PHYSICS 221 SPRING FINAL EXAM: May 6, :00pm - 9:00pm

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1 PHYSICS 221 SPRING 2014 FINAL EXAM: May 6, :00pm - 9:00pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions, each worth 4 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, palmtops, pocket computers, PDA or e-book readers) are NOT permitted. Wireless devices 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 that you already used for the first midterm exam. Bubble answers on the bubble sheet for this exam. 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, Drs. Kai-Ming Ho, Eli Rosenberg, and Kerry Whisnant

2 55) In the figure on the right, a diatomic ideal gas is going clockwise through a cyclic process. Which of the following statements about the work done in the process is correct? A) It is equal to the area under the curve abc. B) It is equal to the area under the curve adc. C) It is zero. D) It is equal to the area enclosed by the curve abcda. E) It is equal to the area under the curve ab minus the area under the curve dc. Solution: The work done for a cycle is always the area enclosed by the cycle on a P-V diagram. Answer: D. 56) A 20-g bullet has an initial speed of 200 m/s. After going through a 2.5 cmthick wall, it s speed is 180 m/s. What is the average force on the bullet as it passes through the wall, in kn? A) 11 B) 30 C) 19 D) 6.4 E) 3.0 Solution: We can use the work-kinetic energy theorem plus the fact that the work done is the average force times the displacement:!ke = W 1 2 m(v 2 f " v 2 i ) = "Fd F = m(v 2 i " v 2 f ) 2d = (0.020)(1802 " ) 2(0.025) = 3.0 #10 3 N. Answer: E. 57) An ice cube at 0 C is placed inside a large metal box at 70 C. Heat is transferred and the ice melts and the box cools until both are at 40 C. Which statement is correct? A. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy is unchanged. B. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy increases. C. The entropy of the ice increases; the entropy of the metal box decreases; the total entropy decreases. D. The entropy of the ice decreases; the entropy of the metal box increases; the total entropy is unchanged. E. The entropy of the ice decreases; the entropy of the metal box is unchanged; the total entropy deceases.

3 Solution: The entropy change is ds = dq/t. For the ice, dq is positive since it is warming up, so ds is positive. For the box, dq is negative since it is giving up heat, so ds is negative. By the second law of thermodynamics the total entropy change cannot be negative; since the heat exchange is occurring at changing temperatures, entropy for the system as a whole is increasing. Answer: B. 58) A 73-kg mass is hung from a massless horizontal beam. The beam is attached to a wall at the left end and supported by a cable on the right end (see diagram). The angle θ is 30 0 and the length of the beam is 2.0 m. If the maximum tension the cable can withstand without breaking is 1000 N, what is the maximum distance x from the wall the mass can be hung, in m, that will not break the cable? A) 0.50 B) 2.0 C) 1.4 D) 1.1 E) 0.70 Solution: We can use the fact that the net torque about the left end of the beam must be zero. Taking CCW torque to be positive, mgx max!t max Lsin! = 0 x max = T Lsin! max = (1000)(2.0)sin30 Answer: C. =1.4 m. mg (73)(9.8) 59) Two boxes are connected by a massless string and are pulled along a frictionless horizontal surface by a horizontal force (see figure). If m A = 2.0 kg, m B = 3.0 kg, and F = 18 N, what is the tension in the string, in N? A) 7.2 B) 18 C) 11 D) 9.0 E) 0 Solution: The acceleration of the system as a whole is F/(m A +m B ) = (18)/(5.0)=3.6 m/s 2. Then T = m A a = (2.0)(3.6) = 7.2 N. Answer: A.

4 60) Water at C with a mass of kg is placed in a freezer whose temperature is kept at 50.0 C. The water eventually turns to ice at 50.0 C. How much heat is absorbed by the freezer, in kj? The specific heat of ice is 2.10 kj/kg C, the specific heat of liquid water is 4.19 kj/kg C, and the latent heat of fusion of water is 334 kj/kg. A) 750 B) 600 C) 450 D) 300 E) 150 Solution: The water must first cool down to 0 0 C, freeze, then cool down to C. The total heat removed from the ice (and therefore absorbed by the freezer) is (using units kj for energy and kg for mass) Q = mc water!t + ml fusion + mc ice!t = m(c water!t + L fusion + c ice!t ) Answer: E. = (0.231)[(4.19)(50.0) (2.10)(50.0)] =150 kj. 61) A Heat Pump with a coefficient of performance 4.0 deposits 8.0 J of heat into a room. How much heat does it remove from the outside air, in J? A) 12 B) 10 C) 8.0 D) 6.0 E) 4.0 Solution: The coefficient of performance for a heat pump is K = Q H / W = Q H /( Q H - Q C ). Therefore Q C = Q H (K-1)/K = (8.0)(4.0-1)/4.0 = 6.0. Answer: D. 62) A 1500-kg car moves eastward with speed 65 mph. It hits an 8000-kg truck that is traveling at 20 mph towards the southwest. If they stick together, what is the speed of the two vehicles immediately after the collision, in mph? A) 12 B) 16 C) 20 D) 24 E) 28 Solution: Momentum is conserved during the collision. In the x direction m c v c! m t v t cos45 = (m c + m t )v x v x = m cv c! m t v t cos45 (1500)(65)! (8000)(20)(0.707) = =!1.6 mph. m c + m t In the y direction 0! m t v t cos45 = (m c + m t )v y v y = 0! m v t t sin 45 = 0! (8000)(20)(0.707) =!11.9 mph. m c + m t Then v = v x 2 + v y 2 =12 m/s. Answer: A.

5 The following information is used in problems 63 and 64. The transverse displacement of wave traveling down a stretched string of mass per unit length 1.0 gm/cm is described by the function y(x,t ) = (5.00 cm) cos[(5.00 rad/m)x + (30.0 rad/s)t ], where x is the position in m and t is the time is seconds. 63) What is the wavelength of this wave, in m? A) 0.05 B) 1.26 C) 0.80 D) 4.80 E) 0.20 Solution: The standard form for a wave is y(x,t) = Acos(kx!!t), where k = 2! / " Then! = 2" / k = 2( ) / (5.00) =1.26 m. Answer: B. 64) What is the tension on the string, in N? A) B) C) 0.60 D) 3.6 E) 6.0 Solution: We have k = 5.00 and ω = The velocity is ω/k and is related to the string tension by (everything is put into standard units)! k = v = T µ or T = µ!! $ # & " k % 2 = ! 30.0 $ # & " 5.00 % = 3.6 N. Answer: D. 65) An object moves in a circular path at constant speed. Compare the directions of the object s velocity and acceleration vectors. A) They are in the same direction. B) They are in the opposite direction. C) They are perpendicular. D) The acceleration vector is zero and has no direction. E) The velocity and acceleration vectors for this motion have no fixed relationship. Solution: The acceleration for uniform circular motion is in the inward radial direction; the velocity is tangential. Therefore they are perpendicular. (Note that if the speed was changing, there would have been a tangential component to the acceleration and the acceleration would not be perpendicular to the velocity.) Answer: C.

6 66) An unknown substance is in a container where the pressure is below the triple-point pressure of the substance. The temperature in the container can be varied. Which states of the substance can be r: observed p = 610 Pa as (0.006 you vary atm), the T = K temperature? A) Either gas, liquid, or solid B) Gas or liquid, but not solid C) Gas or solid, but not liquid D) Liquid or solid, but not gas E) Only gas p solid liquid Critical point Solution: The triple point on a p-t diagram is the point where a substance can have all three forms solid, liquid, and gas. Below the triple point, a liquid is not possible. Answer: C. Triple point = T 67) A baseball is hit with initial speed 40 m/s at an angle 30 0 above the horizontal. Its initial height is 1.0 m above the ground. A 20-m high fence is 130 m away. How high up on the fence does the ball hit, in m? Neglect air resistance, gas T A) The baseball hits the ground before reaching the fence. B) 7.0 C) 12 D) 17 E) The baseball goes over the fence. Solution: The equations for projectile motion are: y = y 0 + v 0 y t! 1 2 gt 2 = y 0 + (v 0 sin!)t! 1 2 gt 2 x = x 0 + v 0 x t = 0 + (v 0 cos!)t = v 0 t cos! We can solve for t using the x equation (the time the ball reaches the fence), then plug that into the y equation (the height on the fence): x t = v 0 cos! = cos30 = 3.75 s y = y 0 + (v 0 sin!)t! 1 2 gt 2 =1.0 + (40)(sin30)(3.75)! 1 2 (9.8)(3.75)2 = 7.0 m. (You may have gotten a slightly different answer if you rounded off your result for the time.) Answer: B.

7 68) A police car siren emits sound that has frequency 2000 Hz and wavelength m. A stationary observer measures a frequency of 1854 Hz. Relative to the observer the police car is moving A) away from the observer at 27 m/s. B) towards from the observer at 27 m/s. C) away from the observer at 25 m/s. D) towards from the observer at 25 m/s. E) in a circle with angular velocity 5825 rad/s. Solution: The formula for the Doppler shift if the source is moving is v f = f 0, where we use the plus sign in the denominator since the observed v + v s frequency is lower, which means that the source is moving away from the observer. The sound wave speed is v =! f = (0.172)(2000) = 344 m/s. Solving for v s, " v s = v f 0 $ # f!1 % " ' = 344$ 2000 & # 1854!1 % ' = 27 m/s. Answer: A. & 69) A bullet is fired horizontally, and at the same instant a second bullet is dropped from rest from the same height. Compare the times it takes for the two bullets to hit the ground. A) The fired bullet hits the ground first. B) The dropped bullet hits the ground first. C) The two bullets hit the ground at the same time. D) The relative times cannot be determined without knowing the masses of the bullets. E) The relative times cannot be determined without knowing the speed of the fired bullet. Solution: The initial heights, initial vertical velocities, and accelerations are the same for each bullet. Therefore the vertical motions are the same and they hit the ground at the same time. Answer: C. The following information is used in problems 70 and 71. Eight moles of a diatomic ideal gas are compressed adiabatically from an initial pressure of 4.0 atm and an initial volume of 50.0 L to a final volume of 25.0 L. 70) What is the initial temperature of the gas, in C? A) 3.0 B) 300 C) 2800 D) 28 E) -3.0

8 Solution: The ideal gas law is pv = nrt. Putting everything in standard units (1 L =.001 m 3, 1 atm = 1.0 x 10 5 N/m 2 ) T = PV nr = (4.0!105 )(0.050) = 301 K = 28 C. (8.0)(8.31) If you used 1.01 x 10 5, you get 304 K = 31 0 C. Answer: D. 71) What is the final pressure of the gas, in atm? A) 2.00 B) 6.50 C) 10.6 D) 8.01 E) 1.50 Solution: The process is adiabatic, so P 1 V 1! = P 2 V 2!, where γ = 1.4 for a diatomic gas. Therefore p 2 = p 1 (V 1 /V 2 )! = (4.0)(0.050 / 0.025) 1.4 =10.6 atm. Answer: C. 72) A large pendulum consists of a massive bob suspended from the ceiling by a very light steel wire. The coefficient of linear expansion of steel is 1.2 x 10-5 /K. At C the period of the pendulum is s. By how much does the period of the pendulum change when it is heated to C, in ms? [You can ignore the mass of the wire.] A) 1.7 B) 4.2 C) 5.0 D) 8.4 E) 0 Solution: The period of the pendulum is T = 2! T = 2! L / g T 0 2! L 0 / g = L = 1+"!T L 0 L / g, so T = T 0 1+"!T = (5.000) 1+ (1.2 "10 #5 )(140) = s. The change in the period is therefore s, or 4.2 ms. Answer: B. 73) A 2.0-kg mass swings through an arc on the end of a 40-cm long massless string (see diagram). The speed of the mass is 4.0 m/s at the bottom of the arc. What is the tension in the string, in N, when the string is horizontal? A) 41 B) 60 C) 33 D) 20 E) 10 Solution: Conservation of energy gives 1 2 mv 2 i + mgy i = 1 2 mv 2 f + mgy f, or 1 2 mv 2 i = 1 2 mv 2 f + mgr, so

9 v f 2 = v i 2! 2gL = (4.0) 2! 2(9.8)(0.40) = 8.16m 2 /s 2. When the string is horizontal, only the tension provides centripetal force, so T = mv f 2 / L = (2.0)(8.16) / (0.40) = 41 N. Answer: A. 74) A 2.5-kg particle is moving in a straight line with speed 2.0 m/s as shown, where d = 0.60 m. The angular momentum of the particle about the point O has magnitude kg-m 2 /s and direction A) 1.2, parallel to the velocity. B) 1.2, out of the page. C) 1.5, opposite to the velocity. D) 3.0, to the left. E) 3.0, into the page. Solution: The angular momentum is mrvsin! = (2.5)(0.60)(2.0)sin90 = 3.0 kg-m 2 /s. By the right-hand rule, the angular momentum is into the page. Answer: E. 75) Two masses are attached to each other by a long, massless string. The string runs over a massless pulley (see diagram). The coefficient of kinetic friction between m 1 and the incline is 0.10 and the coefficient of static friction is If there is no slipping of the string on the pulley, m 1 = 3.0 kg, m 2 = 1.0 kg, and α = 35 0, what is the acceleration of m 2, in m/s 2? A) 0 B) 2.4, down C) 9.8, down D) 2.4, up E) 1.2, up Solution: In this case we choose the string as the x-axis, with motion to the right and down as positive. Treating the two masses as one system, we can ignore the tension in the string, which is internal to the system. To determine which direction friction acts, we look at the other forces in the x direction: m 2 g! m 1 gsin! = (1.0)(9.8)! (3.0)(9.8)sin35 =!7.1 N. Therefore the system wants to move to the left (m 1 down the incline, with m 2 going up), and friction acts up the incline. The next question is whether or not static friction is enough to hold the objects in place; the static friction force is f S = µ S N = µ S m 1 gcos! = (0.20)(3.0)(9.8)cos35 = 4.8 N.

10 This is not enough to stop the motion, so the m 1 slides down the incline and m 2 moves up. Friction is therefore kinetic and has magnitude f k = µ k N = µ k m 1 gcos! = (0.10)(3.0)(9.8)cos35 = 2.4 N. The net force on the system is therefore = N (down the incline). The acceleration is then a = F net / (m 1 + m 2 ) =!4.7 / 4.0 =!1.2 m/s 2. Answer: E. 76) A kg mass on a spring is displaced 0.25 m from its equilibrium position and begins to oscillate with simple harmonic motion about equilibrium. The spring constant of the spring is k = 200 N/m. What is the speed of the mass when it passes through its equilibrium position, in m/s? A) B) 5.0 C) 0.50 D) 2.5 E) 0.25 Solution: The standard simple harmonic motion equation is x = Acos(!t +"), where! = k / m = 200 / = 20 rad/s, and so the velocity is v = dx / dt =!A! sin(!t +"). The maximum velocity is the constant in front of the sine function in the velocity, so v max = A! = (0.25)(20) = 5.0 m/s. Answer: B. 77) Three forces act on a 2.5-kg, uniform, cylindrical wheel as shown, where r = 40 cm and R = 80 cm. If the forces are F 1 = 20 N, F 2 = 10 N, and F 3 = 25 N, through how many complete revolutions does the wheel rotate in 1.7 s if the wheel is initially at rest? A) 1.4 B) 5.7 C) 8.2 D) 2.9 E) 4.1 Solution: The angle it rotates through is!! = " 0 t #t 2 = #t 2 = 1 2 #t 2, so we need the angular acceleration. We can find that using the torque and moment of inertia,! = " / I, where I = 1 2 mr2 for a uniform cylinder. Choosing CCW torque as positive, the net torque is! net =! r i F i = " (0.40)(20)+ (0.80)(10)" (0.80)(25) = 20 N-m where in each case the force is perpendicular to the radial vector, so all angles are Then! = 2" net mr = 2(20) 2 (2.5)(0.80) = 25 2 rad/s2. Plugging into the angle equation,

11 !! = 1 2 "t 2 = 1 2 (25)(1.7)2 = 36.1 rad. We just divide this by 2π to get the number of revolutions, 5.7 rev. Answer: B.

12 Laboratory final exam 78) To estimate the height of a cliff, Sam drops a stone that can be clearly heard as it hits the water at the bottom of the cliff. Using a stopwatch, he measures the time that the stone spends in the air. After several attempts, he computes the average value and standard uncertainty of his measurements: Thus, the estimated height of the cliff is: t = ( 2.3± 0.1 ) s m/s 2 gt 2.3 s 26 m ( )( ) 2 h= = = 2 2 What is the uncertainty in the height due to the uncertainty in the time? A. ± 0.5 m B. ± 1 m C. ± 2 m D. ± 3 m E. ± 4 m The minimum and maximum values of the time are t min = 2.2 s and t max = 2.4 s. With these values, we obtain h min = 24 m and h min = 28 m. Thus, h = (26±2) m. Alternatively, h g( t t) gt 2 gt t = ±Δ = ± Δ + gδt The last term is negligible if Δt << t, so h = gt ± gtδ t 2 2 gtδ t = 9.81 m/s 2.3 s 0.1 s = 2 m Plugging in the given values, ( )( )( ) 2 Answer: C.

13 79) In one of this semester s experiments, a cart was pulled on a track as shown in the figure below. The force probe was fastened to the body of the cart, and the whole system (cart + probe) was pulled through a string attached to the probe s hook. Force probe Pulley Cart Hanging mass In any experiment, it is very important to know exactly what each sensor measures. In this setup, which of the following is the best description of the force measured by the probe when the system is in motion? A. The net force on the cart B. The net force on the cart and probe system C. The weight of the hanging mass D. The tension in the string E. The friction between the cart and the track The probe measures the force exerted on it, and it is the string that is attached to it. Note that this tension is smaller than the weight of the hanging mass when the system is in motion and has an acceleration. For a hanging mass of mass m, Newton s second law gives mg T = ma Answer: D. T = mg ma< mg

14 80) A motion detector like the ones we used in the labs is placed on the floor, facing up. A ball is thrown vertically up right above the detector, and its position is measured while it is in the air (trajectory shown in the figure to the right). Remember that these detectors measure the distance of the object to the detector, so the positive x axis is as shown in the figure. +x The detector sends the position and time data to the computer, where the software uses a numerical derivative to obtain the corresponding velocity v x as a function of time. Which of the plots below is closest to the graph you will obtain for the motion of this ball? Motion detector vx vx 0 t 0 t A B vx vx 0 t 0 t C D vx 0 t E Answer: A.

15 81) In the setup shown in the figure to the right, the syringe is slowly pushed in while the pressure in the sample of gas is measured with a pressure sensor. The whole system is submerged in icy water, so the temperature in the gas remains constant at 273K. The graphs below shows the volume markings in the syringe as the piston is pushed versus the pressure, and versus the inverse of the pressure. The latter is a linear graph, as expected. How many moles of gas are there in the sample? A. 0.2 moles B. 0.4 moles C. 2 moles D. 4 moles E. It cannot be determined unless we know which gas this is. The observed dependence seems to indicate that this gas behaves as an ideal gas for this range of pressure, volume and temperature. For an ideal gas. pv = nrt should be verified. Note that V is the entire volume of the sample, V = V syringe + V bottle and tubes ( syr b+ t ) p V + V = nrt V syr nrt = V p b+ t The last equation corresponds to the linear behavior observed in the second graph. The slope of the linear fit (m) should thus be equal to nrt 5 3 m = ml kpa = 908 m Pa = 908 J nrt = m m 908 J n = = = 0.4 mol RT J 8.31 ( 273 K) mol K Answer: B.

16 55 D 64 D 73 A 56 E 65 C 74 E 57 B 66 C 75 E 58 C 67 B 76 B 59 A 68 A 77 B 60 E 69 C 78 C 61 D 70 D 79 D 62 A 71 C 80 A 63 B 72 B 81 B