Sample Test 3. STUDENT NAME: STUDENT id #:

Similar documents
Sample Test 3. STUDENT NAME: STUDENT id #:

CHAPTER 6 WORK AND ENERGY

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1

f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied

GENERAL PHYSICS PH 221-1D (Dr. S. Mirov) Test 4 (Sample) ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

Kinetics of Particles. Chapter 3

Chapter 5: Force and Motion I-a

Solution to HW14 Fall-2002

Sample Test 2. GENERAL PHYSICS PH 221-3A (Dr. S. Mirov) Test 2 (10/10/07) ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

Study Guide Physics Pre-Comp 2013

Conceptual Dynamics SDC. An Interactive Text and Workbook. Kirstie Plantenberg Richard Hill. Better Textbooks. Lower Prices.

Work, Energy, and Power

Yeu-Sheng Paul Shiue, Ph.D 薛宇盛 Professor and Chair Mechanical Engineering Department Christian Brothers University 650 East Parkway South Memphis, TN

Q x = cos 1 30 = 53.1 South

Honors Physics Final Review Summary

Phys101 First Major-131 Zero Version Coordinator: Dr. A. A. Naqvi Wednesday, September 25, 2013 Page: 1

Phy 213: General Physics III 6/14/2007 Chapter 28 Worksheet 1

Chapter 3 Kinematics in Two Dimensions; Vectors

i-clicker i-clicker Newton s Laws of Motion First Exam Coming Up! Components of Equation of Motion

12/01/10. STUDENT NAME: STUDENT id #: NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.

= m. Suppose the speed of a wave on a string is given by v = Κ τμ

Example 1. A robot has a mass of 60 kg. How much does that robot weigh sitting on the earth at sea level? Given: m. Find: Relationships: W

CHAPTER 8b Static Equilibrium Units

PHYS 219 Spring semester Lecture 02: Coulomb s Law how point charges interact. Ron Reifenberger Birck Nanotechnology Center Purdue University

WYSE Academic Challenge Sectional Physics 2007 Solution Set

Q1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?

1 Course Notes in Introductory Physics Jeffrey Seguritan

Study Guide: PS. 10 Motion, Forces, Work & Simple Machines DESCRIBING MOTION SPEED

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

AP Physics Kinematic Wrap Up

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle?

14. Which shows the direction of the centripetal force acting on a mass spun in a vertical circle?

Physic 231 Lecture 12

Fundamental Concepts in Structural Plasticity

NGSS High School Physics Domain Model

Chapter 9 Vector Differential Calculus, Grad, Div, Curl

205MPa and a modulus of elasticity E 207 GPa. The critical load 75kN. Gravity is vertically downward and the weight of link 3 is W3

Equilibrium of Stress

ENGI 4430 Parametric Vector Functions Page 2-01

CHAPTER 6 -- ENERGY. Approach #2: Using the component of mg along the line of d:

Figure 1a. A planar mechanism.

Q1. In figure 1, Q = 60 µc, q = 20 µc, a = 3.0 m, and b = 4.0 m. Calculate the total electric force on q due to the other 2 charges.

PHYSICS LAB Experiment 10 Fall 2004 ROTATIONAL DYNAMICS VARIABLE I, FIXED

2015 Regional Physics Exam Solution Set

Q1. A) 48 m/s B) 17 m/s C) 22 m/s D) 66 m/s E) 53 m/s. Ans: = 84.0 Q2.

Phys102 Final-061 Zero Version Coordinator: Nasser Wednesday, January 24, 2007 Page: 1

Lecture 5: Equilibrium and Oscillations

EXAM #1 PHYSICAL SCIENCE 103 FALLF, 2017

SPH3U1 Lesson 06 Kinematics

Surface and Contact Stress

Phys102 Second Major-102 Zero Version Coordinator: Al-Shukri Thursday, May 05, 2011 Page: 1

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Physics 321 Solutions for Final Exam

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

EXAM #1 PHYSICAL SCIENCE 103 Spring, 2016

. (7.1.1) This centripetal acceleration is provided by centripetal force. It is directed towards the center of the circle and has a magnitude

CHAPTER 4 Dynamics: Newton s Laws of Motion /newtlaws/newtltoc.html

Faculty of Engineering and Department of Physics Engineering Physics 131 Midterm Examination February 27, 2006; 7:00 pm 8:30 pm

Fall 2013 Physics 172 Recitation 3 Momentum and Springs

Chapter 10. Simple Harmonic Motion and Elasticity. Example 1 A Tire Pressure Gauge

Torque and Angular Momentum

GAUSS' LAW E. A. surface

Introduction: A Generalized approach for computing the trajectories associated with the Newtonian N Body Problem

CLASS XI SET A PHYSICS

SAMPLE USEFUL INFORMATION: Acceleration due to gravity, gg 9.80 m s 2. Avogadro s number, NN AA particles

making triangle (ie same reference angle) ). This is a standard form that will allow us all to have the X= y=

"NEET / AIIMS " SOLUTION (6) Avail Video Lectures of Experienced Faculty.

CHAPTER 1 -- MATH REVIEW

ES201 - Examination 2 Winter Adams and Richards NAME BOX NUMBER

PHYS 314 HOMEWORK #3

4) What is the magnitude of the net electric field at the center of the square?

Lecture 7 Further Development of Theory and Applications

Chapter 32. Maxwell s Equations and Electromagnetic Waves

STUDENT NAME: STUDENT id #: NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.

Rigid Body Dynamics (continued)

Physics UCSB TR 2:00-3:15 lecture Final Exam Wednesday 3/17/2010

ENGINEERING PHYSICS 1A By Dr. Z. Birech Department of Physics, University of Nairobi

Chapter 2 GAUSS LAW Recommended Problems:

( ) + θ θ. ω rotation rate. θ g geographic latitude - - θ geocentric latitude - - Reference Earth Model - WGS84 (Copyright 2002, David T.

1.2.1 Vectors. 1 P age. Examples What is the reference vector angle for a vector that points 50 degrees east of south?

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET

Hooke s Law (Springs) DAVISSON. F A Deformed. F S is the spring force, in newtons (N) k is the spring constant, in N/m

Work, Energy, and Power. AP Physics C

i-clicker!! x 2 lim Lecture 3 Motion in 2- and 3-Dimensions lim REVIEW OF 1-D MOTION

Lecture 2: Single-particle Motion

PHYSICS Unit 3 Trial Examination

Fundamental concept of metal rolling

Physics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1

Chapter 6. Dielectrics and Capacitance

GENERAL PHYSICS PH 221-1D (Dr. S. Mirov) Test 4 (04/29/13) ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.

39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution

ASTRODYNAMICS. o o o. Early Space Exploration. Kepler's Laws. Nicolaus Copernicus ( ) Placed Sun at center of solar system

Aircraft Performance - Drag

2015 Iredell-Statesville Schools Physics

Assume that the water in the nozzle is accelerated at a rate such that the frictional effect can be neglected.

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial

Being able to connect displacement, speed, and acceleration is fundamental to working

Professional Development. Implementing the NGSS: High School Physics

Transcription:

GENERAL PHYSICS PH -3A (Dr. S. Mirv) Test 3 (/7/07) ke Sample Test 3 STUDENT NAME: STUDENT id #: ------------------------------------------------------------------------------------------------------------------------------------------- ALL QUESTIONS ARE WORTH 0 POINTS. WORK OUT FIVE PROBLEMS. NOTE: Clearl write ut slutins and answers (circle the answers) b sectin r each part (a., b., c., etc.) Imprtant Frmulas:. Mtin alng a straight line with a cnstant acceleratin v aver. speed = [dist. taken]/[time trav.]=s/t; v aver.vel. = /t; v ins =d/t; a aver. = v aver. vel. /t; a = dv/t; v = v + at; = /(v +v)t; = v t + / at ; v = v + a (i =0 at t =0). Free all mtin (with psitive directin ) g = 9.80 m/s ; = v aver. t v aver. = (v+v )/; v = v - gt; = v t - / g t ; v = v g (i =0 at t =0) 3. Mtin in a plane v = v cs; v = v sin; = v t+ / a t ; = v t + / a t ; v = v + at; v = v + at; 4. Prjectile mtin (with psitive directin ) v = v = v cs; = v t; ; ma = ( v sin cs)/g = (v sin)/g r in = in ; v = v - gt = v sin - gt; = v t - / gt ; 5. Unirm circular Mtin a=v /r, T=r/v 6. Relative mtin v P A v P B v B A a PA a PB 7. Cmpnent methd vectr additin

A = A + A ; A = A + A and A = A + A ; A A A ; = tan - A /A ; The scalar prduct A a b = a b c s a b ( a ˆ ˆ ˆ) ( ˆ ˆ ˆ i a j a zk b i b j b zk ) a b = a b a b a zb z The vectr prduct a b ( a ˆ ˆ ˆ) ( ˆ ˆ ˆ i a j a zk b i b j b zk ) iˆ ˆj kˆ a a z a ˆ ˆ a a z ˆ a a b b a a a a z i j k b b z b b b z b b b b z ( a b b a ) iˆ ( a b b a ) ˆj ( a b b a ) kˆ z z z z. Secnd Newtn s Law ma=f net ;. Kinetic rictin k = k N; 3. Static ti rictin s = s N; 4. Universal Law Gravitatin: F=GMm/r ; G=6.670 - Nm /kg ; 5. Drag ceicient D C A v 6. Terminal speed v t mg C A 7. Centripetal rce: F c =mv /r 8. Speed the satellite in a circular rbit: v =GM E /r 9. The wrk dne b a cnstant rce acting n an bject: W F d cs F d 0. Kinetic energ: K mv. Ttal mechanical energ: E=K+U. The wrk-energ therem: W=K -K ; W nc =K+U=E -E 3. The principle cnservatin mechanical energ: when W nc =0, E =E 4. Wrk dne b the gravitatinal rce: W m g d cs g

. Wrk dne in Liting and Lwering the bject: K K K W W ; i K K ; W W i a g i a g. Spring Frce: F k ( H k's law ) 3. Wrk dne b a spring rce: W k k ; i 0 and ; W k s i i s 4. Wrk dne b a variable rce: W F ( ) d W d W 5. Pwer: P avg ; P ; P F v cs F v t d t 6. Ptential energ: U W ; U F ( ) d 7. Gravitatinal Ptential Energ: U m g ( ) m g ; i 0 a n d U 0 ; U ( ) m g i i i i i 8. Elastic ptential Energ: U ( ) k 9. Ptential energ curves: du ( ) F ( ) ; K ( ) E m e c U ( ) d 0. Wrk dne n a sstem b an eternal rce: F ric ti n is n t in v lv e d W h e n k in e tic ric ti n rc e a c ts w ith in th e s ste m E d th k W E K U m e c W E E m e c th. Cnservatin energ: W E E E E m e c th r islated sstem (W =0) E E E 0 in t m e c th in t E. Pwer: P avg ; t P de d t ; 3. Center mass: r m r n i i M i 4. Newtns Secnd Law r a sstem particles: F net M a

. Linear Mmentum and Newtn s Secnd law r a sstem particles: P M v and F net dp dt t. Cllisin and impulse: J F ( t ) d t; J F t; t i avg when a stream bdies with mass m and n n m speed v, cllides with a bd whse psitin is ied F avg p m v v t t t Impulse-Linear Mmentum Therem: p p i J 3. Law Cnservatin Linear mmentum: P P r clsed, islated sstem 4. Inelastic cllisin in ne dimensin: p p p p i i i 5. Mtin the Center Mass: The center mass a clsed, islated sstem tw clliding bdies is nt aected b a cllisin. 6. Elastic Cllisin in One Dimensin: m m v v ; v m v i i m m m m 7. Cllisin in Tw Dimensins: p p p p ; p p p p i i i i 8. Variable-mass sstem: Rv rel v v v M a (irst rcket equatin) i r e l M i ln (secnd rcket equatin) M S 9. Angular Psitin: (radian m easure) r 0. Angular Displacement: (p sitiv e r c u n terclc k w ise r tatin ) d. Angular velcit and speed: avg ; (p sitive r c u n tercl ck w ise rtati n ) t d t. Angular acceleratin: avg ; t d d t

t ( ) t. angular acceleratin: t t ( ) t t. Linear and angular variables related: v r s r ; v r ; a t r ; a r r ; T r v 3. Rtatinal Kinetic Energ and Rtatinal Inertia: K I ; I m iri r bd as a sstem discrete particles; I r d m r a bd w ith c n tin u u sl distributed m ass. 4. The parallel aes therem: I I M h 5. Trque: rf r F rf sin t 6. Newtn s secnd law in angular rm: net I 7. Wrk and Rtatinal Kinetic Energ: W d ; W ( ) r cnst; i i dw P K K K I I W dt ; i i w rk e n e rg th e re m r r ta tin g b d ie s v K I m v 8. Rlling bdies: a R a R g sin r rlling sm thl dw n the ram p I / M R 9. Trque as a vectr: r F ; rf sin rf r F

. Angular Mmentum a particle:. Newtn s Secnd law in Angular Frm: net l r p m( r v); l rmvsin rp rmv r p r mv 3. Angular mmentum a sstem particles: 4. Angular Mmentum a Rigid idbd: L I dl dt L net et n l i i dl dt 5. Cnservatin Angular Mmentum: L L (islated sstem) 6. Static equilibrium: F net 0; 0 7. Elastic Mduli: stress=mdulus strain net i i all the rces lie in plane F 0; F 0; 0 net, net, net, z F L 8. Tensin and Cmpressin: E, E is the Yung's mdulus A L F L 9. Shearing: G, G is the shear mdulus A L 0. Hdraulic Stress: p B V, B is the bulk mdulus V

. At the same instant that a 0.50-kg ball is drpped rm 5m abve Earth, a secnd ball, with a mass 0.5 kg, is thrwn straight upward rm Earth's surace with an initial speed 5m/s. The mve alng nearb lines and pass each ther withut clliding. + What is the height abve Earth's surace the center mass the tw-ball sstem at the end.0 s? m v =0 =5m v =5 m/s m 5.4m 0.4m t=0 t=.0s ( a) Find inal crdinate the ball ater.0 s. gt 9.8 5 0 t 5 5.4m (b) Find inal crdinate the ball ater.0 s gt 9.8 vt 5 0.4m (c) Find crdinate the center mass tw balls m m (0.5)(5.4) (0.5)(0.4) 7.m m m (0.5 0.5)

. Tw skaters with masses 00 kg and 60 kg, respectivel, stand 0.0 m apart; each hlds ne end a piece rpe. I the pull themselves alng the rpe until the meet, hw ar des each skater travel? (Neglect rictin) 00kg 0m 60kg 0 + ( a) Since tw scaters represent an islated sstem and initiall theu are at rest their center mass will be at rest and the will meet at the center mass m m m m (00)(0) (60)(0) 006 0 3.75m 3.8m (b) The 00 kg skater mves 3.75 m (c) The 60 kg skater mves 0-3.75=6.5m=6.m

3. A kg blck wd rests n a lng tabletp. A 5 g bullet mving hrizntall with a speed 50 m/s is sht int the blck and sticks in it. The blck then slides 70 cm alng the table and stps. (a) ind the speed the blck just ater impact. (b) ind the rictin rce between blck and table. v v 0 =0 m m.7 m (a) Cnsider inelastic bullet-blck cllisin. At the instant cllisn bullet-blck sstem is islated, hence, we can use law cnservatin linear mmentum t describe the cllisin mv mv 0 0 mv 0mv 0 (0.005)(50) ()(0) ( m m ) v ; v 0.374 4 0 m/ s ( m m ) (0.005 ) (b) Calculate wrk dne b the net rce n a blck-bullet sstem during its mtin alng the table There are 3 rces acting n ur sstem: Weight, Nrmal reactin, and kineti rictin. Onl kinetic rictin rce will perrm nn zer wrk. W (c) Mtin a blck-bullet sstem ater the cllisin can be described b a wrk-energ therem net S ( m m ) v ( m m ) v (0005 (0.005 )(0374) )(0.374) net i k 0 k 50 W K K S N S (.70) k

4. The impact the head a gl club n a gl ball can be apprimatel regarded as an elastic cllisin. The mass the head the gl club is 0.5 kg. The speed the club bere the cllisin is 46 m/s. The ball acquires a speed 70 m/s ater the cllisin. The gl club and a ball are in cntact r abut 0.5 ms. a) What must be the mass the ball? b) What is the average rce eerted b the club n the ball? (a) Fr elastic gl club-ball cllisin v b mc m m mv (0.5)(46) c ci mb v m 0.5 0.047 4.7 0 b c kg 70 c b v ci P P P (0.047)(70) 0 ( ) 6600 6.6 0 t t 0.00050005 b bi bfavg N N

5.

6. Je is painting the lr his basement using a paint rller. The rller has a mass.4 kg and a radius 3.8 cm. In rlling the rller acrss the lr Je applies a rce F=6N directed at an angle 35% as shwn. Ignring the mass the rller handle, what is the magnitude the angular acceleratin the rller? F 35 N (a) nd Newtn law ais: s Fsin ma ( b ) nd Newtn Law r rtatin ti abut : R I (c) recall the relatinship between a and : a R; I ( d) substitute eq.(b) and (c) in (a): Fsin m R ; R F sin F sin F sin (6)sin 35 I mr mr mr 3mR 3(4)(0 3(.4)(0.038) 038) R rad 67 s s s mg F

7. A hp (I h =MR ), a unirm disk (I d =/MR ), and a unirm sphere (I s =/5MR ), all with the same mass and uter radius, start with the same speed and rll withut sliding up identical inclines. Rank the bjects accrding t hw high the g, least t greatest. ( a)during rlling up the incline mtin the mechanical energ the hp-earth, disk-earth, and sphere-earth sstems is cnserved since )rictinal rce des nt transer an energ t thermal energ because the bjects d nt slide; ) nrmal rce is perpendicular t the pass; 3) gravitatinal rce is a cnservative ne. I Mv 0 0 Mgh ( b) secnd term is the same r all ur bjects, is als the same, the larger I the larger maimum height h reached b the bject. rm least t greatest t in terms height ht attained: sphere, disk, hp.

9. A unirm seesaw length 6 m has tw ungsters weights w =700N and w =400 N sitting n the ends. Find the prper lcatin the pivt r the seesaw t be just in balance, i (a) the weight the seesaw can be ignred (b) the seesaw weighs Mg=300N 6 m w Mg ( a) Use nd cnditin equilibrium: W W(6 ) 0; 6W 6(400).8m W W 700 400 w ( b )Use nd cnditin equilibrium: WMg(3 ) W(6 ) 0; 6W 3Mg 6(400) 3(300) W W Mg 700 400 300 36.36 m