PHYSICS 221 SPRING 2013

FINAL EXAM: May 7, 2013 2:15pm - 4:15pm 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 55-81 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

55) A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s2. After 4.00 seconds the motor turns off, but the rocket continues to coast upward with no appreciable air resistance. What maximum elevation does the rocket reach, in m? You may assume that the acceleration of gravity remains constant.

A) 320 B) 160 C) 327 D) 408 E) 487 Solution: The distance traveled during acceleration is y = at2 / 2 = (20.0)(4.0)2 / 2 =160 m. The speed at that point is vy = at = (20.0)(4.0) = 80 m/s. The rocket then coasts under gravity; the distance traveled during this part of the motion is given by vyf

2 = vy02 + 2a!y , or

!y =vyf

2 " vy02

2a=

(0)2 " (80.0)2

2("9.8)= 327 m , where the velocity at the top is zero and the

acceleration is -9.8 m/s2. The total distance to the top is therefore 487 m. Answer: E. 56) A 0.60 kg rock is swung in a circular path in a vertical plane on a 0.30 m

length string. At the top of the path, the angular velocity is 12.0 rad/s. What is the tension in the string, in N, at that point?

A) 5.9 B) 15 C) 20 D) 26 E) 32 Solution: For circular motion, at the top both the tension and gravity act down, and must combine to give the centripetal acceleration. Thus T + mg = mω2R, so T = m(ω2R-g) = (0.60)[(12.0)2(0.30)−9.8] = 20 N. Answer: C.

57) Tom and Jerry throw identical rocks off a tall building at the same time. The ground near the building is flat. Tom throws his rock straight downward. Jerry throws his rock at an initial angle 30o below the horizontal. Jerry throws the rock with a speed twice that of Tom's rock. If air resistance is negligible, which rock hits the ground first?

A) Tom’s rock B) Jerry’s rock C) They hit at the same time D) It is impossible to determine without more information E) It depends on the height of the building

Solution: The equation that determines the vertical motion is .

The rock with the most negative v0y will hit the ground first. If the speed of Tom’s rock is v, then v0y = -v. The vertical component of the velocity of Jerry’s rock is v0y = -(2v)sin30o = -2v(0.5) = -v, i.e., the same. Therefore the two rocks will hit the ground at the same time. Answer: C. 58) A series of weights connected by very light cords are

given an upward acceleration of 4.00 m/s2 by a pulling force P, as shown in the figure. A, B, and C are the tensions in the connecting cords. What is the SMALLEST of the three tensions, A, B, and C, in N?

A) 483 B) 196 C) 621 D) 276 E) 80.0 Solution: The force P must pull all four masses, the tension C must pull only the bottom three masses, B the bottom two, and A only the bottom mass. Since the accelerations are all the same, A will be the smallest. The net force on the 20.00-kg mass is A – mg = ma, or A = m(g+a) = (20.00)(9.81+4.00) = 276 N. Answer: D.

y = y0 + v0 yt !12gt2

59) A 2.0 kg box sits on an incline that makes an angle of 20o with respect to the horizontal (see figure). The coefficient of static friction between the incline and the box is µS = 0.30. What is the horizontal force, F, required to start the box moving up the incline, in N?

A) 12 B) 15 C) 20 D) 26 E) 32 Solution: The net force up the incline is Fcos20o - mgsin20o – f, where f is the friction force; friction acts down the incline, opposing the motion. The net force is zero when the box just starts to move. Also, when the box is about to move, friction has its maximum value of f = µs N. The normal force is determined from the net force perpendicular to the incline, N – Fsin20o – mgcos20o = 0. Then

Answer: B.

60) A 2000-kg car is initially moving with speed 30.0 m/s. The driver slams on his

brakes and skids for 5.00 s. If the coefficient of kinetic friction between the tires and the road is 0.400, how much work was done by friction in slowing down the car, in J?

A) −7.92x105

B) −3.96x104 C) −2.38x106 D) −3.17x105 E) 0

Solution: The velocity of the car is v = v0 + at, and the acceleration is a = -F/m = -(µkN)/m = -(µkmg)/m = -µkg. Therefore the final velocity is v = 30.0 – (0.400)(5.00)(9.81) = 10.4 m/s. The work done is equal to the change in kinetic energy,

Answer: A.

Fcos20o !mgsin20o = f = µsN = µs (F sin20o +mgcos20o )

F =mg sin20o +µs cos20o

cos20o !µs sin20o =15 N.

W =12mvf

2 !12mvi

2 =12m vf

2 ! vi2( ) = 1

2(2000) (10.4)2 ! (30.0)2"# $%= !7.92x105 J.

F  20o  

61) Swimmers at a water park have the choice of two frictionless water slides. Both slides have the same vertical height, but one slide is straight and the other is curved (see figure). How does the speed v1 of a swimmer reaching the end of slide 1 compare with v2, the speed of a swimmer reaching the end of slide 2?

A) v1 = v2 B) v1 > v2 C) v1 < v2 D) It can’t be determined without knowing the curvature of slide 2 E) It can’t be determined without knowing the relative mass of the swimmers Solution: Gravitational potential energy is converted to kinetic energy, so mgh = mv2/2, or

v2= 2gh. This doesn't depend on the mass or shape of the slide, only

the height. Since they start from the same height, they have the same speed at the bottom. Answer: A.

62) A 2000-kg car moves eastward with speed 65 mph. It hits a 6000-kg truck

that is traveling at 20 mph towards the northeast. They bounce off of each other, with the truck heading directly to the east at 15 mph after the collision. What is the direction of the car after the collision?

A) 11o south of east B) 11o north of east C) directly east D) 34o north of east E) 4.0o north of east Solution: Momentum must be conserved. We take east to be the positive x direction and north to be the positive y direction. In the x direction we have

In the y direction we have

mcarv0,car +mtruckv0,truckcos45o =mcarvxf ,car +mtruckv f(2000)(65)+ (6000)(20)(0.707) = (2000)vxf ,car + (6000)(15)vxf ,car = 62.4 mph.

Then the angle is arctan(42.4/62.4) = 34o in the first quadrant, which is 34o north of east. Answer: D. 63) A tire is rolling along a road, without slipping, with a velocity v. A piece of

tape is attached to the tire. When the tape is opposite the road (at the top of the tire), its velocity with respect to the road is

A) undeterminable without knowing the radius of the tire B) 0 C) v D) 1.5v E) 2v Solution:The velocity of a point on the tire is equal to the velocity of the CM of the tire, v, plus the velocity of that point relative to the CM, which for a point on the top is also v, in the same direction as the tire velocity. Answer: E. 64) A dentist’s drill has an initial angular speed of 2.15 x 104 rad/s. It undergoes

constant angular acceleration for 2.50 s and reaches an angular speed of 6.75 x 104 rad/s. How many revolutions does the drill bit make during this period?

A) 8.55 x 103 B) 1.77 x 104 C) 3.54 x 104 D) 1.11 x 105 E) 2.23 x 105 Solution: The easiest way to solve this is by using ωavg = (ωi+ωf)/2 for constant acceleration, or ωavg = (2.15 x 104 + 6.75 x 104)/2 = 4.45 x 104. Then θ = ωavg t = (4.45 x 104)(2.50) = 1.11 x 105 rad. To convert to revolutions, we must divide θ by 2π, to get 1.11 x 105 rad/(2π) = 1.77 x 104 revolutions. You could also find the angular acceleration α from Δω/Δt and then use Δθ = ω0t + αt2/2. Answer: B.

0+mtruckv0,truck sin 45o =mcarvyf ,car + 0(6000)(20)(0.707) = (2000)vyf ,carvyf ,car = 42.4 mph.

65) A uniform 300-kg beam, 6.00 m long, is freely pivoted at P, as shown in the figure. The beam is supported in a horizontal position by a light strut, 5.00 m long, which is freely pivoted at Q and is loosely pinned to the beam at R. A load of mass is suspended from the end of the beam at S. A maximum compression of 23,000 N in the strut is permitted, due to safety. The maximum mass M of the load is what value, in kg?

A) 554 B) 789 C) 1020 D) 1090 E) 1320 Solution: The strut makes angle of 53.1o with respect to the beam. The CCW torque of the strut must cancel the CW torques of the weight of the beam and the weight of the mass. If the compression of the strut is F, the strut exerts a force along the direction of the strut of F. Taking torques about the left end of the beam, we don’t need to know the force exerted by the wall on the beam, and we get

Answer: B. 66) The Eiffel Tower, constructed in 1889, is a lattice work structure made of

iron. On a 22° C day the tower is 301 m high. How much smaller, in cm, is the tower on a day when the temperature is only 0.0° C? The coefficient of linear expansion of iron is 1.2x10-5/K.

A) 14 B) 4.4 C) 3.0x10-2 D) 1.1x102 E) 8.0 Solution: The change in length is ΔL = αLΔT = (1.2x10-5)(301)(−22) = −0.080 m = -8.0 cm. Answer: E.

F L2

sin53.1o =mg L2+MgL

M =F sin53.1o

2g!m2=

(23000)(0.800)2(9.8)

!300

2= 789 kg.

67) A 1.50 m long tube closed at one end is filled with air. The speed of sound in air is 343 m/s. Which one of the following is a resonant frequency of this tube, in Hz?

A) 114 B) 76 C) 143 D) 286 E) 229 Solution: For a tube closed at one end, the frequencies are f = noddv/(4L) = nodd(343)/(6.00) = 57, 172, 286, 400,… Hz, so the fifth harmonic (nodd = 5) is in the list. Answer: D. 68) A 100 g block of a metal at 20.00° C is placed into an insulated container

with 400 g of water at a temperature of 0.00° C. The temperature of the metal and the water ends up at 2.00° C. What is the specific heat of this metal, in J/kg-K? The specific heat of water is 4.19x103 J/kg-K.

A) 1.16x102 B) 9.32x102 C) 1.86x103 D) 2.36x104 E) 1.51x105 Solution: The heat gained by the water is equal to the heat lost by the metal, so mmcm(20.00-2.00) = mwcw(2.00-0.00), or cm = mwcw/(9mm) = (400)(4190)/(900) = 1862 J/kg-K. Answer: C. 69) Water near the surface of a tropical ocean has a temperature of 25° C.

Water 700 m below the surface has a temperature of 7° C. What is the maximum efficiency of a heat engine which uses the warm water as a hot reservoir and the cool water as cold reservoir?

A) 94% B) Not enough information is given to know C) 72% D) 28% E) 6.0% Solution: The maximum possible efficiency is ε = (TH −− TC)/TH, where the temperatures must be in K, so ε = (298 − 280)/298 = 0.060, or 6.0%. Answer: E.

70) A pendulum consists of a massive bob suspended from the ceiling by a massless string. To double the period of the pendulum, you should

A) Quadruple the length of the string B) Double the length of the string C) Double the mass of the pendulum bob D) Halve the length of the string E) Halve the mass of the pendulum bob Solution: The period of a pendulum is T = 2! L / g , so to double T you must quadruple L. Answer: A. 71) A transverse wave is described by y(x, t) = (2.0 cm) sin (3.2 m!1) x ! (4.3m/s)t( )"# $%

What is the frequency of the wave, in Hz? A) 2.0 B) 2.2 C) 3.2 D) 4.3 E) 1.3 Solution: The standard form is y(x, t) = A sin k x ! vt( )"# $%= A sin kx !!t[ ] . The

frequency is f = !2"

=kv2"

=(3.2)(4.3)

2"= 2.2 Hz. Answer: B.

72) A Carnot engine takes heat in from a reservoir at 100oC and discards heat to

a reservoir at 10oC. If the engine does 140 J of work per cycle, how much heat does it take in from the hot reservoir per cycle, in J?

A) 140 B) 156 C) 185 D) 580 E) 1080

Solution: For a Carnot engine, TH !TCTH

=QH !QC

QH

=WQH

, where the temperatures

must be expressed in K, so QH =WTHTH !TC

=140(373)373! 283

= 580 J. Answer: D.

73) The average speed of a molecule in a particular sample of an ideal monatomic gas is 380 m/s. If the gas is in a 1.20-m3 container at STP, what is the total mass of the gas, in kg?

A) 1.68 B) 2.14 C) 2.52 D) 3.26 E) 5.79

Solution: The average speed (not rms speed) is given by vavg2 =

8kT!m

=8RT!M

,

where M is the molar mass. Also, RT = PV/n, so vavg2 =

8PV!Mn

, and the total mass is

nM =8PV!vavg

2 =8(1.013x105)(1.20)

3.14(380)2 = 2.14 kg. Answer: B.

74) For the pV diagram shown, what is

the work done by the gas for one cycle (abca), in L-atm? Process bc is isothermal.

A) −0.30 B) −0.25 C) 0 D) 0.25 E) 0.30 Solution: The work done during ab is zero, since the volume is constant. The work done during bc is W = nRT ln(Vc/Vb) = p1V1 ln(Vc/Vb), since bc is isothermal. Also, pbVb = pcVc, so Vc = 0.80 L (=Va also), and therefore Wbc = (2.0)(0.2) ln(4) = 0.55 L-atm. The work done during the isobaric process ca is just p(Va – Vc) = (0.50)(0.20 – 0.80) = − 0.30 L-atm, so the total work done is 0.25 L-atm. Answer: D.

75) When an ideal gas is allowed to expand isothermally from volume V1 to a larger volume V2, the gas does an amount of work equal to W12. If the same ideal gas is allowed to expand adiabatically from volume V1 to the same larger volume V2, the gas does an amount of work that is

A) greater than W12 B) equal to W12 C) less than W12 D) either A, B, or C, depending on the ratio V2/V1 E) either A, B, or C, depending on the final temperature

Solution: An adiabatic process is described by a curve on the pV diagram that falls off more steeply than an isothermal. Therefore is both processes start at the same pressure and volume and end at the same volume, the adiabatic process ends on a lower pressure, and hence has a smaller area under the curve, and therefore less work. Answer: C. 76) You have 1.00 mol of an ideal monatomic gas and 1.00 mol of an ideal

diatomic gas whose molecules can rotate. The gases are in identical fixed containers and initially both gases are at the same temperature. If the same amount of heat flows into each gas, which gas will undergo the greatest increase in temperature?

A) The monatomic gas B) The diatomic gas C) They have the same increase in temperature D) The answer depends on the molar masses on the gases E) The answer depends on the value of the initial temperature Solution: The heat added to change the temperature at constant volume is Q = nCVΔT. Since CV is smaller for the monatomic gas (3R/2 versus 5R/2 for a diatomic gas), its temperature change will be larger for the same amount of heat. Answer: A. 77) A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will measure a frequency f and wavelength λ for this sound such that A) f > 1000 Hz and λ > 0.343 m B) f > 1000 Hz and λ < 0.343 m C) f > 1000 Hz and λ = 0.343 m D) f = 1000 Hz and λ > 0.343 m E) f = 1000 Hz and λ < 0.343 m

Solution: The listener is moving towards from the source, so that tends to increase the frequency. The wavelength is affected by the source motion; since the source is not moving, the wavelength is unchanged. Answer: C. 78) The side L of a square is measured with an ordinary plastic ruler. The result, along with the estimated uncertainty, is:

L = 50.0 ± 0.5 mm

Which of the following is the area of the square with the corresponding propagated uncertainty? A. 2500 ± 0.3 mm2 B. 2500 ± 0.5 mm2 C. 2500 ± 1 mm2 D. 2500 ± 2 mm2 E. 2500 ± 50 mm2

Solution: We can write the length of a side as 50.0(1.00± 0.01) mm, so the fractional error is 1%. The area is then (50.0)2 (1.00± 0.01)2 mm2, with fractional error 2%, or 2500(1.00± 0.02) = 2500± 50mm2. Answer: E.

 79)  A 300-g mass oscillates at the end of vertical spring. The vertical position of the object is measured with a motion detector. The data is shown in the figure below:

 

 Use the data to estimate the spring constant k of the spring. A. 14 N/m B. 33 N/m C. 62 N/m D. 95 N/m E. 120 N/m

 

Solution: The spring constant is k = mω2 = m(2π/T)2. From the graph we estimate the period to be 0.6 s, so k = (0.300)( 2π/0.6)2 = 33 N/m. Answer: B.  80) A cart is pulled by a string on a track. The motion of the cart is monitored with a motion detector like the ones used in the labs, and the tension in the string is measured by a force probe mounted on the cart (see figure below). Using the data of position versus time produced by the motion detector, our software can compute numerical derivatives of position, so we also have velocity and acceleration as a function of time.

 

 

 

Shown below is a graph showing the tension in the string versus the acceleration of the cart. Based on this data, what is the mass of the cart + force probe system?

 

A. 1.2 kg B. 2.6 kg C. 3.2 kg D. 3.8 kg E. 4.4 kg

 

 

 

 

               

Motion  detector  

T  

Force  probe  

Cart  

Tension  (N)  

Acceleration  (m/s2)  

Solution: Since F = ma, where F is the tension, so mass is the slope of the F versus a curve. Estimating the slope from the graph to be !T /!a = 0.6 / 0.5=1.2kg. Answer: A.  81) A flask containing air from the room is closed with a tightly-fitting rubber stop. The system is then slowly cooled down by placing it in a cold bath. Both the temperature and the pressure inside the flask are measured throughout the process. Which of the following graphs is closest to the graph you will obtain with that data?  

 

Solution: Since pV = nRT, if the volume and number of moles remains constant (the flask does not change size and the stopper keeps all of the gas inside), then the pressure and temperature must be proportional. This gives a straight line with positive slope for p versus T. Answer: B.    

p  

T  

A  

p  

T  

B  

p  

T  

D  

p  

T  

E  

p  

T  

C  

55    E     64    B     73    B  

56    C     65    B     74    D  

57    C       66    E     75    C  

58    D     67    D     76    A  

59    B     68    C     77    C  

60    A     69    E     78    E  

61    A     70    A     79    B  

62    D     71    B     80    A  

63    E     72    D     81    B