PH1104/PH114S MECHANICS

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "PH1104/PH114S MECHANICS"

Transcription

1 PH04/PH4S MECHANICS SEMESTER I EXAMINATION SOLUTION MULTIPLE-CHOICE QUESTIONS. (B) For freely falling bodies, the equation v = gh holds. v is proportional to h, therefore v v = h h = h h =.. (B).5i + 0.3j From the scalar product of vectors A and B, we get A B =.00x b y B = 3.00 x b = y B..00 Because the angle between A and B is 45.0, the scalar product can also be written as A B cos 45.0 = x b + y B = 3.00 Substituting x b from the previous equation, y B = 0.3 or y B =.5 Choose y B = 0.3 and x b =.5 because the x-component of vector B is positive. 3. (B) 390N The object does not move perpendicularly to the ground, therefore F y = 0 F sin 30 mg + N = 0 N = 390N 4. (C) 5N Forces on the upper block: F T mg =.3 () Forces on the lower block: T T mg =.3 () () + () and inserting T = m rope g: Back substitution to (): F = (m rope + m)g T = (m rope + m)g +.3 = 5N 5. (A) 4. x 0 3 N

2 In the plane s frame of reference, the pilot is experiencing a fictitious centrifugal force and not moving perpendicular to the floor, F y = 0 N = mg + F centrifugal = N 6. (B) 00 m/s All the kinetic energy after the collision between the bullet and the block is converted into the spring s potential energy mv = kx v =.ms From conservation of momentum in the event of collision we know, m bullet v bullet = (m block + m bullet )v v bullet = 00m/s 7. (D).67R The race car leaves the track when there is no normal force from the track onto the car. N = 0 mv R = mg sin θ v = gr sin θ Since there is no external work on the system, conservation of energy yields KE + PE = KE + PE 0 + mg(r) = mgr sin θ + mgr( + sin θ) sin θ = 3 The height of the car is h = R( + sin θ) =.67R 8. (C) the momentum change of the lighter fragment is exactly the same as the momentum change of the heavier fragment. We can apply the conservation of momentum here, inserting zero to the initial momentum as the two fragments were not moving relative to each other. 0 = p + p p = p p = p 9. (C) 3.67 m/s The work done on the object W = U = U(.00) U(5.00) = 56 5 The kinetic energy of the object when it is located at x = 5.00m. KE = KE + W mv = mv v = 3.67m/s

3 0. (C).7cm The initial condition when the block is held in place against the spring F = kx k = 74 x () Apply conservation of energy at the initial condition and at the time the block is released kx = mv 74x =.37(.) x =.7 0 m. (A) 0.7m/s Tangential acceleration a t = αr = 0.m/s Radial acceleration a r = ω r = (αθ)r = 0.7m/s Total linear acceleration a = a r + a t = 0.7m/s. (C) 3gl Use conservation of energy, i.e. all the gravitational potential energy is converted into rotational kinetic energy Iω = Mg h Since the rod is rotating around one of its end, we know I = 3 ML and ω = v tip L h is the center of mass change of height, which is equal to L for thin enough rod (shown in the figure on the right) We then obtain v tip = 3gL 3. (C).76 x 0-3 s Simply apply the conservation of angular momentum law Iω = I ω Assume the shape remains the same (spherical), therefore I can be expressed as a function of its radius, i.e. I(r) = kmr. kmr ω = kmr ω ω = ( r r ) ω The period is T = π ω = T (r r ) = s

4 4. (A) Axis Moment of inertia is proportional to radius (it is more difficult to rotate a body that is farther away from the axis) 5. (D) Search for gyroscopic stabilization and tail rotor for more information. 6. (B) 3g Surface gravity is proportional to the body s mass and inversely proportional to its radius squared g M and g R g = 3 M M ( R ) g = 3g 3 R 7. (A).99 x 0 4 kg We have two equations with two unknowns (R and M). At the surface, G Mm = 389N () R At.86 x 0 4 km above the planet s surface, Mm G (R = 4.3N () km) Dividing () by () (R km) R = 389 R = 600km 4.3 Inserting G = 6.67 x 0 - N m kg -, m = 75.0 kg, and R = 6. x 0 6 m, we get 389N R M = = kg Gm 8. (A) (a)7. x 0 5 kg, (b).9 x 0 30 kg The surface gravity of a planet is related to its mass(m) and radius(r) but not its rotation period (.3 hours is not needed) g = Gm m = kg r The dynamics of the planet s revolution is governed by the circular motion equation F centripetal = F gravity mω R = GMm R M is the mass of the star, R is the orbital radius of the planet, ω = π where T = 40 earth days. T M = 4π R 3 G T = kg 9. (A).8 x 0 4 kg

5 From the orbital speed equation v orbit = GM, we can represent the radius of the planet in R+h terms of the known variables (G, v orbit, h) and the mass. R = GM h v orbit We can then substitute this R into the equation for escape velocity ( v escape v orbital v e = GM R h ) M = vescape G 0. (E) 6.4 x 0 8 m/s The magnitude of acceleration due to gravity from each mass a = GM R = m/s Since the masses lie on the corners of an equilateral triangle, we know the angle between the two acceleration vectors is 60. Therefore, the total acceleration is a total = a ( + cos 60 ) = m/s. #a Since the plane is only moving horizontally, the kinematics equation for the vertical position of the bomb is When the bomb hits the ground, y = h gt y = 0 t = h g For the x(horizontal) coordinate, the bomb has the same constant velocity as the plane. Simply substitute the time t into the kinematics equation for movement constant velocity. x = vt = v h g #b (a) Viewed in a frame of reference that is moving with the cart, the kg box will experience a fictitious force to the opposite direction from the acceleration. The magnitude of this fictitious force is F = ma where m is the mass of the box and a is the acceleration of the cart. The box must remain stationary in the moving frame of reference for it to not fall (in the non-moving frame of reference: moving forward together with the cart and not moving vertically).

6 Newton s equation for the box in this frame of reference is then: F x = 0 N = F N = ma We know that normal force affects the maximum magnitude of static friction, i.e. f s,max = μ s N. From these two relations, we can think of finding the minimum acceleration as finding the minimum value of maximum static friction, i.e. when the maximum static friction is just enough to counter the box s weight. (b) 0N. f s,max = mg μ s ma = mg a = g μ s = 6m/s (c) 0N. Frictional force always resists the tendency of movement, but never creates movement. Here, static frictional force only nullify gravity and will not exceed the weight. (d) a g μ s, f s,max = μ s m(a) μ s m( g μ s ) = mg The box will not fall #a On the radial direction(r) the string is always tense and will not increase or decrease in length: a r = 0 On the tangential direction(t): F t = ma t a t = g sin θ (i) (ii) (iii) The acceleration of ball A after it is released a A = a t = 9.8 sin 30 = 4.9m/s The acceleration of ball A just before colliding with ball B a A = a t = 9.8 sin 0 = 0m/s Ball A stops, and ball B moves with the same velocity as ball A upon the impact. Ball B then instantaneously collides with C, and C with D. Finally, D moves upward as high as A initially. This happens because the collision is elastic, hence kinetic energy is conserved. One example is perfectly non-elastic collision where all the balls stick and move together. Linear momentum is conserved if the collision is non-elastic, but energy is not. m A v A = (m A + m B + m C + m D )v m Av A (m A + m B + m C + m D )v

7 #b Because the snow is frictionless, there is no energy loss along the ramp. From conservation of energy: v = g h = 9.8 (5 3.0) = m/s The rest of the steps required to find the touchdown point is simply the kinematics of parabolic motion. We know that she launches off from the ramp at an angle of 30 from the ground, so The uptime is t up = 9.8 =.s v x = cos 30 = 8m/s and v y = sin 30 = m/s The maximum height and the downtime are h max = = 9.m and t down = h max g = (9.) 9.8 =.4s The distance to her touchdown point is then x = v x (t up + t down ) = 8(. +.4) = 45m #3a (i) (ii) There is slipping between the spool and the incline. The string is pulling the spool to rotate backwards while the spool itself is moving forward down due to gravity. Translationally, the spool is not moving perpendicularly to the incline, so a = 0. Parallel to the incline, the ball is moving because of the forces F = ma ma = mg sin θ T a = g sin θ T m. (iii) Substituting α = a r Rotationally the spool is only rotating because of the tension: Iα = Tr α = Tr I into the equations from (ii), we get a = g sin θ Ia mr a ( + I mr) = g sin θ g sin θ a = + I mr #3b The static friction resists the tendency of movement which is downward. Therefore, the direction of the static frictional force is upward parallel to the incline. Since the spool is stationary, we know

8 F = 0 T = mg sin θ f () τ = 0 fr Tr = 0 T = fr r () Equating T from () and () mg sin θ f = fr r f() = mgr sin θ f = r mg sin θ To find the minimum coefficient of static friction we take the maximum static frictional force to be equal with the frictional force we have found, i.e. f s,max = μ s N = r mg sin θ r mg sin θ The normal force can be expressed in terms of m, g, and θ by considering the forces along the perpendicular direction F = 0 N = mg cos θ μ s mg cos θ = μ s = r mg sin θ tan θ

Exam 3 Practice Solutions

Exam 3 Practice Solutions Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at

More information

= o + t = ot + ½ t 2 = o + 2

= o + t = ot + ½ t 2 = o + 2 Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the

More information

PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011

PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011 PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this

More information

PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009

PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively.

More information

Name (please print): UW ID# score last first

Name (please print): UW ID# score last first Name (please print): UW ID# score last first Question I. (20 pts) Projectile motion A ball of mass 0.3 kg is thrown at an angle of 30 o above the horizontal. Ignore air resistance. It hits the ground 100

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Common Quiz Mistakes / Practice for Final Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A ball is thrown directly upward and experiences

More information

Concept Question: Normal Force

Concept Question: Normal Force Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical

More information

= y(x, t) =A cos (!t + kx)

= y(x, t) =A cos (!t + kx) A harmonic wave propagates horizontally along a taut string of length L = 8.0 m and mass M = 0.23 kg. The vertical displacement of the string along its length is given by y(x, t) = 0. m cos(.5 t + 0.8

More information

. d. v A v B. e. none of these.

. d. v A v B. e. none of these. General Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibrium Oct. 28, 2009 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show the formulas you use, the essential

More information

Distance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:

Distance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is: Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v =

More information

Physics I (Navitas) FINAL EXAM Fall 2015

Physics I (Navitas) FINAL EXAM Fall 2015 95.141 Physics I (Navitas) FINAL EXAM Fall 2015 Name, Last Name First Name Student Identification Number: Write your name at the top of each page in the space provided. Answer all questions, beginning

More information

TOPIC D: ROTATION EXAMPLES SPRING 2018

TOPIC D: ROTATION EXAMPLES SPRING 2018 TOPIC D: ROTATION EXAMPLES SPRING 018 Q1. A car accelerates uniformly from rest to 80 km hr 1 in 6 s. The wheels have a radius of 30 cm. What is the angular acceleration of the wheels? Q. The University

More information

Circular Motion, Pt 2: Angular Dynamics. Mr. Velazquez AP/Honors Physics

Circular Motion, Pt 2: Angular Dynamics. Mr. Velazquez AP/Honors Physics Circular Motion, Pt 2: Angular Dynamics Mr. Velazquez AP/Honors Physics Formulas: Angular Kinematics (θ must be in radians): s = rθ Arc Length 360 = 2π rads = 1 rev ω = θ t = v t r Angular Velocity α av

More information

Your Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon

Your Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon 1 Your Name: PHYSICS 101 MIDTERM October 26, 2006 2 hours Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon Problem Score 1 /13 2 /20 3 /20 4

More information

Practice Test for Midterm Exam

Practice Test for Midterm Exam A.P. Physics Practice Test for Midterm Exam Kinematics 1. Which of the following statements are about uniformly accelerated motion? Select two answers. a) If an object s acceleration is constant then it

More information

PHYSICS 220. Lecture 15. Textbook Sections Lecture 15 Purdue University, Physics 220 1

PHYSICS 220. Lecture 15. Textbook Sections Lecture 15 Purdue University, Physics 220 1 PHYSICS 220 Lecture 15 Angular Momentum Textbook Sections 9.3 9.6 Lecture 15 Purdue University, Physics 220 1 Last Lecture Overview Torque = Force that causes rotation τ = F r sin θ Work done by torque

More information

Physics 5A Final Review Solutions

Physics 5A Final Review Solutions Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone

More information

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved: 8) roller coaster starts with a speed of 8.0 m/s at a point 45 m above the bottom of a dip (see figure). Neglecting friction, what will be the speed of the roller coaster at the top of the next slope,

More information

Review questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.

Review questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. Review questions Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the

More information

Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 2

Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 2 Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 1 3 problems from exam 2 6 problems 13.1 14.6 (including 14.5) 8 problems 1.1---9.6 Go through the

More information

Classical Mechanics. FIG. 1. Figure for (a), (b) and (c). FIG. 2. Figure for (d) and (e).

Classical Mechanics. FIG. 1. Figure for (a), (b) and (c). FIG. 2. Figure for (d) and (e). Classical Mechanics 1. Consider a cylindrically symmetric object with a total mass M and a finite radius R from the axis of symmetry as in the FIG. 1. FIG. 1. Figure for (a), (b) and (c). (a) Show that

More information

PSI AP Physics I Work and Energy

PSI AP Physics I Work and Energy PSI AP Physics I Work and Energy Multiple-Choice questions 1. A driver in a 2000 kg Porsche wishes to pass a slow moving school bus on a 4 lane road. What is the average power in watts required to accelerate

More information

Lecture 13 REVIEW. Physics 106 Spring What should we know? What should we know? Newton s Laws

Lecture 13 REVIEW. Physics 106 Spring What should we know? What should we know? Newton s Laws Lecture 13 REVIEW Physics 106 Spring 2006 http://web.njit.edu/~sirenko/ What should we know? Vectors addition, subtraction, scalar and vector multiplication Trigonometric functions sinθ, cos θ, tan θ,

More information

Physics 53 Summer Final Exam. Solutions

Physics 53 Summer Final Exam. Solutions Final Exam Solutions In questions or problems not requiring numerical answers, express the answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s

More information

Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!

Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm! Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm! Final Exam Review December 9, 009 3 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector

More information

Honors Physics Review

Honors Physics Review Honors Physics Review Work, Power, & Energy (Chapter 5) o Free Body [Force] Diagrams Energy Work Kinetic energy Gravitational Potential Energy (using g = 9.81 m/s 2 ) Elastic Potential Energy Hooke s Law

More information

PH1104/PH114S - MECHANICS

PH1104/PH114S - MECHANICS PH04/PH4S - MECHANICS FAISAN DAY FALL 06 MULTIPLE CHOICE ANSWES. (E) the first four options are clearly wrong since v x needs to change its sign at a moment during the motion and there s no way v x could

More information

11. (7 points: Choose up to 3 answers) What is the tension,!, in the string? a.! = 0.10 N b.! = 0.21 N c.! = 0.29 N d.! = N e.! = 0.

11. (7 points: Choose up to 3 answers) What is the tension,!, in the string? a.! = 0.10 N b.! = 0.21 N c.! = 0.29 N d.! = N e.! = 0. A harmonic wave propagates horizontally along a taut string of length! = 8.0 m and mass! = 0.23 kg. The vertical displacement of the string along its length is given by!!,! = 0.1!m cos 1.5!!! +!0.8!!,

More information

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational

More information

Physics 123 Quizes and Examinations Spring 2002 Porter Johnson

Physics 123 Quizes and Examinations Spring 2002 Porter Johnson Physics 123 Quizes and Examinations Spring 2002 Porter Johnson Physics can only be learned by thinking, writing, and worrying. -David Atkinson and Porter Johnson (2002) There is no royal road to geometry.

More information

Random sample problems

Random sample problems UNIVERSITY OF ALABAMA Department of Physics and Astronomy PH 125 / LeClair Spring 2009 Random sample problems 1. The position of a particle in meters can be described by x = 10t 2.5t 2, where t is in seconds.

More information

End-of-Chapter Exercises

End-of-Chapter Exercises End-of-Chapter Exercises Exercises 1 12 are conceptual questions that are designed to see if you have understood the main concepts of the chapter. 1. Figure 11.21 shows four different cases involving a

More information

(1) +0.2 m/s (2) +0.4 m/s (3) +0.6 m/s (4) +1 m/s (5) +0.8 m/s

(1) +0.2 m/s (2) +0.4 m/s (3) +0.6 m/s (4) +1 m/s (5) +0.8 m/s 77777 77777 Instructor: Biswas/Ihas/Whiting PHYSICS DEPARTMENT PHY 2053 Exam 2, 120 minutes November 13, 2009 Name (print, last first): Signature: On my honor, I have neither given nor received unauthorized

More information

What is the initial velocity (magnitude and direction) of the CM? Ans: v CM (0) = ( 7 /2) v 0 ; tan 1 ( 3 /2) 41 above horizontal.

What is the initial velocity (magnitude and direction) of the CM? Ans: v CM (0) = ( 7 /2) v 0 ; tan 1 ( 3 /2) 41 above horizontal. Reading: Systems of Particles, Rotations 1, 2. Key concepts: Center of mass, momentum, motion relative to CM, collisions; vector product, kinetic energy of rotation, moment of inertia; torque, rotational

More information

Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)

Kinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction) Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position

More information

Topic 1: Newtonian Mechanics Energy & Momentum

Topic 1: Newtonian Mechanics Energy & Momentum Work (W) the amount of energy transferred by a force acting through a distance. Scalar but can be positive or negative ΔE = W = F! d = Fdcosθ Units N m or Joules (J) Work, Energy & Power Power (P) the

More information

Chapter 9-10 Test Review

Chapter 9-10 Test Review Chapter 9-10 Test Review Chapter Summary 9.2. The Second Condition for Equilibrium Explain torque and the factors on which it depends. Describe the role of torque in rotational mechanics. 10.1. Angular

More information

24/06/13 Forces ( F.Robilliard) 1

24/06/13 Forces ( F.Robilliard) 1 R Fr F W 24/06/13 Forces ( F.Robilliard) 1 Mass: So far, in our studies of mechanics, we have considered the motion of idealised particles moving geometrically through space. Why a particular particle

More information

Newton s Laws of Motion

Newton s Laws of Motion Chapter 4 Newton s Second Law: in vector form Newton s Laws of Motion σ റF = m റa in component form σ F x = ma x σ F y = ma y in equilibrium and static situations a x = 0; a y = 0 Strategy for Solving

More information

AP Physics C. Momentum. Free Response Problems

AP Physics C. Momentum. Free Response Problems AP Physics C Momentum Free Response Problems 1. A bullet of mass m moves at a velocity v 0 and collides with a stationary block of mass M and length L. The bullet emerges from the block with a velocity

More information

AP practice ch 7-8 Multiple Choice

AP practice ch 7-8 Multiple Choice AP practice ch 7-8 Multiple Choice 1. A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the spool? "Average radius" allows us to

More information

Physics 2211 M Quiz #2 Solutions Summer 2017

Physics 2211 M Quiz #2 Solutions Summer 2017 Physics 2211 M Quiz #2 Solutions Summer 2017 I. (16 points) A block with mass m = 10.0 kg is on a plane inclined θ = 30.0 to the horizontal, as shown. A balloon is attached to the block to exert a constant

More information

Do not fill out the information below until instructed to do so! Name: Signature: Student ID: Section Number:

Do not fill out the information below until instructed to do so! Name: Signature: Student ID:   Section Number: Do not fill out the information below until instructed to do so! Name: Signature: Student ID: E-mail: Section Number: Formulae are provided on the last page. You may NOT use any other formula sheet. You

More information

11th Grade. Review for General Exam-3. decreases. smaller than. remains the same

11th Grade. Review for General Exam-3. decreases. smaller than. remains the same 1. An object is thrown horizontally with a speed of v from point M and hits point E on the vertical wall after t seconds as shown in the figure. (Ignore air friction.). Two objects M and S are thrown as

More information

Force. The cause of an acceleration or change in an object s motion. Any kind of a push or pull on an object.

Force. The cause of an acceleration or change in an object s motion. Any kind of a push or pull on an object. Force The cause of an acceleration or change in an object s motion. Any kind of a push or pull on an object. Forces do not always give rise to motion. Forces can be equal and opposite. Force is a vector

More information

PHY2020 Test 2 November 5, Name:

PHY2020 Test 2 November 5, Name: 1 PHY2020 Test 2 November 5, 2014 Name: sin(30) = 1/2 cos(30) = 3/2 tan(30) = 3/3 sin(60) = 3/2 cos(60) = 1/2 tan(60) = 3 sin(45) = cos(45) = 2/2 tan(45) = 1 sin(37) = cos(53) = 0.6 cos(37) = sin(53) =

More information

The net force on a moving object is suddenly reduced to zero. As a consequence, the object

The net force on a moving object is suddenly reduced to zero. As a consequence, the object The net force on a moving object is suddenly reduced to zero. As a consequence, the object (A) stops abruptly (B) stops during a short time interval (C) changes direction (D) continues at a constant velocity

More information

Class XI Chapter 7- System of Particles and Rotational Motion Physics

Class XI Chapter 7- System of Particles and Rotational Motion Physics Page 178 Question 7.1: Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie

More information

St. Joseph s Anglo-Chinese School

St. Joseph s Anglo-Chinese School Time allowed:.5 hours Take g = 0 ms - if necessary. St. Joseph s Anglo-Chinese School 008 009 First Term Examination Form 6 ASL Physics Section A (40%) Answer ALL questions in this section. Write your

More information

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics Physics 111.6 MIDTERM TEST #2 November 15, 2001 Time: 90 minutes NAME: STUDENT NO.: (Last) Please Print (Given) LECTURE SECTION

More information

Force, Energy & Periodic Motion. Preparation for unit test

Force, Energy & Periodic Motion. Preparation for unit test Force, Energy & Periodic Motion Preparation for unit test Summary of assessment standards (Unit assessment standard only) In the unit test you can expect to be asked at least one question on each sub-skill.

More information

AP Mechanics Summer Assignment

AP Mechanics Summer Assignment 2012-2013 AP Mechanics Summer Assignment To be completed in summer Submit for grade in September Name: Date: Equations: Kinematics (For #1 and #2 questions: use following equations only. Need to show derivation

More information

Solution to Problem. Part A. x m. x o = 0, y o = 0, t = 0. Part B m m. range

Solution to Problem. Part A. x m. x o = 0, y o = 0, t = 0. Part B m m. range PRACTICE PROBLEMS: Final Exam, December 4 Monday, GYM, 6 to 9 PM Problem A Physics Professor did a daredevil stunt in his spare time. In the figure below he tries to cross a river from a 53 ramp at an

More information

Physics 23 Exam 2 March 3, 2009

Physics 23 Exam 2 March 3, 2009 Use the following to answer question 1: A stationary 4-kg shell explodes into three pieces. Two of the fragments have a mass of 1 kg each and move along the paths shown with a speed of 10 m/s. The third

More information

Lecture 10. Example: Friction and Motion

Lecture 10. Example: Friction and Motion Lecture 10 Goals: Exploit Newton s 3 rd Law in problems with friction Employ Newton s Laws in 2D problems with circular motion Assignment: HW5, (Chapter 7, due 2/24, Wednesday) For Tuesday: Finish reading

More information

Broward County Schools AP Physics 1 Review

Broward County Schools AP Physics 1 Review Broward County Schools AP Physics 1 Review 1 AP Physics 1 Review 1. The Basics of the Exam Important info: No penalty for guessing. Eliminate one or two choices and then take a shot. Multi-select questions

More information

AAPT UNITED STATES PHYSICS TEAM AIP 2017

AAPT UNITED STATES PHYSICS TEAM AIP 2017 2017 F = ma Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 2017 2017 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Use g = 10 N/kg throughout this

More information

Name SOLUTION Student ID Score Speed of blocks is is decreasing. Part III. [25 points] Two blocks move on a frictionless

Name SOLUTION Student ID Score Speed of blocks is is decreasing. Part III. [25 points] Two blocks move on a frictionless Name SOLUTION Student ID Score last first Speed of blocks is is decreasing. Part III. [25 points] Two blocks move on a frictionless v o incline with initial speed v o, as shown, while a hand pushes with

More information

Chapter Work, Energy and Power. Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Ans: (a)

Chapter Work, Energy and Power. Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Ans: (a) Chapter Work, Energy and Power Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Q2. A bullet of mass 10g leaves a rifle at an initial velocity of

More information

Version A (01) Question. Points

Version A (01) Question. Points Question Version A (01) Version B (02) 1 a a 3 2 a a 3 3 b a 3 4 a a 3 5 b b 3 6 b b 3 7 b b 3 8 a b 3 9 a a 3 10 b b 3 11 b b 8 12 e e 8 13 a a 4 14 c c 8 15 c c 8 16 a a 4 17 d d 8 18 d d 8 19 a a 4

More information

2. What would happen to his acceleration if his speed were half? Energy The ability to do work

2. What would happen to his acceleration if his speed were half? Energy The ability to do work 1. A 40 kilogram boy is traveling around a carousel with radius 0.5 meters at a constant speed of 1.7 meters per second. Calculate his centripetal acceleration. 2. What would happen to his acceleration

More information

1. (P2.1A) The picture below shows a ball rolling along a table at 1 second time intervals. What is the object s average velocity after 6 seconds?

1. (P2.1A) The picture below shows a ball rolling along a table at 1 second time intervals. What is the object s average velocity after 6 seconds? PHYSICS FINAL EXAM REVIEW FIRST SEMESTER (01/2017) UNIT 1 Motion P2.1 A Calculate the average speed of an object using the change of position and elapsed time. P2.1B Represent the velocities for linear

More information

AAPT UNITED STATES PHYSICS TEAM AIP 2016

AAPT UNITED STATES PHYSICS TEAM AIP 2016 216 F = ma Exam 1 AAPT UNITED STATES PHYSICS TEAM AIP 216 216 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Use g = 1 N/kg throughout this contest.

More information

Show all work in answering the following questions. Partial credit may be given for problems involving calculations.

Show all work in answering the following questions. Partial credit may be given for problems involving calculations. Physics 3210, Spring 2018 Final Exam Name: Signature: UID: Please read the following before continuing: Show all work in answering the following questions. Partial credit may be given for problems involving

More information

Mechanics II. Which of the following relations among the forces W, k, N, and F must be true?

Mechanics II. Which of the following relations among the forces W, k, N, and F must be true? Mechanics II 1. By applying a force F on a block, a person pulls a block along a rough surface at constant velocity v (see Figure below; directions, but not necessarily magnitudes, are indicated). Which

More information

Chapter 10. Rotation

Chapter 10. Rotation Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised

More information

Department of Physics

Department of Physics Department of Physics PHYS101-051 FINAL EXAM Test Code: 100 Tuesday, 4 January 006 in Building 54 Exam Duration: 3 hrs (from 1:30pm to 3:30pm) Name: Student Number: Section Number: Page 1 1. A car starts

More information

PHYSICS 111 SPRING EXAM 2: March 7, 2017; 8:15-9:45 pm

PHYSICS 111 SPRING EXAM 2: March 7, 2017; 8:15-9:45 pm PHYSICS 111 SPRING 017 EXAM : March 7, 017; 8:15-9:45 pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 0 multiple-choice questions plus 1 extra credit question, each

More information

Friction is always opposite to the direction of motion.

Friction is always opposite to the direction of motion. 6. Forces and Motion-II Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:

More information

P211 Spring 2004 Form A

P211 Spring 2004 Form A 1. A 2 kg block A traveling with a speed of 5 m/s as shown collides with a stationary 4 kg block B. After the collision, A is observed to travel at right angles with respect to the initial direction with

More information

Newton s Gravitational Law

Newton s Gravitational Law 1 Newton s Gravitational Law Gravity exists because bodies have masses. Newton s Gravitational Law states that the force of attraction between two point masses is directly proportional to the product of

More information

Forces Part 1: Newton s Laws

Forces Part 1: Newton s Laws Forces Part 1: Newton s Laws Last modified: 13/12/2017 Forces Introduction Inertia & Newton s First Law Mass & Momentum Change in Momentum & Force Newton s Second Law Example 1 Newton s Third Law Common

More information

Name & Surname:... No:... Class: 11 /...

Name & Surname:... No:... Class: 11 /... METU D. F. HIGH SCHOOL 2017-2018 ACADEMIC YEAR, 1 st SEMESTER GRADE 11 / PHYSICS REVIEW FOR GENERAL EXAM-3 UNIFORMLY ACCELERATED MOTION IN TWO DIMENSIONS, ENERGY, IMPULSE & MOMENTUM & TORQUE DECEMBER 2017

More information

Practice Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question.

Practice Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question. Practice Test 3 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During

More information

Write your name legibly on the top right hand corner of this paper

Write your name legibly on the top right hand corner of this paper NAME Phys 631 Summer 2007 Quiz 2 Tuesday July 24, 2007 Instructor R. A. Lindgren 9:00 am 12:00 am Write your name legibly on the top right hand corner of this paper No Books or Notes allowed Calculator

More information

On my honor, I have neither given nor received unauthorized aid on this examination.

On my honor, I have neither given nor received unauthorized aid on this examination. Instructor(s): Profs. D. Reitze, H. Chan PHYSICS DEPARTMENT PHY 2053 Exam 2 April 2, 2009 Name (print, last first): Signature: On my honor, I have neither given nor received unauthorized aid on this examination.

More information

Extra Circular Motion Questions

Extra Circular Motion Questions Extra Circular Motion Questions Elissa is at an amusement park and is driving a go-cart around a challenging track. Not being the best driver in the world, Elissa spends the first 10 minutes of her go-cart

More information

Static Equilibrium, Gravitation, Periodic Motion

Static Equilibrium, Gravitation, Periodic Motion This test covers static equilibrium, universal gravitation, and simple harmonic motion, with some problems requiring a knowledge of basic calculus. Part I. Multiple Choice 1. 60 A B 10 kg A mass of 10

More information

Physics 2210 Homework 18 Spring 2015

Physics 2210 Homework 18 Spring 2015 Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle

More information

Final Exam April 26, 2016

Final Exam April 26, 2016 PHYS 050 Spring 016 Name: Final Exam April 6, 016 INSTRUCTIONS: a) No books or notes are permitted. b) You may use a calculator. c) You must solve all problems beginning with the equations on the Information

More information

SAPTARSHI CLASSES PVT. LTD.

SAPTARSHI CLASSES PVT. LTD. SAPTARSHI CLASSES PVT. LTD. NEET/JEE Date : 13/05/2017 TEST ID: 120517 Time : 02:00:00 Hrs. PHYSICS, Chem Marks : 360 Phy : Circular Motion, Gravitation, Che : Halogen Derivatives Of Alkanes Single Correct

More information

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.

More information

PHYSICS. Chapter 8 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.

PHYSICS. Chapter 8 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc. PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 8 Lecture RANDALL D. KNIGHT Chapter 8. Dynamics II: Motion in a Plane IN THIS CHAPTER, you will learn to solve problems about motion

More information

Phys 270 Final Exam. Figure 1: Question 1

Phys 270 Final Exam. Figure 1: Question 1 Phys 270 Final Exam Time limit: 120 minutes Each question worths 10 points. Constants: g = 9.8m/s 2, G = 6.67 10 11 Nm 2 kg 2. 1. (a) Figure 1 shows an object with moment of inertia I and mass m oscillating

More information

Exam II Difficult Problems

Exam II Difficult Problems Exam II Difficult Problems Exam II Difficult Problems 90 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Two boxes are connected to each other as shown. The system is released

More information

Motion. Argument: (i) Forces are needed to keep things moving, because they stop when the forces are taken away (evidence horse pulling a carriage).

Motion. Argument: (i) Forces are needed to keep things moving, because they stop when the forces are taken away (evidence horse pulling a carriage). 1 Motion Aristotle s Study Aristotle s Law of Motion This law of motion was based on false assumptions. He believed that an object moved only if something was pushing it. His arguments were based on everyday

More information

4) Vector = and vector = What is vector = +? A) B) C) D) E)

4) Vector = and vector = What is vector = +? A) B) C) D) E) 1) Suppose that an object is moving with constant nonzero acceleration. Which of the following is an accurate statement concerning its motion? A) In equal times its speed changes by equal amounts. B) In

More information

Physics 6A Winter 2006 FINAL

Physics 6A Winter 2006 FINAL Physics 6A Winter 2006 FINAL The test has 16 multiple choice questions and 3 problems. Scoring: Question 1-16 Problem 1 Problem 2 Problem 3 55 points total 20 points 15 points 10 points Enter the solution

More information

2010 F=ma Solutions. that is

2010 F=ma Solutions. that is 2010 F=ma Solutions 1. The slope of a position vs time graph gives the velocity of the object So you can see that the position from B to D gives the steepest slope, so the speed is the greatest in that

More information

PRACTICE TEST for Midterm Exam

PRACTICE TEST for Midterm Exam South Pasadena AP Physics PRACTICE TEST for Midterm Exam FORMULAS Name Period Date / / d = vt d = v o t + ½ at 2 d = v o + v 2 t v = v o + at v 2 = v 2 o + 2ad v = v x 2 + v y 2 = tan 1 v y v v x = v cos

More information

1. A train moves at a constant velocity of 90 km/h. How far will it move in 0.25 h? A. 10 km B km C. 25 km D. 45 km E. 50 km

1. A train moves at a constant velocity of 90 km/h. How far will it move in 0.25 h? A. 10 km B km C. 25 km D. 45 km E. 50 km Name: Physics I Mid Term Exam Review Multiple Choice Questions Date: Mr. Tiesler 1. A train moves at a constant velocity of 90 km/h. How far will it move in 0.25 h? A. 10 km B. 22.5 km C. 25 km D. 45 km

More information

General Physics (PHY 2130)

General Physics (PHY 2130) General Physics (PHY 130) Lecture 0 Rotational dynamics equilibrium nd Newton s Law for rotational motion rolling Exam II review http://www.physics.wayne.edu/~apetrov/phy130/ Lightning Review Last lecture:

More information

Announcements Oct 27, 2009

Announcements Oct 27, 2009 Announcements Oct 7, 009 1. HW 14 due tonight. Reminder: some of your HW answers will need to be written in scientific notation. Do this with e notation, not with x signs. a. 6.57E33 correct format b.

More information

PHYSICS 221 SPRING EXAM 2: March 31, 2016; 8:15pm 10:15pm

PHYSICS 221 SPRING EXAM 2: March 31, 2016; 8:15pm 10:15pm PHYSICS 221 SPRING 2016 EXAM 2: March 31, 2016; 8:15pm 10:15pm Name (printed): Recitation Instructor: Section # Student ID# INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit

More information

December 2015 Exam Review July :39 AM. Here are solutions to the December 2014 final exam.

December 2015 Exam Review July :39 AM. Here are solutions to the December 2014 final exam. December 2015 Exam Review July-15-14 10:39 AM Here are solutions to the December 2014 final exam. 1. [5 marks] A soccer ball is kicked from the ground so that it is projected at an initial angle of 39

More information

Today s lecture. WEST VIRGINIA UNIVERSITY Physics

Today s lecture. WEST VIRGINIA UNIVERSITY Physics Today s lecture Review of chapters 1-14 Note: I m taking for granted that you ll still know SI/cgs units, order-of-magnitude estimates, etc., so I m focusing on problems. Velocity and acceleration (1d)

More information

Practice Problems for Exam 2 Solutions

Practice Problems for Exam 2 Solutions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 008 Practice Problems for Exam Solutions Part I Concept Questions: Circle your answer. 1) A spring-loaded toy dart gun

More information

Physics 101 Fall 2006: Final Exam Free Response and Instructions

Physics 101 Fall 2006: Final Exam Free Response and Instructions Last Name: First Name: Physics 101 Fall 2006: Final Exam Free Response and Instructions Print your LAST and FIRST name on the front of your blue book, on this question sheet, the multiplechoice question

More information

Written Homework problems. Spring (taken from Giancoli, 4 th edition)

Written Homework problems. Spring (taken from Giancoli, 4 th edition) Written Homework problems. Spring 014. (taken from Giancoli, 4 th edition) HW1. Ch1. 19, 47 19. Determine the conversion factor between (a) km / h and mi / h, (b) m / s and ft / s, and (c) km / h and m

More information

Physics 121, Sections 1 and 2, Winter 2011 Instructor: Scott Bergeson Exam #3 April 16 April 21, 2011 RULES FOR THIS TEST:

Physics 121, Sections 1 and 2, Winter 2011 Instructor: Scott Bergeson Exam #3 April 16 April 21, 2011 RULES FOR THIS TEST: Physics 121, Sections 1 and 2, Winter 2011 Instructor: Scott Bergeson Exam #3 April 16 April 21, 2011 RULES FOR THIS TEST: This test is closed book. You may use a dictionary. You may use your own calculator

More information