2. 9/2/2015 Physics 201, Spring 20111 Physics 201: Review Final Exam: Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272 The exam will cover chapters 1 – 14 The exam will have about 30 multiple choice questions Consultations hours the same as before. Another review sessions will be held by your TA’s at the discussion session

3. 9/2/2015 Physics 201, Spring 20112 Problem Solving l Read and understand the problem statement completely: çOften this is helped by a diagram showing the relationships of the objects. çBe sure you understand what is wanted  Be sure you understand what information is available to you (or can be found from the available information) l Translate the situation described to physics concepts. çBe alert for clues regarding the choice of relationships (e.g. conservation of energy, conservation of momentum, rotational or linear motion, …….) çBe alert for detail that would qualify the use of some concepts (e.g. friction affecting conservation of energy.) çIf there are other “unknowns” involved how can you find them (or eliminate them).

4. 9/2/2015 Physics 201, Spring 20113 Problem solving….. l After choosing the appropriate relationship between the concepts (equation), find the target quantities -- at first algebraically, then substitute numbers at the end. l Check çDoes the answer make sense? çAre the units consistent? çOften with multiple choice questions you can round the numerical quantities and check the final choice without the calculator. l Techniques and Hints çBe clear and organized -- neat in your solution.You must be able to read and understand your own notes! çGo through the exam completely at first and complete those questions that you are confident in solving. Then return to the others.

5. Kinematics: 9/2/2015 Physics 201, Spring 20114 Chapters 2, 3 (linear) 9 (rotational)

6. Dynamics: 9/2/2015 Physics 201, Spring 20115 Chapters 4, 5, (the 2 nd law) 6, 7, (energy and work) 10, 11 (rotational, gravity)

7. No net force, No net torque: 9/2/2015 Physics 201, Spring 20116 Chapters 8 (linear: consequence of the 2 nd law) 10 (rotational)

8. Statics: Stationary balance 9/2/2015 Physics 201, Spring 20117 Chapters 12 (static equilibrium, elasticity) Fluids Archimedes’ Principle : A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid.

9. Oscillations: 9/2/2015 Physics 201, Spring 20118 Resonance frequency:

10. 9/2/2015 Physics 201, Spring 20119 Question (Chapt 2) An European sports car dealer claims that his product will accelerate at a constant rate from rest to a speed of 100 km/hr in 8s. What is the speed after first 5 s of acceleration? 17.4 m/s 53.2 m/s 44.4 m/s 34.7 m/s 28.7 m/s

11. 9/2/2015 Physics 201, Spring 201110 Two gliders of unequal mass m A <m B are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right. Let F hA represent the magnitude of the force of the hand on the glider A. Let F BA represent the magnitude of the force exerted by the glider A on the glider B. Which one of the following is true? F hA < F BA F hA = F BA F hA > F BA Question (Chapt 4) Newton’s Second Law: Net external, F hA -F BA, is causing block A to accelerate to the right. F BA = F AB < F hA

12. 9/2/2015 Physics 201, Spring 201111 Two gliders of unequal mass m A <m B are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right. Let F hA represent the magnitude of the force of the hand on the glider A. Let F BA represent the magnitude of the force exerted by the glider A on the glider B. Which one of the following is true? F BA < F AB F BA = F AB F BA > F AB Question, Continued Newton’s Third Law

13. 9/2/2015 Physics 201, Spring 201112 Two gliders of unequal mass m A <m B are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right. How does the net force on glider B (F B ) compare to the magnitude of the net force on glider A (F A )? F B < F A F B = F A F B > F A Question, Continued

14. 12/10/07 Physics 103, Fall 2007, U. Wisconsin13 Question (Chapt. 6) How much power is needed to lift a 75-kg student vertically upward at a constant speed of 0.33 m/s? 25 W 12.5 W 243 W 115 W 230 W

15. 12/10/07 Physics 103, Fall 2007, U. Wisconsin14 A block is placed on a planar sheet that is pivoted at one end. The free side of the sheet is then raised very slowly, as shown. When the sheet is first raised, friction between block and sheet keeps the block from moving. At a certain angle, however, the block begins to slide down the inclined sheet. If the sheet is kept at this angle, will the acceleration of the block be zero, constant, or neither? Friction

16. 9/2/2015 Physics 201, Spring 201115 A moving object collides with an object initially at rest. Question (Chapt 8) Is it possible for both objects to be at rest after the collision? çYes çNo Can one of them be at rest after the collision? çYes çNo If the objects stick together after the collision, is the kinetic energy conserved? çYes çNo

17. Elastic collision (Q4) l A block of mass m moving at to the right with speed v hits a block of mass M that is at rest. If the surface is frictionless and the collision is elastic, what are the final velocities of the two blocks? 9/2/2015 Physics 201, Spring 201116 m M v

18. Elastic collision l A block of mass m moving at to the right with speed v hits a block of mass M that is at rest. If the surface is frictionless and the collision is elastic, what are the final velocities of the two blocks? In the center of mass frame, velocities reverse after an elastic collision 9/2/2015 Physics 201, Spring 201117 m M v m M v-v CM -v CM m M -(v- v CM ) v CM v CM = mv/(m+M)

19. Elastic collision Now find velocity of each block in lab frame: Velocity of m = v CM - (v-v CM ) = 2v CM – v = (m-M)v/(m+M) Velocity of M = 2v CM = 2mv/(m+M) 9/2/2015 Physics 201, Spring 201118 m M v v CM = mv/(m+M)

20. 9/2/2015 Physics 201, Spring 201119 A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the string? The tension increases to four times its original value. The tension increases to two times its original value. The tension is unchanged. The tension reduces to one half its original value. The tension reduces to one fourth its original value. Question (Chapt 9)

21. 9/2/2015 Physics 201, Spring 201120 Concepts Is there a net force acting on the system? çYes çNo Yes, the direction of velocity is changing. Centripetal acceleration is provided by the tension in the string. The centripetal acceleration is different in the two cases presented, therefore, the tension will be different Note that the radius has not changed in the two conditions Note also that angular velocity is given (in words).

22. 9/2/2015 Physics 201, Spring 201121 Solution Tension goes up by a factor of 4! Centripetal force for the two situations: Need to write in terms of change to angular velocity because that is what is specified Some of the quantities are not given but we are comparing situations, I.e., take ratios and cancel common factors!

23. 12/10/07 Physics 103, Fall 2007, U. Wisconsin22 The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which case is the torque on the nut the biggest? Case 1 Case 2 Case 3 Question (Chapt. 9)  = F d sin  Longest lever arm, d at 90 o angle

24. 9/2/2015 Physics 201, Spring 201123 Question (Chapt 12) A sign of mass M is hung 1 m from the end of a 4 m long uniform beam of mass m, as shown in the diagram. The beam is hinged at the wall. What is the tension in the guy wire? Determine the tension T, and the contact force F at the hinge. SIGN wire   1 m 3 m

25. 9/2/2015 Physics 201, Spring 201124 What are the concepts involved? Is there a net force acting on the system? çYes çNo Is there a net torque acting on the system? çYes çNo Draw the free body diagram. How many forces are acting on the system? ç2 ç3 ç4 ç5 What is the direction of the contact force at the hinge between the wall and the beam ? çVertical çHorizontal çIt has both vertical and horizontal components SIGN wire   1 m mg, Mg, tension, force from hinge

26. 9/2/2015 Physics 201, Spring 201125 Solution mg Mg T FyFy FxFx 30 0 Hint: Choose axis of rotation at support because F x & F y are not known 2m 3m Forces Torques

27. 12/10/07 Physics 103, Fall 2007, U. Wisconsin26 A rock is thrown straight up from the Earth’s surface. Which one of the following statements concerning the net force acting on the rock at the top of its path is true? It is equal to zero for an instant. It is equal to the force used to throw it up but in opposite direction It is equal to the weight of the rock Its direction changes from up to down Its magnitude is equal to the sum of the force used to throw it up and its weight Motion in Gravity (Chapt. 11)

28. 9/2/2015 Physics 201, Spring 201127 Question (Chapt 12) A mass of 100 tons (10 5 kg) is lifted on a steel rod two cm in diameter and 10 m in length. (Young’s modulus of steel is 210 10 9 N/m 2 ) (a) How long does the rod stretch?

29. 9/2/2015 Physics 201, Spring 201128 A mass of 100 tons (10 5 kg) is lifted on a steel rod two cm in diameter and 10 m in length. (Young’s modulus of steel is 210 10 9 N/m 2 ) (a) How long does the rod stretch? F = force A = area of rod L = length of rod ΔL = change of length of the rod

30. 9/2/2015 Physics 201, Spring 201129 1)The pressure on the roof of a tall building is 0.985 × 10 5 Pa and the pressure on the ground is 1.000 × 10 5 Pa. The density of air is 1.29 kg/m 3. What is the height of the building? A.100 m B.118 m C.135 m D.114 m E.None of the above Question (Chapt 13)

31. 9/2/2015 Physics 201, Spring 201130 Question (Chapt 13) A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 m/s in the 2.0-cm section? (fuel density = 1.2 kg/m 3.) ? Velocity is faster in constricted section because mass flow is conserved (mass that flows into constriction must also flow out). Pressure drops because of Bernoulli principle: (applies to incompressible, frictionless fluid)

32. 9/2/2015 Physics 201, Spring 201131 Calculate velocity in constriction Volume flow rate: ΔV/Δt = A Δx/Δt = Av (m 3 /s) Continuity: A 1 v 1 = A 2 v 2 i.e., mass that flows in must then flow out Fluid flow without friction

33. 9/2/2015 Physics 201, Spring 201132 Question, continued A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 m/s in the 2.0-cm section? (fuel density = 1.2 kg/m 3.) ? 70 Pa 85 Pa 100 Pa 115 Pa 81 Pa

34. 9/2/2015 Physics 201, Spring 201133 Question (Chapt 13) The water level in identical bowls, A and B, is exactly the same. A contains only water; B contains ice as well as water. When we weigh the bowls, we find that W A < W B W A = W B W A > W B W A < W B if the volume of the ice cubes is greater than one-ninths the volume of the water. Eureka! Archimedes Principle. Weight of the water displaced = Bouyant Force

35. 9/2/2015 Physics 201, Spring 201134 1)A block of aluminum (density 3041 kg/m 3 ) is lifted very slowly but at constant speed from the bottom of a tank filled with water. If it is a cube 20 cm on each side, the tension in the cord is: A.160 N B. 4 N C. 80 N D. 8 N E.None of the above T FbFb W Question (Chapt 13)

36. 9/2/2015 Physics 201, Spring 201135 Question (Chapt 13) A wind with velocity 10 m/s is blowing through a wind generator with blade radius 5.0 meters. What is the maximum power output if 30% of the wind’s energy can be extracted? (air density = 1.25 kg/m 3.) 7.2 kW 14.7 kW 21.3 kW 29.4 kW 39.6 kW

37. 9/2/2015 Physics 201, Spring 201136 Firemen connect a hose (8 cm in diameter) to a fire hydrant. When the nozzle is open, the pressure in the hose is 2.35 atm. (1 atm. = 10 5 Pa). The firemen hold the nozzle at the same height of the hydrant and at 45 o to the horizontal. The stream of water just barely reaches a window 10 m above them. The diameter of the nozzle is about: A.8 cm B.6 cm C.4 cm D.2 cm E.None of the above 10m Point 1 Point 2 Point 3 Question (Chapt 13)

38. 9/2/2015 Physics 201, Spring 201137 Question (Chapt 14) At t=0, a 795-g mass at rest on the end of a horizontal spring (k=127 N/m) is struck by a hammer, giving it an initial speed of 2.76 m/s. The position of the mass is described by, with What is period of the motion? period = 2π/ω What is the frequency of the motion? What is the maximum acceleration? What is the total energy? 0.497 s 2.01 Hz 34.9 m/s 2 3.03 J

39. 9/2/2015 Physics 201, Spring 201138 Question (Chapt 14) The amplitude of a system moving with simple harmonic motion is doubled. The total energy will then be 4 times larger 2 times larger the same as it was half as much quarter as much