Chapter 3 Kinematics in Two Dimensions

Similar documents
Kinematics in two dimensions

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

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

KINEMATICS IN ONE DIMENSION

IB Physics Kinematics Worksheet

Kinematics in two Dimensions

s in boxe wers ans Put

Main Ideas in Class Today

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

Physics Notes - Ch. 2 Motion in One Dimension

Equations of motion for constant acceleration

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

One-Dimensional Kinematics

Q2.4 Average velocity equals instantaneous velocity when the speed is constant and motion is in a straight line.

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

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

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

Physics Unit Workbook Two Dimensional Kinematics

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

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

Displacement ( x) x x x

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

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

Chapter 12: Velocity, acceleration, and forces

1. VELOCITY AND ACCELERATION


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

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

Physics for Scientists and Engineers. Chapter 2 Kinematics in One Dimension

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

SOLUTIONS TO CONCEPTS CHAPTER 3

DEVIL PHYSICS BADDEST CLASS ON CAMPUS IB PHYSICS

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

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

Review Equations. Announcements 9/8/09. Table Tennis

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

University Physics with Modern Physics 14th Edition Young TEST BANK

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

INSTANTANEOUS VELOCITY

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

x i v x t a dx dt t x

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

AP Calculus BC Chapter 10 Part 1 AP Exam Problems

Two Dimensional Dynamics

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

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

Two Dimensional Dynamics

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

SPH3U: Projectiles. Recorder: Manager: Speaker:

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

Today: Falling. v, a

Check in: 1 If m = 2(x + 1) and n = find y when. b y = 2m n 2

A B C D September 25 Exam I Physics 105. Circle the letter of the single best answer. Each question is worth 1 point

Ex: An object is released from rest. Find the proportion of its displacements during the first and second seconds. y. g= 9.8 m/s 2

Topic 1: Linear motion and forces

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

Motion along a Straight Line

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

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

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

NEWTON S SECOND LAW OF MOTION

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 149: Lecture 9

LAB 05 Projectile Motion

Integration of the equation of motion with respect to time rather than displacement leads to the equations of impulse and momentum.

Lab #2: Kinematics in 1-Dimension

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

Course II. Lesson 7 Applications to Physics. 7A Velocity and Acceleration of a Particle

LAB # 2 - Equilibrium (static)

RECTILINEAR MOTION. Contents. Theory Exercise Exercise Exercise Exercise Answer Key

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

10.6 Parametric Equations

Version 053 Midterm 1 OConnor (05141) 1

Conceptual Physics Review (Chapters 2 & 3)

1. The 200-kg lunar lander is descending onto the moon s surface with a velocity of 6 m/s when its retro-engine is fired. If the engine produces a

4.5 Constant Acceleration

3.6 Derivatives as Rates of Change

Q.1 Define work and its unit?

Variable acceleration, Mixed Exercise 11

MEI Mechanics 1 General motion. Section 1: Using calculus

Two Coupled Oscillators / Normal Modes

I. OBJECTIVE OF THE EXPERIMENT.

SPH3U1 Lesson 03 Kinematics

Chapters 6 & 7: Trigonometric Functions of Angles and Real Numbers. Divide both Sides by 180

Chapter 2. Motion along a straight line

EF 151 Exam #2 - Spring, 2014 Page 1 of 6

15. Vector Valued Functions

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

Velocity is a relative quantity

Questions 1 and 2 refer to the graph below. The graph is a displacement-time graph for a runner. Displacement / m. Time / s

2. What is the displacement of the bug between t = 0.00 s and t = 20.0 s? A) cm B) 39.9 cm C) cm D) 16.1 cm E) +16.

MA Study Guide #1

Let us start with a two dimensional case. We consider a vector ( x,

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)

RELATIVE MOTION. Contents. Theory 01. Exercise Exercise Exercise Exercise Answer Key 13.

Summary:Linear Motion

1. Kinematics I: Position and Velocity

Dynamics. Option topic: Dynamics

v 1 =4 m/s v 2 =0 m 1 =0.5kg m 2 Momentum F (N) t (s) v 0y v x

Transcription:

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 moions of a projecile are compleel independen of one anoher, and can be described b consan eloci in he horizonal direcion, and free fall in he erical direcion. Since he wo-dimensional moion described in his chaper inoles onl consan acceleraions, we ma use he kinemaic equaions. QUICK REFERENCE Imporan Terms projecile an objec ha is projeced b a force and coninues o moe b is own ineria range of a projecile he horizonal disance beween he launch poin of a projecile and where i reurns o is launch heigh rajecor he pah followed b a projecile Equaions and Smbols Horizonal direcion: ( o o + a o + o + ) a + a Verical direcion: ( o o + a o + o + ) a + a For a projecile near he surface of he earh: a 0, is consan, and a g 0 m/s. DISCUSSION OF SELECTED SECTIONS 37

3.3 Projecile Moion Projecile moion resuls when an objec is hrown eiher horizonall hrough he air or a an angle relaie o he ground. In boh cases, he objec moes hrough he air wih a consan horizonal eloci, and a he same ime is falling freel under he influence of grai. In oher words, he projeced objec is moing horizonall and ericall a he same ime, and he resuling pah of he projecile, called he rajecor, has a parabolic shape. For his reason, projecile moion is considered o be wo-dimensional moion. The moion of a projecile can be broken down ino consan eloci and zero acceleraion in he horizonal direcion, and a changing erical eloci due o he acceleraion of grai. Le s label an quani in he horizonal direcion wih he subscrip, and an quani in he erical direcion wih he subscrip. If we fire a cannonball from a cannon on he ground poining up a an angle θ, he ball will follow a parabolic pah and we can draw he ecors associaed wih he moion a each poin along he pah: A each poin, we can draw he horizonal eloci ecor, he erical eloci ecor, and he erical acceleraion ecor g, which is simpl he acceleraion due o grai. Noice ha he lengh of he horizonal eloci and he acceleraion due o grai ecors do no change, since he are consan. The erical eloci decreases as he ball rises and increases as he ball falls. The moion of he ball is smmeric, ha is, he elociies and acceleraion of he ball on he wa up is he same as on he wa down, wih he erical eloci being zero a he op of he pah and reersing is direcion a his poin. 38

A an poin along he rajecor, he eloci ecor is he ecor sum of he horizonal and erical eloci ecors, ha is, +. θ B he Phagorean heorem, and cosq sinq q an - + æ ç è ö ø In boh he horizonal and erical cases, he acceleraion is consan, being zero in he horizonal direcion and 0 m/s downward in he erical direcion, and herefore we can use he kinemaic equaions o describe he moion of a projecile. Kinemaic Equaions for a Projecile Horizonal moion Verical moion a 0 a g - 0 m/s o + g g o + Noice he minus sign in he equaions in he righ column. Since he acceleraion g and he iniial erical eloci o are in opposie direcions, we mus gie one of hem a negaie sign, and here we e chosen o make g negaie. Remember, he horizonal eloci of a projecile is consan, bu he erical eloci is changed b grai. 39

Eample A golf ball resing on he ground is sruck b a golf club and gien an iniial eloci of 50 m/s a an angle of 30º aboe he horizonal. The ball heads oward a fence meers high a he end of he golf course, which is 00 meers awa from he poin a which he golf ball was sruck. Neglec an air resisance ha ma be acing on he golf ball. 30º 50 m/s m 00 m (a) Calculae he ime i akes for he ball o reach he plane of he fence. (b) Will he ball hi he fence or pass oer i? Jusif our answer b showing our calculaions. (c) On he aes below, skech a graph of he erical eloci of he golf ball s. ime. Be sure o label all significan poins on each ais. (m/s) (s) 40

Soluion: (a) The ime i akes for he ball o reach he plane of he fence can be found b cosq ( 50 m / s) 00m 4.6 s cos30 (b) To deermine wheher or no he golf ball will srike he fence we need o find he ball s erical posiion a he ime when he ball is a 00 m, ha is, a 4.6 seconds. o + g sin 30 + g ( 50 m / s) sin 30( 4.6 s) + (-0 m / s )( 4.6 s) 9.7 m Thus, he ball will srike he fence, since he ball is a a heigh of less han m when i reaches he plane of he fence. (c) The -componen of he ball s eloci is iniiall sin 30 (50 m/s) sin 30 5 m/s. So he erical speed would begin a 5 m/s on he erical ais, and decrease wih a negaie slope of 0 m/s, crossing he ime ais when he erical eloci is zero, ha is, when he ball has reached is maimum heigh. We can find his ime b using he equaion 0 0 sin30 g + g sin30 + g ( 50 m / s) sin30.5 s 0 m / s The ball s erical eloci is negaie (downward) afer.5 s, unil i srikes he fence a 4.6 s. (m/s) 5 m/s.5 s 4.6 s (s) 4

CHAPTER 3 REVIEW QUESTIONS For each of he muliple choice quesions below, choose he bes answer. Unless oherwise noed, use g 0 m/s and neglec air resisance.. Which of he following is NOT rue of a projecile launched from he ground a an angle? (A) The horizonal eloci is consan (B) The erical acceleraion is upward during he firs half of he fligh, and downward during he second half of he fligh. (C) The horizonal acceleraion is zero. (D) The erical acceleraion is 0 m/s (E) The ime of fligh can be found b horizonal disance diided b horizonal eloci.. A projecile is launched horizonall from he edge of a cliff 0 m high wih an iniial speed of 0 m/s. Wha is he horizonal disance he projecile raels before sriking he leel ground below he cliff? (A) 5 m (B) 0 m (C) 0 m (D) 40 m (E) 60 m 3. A projecile is launched from leel ground wih a eloci of 40 m/s a an angle of 30 from he ground. Wha will be he erical componen of he projecile s eloci jus before i srikes he ground? (sin 30 0.5, cos 30 0.87) (A) 0 m/s (B) 0 m/s (C) 30 m/s (D) 35 m/s (E) 40 m/s Free Response Problem Direcions: Show all work in working he following quesion. The quesion is worh 0 poins, and he suggesed ime for answering he quesion is abou 0 minues. The pars wihin a quesion ma no hae equal weigh.. (0 poins) Two planear eplorers land on an unchared plane and decide o es he range of cannon he brough along. When he fire a cannonball wih a speed of 00 m/s a an angle of 5 from he horizonal ground, he find ha he cannonball follows a parabolic pah and akes 0 seconds o reurn o he ground. (a) Deermine he acceleraion due o grai on his unchared plane. (b) Deermine he maimum heigh aboe he leel ground he cannonball reaches. 4

(c) One of he asronaus eclaims ha he cannonball mus hae landed oer a mile awa! Is he asronau righ? Jusif our answer ( mile 600 m). (d) The asronaus hen fire anoher idenical cannonball a 00 m/s a an angle of 75 o he horizonal ground. Will he cannonball rael a horizonal range which is less han, greaer han, or equal o he horizonal range for a 5 launch angle? less han greaer han equal o Jusif our answer. 43

ANSWERS AND EXPLANATIONS TO CHAPTER 3 REVIEW QUESTIONS Muliple Choice. B Since he erical acceleraion is due o grai, i is alwas downward.. C Firs we find he ime of fligh, which can be calculaed from he heigh: ( 0 m) g, so s g 0 m / s ( 0 m / s)( s) m Then, 0 3. B Neglecing air resisance, he componen of he eloci of he projecile jus before i lands is equal o he componen of he eloci when i is firs fired: 40 m / s sin 30 0 m / ( ) s 44

Free Response Quesion Soluion (a) 3 poins Since i akes 0 s o reurn o he ground, i akes 5s o reach maimum heigh, a which poin he erical eloci 0. Thus, - g g o o o sin 5 ( 00m / s) sin 5 5s 8.5m / s (b) 3 poins ma g ( 8.5m / s )( 5s) 05m (c) poins o cos 5 ( 00m / s) cos5( 0 s) 906m, and so i lands less han a mile awa from where i was launched. (d) poins Two launch angles which are complemenar, i.e., whose sum is 90, will produce he same horizonal range for a paricular iniial eloci. The complemen of 5 is 65. Since he new launch angle is greaer han 65, he horizonal componen of he eloci for he 75 launch angle will be less han ha of a 65 (and 5 ) launch angle, and herefore he horizonal range will be less for he 75 launch angle. 45