There is negligible ir resistnce. wire 100 kg Fig. 1.1 (i) lculte the ccelertion of the cylinder when the
More informationIn-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Physics Physics 8T Fll Term 4 In-Clss Problems nd 3: Projectile Motion Solutions We would like ech group to pply the problem solving strtegy with the
More informationYour Thoughts. Mechanics Lecture 16, Slide 1
Your Thoughts I get dizzy with ll the equtions being shifted, spun nd switched so much in the pre-lectures. If the pre-lectures for, 3 nd 4 re like tht, I m pretty worried. Are we going to be rcing spheres,
More informationPhET INTRODUCTION TO MOTION
IB PHYS-1 Nme: Period: Dte: Preprtion: DEVIL PHYSICS BADDEST CLASS ON CAMPUS PhET INTRODUCTION TO MOTION 1. Log on to computer using your student usernme nd pssword. 2. Go to https://phet.colordo.edu/en/simultion/moing-mn.
More informationEunil Won Dept. of Physics, Korea University 1. Ch 03 Force. Movement of massive object. Velocity, acceleration. Force. Source of the move
Eunil Won Dept. of Phsics, Kore Uniersit 1 Ch 03 orce Moement of mssie object orce Source of the moe Velocit, ccelertion Eunil Won Dept. of Phsics, Kore Uniersit m ~ 3.305 m ~ 1.8 m 1.8 m Eunil Won Dept.
More informationLecture 5. Today: Motion in many dimensions: Circular motion. Uniform Circular Motion
Lecture 5 Physics 2A Olg Dudko UCSD Physics Tody: Motion in mny dimensions: Circulr motion. Newton s Lws of Motion. Lws tht nswer why questions bout motion. Forces. Inerti. Momentum. Uniform Circulr Motion
More informationProblems (Motion Relative to Rotating Axes)
1. The disk rolls without slipping on the roblems (Motion Reltie to Rotting xes) horizontl surfce, nd t the instnt represented, the center O hs the elocity nd ccelertion shown in the figure. For this instnt,
More information= 40 N. Q = 60 O m s,k
Multiple Choice ( 6 Points Ech ): F pp = 40 N 20 kg Q = 60 O m s,k = 0 1. A 20 kg box is pulled long frictionless floor with n pplied force of 40 N. The pplied force mkes n ngle of 60 degrees with the
More informationAP Physics 1. Slide 1 / 71. Slide 2 / 71. Slide 3 / 71. Circular Motion. Topics of Uniform Circular Motion (UCM)
Slide 1 / 71 Slide 2 / 71 P Physics 1 irculr Motion 2015-12-02 www.njctl.org Topics of Uniform irculr Motion (UM) Slide 3 / 71 Kinemtics of UM lick on the topic to go to tht section Period, Frequency,
More informationMathematics Extension 1
04 Bored of Studies Tril Emintions Mthemtics Etension Written by Crrotsticks & Trebl. Generl Instructions Totl Mrks 70 Reding time 5 minutes. Working time hours. Write using blck or blue pen. Blck pen
More informationPhysics 207 Lecture 7
Phsics 07 Lecture 7 Agend: Phsics 07, Lecture 7, Sept. 6 hpter 6: Motion in (nd 3) dimensions, Dnmics II Recll instntneous velocit nd ccelertion hpter 6 (Dnmics II) Motion in two (or three dimensions)
More informationTime : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A
Time : hours 0 - Mthemtics - Mrch 007 Mrks : 100 Pg - 1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new
More informationExam 1: Tomorrow 8:20-10:10pm
x : Toorrow 8:0-0:0p Roo Assignents: Lst Ne Roo A-D CCC 00 -J CS A0 K- PUGH 70 N-Q LI 50 R-S RY 30 T-Z W 00 redown o the 0 Probles teril # o Probles Chpter 4 Chpter 3 Chpter 4 6 Chpter 5 3 Chpter 6 5 Crib
More informationKEY. Physics 106 Common Exam 1, Spring, 2004
Physics 106 Common Exm 1, Spring, 2004 Signture Nme (Print): A 4 Digit ID: Section: Instructions: Questions 1 through 10 re multiple-choice questions worth 5 points ech. Answer ech of them on the Scntron
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationPROBLEM 11.3 SOLUTION
PROBLEM.3 The verticl motion of mss A is defined by the reltion x= 0 sin t+ 5cost+ 00, where x nd t re expressed in mm nd seconds, respectively. Determine () the position, velocity nd ccelertion of A when
More informationHW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy.
HW Solutions # 5-8.01 MIT - Prof. Kowlski Friction, circulr dynmics, nd Work-Kinetic Energy. 1) 5.80 If the block were to remin t rest reltive to the truck, the friction force would need to cuse n ccelertion
More informationMathematics of Motion II Projectiles
Chmp+ Fll 2001 Dn Stump 1 Mthemtics of Motion II Projectiles Tble of vribles t time v velocity, v 0 initil velocity ccelertion D distnce x position coordinte, x 0 initil position x horizontl coordinte
More informationKINEMATICS OF RIGID BODIES
KINEMTICS OF RIGID ODIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. Description
More informationPage 1. Motion in a Circle... Dynamics of Circular Motion. Motion in a Circle... Motion in a Circle... Discussion Problem 21: Motion in a Circle
Dynics of Circulr Motion A boy ties rock of ss to the end of strin nd twirls it in the erticl plne. he distnce fro his hnd to the rock is. he speed of the rock t the top of its trectory is. Wht is the
More informationPHYS Summer Professor Caillault Homework Solutions. Chapter 2
PHYS 1111 - Summer 2007 - Professor Cillult Homework Solutions Chpter 2 5. Picture the Problem: The runner moves long the ovl trck. Strtegy: The distnce is the totl length of trvel, nd the displcement
More informationHIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)
HIGHER SCHOOL CERTIFICATE EXAMINATION 999 MATHEMATICS UNIT (ADDITIONAL) Time llowed Three hours (Plus 5 minutes reding time) DIRECTIONS TO CANDIDATES Attempt ALL questions ALL questions re of equl vlue
More informationPHYSICS 211 MIDTERM I 22 October 2003
PHYSICS MIDTERM I October 3 Exm i cloed book, cloed note. Ue onl our formul heet. Write ll work nd nwer in exm booklet. The bck of pge will not be grded unle ou o requet on the front of the pge. Show ll
More informationDYNAMICS. Kinematics of Rigid Bodies VECTOR MECHANICS FOR ENGINEERS: Tenth Edition CHAPTER
Tenth E CHTER 15 VECTOR MECHNICS FOR ENGINEERS: YNMICS Ferdinnd. eer E. Russell Johnston, Jr. hillip J. Cornwell Lecture Notes: rin. Self Cliforni olytechnic Stte Uniersity Kinemtics of Rigid odies 013
More informationAnswers to selected problems from Essential Physics, Chapter 3
Answers to selected problems from Essentil Physics, Chpter 3 1. FBD 1 is the correct free-body dirm in ll five cses. As fr s forces re concerned, t rest nd constnt velocity situtions re equivlent. 3. ()
More informationPhys101 Lecture 4,5 Dynamics: Newton s Laws of Motion
Phys101 Lecture 4,5 Dynics: ewton s Lws of Motion Key points: ewton s second lw is vector eqution ction nd rection re cting on different objects ree-ody Digrs riction Inclines Ref: 4-1,2,3,4,5,6,7,8,9.
More informationVersion 001 HW#6 - Circular & Rotational Motion arts (00223) 1
Version 001 HW#6 - Circulr & ottionl Motion rts (00223) 1 This print-out should hve 14 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Circling
More informationSolutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16
CHAPTER 16 1. The number of electrons is N = Q/e = ( 30.0 10 6 C)/( 1.60 10 19 C/electrons) = 1.88 10 14 electrons.. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces,
More information8A Review Solutions. Roger Mong. February 24, 2007
8A Review Solutions Roer Mon Ferury 24, 2007 Question We ein y doin Free Body Dirm on the mss m. Since the rope runs throuh the lock 3 times, the upwrd force on the lock is 3T. (Not ecuse there re 3 pulleys!)
More informationKinematics in Two-Dimensions
Slide 1 / 92 Slide 2 / 92 Kinemtics in Two-imensions www.njctl.org Slide 3 / 92 How to Use this File ch topic is composed of brief direct instruction There re formtie ssessment questions fter eer topic
More informationIntroduction to Mechanics Practice using the Kinematics Equations
Introduction to Mechnics Prctice using the Kinemtics Equtions Ln Sheridn De Anz College Jn 24, 2018 Lst time finished deriing the kinemtics equtions some problem soling prctice Oeriew using kinemtics equtions
More informationSimple Harmonic Motion I Sem
Simple Hrmonic Motion I Sem Sllus: Differentil eqution of liner SHM. Energ of prticle, potentil energ nd kinetic energ (derivtion), Composition of two rectngulr SHM s hving sme periods, Lissjous figures.
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More informationStudent Session Topic: Particle Motion
Student Session Topic: Prticle Motion Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position, velocity or ccelertion my be
More informationWhat determines where a batted baseball lands? How do you describe
MTIN IN TW R THREE DIMENIN 3 LEARNING GAL studing this chpter, ou will lern:?if cr is going round cure t constnt speed, is it ccelerting? If so, in wht direction is it ccelerting? Wht determines where
More informationINTRODUCTION. The three general approaches to the solution of kinetics problems are:
INTRODUCTION According to Newton s lw, prticle will ccelerte when it is subjected to unblnced forces. Kinetics is the study of the reltions between unblnced forces nd the resulting chnges in motion. The
More information16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings When, in the cse of tilted coordinte system, you brek up the
More information1 Which of the following summarises the change in wave characteristics on going from infra-red to ultraviolet in the electromagnetic spectrum?
Which of the following summrises the chnge in wve chrcteristics on going from infr-red to ultrviolet in the electromgnetic spectrum? frequency speed (in vcuum) decreses decreses decreses remins constnt
More informationPREVIOUS EAMCET QUESTIONS
CENTRE OF MASS PREVIOUS EAMCET QUESTIONS ENGINEERING Two prticles A nd B initilly t rest, move towrds ech other, under mutul force of ttrction At n instnce when the speed of A is v nd speed of B is v,
More informationMathematics Extension 2
00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Etension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors my be used A tble of stndrd
More informationMathematics Extension Two
Student Number 04 HSC TRIAL EXAMINATION Mthemtics Etension Two Generl Instructions Reding time 5 minutes Working time - hours Write using blck or blue pen Bord-pproved clcultors my be used Write your Student
More informationA Case Study on Simple Harmonic Motion and Its Application
Interntionl Journl of Ltest Engineering nd Mngement Reserch IJLEMR ISSN: 55-87 Volume 0 - Issue 08 August 07 PP. 5-60 A Cse Stud on Simple Hrmonic Motion nd Its Appliction Gowri.P, Deepik.D, Krithik.S
More informationPHYSICS. (c) m D D m. v vt. (c) 3. Width of the river = d = 800 m m/s. 600m. B 800m v m,r. (b) 1000m. v w
PHYSICS T. T 8 k/hr = /s, ' Hz, ' 7 Hz T. r D D D,. r D D. D. D.... g/c. Width o the rier = d = 8 w /s B 8,r P O. sin t cos., on the copring this eqution, with sin t cos k =, k =. k. /s w Q. =, 9 9 = 8,
More informationC D o F. 30 o F. Wall String. 53 o. F y A B C D E. m 2. m 1. m a. v Merry-go round. Phy 231 Sp 03 Homework #8 Page 1 of 4
Phy 231 Sp 3 Hoework #8 Pge 1 of 4 8-1) rigid squre object of negligible weight is cted upon by the forces 1 nd 2 shown t the right, which pull on its corners The forces re drwn to scle in ters of the
More informationInstantaneous Rate of Change of at a :
AP Clculus AB Formuls & Justiictions Averge Rte o Chnge o on [, ]:.r.c. = ( ) ( ) (lger slope o Deinition o the Derivtive: y ) (slope o secnt line) ( h) ( ) ( ) ( ) '( ) lim lim h0 h 0 3 ( ) ( ) '( ) lim
More informationPhysics 319 Classical Mechanics. G. A. Krafft Old Dominion University Jefferson Lab Lecture 2
Physics 319 Clssicl Mechnics G. A. Krfft Old Dominion University Jefferson Lb Lecture Undergrdute Clssicl Mechnics Spring 017 Sclr Vector or Dot Product Tkes two vectors s inputs nd yields number (sclr)
More informationYear 12 Mathematics Extension 2 HSC Trial Examination 2014
Yer Mthemtics Etension HSC Tril Emintion 04 Generl Instructions. Reding time 5 minutes Working time hours Write using blck or blue pen. Blck pen is preferred. Bord-pproved clcultors my be used A tble of
More informationThe Atwood Machine OBJECTIVE INTRODUCTION APPARATUS THEORY
The Atwood Mchine OBJECTIVE To derive the ening of Newton's second lw of otion s it pplies to the Atwood chine. To explin how ss iblnce cn led to the ccelertion of the syste. To deterine the ccelertion
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationFirst, we will find the components of the force of gravity: Perpendicular Forces (using away from the ramp as positive) ma F
1.. In Clss or Homework Eercise 1. An 18.0 kg bo is relesed on 33.0 o incline nd ccelertes t 0.300 m/s. Wht is the coeicient o riction? m 18.0kg 33.0? 0 0.300 m / s irst, we will ind the components o the
More information1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation
1 1.1. Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem. 1.1.1. Overview
More informationMathematics Extension 2
S Y D N E Y B O Y S H I G H S C H O O L M O O R E P A R K, S U R R Y H I L L S 005 HIGHER SCHOOL CERTIFICATE TRIAL PAPER Mthemtics Extension Generl Instructions Totl Mrks 0 Reding Time 5 Minutes Attempt
More informationa) mass inversely proportional b) force directly proportional
1. Wht produces ccelertion? A orce 2. Wht is the reltionship between ccelertion nd ) mss inersely proportionl b) orce directly proportionl 3. I you he orce o riction, 30N, on n object, how much orce is
More informationKINEMATICS OF RIGID BODIES
KINEMTICS OF RIGI OIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. escription
More informationDynamics Applying Newton s Laws Accelerated Frames
Dynmics Applying Newton s Lws Accelerted Frmes Ln heridn De Anz College Oct 18, 2017 Lst time Circulr motion nd force Centripetl force Exmples Non-uniform circulr motion Overview one lst circulr motion
More informationAP Physics C - Mechanics. Introduction. Sliding Blocks. Slide 1 / 139 Slide 2 / 139. Slide 3 / 139. Slide 4 / 139. Slide 5 / 139.
Slide 1 / 139 Slide 2 / 139 P Physics C - Mechnics Dynmics - pplictions of Newtons Lws 2015-12-03 www.njctl.org Slide 3 / 139 Slide 4 / 139 Tble of Contents Click on the topic to go to tht section Introduction
More informationSPECIALIST MATHEMATICS
Victorin Certificte of Eduction 006 SUPERVISOR TO ATTACH PROCESSING LABEL HERE STUDENT NUMBER Letter Figures Words SPECIALIST MATHEMATICS Written exmintion Mondy 30 October 006 Reding time: 3.00 pm to
More informationChapter 4. (a) (b) (c) rocket engine, n r is a normal force, r f is a friction force, and the forces labeled mg
Chpter 4 0. While the engines operte, their totl upwrd thrust eceeds the weight of the rocket, nd the rocket eperiences net upwrd fce. his net fce cuses the upwrd velocit of the rocket to increse in mgnitude
More informationpivot F 2 F 3 F 1 AP Physics 1 Practice Exam #3 (2/11/16)
AP Physics 1 Prctice Exm #3 (/11/16) Directions: Ech questions or incomplete sttements below is followed by four suggested nswers or completions. Select one tht is best in ech cse nd n enter pproprite
More informationOn the diagram below the displacement is represented by the directed line segment OA.
Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples
More informationStudy Guide Final Exam. Part A: Kinetic Theory, First Law of Thermodynamics, Heat Engines
Msschusetts Institute of Technology Deprtment of Physics 8.0T Fll 004 Study Guide Finl Exm The finl exm will consist of two sections. Section : multiple choice concept questions. There my be few concept
More information