# WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

Size: px
Start display at page:

Download "WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x"

Transcription

1 WEEK-3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, (a) The horizonal componen of he projecile s velociy remains consan hroughou he moion, since he acceleraion a x in he horizonal direcion is zero (a x = 0 m/s ). The verical componen v y, however, changes as he projecile moves. This componen is greaes a poin 1, decreases o zero a poin a he op of he rajecory, and hen increases o a magniude less han ha a poin 1 as he projecile approaches poin (c) The acceleraion due o graviy is he same for boh balls, despie he fac ha hey have differen velociies. 4. (d) The acceleraion of a projecile is he same a all poins on he rajecory. I poins downward, oward he earh, and has a magniude of 9.80 m/s. 6. (c) The ime for a projecile o reach he ground depends only on he y componen (or verical componen) of is variables, i.e., y, v 0y, and a y. These variables are he same for boh balls. The fac ha Ball 1 is moving horizonally a he op of is rajecory does no play a role in he ime i akes for i o reach he ground. 14. (c) The velociy v PC of he passenger relaive o he car is given by v PC = v PB + v BC, according o he subscriping mehod discussed in Secion 3.4. However, he las erm on he righ of his equaion is given by v BC = v BG + v GC. So, v PC = v PB + v BG + v GC = + m/s + 16 m/s + ( 1 m/s) = +6 m/s.

2 Ch. 3: Problems 9, 37, 41 & REASONING The verical displacemen y of he ball depends on he ime ha i is in he air before being caugh. These variables depend on he y-direcion daa, as indicaed in he able, where he +y direcion is "up." y-direcion Daa y a y v y v 0y? 9.80 m/s 0 m/s? Since only wo variables in he y direcion are known, we canno deermine y a his poin. Therefore, we examine he daa in he x direcion, where +x is aken o be he direcion from he picher o he cacher. x-direcion Daa x a x v 0x m 0 m/s m/s? Since his able conains hree known variables, he ime can be evaluaed by using an equaion of kinemaics. Once he ime is known, i can hen be used wih he y-direcion daa, along wih he appropriae equaion of kinemaics, o find he verical displacemen y. SOLUTION Using he x-direcion daa, Equaion 3.5a can be employed o find he ime ha he baseball is in he air: Solving for gives ( ) x= v + a = v since a = 0 m/s x x x x x m = = = s v m/s 0x The displacemen in he y direcion can now be evaluaed by using he y-direcion daa able and he value of = s. Using Equaion 3.5b, we have y 1 1 = v + a = ( 0 m/s)( s) + ( 9.80 m/s )( s) = m 0 y y The disance ha he ball drops is given by he magniude of his resul, so Disance = m.

3 37. SSM REASONING a. The drawing shows he iniial velociy v 0 of he package when i is released. The iniial speed of he package is 97.5 m/s. The componen of is displacemen along he ground is labeled as x. The daa for he x direcion are indicaed in he daa able below. v y +y +x x-direcion Daa x a x v 0x? 0 m/s +(97.5 m/s) cos 50.0 = +6.7 m/s Since only wo variables are known, i is no possible o deermine x from he daa in his able. A value for a hird variable is needed. We know ha he ime of fligh is he same for boh he x and y moions, so le s now look a he daa in he y direcion. y-direcion Daa x y a y v y v 0y 73 m 9.80 m/s +(97.5 m/s) sin 50.0 = m/s? Noe ha he displacemen y of he package poins from is iniial posiion oward he ground, so is value is negaive, i.e., y = 73 m. The daa in his able, along wih he appropriae equaion of kinemaics, can be used o find he ime of fligh. This value for can, in urn, be used in conjuncion wih he x-direcion daa o deermine x. b. The drawing a he righ shows he velociy of he package jus before impac. The angle ha he velociy makes wih respec o he ground can be found from he inverse 1 angen funcion as θ = an ( v / v y x). Once he ime has been found in par (a), he values of v y and can be deermined from he daa in he ables and he appropriae equaions of kinemaics. v y +y θ v +x

4 SOLUTION a. To deermine he ime ha he package is in he air, we will use Equaion 3.5b 1 ( y v a 0 y y ) = + and he daa in he y-direcion daa able. Solving his quadraic equaion for he ime yields 1 ( ay ) ( )( ) 1 v ± v 4 a y 0y 0y y = ( 74.7 m/s) ( 74.7 m/s) 4( 1 )( 9.80 m/s )( 73 m) ± = = 6.78 s and.0 s 9.80 m/s 1 ( )( ) We discard he firs soluion, since i is a negaive value and, hence, unrealisic. The displacemen x can be found using =.0 s, he daa in he x-direcion daa able, and Equaion 3.5a: b. The angle θ ha he velociy of he package makes wih respec o he ground is given by 1 θ = an v / v. Since here is no acceleraion in he x direcion (a x = 0 m/s ), is he ( y x) same as v 0x, so ha = v 0x = +6.7 m/s. Equaion 3.3b can be employed wih he y-direcion daa o find v y : Therefore, ( )( ) v = v + a= m/s m/s.0 s = 141 m/s y 0 y y v 1 y m/s = an = an = 66.0 v m/s x θ where he minus sign indicaes ha he angle is 66.0 below he horizonal.

5 41. REASONING The speed v of he soccer ball jus before he goalie caches i is given by v = v + v, where and are he x and y componens of he final velociy of he ball. x y The daa for his problem are (he +x direcion is from he kicker o he goalie, and he +y direcion is he up direcion): x-direcion Daa x a x v 0x m 0 m/s? +(16.0 m/s) cos 8.0 = m/s y-direcion Daa y a y v y v 0y 9.80 m/s? +(16.0 m/s) sin 8.0 = m/s Since here is no acceleraion in he x direcion (a x = 0 m/s ), remains he same as v 0x, so = v 0x = m/s. The ime ha he soccer ball is in he air can be found from he x-direcion daa, since hree of he variables are known. Wih his value for he ime and he y-direcion daa, he y componen of he final velociy can be deermined. SOLUTION Since a x = 0 m/s, he ime can be calculaed from Equaion 3.5a as x m = = = 1.19 s. The value for v v m/s y can now be found by using Equaion 3.3b 0x wih his value of he ime and he y-direcion daa: ( )( ) v = v + a= m/s m/s 1.19 s = 4.15 m/s y 0 y y The speed of he ball jus as i reaches he goalie is x y ( ) ( ) v = v + v = m/s m/s = 14.7 m/s

6 71. SSM REASONING Once he diver is airborne, he moves in he x direcion wih consan velociy while his moion in he y direcion is acceleraed (a he acceleraion due o graviy). Therefore, he magniude of he x componen of his velociy remains consan a 1.0 m/s for all imes. The magniude of he y componen of he diver's velociy afer he has fallen hrough a verical displacemen y can be deermined from Equaion 3.6b: v = v + a y. y 0 y y Since he diver runs off he plaform horizonally, v 0 y = 0 m/s. Once he x and y componens of he velociy are known for a paricular verical displacemen y, he speed of he diver can be obained from v = v + v. x y SOLUTION For convenience, we will ake downward as he posiive y direcion. Afer he diver has fallen 10.0 m, he y componen of his velociy is, from Equaion 3.6b, v = v + a y = 0 + (9.80 m / s )( m) = m / s y 0 y y Therefore, v = v + v = (1.0 m / s) + ( 14.0 m / s) = 14.1 m / s x y Ch. 4: FOC 1, 3, 5, (b) If only one force acs on he objec, i is he ne force; hus, he ne force mus be nonzero. Consequenly, he velociy would change, according o Newon s firs law, and could no be consan. 3. (e) Newon s firs law saes ha an objec coninues in a sae of res or in a sae of moion a a consan speed along a sraigh line, unless compelled o change ha sae by a ne force. All hree saemens are consisen wih he firs law. 5. (c) Newon s second law gives he answer direcly, provided he ne force is calculaed by vecor addiion of he wo given forces. The direcion of he ne force gives he direcion of he acceleraion. 8. (b) Newon s hird law indicaes ha Paul and Tom apply forces of equal magniude o each oher. According o Newon s second law, he magniude of each of hese forces is he mass imes he magniude of he acceleraion. Thus, we have m Paul a Paul = m Tom a Tom, or m Paul /m Tom = a Tom /a Paul.

7 Ch. 4: Problems 3, 5 & REASONING In each case, we will apply Newon s second law. Remember ha i is he ne force ha appears in he second law. The ne force is he vecor sum of boh forces. SOLUTION a. We will use Newon s second law, ΣF x, o find he force F. Taking he posiive x direcion o be o he righ, we have so F F 1 F = (3.0 kg)(+5.0 m/s ) (+9.0 N) = + 6N b. Applying Newon s second law again gives F F 1 = (3.0 kg)( 5.0 m/s ) (+9.0 N) = 4 N c. An applicaion of Newon s second law gives F F 1 = (3.0 kg)(0 m/s ) (+9.0 N) = 9.0 N 5. SSM REASONING The magniude ΣF of he ne force acing on he kayak is given by Newon s second law as Σ F = ma (Equaion 4.1), where m is he combined mass of he person and kayak, and a is heir acceleraion. Since he iniial and final velociies, v 0 and v, and he displacemen x are known, we can employ one of he equaions of kinemaics from Chaper o find he acceleraion. SOLUTION Solving Equaion.9 ( v v ax) = 0 + from he equaions of kinemaics for he acceleraion, we have v v0 a = x Subsiuing his resul ino Newon s second law gives ( ) ( ) ( ) v v m/s 0 m/s Σ F = ma = m = 73 kg = 3 N x ( 0.41 m)

8 16. REASONING Since here is only one force acing on he man in he horizonal direcion, i is he ne force. According o Newon s second law, Equaion 4.1, he man mus accelerae under he acion of his force. The facors ha deermine his acceleraion are (1) he magniude and () he direcion of he force exered on he man, and (3) he mass of he man. When he woman exers a force on he man, he man exers a force of equal magniude, bu opposie direcion, on he woman (Newon s hird law). I is he only force acing on he woman in he horizonal direcion, so, as is he case wih he man, she mus accelerae. The facors ha deermine her acceleraion are (1) he magniude and () he direcion of he force exered on her, and (3) he her mass. SOLUTION a. The acceleraion of he man is, according o Equaion 4.1, equal o he ne force acing on him divided by his mass. a man ΣF 45 N = = = m 8 kg 0.55 m / s (due eas) b. The acceleraion of he woman is equal o he ne force acing on her divided by her mass. a woman ΣF 45 N = = = m 48 kg 0.94 m / s (due wes)

### and v y . The changes occur, respectively, because of the acceleration components a x and a y

Week 3 Reciaion: Chaper3 : Problems: 1, 16, 9, 37, 41, 71. 1. A spacecraf is raveling wih a veloci of v0 = 5480 m/s along he + direcion. Two engines are urned on for a ime of 84 s. One engine gives he

### IB Physics Kinematics Worksheet

IB Physics Kinemaics Workshee Wrie full soluions and noes for muliple choice answers. Do no use a calculaor for muliple choice answers. 1. Which of he following is a correc definiion of average acceleraion?

### KINEMATICS IN ONE DIMENSION

KINEMATICS IN ONE DIMENSION PREVIEW Kinemaics is he sudy of how hings move how far (disance and displacemen), how fas (speed and velociy), and how fas ha how fas changes (acceleraion). We say ha an objec

### 1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under

### Physics 101 Fall 2006: Exam #1- PROBLEM #1

Physics 101 Fall 2006: Exam #1- PROBLEM #1 1. Problem 1. (+20 ps) (a) (+10 ps) i. +5 ps graph for x of he rain vs. ime. The graph needs o be parabolic and concave upward. ii. +3 ps graph for x of he person

### 2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance

Ch: Moion along a sraigh line Moion Posiion and Displacemen Average Velociy and Average Speed Insananeous Velociy and Speed Acceleraion Consan Acceleraion: A Special Case Anoher Look a Consan Acceleraion

### x(m) t(sec ) Homework #2. Ph 231 Introductory Physics, Sp-03 Page 1 of 4

Homework #2. Ph 231 Inroducory Physics, Sp-03 Page 1 of 4 2-1A. A person walks 2 miles Eas (E) in 40 minues and hen back 1 mile Wes (W) in 20 minues. Wha are her average speed and average velociy (in ha

### Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

Ground Rules PC11 Fundamenals of Physics I Lecures 3 and 4 Moion in One Dimension A/Prof Tay Seng Chuan 1 Swich off your handphone and pager Swich off your lapop compuer and keep i No alking while lecure

### Physics 221 Fall 2008 Homework #2 Solutions Ch. 2 Due Tues, Sept 9, 2008

Physics 221 Fall 28 Homework #2 Soluions Ch. 2 Due Tues, Sep 9, 28 2.1 A paricle moving along he x-axis moves direcly from posiion x =. m a ime =. s o posiion x = 1. m by ime = 1. s, and hen moves direcly

### Displacement ( x) x x x

Kinemaics Kinemaics is he branch of mechanics ha describes he moion of objecs wihou necessarily discussing wha causes he moion. 1-Dimensional Kinemaics (or 1- Dimensional moion) refers o moion in a sraigh

### 1. The graph below shows the variation with time t of the acceleration a of an object from t = 0 to t = T. a

Kinemaics Paper 1 1. The graph below shows he ariaion wih ime of he acceleraion a of an objec from = o = T. a T The shaded area under he graph represens change in A. displacemen. B. elociy. C. momenum.

### Physics 180A Fall 2008 Test points. Provide the best answer to the following questions and problems. Watch your sig figs.

Physics 180A Fall 2008 Tes 1-120 poins Name Provide he bes answer o he following quesions and problems. Wach your sig figs. 1) The number of meaningful digis in a number is called he number of. When numbers

### In this chapter the model of free motion under gravity is extended to objects projected at an angle. When you have completed it, you should

Cambridge Universiy Press 978--36-60033-7 Cambridge Inernaional AS and A Level Mahemaics: Mechanics Coursebook Excerp More Informaion Chaper The moion of projeciles In his chaper he model of free moion

### PHYSICS 220 Lecture 02 Motion, Forces, and Newton s Laws Textbook Sections

PHYSICS 220 Lecure 02 Moion, Forces, and Newon s Laws Texbook Secions 2.2-2.4 Lecure 2 Purdue Universiy, Physics 220 1 Overview Las Lecure Unis Scienific Noaion Significan Figures Moion Displacemen: Δx

### 0 time. 2 Which graph represents the motion of a car that is travelling along a straight road with a uniformly increasing speed?

1 1 The graph relaes o he moion of a falling body. y Which is a correc descripion of he graph? y is disance and air resisance is negligible y is disance and air resisance is no negligible y is speed and

### Kinematics Vocabulary. Kinematics and One Dimensional Motion. Position. Coordinate System in One Dimension. Kinema means movement 8.

Kinemaics Vocabulary Kinemaics and One Dimensional Moion 8.1 WD1 Kinema means movemen Mahemaical descripion of moion Posiion Time Inerval Displacemen Velociy; absolue value: speed Acceleraion Averages

### PHYSICS 149: Lecture 9

PHYSICS 149: Lecure 9 Chaper 3 3.2 Velociy and Acceleraion 3.3 Newon s Second Law of Moion 3.4 Applying Newon s Second Law 3.5 Relaive Velociy Lecure 9 Purdue Universiy, Physics 149 1 Velociy (m/s) The

### MEI Mechanics 1 General motion. Section 1: Using calculus

Soluions o Exercise MEI Mechanics General moion Secion : Using calculus. s 4 v a 6 4 4 When =, v 4 a 6 4 6. (i) When = 0, s = -, so he iniial displacemen = - m. s v 4 When = 0, v = so he iniial velociy

### Chapter 3 Kinematics in Two Dimensions

Chaper 3 KINEMATICS IN TWO DIMENSIONS PREVIEW Two-dimensional moion includes objecs which are moing in wo direcions a he same ime, such as a projecile, which has boh horizonal and erical moion. These wo

### t A. 3. Which vector has the largest component in the y-direction, as defined by the axes to the right?

Ke Name Insrucor Phsics 1210 Exam 1 Sepember 26, 2013 Please wrie direcl on he exam and aach oher shees of work if necessar. Calculaors are allowed. No noes or books ma be used. Muliple-choice problems

### Unit 1 Test Review Physics Basics, Movement, and Vectors Chapters 1-3

A.P. Physics B Uni 1 Tes Reiew Physics Basics, Moemen, and Vecors Chapers 1-3 * In sudying for your es, make sure o sudy his reiew shee along wih your quizzes and homework assignmens. Muliple Choice Reiew:

### Brock University Physics 1P21/1P91 Fall 2013 Dr. D Agostino. Solutions for Tutorial 3: Chapter 2, Motion in One Dimension

Brock Uniersiy Physics 1P21/1P91 Fall 2013 Dr. D Agosino Soluions for Tuorial 3: Chaper 2, Moion in One Dimension The goals of his uorial are: undersand posiion-ime graphs, elociy-ime graphs, and heir

### Giambattista, Ch 3 Problems: 9, 15, 21, 27, 35, 37, 42, 43, 47, 55, 63, 76

Giambaisa, Ch 3 Problems: 9, 15, 21, 27, 35, 37, 42, 43, 47, 55, 63, 76 9. Sraeg Le be direced along he +x-axis and le be 60.0 CCW from Find he magniude of 6.0 B 60.0 4.0 A x 15. (a) Sraeg Since he angle

### v x + v 0 x v y + a y + v 0 y + 2a y + v y Today: Projectile motion Soccer problem Firefighter example

Thurs Sep 10 Assign 2 Friday SI Sessions: Moron 227 Mon 8:10-9:10 PM Tues 8:10-9:10 PM Thur 7:05-8:05 PM Read Read Draw/Image lay ou coordinae sysem Wha know? Don' know? Wan o know? Physical Processes?

### !!"#"\$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

"#"\$%&#'()"#&'(*%)+,&',-)./)1-*) #\$%&'()*+,&',-.%,/)*+,-&1*#\$)()5*6\$+\$%*,7&*-'-&1*(,-&*6&,7.\$%\$+*&%'(*8\$&',-,%'-&1*(,-&*6&,79*(&,%: ;..,*&1\$&\$.\$%&'()*1\$\$.,'&',-9*(&,%)?%*,('&5

### 4.5 Constant Acceleration

4.5 Consan Acceleraion v() v() = v 0 + a a() a a() = a v 0 Area = a (a) (b) Figure 4.8 Consan acceleraion: (a) velociy, (b) acceleraion When he x -componen of he velociy is a linear funcion (Figure 4.8(a)),

### Suggested Practice Problems (set #2) for the Physics Placement Test

Deparmen of Physics College of Ars and Sciences American Universiy of Sharjah (AUS) Fall 014 Suggesed Pracice Problems (se #) for he Physics Placemen Tes This documen conains 5 suggesed problems ha are

### Motion along a Straight Line

chaper 2 Moion along a Sraigh Line verage speed and average velociy (Secion 2.2) 1. Velociy versus speed Cone in he ebook: fer Eample 2. Insananeous velociy and insananeous acceleraion (Secions 2.3, 2.4)

### Physics 20 Lesson 5 Graphical Analysis Acceleration

Physics 2 Lesson 5 Graphical Analysis Acceleraion I. Insananeous Velociy From our previous work wih consan speed and consan velociy, we know ha he slope of a posiion-ime graph is equal o he velociy of

### Phys 221 Fall Chapter 2. Motion in One Dimension. 2014, 2005 A. Dzyubenko Brooks/Cole

Phys 221 Fall 2014 Chaper 2 Moion in One Dimension 2014, 2005 A. Dzyubenko 2004 Brooks/Cole 1 Kinemaics Kinemaics, a par of classical mechanics: Describes moion in erms of space and ime Ignores he agen

### Lab #2: Kinematics in 1-Dimension

Reading Assignmen: Chaper 2, Secions 2-1 hrough 2-8 Lab #2: Kinemaics in 1-Dimension Inroducion: The sudy of moion is broken ino wo main areas of sudy kinemaics and dynamics. Kinemaics is he descripion

### Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

Lecure - Kinemaics in One Dimension Displacemen, Velociy and Acceleraion Everyhing in he world is moving. Nohing says sill. Moion occurs a all scales of he universe, saring from he moion of elecrons in

### SPH3U: Projectiles. Recorder: Manager: Speaker:

SPH3U: Projeciles Now i s ime o use our new skills o analyze he moion of a golf ball ha was ossed hrough he air. Le s find ou wha is special abou he moion of a projecile. Recorder: Manager: Speaker: 0

### a 10.0 (m/s 2 ) 5.0 Name: Date: 1. The graph below describes the motion of a fly that starts out going right V(m/s)

Name: Dae: Kinemaics Review (Honors. Physics) Complee he following on a separae shee of paper o be urned in on he day of he es. ALL WORK MUST BE SHOWN TO RECEIVE CREDIT. 1. The graph below describes he

### Dynamics. Option topic: Dynamics

Dynamics 11 syllabusref Opion opic: Dynamics eferenceence In his cha chaper 11A Differeniaion and displacemen, velociy and acceleraion 11B Inerpreing graphs 11C Algebraic links beween displacemen, velociy

### Kinematics in two dimensions

Lecure 5 Phsics I 9.18.13 Kinemaics in wo dimensions Course websie: hp://facul.uml.edu/andri_danlo/teaching/phsicsi Lecure Capure: hp://echo36.uml.edu/danlo13/phsics1fall.hml 95.141, Fall 13, Lecure 5

### k 1 k 2 x (1) x 2 = k 1 x 1 = k 2 k 1 +k 2 x (2) x k series x (3) k 2 x 2 = k 1 k 2 = k 1+k 2 = 1 k k 2 k series

Final Review A Puzzle... Consider wo massless springs wih spring consans k 1 and k and he same equilibrium lengh. 1. If hese springs ac on a mass m in parallel, hey would be equivalen o a single spring

### Best test practice: Take the past test on the class website

Bes es pracice: Take he pas es on he class websie hp://communiy.wvu.edu/~miholcomb/phys11.hml I have posed he key o he WebAssign pracice es. Newon Previous Tes is Online. Forma will be idenical. You migh

### Physics 101: Lecture 03 Kinematics Today s lecture will cover Textbook Sections (and some Ch. 4)

Physics 101: Lecure 03 Kinemaics Today s lecure will coer Texbook Secions 3.1-3.3 (and some Ch. 4) Physics 101: Lecure 3, Pg 1 A Refresher: Deermine he force exered by he hand o suspend he 45 kg mass as

### Guest Lecturer Friday! Symbolic reasoning. Symbolic reasoning. Practice Problem day A. 2 B. 3 C. 4 D. 8 E. 16 Q25. Will Armentrout.

Pracice Problem day Gues Lecurer Friday! Will Armenrou. He d welcome your feedback! Anonymously: wrie somehing and pu i in my mailbox a 111 Whie Hall. Email me: sarah.spolaor@mail.wvu.edu Symbolic reasoning

### AP Calculus BC Chapter 10 Part 1 AP Exam Problems

AP Calculus BC Chaper Par AP Eam Problems All problems are NO CALCULATOR unless oherwise indicaed Parameric Curves and Derivaives In he y plane, he graph of he parameric equaions = 5 + and y= for, is a

### Farr High School NATIONAL 5 PHYSICS. Unit 3 Dynamics and Space. Exam Questions

Farr High School NATIONAL 5 PHYSICS Uni Dynamics and Space Exam Quesions VELOCITY AND DISPLACEMENT D B D 4 E 5 B 6 E 7 E 8 C VELOCITY TIME GRAPHS (a) I is acceleraing Speeding up (NOT going down he flume

### Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

Chaper 2 Newonian Mechanics Single Paricle In his Chaper we will review wha Newon s laws of mechanics ell us abou he moion of a single paricle. Newon s laws are only valid in suiable reference frames,

### Q2.1 This is the x t graph of the motion of a particle. Of the four points P, Q, R, and S, the velocity v x is greatest (most positive) at

Q2.1 This is he x graph of he moion of a paricle. Of he four poins P, Q, R, and S, he velociy is greaes (mos posiive) a A. poin P. B. poin Q. C. poin R. D. poin S. E. no enough informaion in he graph o

### Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration

PHYS 54 Tes Pracice Soluions Spring 8 Q: [4] Knowing ha in he ne epression a is acceleraion, v is speed, is posiion and is ime, from a dimensional v poin of view, he equaion a is a) incorrec b) correc

### Conceptual Physics Review (Chapters 2 & 3)

Concepual Physics Review (Chapers 2 & 3) Soluions Sample Calculaions 1. My friend and I decide o race down a sraigh srech of road. We boh ge in our cars and sar from res. I hold he seering wheel seady,

### Chapter 12: Velocity, acceleration, and forces

To Feel a Force Chaper Spring, Chaper : A. Saes of moion For moion on or near he surface of he earh, i is naural o measure moion wih respec o objecs fixed o he earh. The 4 hr. roaion of he earh has a measurable

### Q.1 Define work and its unit?

CHP # 6 ORK AND ENERGY Q.1 Define work and is uni? A. ORK I can be define as when we applied a force on a body and he body covers a disance in he direcion of force, hen we say ha work is done. I is a scalar

### SOLUTIONS TO CONCEPTS CHAPTER 3

SOLUTIONS TO ONEPTS HPTER 3. a) Disance ravelled = 50 + 40 + 0 = 0 m b) F = F = D = 50 0 = 30 M His displacemen is D D = F DF 30 40 50m In ED an = DE/E = 30/40 = 3/4 = an (3/4) His displacemen from his

### Lecture 4 Kinetics of a particle Part 3: Impulse and Momentum

MEE Engineering Mechanics II Lecure 4 Lecure 4 Kineics of a paricle Par 3: Impulse and Momenum Linear impulse and momenum Saring from he equaion of moion for a paricle of mass m which is subjeced o an

### MOMENTUM CONSERVATION LAW

1 AAST/AEDT AP PHYSICS B: Impulse and Momenum Le us run an experimen: The ball is moving wih a velociy of V o and a force of F is applied on i for he ime inerval of. As he resul he ball s velociy changes

### MEI STRUCTURED MATHEMATICS 4758

OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions Thursday 5 JUNE 006 Afernoon

### 3.6 Derivatives as Rates of Change

3.6 Derivaives as Raes of Change Problem 1 John is walking along a sraigh pah. His posiion a he ime >0 is given by s = f(). He sars a =0from his house (f(0) = 0) and he graph of f is given below. (a) Describe

### x i v x t a dx dt t x

Physics 3A: Basic Physics I Shoup - Miderm Useful Equaions A y A sin A A A y an A y A A = A i + A y j + A z k A * B = A B cos(θ) A B = A B sin(θ) A * B = A B + A y B y + A z B z A B = (A y B z A z B y

### Ch.1. Group Work Units. Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Ch.. Group Work Unis Coninuum Mechanics Course (MMC) - ETSECCPB - UPC Uni 2 Jusify wheher he following saemens are rue or false: a) Two sreamlines, corresponding o a same insan of ime, can never inersec

### Physics Notes - Ch. 2 Motion in One Dimension

Physics Noes - Ch. Moion in One Dimension I. The naure o physical quaniies: scalars and ecors A. Scalar quaniy ha describes only magniude (how much), NOT including direcion; e. mass, emperaure, ime, olume,

### PHYS 100: Lecture 2. Motion at Constant Acceleration. Relative Motion: Reference Frames. x x = v t + a t. x = vdt. v = adt. x Tortoise.

a PHYS 100: Lecure 2 Moion a Consan Acceleraion a 0 0 Area a 0 a 0 v ad v v0 a0 v 0 x vd 0 A(1/2)( v) Area v 0 v v-v 0 v 0 x x v + a 1 0 0 2 0 2 Relaive Moion: Reference Frames x d Achilles Toroise x Toroise

### SPH3U1 Lesson 03 Kinematics

SPH3U1 Lesson 03 Kinemaics GRAPHICAL ANALYSIS LEARNING GOALS Sudens will Learn how o read values, find slopes and calculae areas on graphs. Learn wha hese values mean on boh posiion-ime and velociy-ime

### Of all of the intellectual hurdles which the human mind has confronted and has overcome in the last fifteen hundred years, the one which seems to me

Of all of he inellecual hurdles which he human mind has confroned and has overcome in he las fifeen hundred years, he one which seems o me o have been he mos amazing in characer and he mos supendous in

### Physics 5A Review 1. Eric Reichwein Department of Physics University of California, Santa Cruz. October 31, 2012

Physics 5A Review 1 Eric Reichwein Deparmen of Physics Universiy of California, Sana Cruz Ocober 31, 2012 Conens 1 Error, Sig Figs, and Dimensional Analysis 1 2 Vecor Review 2 2.1 Adding/Subracing Vecors.............................

### Applications of the Basic Equations Chapter 3. Paul A. Ullrich

Applicaions of he Basic Equaions Chaper 3 Paul A. Ullrich paullrich@ucdavis.edu Par 1: Naural Coordinaes Naural Coordinaes Quesion: Why do we need anoher coordinae sysem? Our goal is o simplify he equaions

### t = x v = 18.4m 44.4m/s =0.414 s.

1 Assuming he horizonal velociy of he ball is consan, he horizonal displacemen is x = v where x is he horizonal disance raveled, is he ime, and v is he (horizonal) velociy Convering v o meers per second,

### Physics 30: Chapter 2 Exam Momentum & Impulse

Physics 30: Chaper 2 Exam Momenum & Impulse Name: Dae: Mark: /29 Numeric Response. Place your answers o he numeric response quesions, wih unis, in he blanks a he side of he page. (1 mark each) 1. A golfer

### Physics 3A: Basic Physics I Shoup Sample Midterm. Useful Equations. x f. x i v x. a x. x i. v xi v xf. 2a x f x i. y f. a r.

Physics 3A: Basic Physics I Shoup Sample Miderm Useful Equaions A y Asin A A x A y an A y A x A = A x i + A y j + A z k A * B = A B cos(θ) A x B = A B sin(θ) A * B = A x B x + A y B y + A z B z A x B =

### Equations of motion for constant acceleration

Lecure 3 Chaper 2 Physics I 01.29.2014 Equaions of moion for consan acceleraion Course websie: hp://faculy.uml.edu/andriy_danylo/teaching/physicsi Lecure Capure: hp://echo360.uml.edu/danylo2013/physics1spring.hml

### CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

### One-Dimensional Kinematics

One-Dimensional Kinemaics One dimensional kinemaics refers o moion along a sraigh line. Een hough we lie in a 3-dimension world, moion can ofen be absraced o a single dimension. We can also describe moion

### 15. Vector Valued Functions

1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,

### 4.6 One Dimensional Kinematics and Integration

4.6 One Dimensional Kinemaics and Inegraion When he acceleraion a( of an objec is a non-consan funcion of ime, we would like o deermine he ime dependence of he posiion funcion x( and he x -componen of

### Physics for Scientists and Engineers I

Physics for Scieniss and Engineers I PHY 48, Secion 4 Dr. Beariz Roldán Cuenya Universiy of Cenral Florida, Physics Deparmen, Orlando, FL Chaper - Inroducion I. General II. Inernaional Sysem of Unis III.

### 9702/1/O/N/02. are set up a vertical distance h apart. M 1 M 2. , it is found that the ball takes time t 1. to reach M 2 ) 2

PhysicsndMahsTuor.com 7 car is ravelling wih uniform acceleraion along a sraigh road. The road has marker poss every 1 m. When he car passes one pos, i has a speed of 1 m s 1 and, when i passes he nex

### Answers, Even-Numbered Problems, Chapter 5

5 he ension in each sring is w (= mg) Answers, Even-Numbered Problems, Chaper 5 54 (a) 540 N (b) The θ = 0 58 (a) (b) 4 53 0 N 4 336 0 N 50 (a) The free-body diagram for he car is given in Figure 50 The

### Position, Velocity, and Acceleration

rev 06/2017 Posiion, Velociy, and Acceleraion Equipmen Qy Equipmen Par Number 1 Dynamic Track ME-9493 1 Car ME-9454 1 Fan Accessory ME-9491 1 Moion Sensor II CI-6742A 1 Track Barrier Purpose The purpose

### Some Basic Information about M-S-D Systems

Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order,

### Summary:Linear Motion

Summary:Linear Moion D Saionary objec V Consan velociy D Disance increase uniformly wih ime D = v. a Consan acceleraion V D V = a. D = ½ a 2 Velociy increases uniformly wih ime Disance increases rapidly

### Constant Acceleration

Objecive Consan Acceleraion To deermine he acceleraion of objecs moving along a sraigh line wih consan acceleraion. Inroducion The posiion y of a paricle moving along a sraigh line wih a consan acceleraion

### Traveling Waves. Chapter Introduction

Chaper 4 Traveling Waves 4.1 Inroducion To dae, we have considered oscillaions, i.e., periodic, ofen harmonic, variaions of a physical characerisic of a sysem. The sysem a one ime is indisinguishable from

### Non-uniform circular motion *

OpenSax-CNX module: m14020 1 Non-uniform circular moion * Sunil Kumar Singh This work is produced by OpenSax-CNX and licensed under he Creaive Commons Aribuion License 2.0 Wha do we mean by non-uniform

### PHYS 100: Lecture 2. Motion at Constant Acceleration. Relative Motion: Reference Frames. x Tortoise. Tortoise. d Achilles. Reference frame = Earth

a PHYS 1: Lecure 2 Moion a Consan Acceleraion a Area = a a v = ad v v = a v x = vd A=(1/2)( v) Area = v v = v-v v x x = v + a 1 2 2 Relaive Moion: Reference Frames x d Achilles Toroise x Toroise Reference

Answers o Homework. x + and y x 5 y To eliminae he parameer, solve for x. Subsiue ino y s equaion o ge y x.. x and y, x y x To eliminae he parameer, solve for. Subsiue ino y s equaion o ge x y, x. (Noe:

### Robotics I. April 11, The kinematics of a 3R spatial robot is specified by the Denavit-Hartenberg parameters in Tab. 1.

Roboics I April 11, 017 Exercise 1 he kinemaics of a 3R spaial robo is specified by he Denavi-Harenberg parameers in ab 1 i α i d i a i θ i 1 π/ L 1 0 1 0 0 L 3 0 0 L 3 3 able 1: able of DH parameers of

### Parametrics and Vectors (BC Only)

Paramerics and Vecors (BC Only) The following relaionships should be learned and memorized. The paricle s posiion vecor is r() x(), y(). The velociy vecor is v(),. The speed is he magniude of he velociy

### d = ½(v o + v f) t distance = ½ (initial velocity + final velocity) time

BULLSEYE Lab Name: ANSWER KEY Dae: Pre-AP Physics Lab Projecile Moion Weigh = 1 DIRECTIONS: Follow he insrucions below, build he ramp, ake your measuremens, and use your measuremens o make he calculaions

### Kinematics in two Dimensions

Lecure 5 Chaper 4 Phsics I Kinemaics in wo Dimensions Course websie: hp://facul.uml.edu/andri_danlo/teachin/phsicsi PHYS.141 Lecure 5 Danlo Deparmen of Phsics and Applied Phsics Toda we are oin o discuss:

### Two Coupled Oscillators / Normal Modes

Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own

### 1. Kinematics I: Position and Velocity

1. Kinemaics I: Posiion and Velociy Inroducion The purpose of his eperimen is o undersand and describe moion. We describe he moion of an objec by specifying is posiion, velociy, and acceleraion. In his

### 3, so θ = arccos

Mahemaics 210 Professor Alan H Sein Monday, Ocober 1, 2007 SOLUTIONS This problem se is worh 50 poins 1 Find he angle beween he vecors (2, 7, 3) and (5, 2, 4) Soluion: Le θ be he angle (2, 7, 3) (5, 2,

### Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence

### Roller-Coaster Coordinate System

Winer 200 MECH 220: Mechanics 2 Roller-Coaser Coordinae Sysem Imagine you are riding on a roller-coaer in which he rack goes up and down, wiss and urns. Your velociy and acceleraion will change (quie abruply),

### Decimal moved after first digit = 4.6 x Decimal moves five places left SCIENTIFIC > POSITIONAL. a) g) 5.31 x b) 0.

PHYSICS 20 UNIT 1 SCIENCE MATH WORKSHEET NAME: A. Sandard Noaion Very large and very small numbers are easily wrien using scienific (or sandard) noaion, raher han decimal (or posiional) noaion. Sandard

### d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or

### 2001 November 15 Exam III Physics 191

1 November 15 Eam III Physics 191 Physical Consans: Earh s free-fall acceleraion = g = 9.8 m/s 2 Circle he leer of he single bes answer. quesion is worh 1 poin Each 3. Four differen objecs wih masses:

### 2002 November 14 Exam III Physics 191

November 4 Exam III Physics 9 Physical onsans: Earh s free-fall acceleraion = g = 9.8 m/s ircle he leer of he single bes answer. quesion is worh poin Each 3. Four differen objecs wih masses: m = kg, m

### A man pushes a 500 kg block along the x axis by a constant force. Find the power required to maintain a speed of 5.00 m/s.

Coordinaor: Dr. F. hiari Wednesday, July 16, 2014 Page: 1 Q1. The uniform solid block in Figure 1 has mass 0.172 kg and edge lenghs a = 3.5 cm, b = 8.4 cm, and c = 1.4 cm. Calculae is roaional ineria abou

### AP CALCULUS BC 2016 SCORING GUIDELINES

6 SCORING GUIDELINES Quesion A ime, he posiion of a paricle moving in he xy-plane is given by he parameric funcions ( x ( ), y ( )), where = + sin ( ). The graph of y, consising of hree line segmens, is

### Welcome Back to Physics 215!

Welcome Back o Physics 215! (General Physics I) Thurs. Jan 19 h, 2017 Lecure01-2 1 Las ime: Syllabus Unis and dimensional analysis Today: Displacemen, velociy, acceleraion graphs Nex ime: More acceleraion