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Physics 201: Review

Prepare for Physics 201 final exam with strategies, hints, and key concepts for solving problems effectively. Covering chapters 1-14.

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Physics 201: Review

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  1. 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 Physics 201, Spring 2011

  2. Problem Solving • Read and understand the problem statement completely: • Often this is helped by a diagram showing the relationships of the objects. • Be sure you understand what is wanted • Be sure you understand what information is available to you (or can be found from the available information) • Translate the situation described to physics concepts. • Be alert for clues regarding the choice of relationships (e.g. conservation of energy, conservation of momentum, rotational or linear motion, …….) • Be alert for detail that would qualify the use of some concepts (e.g. friction affecting conservation of energy.) • If there are other “unknowns” involved how can you find them (or eliminate them). Physics 201, Spring 2011

  3. Problem solving….. • After choosing the appropriate relationship between the concepts (equation), find the target quantities -- at first algebraically, then substitute numbers at the end. • Check • Does the answer make sense? • Are the units consistent? • Often with multiple choice questions you can round the numerical quantities and check the final choice without the calculator. • Techniques and Hints • Be clear and organized -- neat in your solution.You must be able to read and understand your own notes! • Go through the exam completely at first and complete those questions that you are confident in solving. Then return to the others. Physics 201, Spring 2011

  4. Kinematics: Chapters 2, 3 (linear) 9 (rotational) Physics 201, Spring 2011

  5. Dynamics: Chapters 4, 5, (the 2nd law) 6, 7, (energy and work) 10, 11 (rotational, gravity) Physics 201, Spring 2011

  6. No net force, No net torque: Chapters 8 (linear: consequence of the 2nd law) 10 (rotational) Physics 201, Spring 2011

  7. Statics: Stationary balance 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. Physics 201, Spring 2011

  8. Oscillations: Resonance frequency: Physics 201, Spring 2011

  9. 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 Physics 201, Spring 2011

  10. Question (Chapt 4) Two gliders of unequal mass mA<mB are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right. Let FhA represent the magnitude of the force of the hand on the glider A. Let FBA represent the magnitude of the force exerted by the glider A on the glider B. Which one of the following is true?      FhA < FBA FhA = FBA FhA > FBA Newton’s Second Law: Net external, FhA-FBA, is causing block A to accelerate to the right. FBA = FAB < FhA Physics 201, Spring 2011

  11. Question , Continued Two gliders of unequal mass mA<mB are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right. Let FhA represent the magnitude of the force of the hand on the glider A. Let FBA represent the magnitude of the force exerted by the glider A on the glider B. Which one of the following is true?      FBA < FAB FBA = FAB FBA > FAB Newton’s Third Law Physics 201, Spring 2011

  12. Question, Continued Two gliders of unequal mass mA<mB 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 (FB) compare to the magnitude of the net force on glider A (FA)?      FB < FA FB = FA FB > FA Physics 201, Spring 2011

  13. 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 Physics 103, Fall 2007, U. Wisconsin

  14. Question (Chapt 8) A moving object collides with an object initially at rest. Is it possible for both objects to be at rest after the collision? • Yes • No Can one of them be at rest after the collision? • Yes • No If the objects stick together after the collision, is the kinetic energy conserved? • Yes • No Physics 201, Spring 2011

  15. Elastic collision (Q4) v • 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? M m Physics 201, Spring 2011

  16. Elastic collision v • 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 M m vCM = mv/(m+M) -vCM v-vCM M m vCM -(v-vCM) M m Physics 201, Spring 2011

  17. Elastic collision v Now find velocity of each block in lab frame: Velocity of m = vCM - (v-vCM) = 2vCM – v = (m-M)v/(m+M) Velocity of M = 2vCM = 2mv/(m+M) M m vCM = mv/(m+M) Physics 201, Spring 2011

  18. Question (Chapt 9) 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. Physics 201, Spring 2011

  19. Concepts Is there a net force acting on the system? • Yes • No 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). Physics 201, Spring 2011

  20. Solution 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! Tension goes up by a factor of 4! Physics 201, Spring 2011

  21. Question (Chapt. 9) 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 • = F d sin θ Longest lever arm, d at 90o angle Physics 103, Fall 2007, U. Wisconsin

  22. wire θ = 30o 1 m SIGN 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. 3 m Physics 201, Spring 2011

  23. wire θ = 30o 1 m SIGN What are the concepts involved? Is there a net force acting on the system? • Yes • No Is there a net torque acting on the system? • Yes • No Draw the free body diagram. How many forces are acting on the system? • 2 • 3 • 4 • 5 mg, Mg, tension, force from hinge What is the direction of the contact force at the hinge between the wall and the beam ? • Vertical • Horizontal • It has both vertical and horizontal components Physics 201, Spring 2011

  24. Fy T 300 Fx mg Mg Solution 3m 2m Forces Torques Hint: Choose axis of rotation at support because Fx & Fy are not known Physics 201, Spring 2011

  25. Motion in Gravity (Chapt. 11) 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  Physics 103, Fall 2007, U. Wisconsin

  26. Question (Chapt 12) A mass of 100 tons (105 kg) is lifted on a steel rod two cm in diameter and 10 m in length. (Young’s modulus of steel is 210 109 N/m2) (a) How long does the rod stretch? Physics 201, Spring 2011

  27. A mass of 100 tons (105 kg) is lifted on a steel rod two cm in diameter and 10 m in length. (Young’s modulus of steel is 210 109 N/m2) (a) How long does the rod stretch? F = force A = area of rod L = length of rod ΔL = change of length of the rod Physics 201, Spring 2011

  28. Question (Chapt 13) • 1) The pressure on the roof of a tall building is 0.985 × 105 Pa and the pressure on the ground is 1.000 × 105 Pa. The density of air is 1.29 kg/m3. What is the height of the building? • A. 100 m • B. 118 m • C. 135 m • 114 m • None of the above Physics 201, Spring 2011

  29. 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/m3.)? 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) Physics 201, Spring 2011

  30. Calculate velocity in constriction Fluid flow without friction • Volume flow rate: ΔV/Δt = A Δx/Δt = Av (m3/s) • Continuity: A1 v1 = A2 v2 • i.e., mass that flows in must then flow out Physics 201, Spring 2011

  31. 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/m3.)? 70 Pa  85 Pa      100 Pa    115 Pa      81 Pa Physics 201, Spring 2011

  32. 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 WA < WB WA = WB WA > WB WA < WBif the volume of the ice cubes is greater than one-ninths the volume of the water. WA < WBif 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 Physics 201, Spring 2011

  33. Question (Chapt 13) Fb W T • 1) A block of aluminum (density 3041 kg/m3) 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 • 8 N • None of the above Physics 201, Spring 2011

  34. 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/m3.) 7.2 kW  14.7 kW      21.3 kW    29.4 kW     39.6 kW Physics 201, Spring 2011

  35. Question (Chapt 13) Point 3 10m Point 2 Point 1 • 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. = 105 Pa). The firemen hold the nozzle at the same height of the hydrant and at 45o 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 • 2 cm • None of the above Physics 201, Spring 2011

  36. 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/s2 3.03 J Physics 201, Spring 2011

  37. 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 Physics 201, Spring 2011

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