PHYSICS 111 SPRING FINAL EXAM: April 30, 2018; 4:30pm - 6:30pm. Name (printed): Recitation Instructor: Section #

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1 PHYSICS 111 SPRING 2018 FINAL EXAM: April 30, 2018; 4:30pm - 6:30pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 30 multiple-choice question, each worth 3 points, for a nominal maximum score of 80 points plus 10 extra credit points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, e-book readers, smart phones) are NOT permitted. Calculators with WiFi technology are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet: Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. You will continue to use the same bubble sheet you already used for Exam 1-3. Bubble answers on the bubble sheet for this exam. Only, if for some reason you are starting a new bubble sheet, write and fill in the bubbles corresponding to: - Your last name, middle initial, and first name. - ««Your ID number (the middle 9 digits on your ISU card) ««- Special codes K to L are your recitation section. Always use two digits (e.g. 01, 09, 11, 13). Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over. Best of luck, Dr. Soeren Prell

2 64. The temperature in your classroom is closest to A) 295 K B) 68 K C) 20 K D) 68 C E) 295 C Room temperature is closest to 295 K, which is about 22 C or 72 F. 65. A large vat contains L of water at 20 C. What volume will this water occupy when it is heated up to 80 C? Water has a volume expansion coefficient of K -1. A) L B) L C) L D) L E) L V 0 + ΔV = V 0 (1+ βδt ) = L ( ( )( 60 K) ) = L ( ) K How much heat is required to increase the temperature of 1.70 moles of Helium gas by 23.0 K at constant pressure? (Treat Helium as an ideal monatomic gas.) A) 391 J B) 812 J C) 487 J D) 346 J E) 751 J Q = nc P ΔT = n 5 2 R ΔT = 1.70 mol 8.31 J/(mol K) ( ) 5 2 ( )( 23.0 K) = 812 J

3 The next two questions pertain to the following situation: An iron mass initially at 0.0 o C is dropped into a large container of liquid nitrogen at 77 K, causing gaseous nitrogen to emerge. After the iron has cooled to 77 K, the iron has transferred 22.0 kj of heat to the nitrogen. The specific heat of iron is 0.45 kj/(kg K) and the latent heat of vaporization of iron is 6340 kj/kg. The latent heat of vaporization of nitrogen is 200 kj/kg. 67. What mass did the iron have? A) 0.74 kg B) 2.3 kg C) 9.1 kg D) 1.6 kg E) 0.25 kg Q = mcδt m = Q cδt = ( 22.0 kj) 0.45 kj/ kg K ( ( ))( 273 K 77 K) = 0.25 kg 68. What total mass of nitrogen was vaporized? A) 0.34 kg B) 0.24 kg C) 0.81 kg D) 0.67 kg E) 0.11 kg Q = ml m = Q L = ( 22.0 kj) = 0.11 kg 200 kj/kg ( )

4 69. The rms speed of a certain sample of carbon dioxide molecules, with a molecular weight of 44 g/mol, is 396 m/s. What is the rms speed of water molecules, with a molecular weight of 18 g/mol, at the same temperature? A) 387 m/s B) 506 m/s C) 253 m/s D) 619 m/s E) 421 m/s v rms = 3kT m = 3RT M 1 M v rms ( H 2 O) = v rms ( CO 2 ) ( ) ( ) M CO 2 M H 2 O = ( 396 m/s) = 619 m/s 70. An ideal gas has a pressure of 2.5 atm, a volume of 1.0 L at a temperature of 30 C. How many molecules are there in this gas? A) B) C) D) E) PV = nrt = NkT N = PV kt = ( 2.5 atm) 1 L J/K ( ) 273 K + 30 K ( )( 10 3 m ) 3 ( ) 273 K + 30 K ( ) ( ) = Pa J/K ( ) = What is the name of an ideal-gas process in which no heat is transferred? A) reversible B) isobaric C) isochoric D) isothermal E) adiabatic A thermal process of an ideal gas in which no heat is transferred is called adiabatic.

5 72. An ideal gas in a heat engine executes the cycle (ABCD) shown in the pressure versus volume graph. The temperature of the gas is maximum p B C A) along the line BC B) at point B C) at point C D) at point D E) along the line CD The temperature is maximum at point C. For an ideal gas, PV = nrt, so T is max, when PV is max, which corresponds to the point furthest from the origin. 73. The figure shows a PV diagram for a cycle of a heat engine for which QH = 59 J. What is the thermal efficiency of the engine? A) 17% B) 8.5% C) 14% D) 34% E) 83% A D V e = W ( area in PV diagram) = = Q H Q H ( ) 200 kpa cm 59 J ( ) 74. The graph in the figure shows the position of a particle as it travels along the x-axis. At what value of t is the speed of the particle equal to 0 m/s? A) 0.0 s B) 1.0 s C) 2.0 s D) 3.0 s E) 4.0 s = 0.17 The velocity (and speed) is equal to the slope of position versus time graph.

6 75. To determine the height of a bridge above the water, a person drops a stone and measures the time it takes for it to hit the water. If the time is 2.3 s, what is the height of the bridge? Neglect air resistance. A) 26 m B) 14 m C) 10 m D) 32 m E) 52 m Δy = 1 2 g Δt ( ) 2 Δt = 1 2 ( 9.8 m/s2 ) 2.3 s ( ) 2 = 26 m 76. When Jeff ran up a hill with a speed of 7.0 m/s, the horizontal component of his velocity vector was 5.1 m/s. What was the vertical component of Jeff's velocity? A) 3.8 m/s B) 3.4 m/s C) 4.3 m/s D) 4.8 m/s E) Not enough information to solve. v = v x 2 + v y 2 v y = v 2 v x 2 = 4.8 m/s 77. A block of mass 2.36 kg slides down a frictionless inclined ramp and experiences no significant air resistance. If the ramp angle is 17.0 above the horizontal and the length of the surface of the ramp is 20.0 m, find the speed of the block as it reaches the bottom of the ramp, assuming it started sliding from rest at the top. A) 7.57 m/s B) 10.7 m/s C) 114 m/s D) 19.6 m/s E) 5.85 m/s h = ( 20.0 m)sin 17 o ( ) = 5.85 m mgh = 1 2 mv2 v = 2gh = 10.7 m

7 78. Two blocks, A and B, are being pulled to the right along a horizontal surface by a horizontal 100-N pull, as shown in the figure. Both of them are moving together at a constant velocity of 2.0 m/s to the right, and both weigh the same. Which of the figures below shows a correct free-body diagram of the horizontal forces acting on the upper block, A? A) B) C) D) E) Box A is moving with constant velocity and therefore the net force on box A is zero. Box B can exert at most one horizontal force on box A, either to the left or to the right. In both cases box A would experience a net force. Thus, no horizontal force is exerted on box A. 79. A 10,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 14,000-kg boxcar. What is the speed of the cars just after the collision? A) Zero. B) 2.10 m/s C) 1.07 m/s D) m/s E) m/s m 1 v 1 = m 1 + m 2 ( )v v = m 1 v 1 m 1 + m 2 = m/s

8 80. In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have so they do not fall if the coefficient of static friction between the passengers and the wall is 0.25? A) 3.0 m/s B) 4.9 m/s C) 8.9 m/s D) 2.3 m/s E) It depends on the mass of the passengers. f s = mg = µ s N; N = mv2 r v = Nr m = gr µ s = 8.9 m/s 81. A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. If the top piece is taken off and removed, what happens to the water level in the tub? A) It does not change. B) It goes down. C) It goes up. D) It cannot be determined without knowing the masses of the two pieces of wood. E) It cannot be determined without knowing the densities of the two pieces of wood. The buoyant force on the two blocks is equal to the weight of the displaced water and equal to the combined weight of the blocks. After the top block is removed, the buoyant force on the single block is equal to the weight of the displaced water and to the weight of the single block. The single block weighs less than the two block together. Therefore, after the top block is removed less water is displaced and the water level in the bathtub drops.

9 82. The figure shows a graph of the position x as a function of time t for a system undergoing simple harmonic motion. Which one of the following graphs represents the velocity of this system as a function of time? A) graph a B) graph b C) graph c D) graph d E) None of the graphs. At t = 0 the system has maximum speed (largest slope of the x versus t curve) and positive velocity (x is increasing).

10 83. Consider a pipe of length L that is open at one end and closed at the other end. What are the wavelengths of the three lowest-pitch tones produced by this pipe? A) 2L, L, L/2 B) 4L, 2L, L C) 2L, L, 2 L/3 D) 4L, 4L/3, 4L/5 E) L, L/3, L/5 The wavelength corresponding to the fundamental (lowest-pitch) frequency of a pipe with one open and one closed end is 4L. The next possible shorter wavelength is 4L/3. The general equation for the wavelength of standing waves in a pipe with one open and one closed end is λ = 4L/n (n = 1, 3, 5, ). 84. The mobile shown in the figure is perfectly balanced, and the horizontal supports have insignificant masses. The mass of fish D is 8.00 grams. What must be the mass of fish B to maintain balance? A) 46.0 g B) 64.0 g C) 82.0 g D) 98.0 g E) 108 g ( 17.5 cm)m D = ( 5.00 cm)m C m C = ( 17.5 cm)m D 5.00 cm ( 15.0 cm) ( m D + m C ) = ( 5.00 cm)m B m B = ( 17.5 cm) ( 8.00 g) = = 28.0 g ( ) ( 5.00 cm) ( 15.0 cm) ( m D + m C ) ( 15.0 cm) ( g) = ( 5.00 cm) ( 5.00 cm) = 108 g

11 85. When a laboratory sample of unknown mass is placed on a vertical spring-scale having a force constant (spring constant) of 467 N/m, the system obeys the equation y = (4.4 cm) cos(33.3 s -1 t). What is the mass of this laboratory sample? A) 60.2 kg B) 22.4 kg C) 3.87 kg D) kg E) kg y = Acos( wt ) = ( 4.4 cm)cos( 33.3 s 1 t) w = 33.3 s 1 ;w = k m m = k 467 N/m = = kg 2 w 33.3 s 1 2 ( ) 86. Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one? A) 16 F B) 4 F C) 2 F D) F/4 E) F F ' = G m 1m 2 = G 2m 1,0 r 2 2r 0 ( )( 2m 2,0 ) ( = G 2m 1,0 )( 2m 2,0 ) = G m 1,0m 2,0 ( ) 2 ( ) 2 2 r 0 2r 0 = F 87. For general projectile motion with no air resistance, the horizontal component of a projectile's velocity A) continuously increases. B) first decreases and then increases. C) continuously decreases. D) remains zero. E) remains a non-zero constant. For general projectile motion with no air resistance, there s no acceleration along the horizontal axis. Therefore the horizontal component of the velocity remains constant. For general projectile motion, the horizontal component of the velocity is non-zero.

12 88. Consider two same-size cubes. One is made of solid iron, the other is made of solid wood. When put in water, the iron block sinks to the ground whereas the wood block floats. Which of the statements regarding the buoyant forces on the two blocks is correct. A) The buoyant force on the iron block is larger. B) The buoyant force on the wood block is larger. The buoyant force on the iron block is zero. C) The buoyant force on the iron block and the buoyant force on the wood block are the same and non-zero. D) The buoyant force on the wood block is larger. The buoyant force on the iron block is non-zero. E) The buoyant force on the iron block and the buoyant force on the wood block are both zero. The buoyant force on the iron block is larger. Since the iron block is fully submerged the amount of displaced water by the iron block is larger than the one by the wood block. 89. A car goes around a circular curve on a horizontal road at constant speed. What is the direction of the friction force on the car due to the road? A) tangent to the curve opposite to the direction of the car's motion B) perpendicular to the curve inward C) tangent to the curve in the forward direction D) perpendicular to the curve outward E) There is no friction on the car because its speed is constant. The friction between the tires and the road provide the centripetal force that lets the car navigate the curve.

13 Laboratory final exam 90. In the collisions experiment, you obtained traces of the motion of a puck like the one shown below: This trace is produced by the sparks generated at a rate of 60 sparks per second. The distance between the first and last point in the trace shown above is measured with a standard metric ruler and found to be 3.40 cm. Due to the limitations of the ruler, we can estimate the uncertainty in the reading to be ± 0.5 mm. You may neglect the uncertainty in the spark generator frequency. What is the uncertainty in the speed of the puck? A. ± 0.5 mm/s B. ± 1 mm/s C. ± 3 mm/s D. ± 5 mm/s E. ± 3 cm/s There are 6 gaps between sparks. Since they come at a rate of 60 sparks per second the time interval between the first and the last spark is 0.1 seconds. The puck moves at a speed of v = 3.4 cm / 0.10 s = 34.0 cm/s. The uncertainty in the distance between the first and the last spark is ± 0.5 mm. Therefore, the uncertainty in the speed is given by Δv = ± 0.5 mm / 0.10 s = ± 5 mm/s.

14 91. In the Forces and Motion experiment, a force probe is fixed to a cart that can roll with or without friction on a track. The whole system (probe + cart) is pulled through a hook screwed onto the probe by a light string that goes through a pulley and has a cylindrical weight attached to the other end (see figure). Motion detector Reflective plate Force probe Pulley Cart Aluminum track Which of the following most accurately describes what the probe measures? A. The net force on the cart + probe system B. The net force on the probe alone C. The tension in the string D. The weight of the cylinder that pulls on the system E. The weight of the cart + probe system Weight The probe measures the tension in the string. To see that A) and B) are incorrect, consider a situation with a large kinetic friction force. The cart would almost not move, but the probe would still measure the tension in the string, which in this case would be close to the weight of the cylinder. On the other hand the acceleration and thus the net force on the cart or cart-and-probe system would be almost zero. Only, if the car and probe system is not being accelerated, the tension in the string is equal to the mass of the cylinder. So, in general D) is incorrect. In no situation, does the probe directly measure the weight of the cart and probe system.

15 92. A ball is thrown across a yard at different angles with the horizontal. The initial velocity v0 is expected to be the same for all throws. For each initial angle q, we measure the horizontal distance Dx covered by the ball. For a projectile that starts and ends at the same height, the theoretical relation between the two quantities is: D x = ( q ) v 2 sin 2 0 g v0 q Dx As expected, the plot of Dx versus sin( 2q ) is linear, so a linear fit was performed. What was the initial velocity of the ball during these throws? A. 1.2 m/s B. 3.7 m/s C. 6.8 m/s D. 11 m/s E. 25 m/s

16 According to the theoretical prediction, the slope of this graph should be equal to 2 ( ) g ( )( ) v0 = slope = 1.38 m 9.8 m/s = 3.7 m/s 2 v 0 g. Therefore,

17 Physics 111 Final Exam KEY 64 A 74 D 84 E 65 A 75 A 85 D 66 B 76 D 86 E 67 E 77 B 87 E 68 E 78 E 88 A 69 D 79 E 89 B 70 D 80 C 90 D 71 E 81 B 91 C 72 C 82 B 92 B 73 A 83 D 93 A