PHYSICS 111 SPRING 2016

FINAL EXAM: May 3, 2016; 2:15pm - 4:15pm 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 64-93 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

64. A ball is thrown straight up, reaches a maximum height, then falls to its initial

height. Which of the following statements about the direction of the velocity and acceleration of the ball as it is going up is correct?

A) Both its velocity and its acceleration point upward. B) Its velocity points upward and its acceleration points downward. C) Its velocity points downward and its acceleration points upward. D) Both its velocity and its acceleration points downward. E) Its velocity points upward and its acceleration is zero.

Solution

As the ball is going upward its velocity points upward. For all objects in free fall the acceleration points downward.

65. A cart with an initial velocity of 5.0 m/s to the right experiences a constant

acceleration of 2.0 m/s2 to the right. What is the cart's displacement during the first 6.0 s of this motion?

A) 10 m B) 55 m C) 66 m D) 80 m E) 91 m

Solution

x = x0 + v0t +

12at 2 ⇒

x = 5.0 m/s( ) 6.0 s( ) + 12

2.0 m/s2( ) 6.0 s( )2 = 66 m

66. Two vectors, of magnitudes 20 mm and 50 mm, are added together. Which one of

the following is a possible value for the magnitude of the resultant?

A) 10 mm B) 20 mm C) 40 mm D) 80 mm E) 90 mm

Solution The magnitude of the sum of two vectors lies between the sum of the magnitudes (parallel vectors) and the difference of the magnitudes (anti-parallel vectors). Thus, the magnitude of a vector that is the sum of two vectors with magnitudes 20 mm and 50 mm has to be between 30 mm and 70 mm.

67. Mary and Debra stand on a snow-covered roof. They both throw snowballs with

the same initial speed, but in different directions. Mary throws her snowball downward, at 30° below the horizontal; Debra throws her snowball upward, at 30° above the horizontal. Which of the following statements are true about just as the snowballs reach the ground below?

A) Debra's snowball has a higher total speed than Mary's snowball. B) Mary's snowball has a higher total speed than Debra's snowball. C) Both snowballs hit the ground with the same total speed. D) Both snowballs hit the ground at the same time. E) Mary's snowball has a higher vertical speed than Debra's snowball. Solution When Debra’s snowball passes through the height again, it has the same velocity as Mary’s snowball at launch. Both snowballs hit the ground with the same total speed, horizontal speed and vertical speed. Debra’s snowball hits the ground later, because her snowball first goes up before it goes down.

68. A block of mass m sits at rest on a rough inclined

ramp that makes an angle θ with the horizontal. What must be true about force of static friction f on the block?

A) f > mg B) f > mg cos θ C) f > mg sin θ D) f = mg cos θ E) f = mg sin θ

Solution Since the block is not moving, the friction force must be equal to the x-component of the weight mg: f = mgsinθ

69. A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the average braking force acting on the car?

A) 690 N B) 550 N C) 410 N D) 340 N E) 220 N

Solution

vf

2 − v02 = 2aΔx→ a = −v0

2

2Δx

F = m a = mv02

2Δx=

1100 kg( ) 27 m/s( )2

2 578 m( ) = 690 N

70. A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring?

A) 41 N/m B) 22 N/m C) 34 N/m D) 46 N/m E) 58 N/m

Solution

F = kΔx⇒ k =FΔx

=30 N( )

0.73 m( ) = 41 N/m

71. In the figure, a ball hangs by a very light string.

What is the minimum speed of the ball at the bottom of its swing (point B) in order for it to reach point A, which is 1.0 m above the bottom of the swing?

A) 2.2 m/s B) 3.1 m/s C) 4.4 m/s D) 4.9 m/s E) 5.2 m/s

Solution

KE0 + PE0 = KEf + PEf ⇒12mv0

2 = mgΔy

v0 = 2gΔy = 2 9.8 m/s2( ) 1.0 m( ) = 4.4 m/s

72. Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a speed of 0.850 m/s. Cart B has a mass of 300 g and moves toward the left with a speed of 1.12 m/s. What is the total momentum of the two-cart system?

A) 0.047 kg · m/s toward the right B) 0.719 kg · m/s toward the right C) 0.719 kg · m/s toward the left D) 0.750 kg · m/s toward the right E) 0.750 kg · m/s toward the left

Solution Let the positive x axis point to the right.

ptotal = mAvA +mBvB

= 0.450 kg( ) 0.850 m/s( ) + −0.300 kg( ) 1.12 m/s( )= 0.0465 kg m/s

73. In the figure, four point masses are placed as shown. Assume that all the numbers in the figure are accurate to two significant figures. What is the x coordinate of the center of mass (or center of gravity) of this arrangement?

A) 2.1 m B) 2.2 m C) 2.3 m D) 2.4 m E) 2.5 m

Solution

xcom =8 kg( ) 0 m( ) + 2 kg( ) 2 m( ) + 4 kg( ) 3 m( ) + 6 kg( ) 5 m( )

8 kg( ) + 2 kg( ) + 4 kg( ) + 6 kg( ) = 2.3 m

74. How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3

rad/s if it has a uniform angular acceleration of 3.45 rad/s2?

A) 4.35 s B) 5.30 s C) 9.57 s D) 10.6 s E) 63.1 s

Solution

ω =ω 0 +αt→ t = ω −ω 0

α=

33.3 rad/s( )− 15.0 rad/s( )3.45 rad/s2( ) = 5.30 s

75. If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration of the oscillator change? A) 2 B) 4 C) It does not change. D) 1/2 E) 1/4

Solution

amax = Aω2 = A 2ω 0( )2 = 4Aω 0 = 4a0,max

76. The figure shows a "snapshot" of a wave at a

given instant of time. The frequency of this wave is 120 Hz. What is the speed of this wave?

A) 24 m/s B) 12 m/s C) 36 m/s D) 48 m/s E) 6 m/s

Solution v = fλ = 120 Hz( ) 0.2 m( ) = 24 m/s

77. As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear a frequency of 65 Hz. What is the speed of the car? The speed of sound in the air is 343 m/s.

A) 26 m/s B) 27 m/s C) 28 m/s D) 29 m/s E) 30 m/s

Solution

fobs =vsound ± vobsvsound ∓ vsource

fsource =vsound

vsound ∓ vsourcefsource ⇒

76 Hz =343 m/s( )

343 m/s( )− vsourcefsource; 65 Hz =

343 m/s( )343 m/s( ) + vsource

fsource

⇒ 7665

=343 m/s( ) + vsource343 m/s( )− vsource

⇒ vsource =11141

343 m/s( ) = 26.8 m/s

78. A 40.0-kg cylinder of solid iron is supported by a string while submerged in

water. What is the tension in the string? The density of iron is 7860 kg/m3 and that of water is 1000 kg/m3.

A) 25 N B) 200 N C) 240 N D) 340 N E) 400 N

Solution

FB +T = mirong; FB = mdispg = ρwaterVg = ρwatermiron

ρiron

g

⇒ T = mirong − FB = mirong 1− ρwater

ρiron

⎛⎝⎜

⎞⎠⎟= 340 N

79. A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a

fluid through a pipe. What is the speed of the fluid when it enters a portion of the pipe that is 3.0 mm in diameter? Treat the fluid as ideal and incompressible.

A) 50 m/s B) 50 cm/s C) 6.0 cm/s D) 22 cm/s E) 25 m/s Solution

A1v1 = A2v2 ⇒ v2 =A1

A2

v1 =π d1 / 2( )2

π d2 / 2( )2 v1 =d1

d2

⎛⎝⎜

⎞⎠⎟

2

v1 = 5000 cm/s = 50 m/s

80. Express a body temperature 98.6°F in Celsius degrees.

A) 37.0°C B) 45.5°C C) 66.6°C D) 72.6°C E) 29.4°C

Solution

T (oC) = 59T (oF)− 32oF( ) = 59 98.6

oF − 32oF( ) = 37oC

81. A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 × 10-6 K-1.

A) 0.33 m B) 0.44 m C) 0.55 m D) 0.66 m E) 0.77 m

Solution ΔL =αLΔT = 11×10−6 /K( ) 1000 m( ) 60 K( ) = 0.66 m

82. If the absolute temperature of an object is tripled, the thermal power radiated by

this object (assuming that its emissivity and size are not affected by the temperature change) will

A) increase by a factor of 3. B) increase by a factor of 9. C) increase by a factor of 18. D) increase by a factor of 27. E) increase by a factor of 81.

Solution ΔQemit

Δt= AeσT 4 = Aeσ 3T0( )4 = 81Aeσ T0( )4 = 81 ΔQemit

Δt⎛⎝⎜

⎞⎠⎟ 0

83. A thermally isolated system is made up of a hot piece of aluminum and a cold

piece of copper, with the aluminum and the copper in thermal contact. The specific heat capacity of aluminum is more than double that of copper. Which object experiences the greater magnitude gain or loss of heat during the time the system takes to reach thermal equilibrium?

A) the aluminum B) the copper C) Neither one; both of them experience the same size gain or loss of heat. D) It is impossible to tell without knowing the masses. E) It is impossible to tell without knowing the volumes.

Solution The system is isolated and no heat is added or removed. The heat that is lost by the higher-temperature aluminum and gained by the lower temperature copper is the same.

84. A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab

technician begins to supply heat to the container at the rate of 1000 J/min. How long after starting will the ice begin to melt, assuming all of the ice has the same temperature? The specific heat of ice is 2090 J/kg · K and the latent heat of fusion of water is 33.5 × 104 J/kg.

A) 23.8 min B) 12.6 min C) 52.1 min D) 63.7 min E) 33.5 min

Solution

t = QP= mciceΔT

P=

0.456 kg( ) 2090 J/(kg K)( ) 25 K( )1000 J/min( ) = 23.8 min

85. A car starts out when the air temperature is 288 K and the absolute (total) air

pressure in the tires is 500 kPa. After driving a while, the temperature of the air in the tires increases to 298 K. What is the pressure in the tires at that point, assuming their volume does not change?

A) 129 kPa B) 483 kPa C) 507 kPa D) 517 kPa E) 532 kPa

Solution P1

T1

= P2

T2

⇒ P2 = P1T2

T1

= 500 kPa( ) 298 K288 K

⎛⎝⎜

⎞⎠⎟ = 517 kPa

86. An ideal gas is compressed isobarically to one-third of its initial volume. The resulting pressure will be A) three times as large as the initial value. B) equal to the initial value. C) more than three times as large as the initial value. D) nine times the initial value. E) impossible to predict on the basis of this data.

Solution

An isobaric process happens at constant pressure.

87. At what temperature is the rms speed of hydrogen molecules, H2, which have a molecular weight of 2.02 g/mole, equal to 2.0 km/s?

A) 17°C B) 34°C C) 51°C D) 68°C E) 72°C

Solution

vrms =3kTm

= 3RTM

⇒ T = Mvrms2

3R=

2.02 ×10−3 kg/mole( ) 2.0 ×103 m/s( )2

3 8.31 J/(mole K)( ) = 324 K = 51oC

88. A gas is taken through the cycle shown in the pV

diagram in the figure. During one cycle, how much work is done by the gas?

A) 0.5 p0V0 B) p0V0 C) 2 p0V0 D) 3 p0V0 E) 4 p0V0

Solution W = W = 0 + 2P0 (4 −1)V0 + 0 +∑ P0 (1− 4)V0 = (2 −1)(4 −1)P0V0 = 3P0V0

89. An ideal Carnot heat engine has an efficiency of 0.600. If it operates between a deep lake with a constant temperature of 294.0 K and a hot reservoir, what is the temperature of the hot reservoir?

A) 735 K B) 470 K C) 526 K D) 784 K E) 651 K Solution

eideal = 1− TcoldThot

⇒ Thot =Tcold

1− eideal=

294.0 K( )1− 0.600

= 735 K

Laboratory final exam

90. A force F is applied at an angle θ with the x axis. We measure both quantities and

obtain the following values and standard uncertainties:

( )( )382 3 N

25 1

F

θ= ±

= ± °

Determine the value of the x component of this force, along with its corresponding standard uncertainty.

( )( )( )( )( )

C. 346 4 N

A. 346 2 N

B. 346 3 N

D. 346 5 N

E. 346 6 N

±

±

±

±

±

Solution

Fx = F cosθ = 382 N( )cos 25o( ) = 346 N

ε Fx( ) = ε F( )cosθ( )2+ Fε cosθ( )( )2

= ε F( )cosθ( )2+ Fε θ( ) sinθ( )2

= 3 N( )cos25o( )2+ 382 N( ) 10( ) π

1800⎛⎝⎜

⎞⎠⎟ sin25o⎛

⎝⎜⎞⎠⎟

2

= 4 N

91. A rod has an initial length of 80 cm at room temperature (29.5°C). The rod is then

warmed up to 96.0°C and slowly allowed to cool down. During the cooling down, both the changes in length and the temperature of the rod are measured many times. The results are shown in the graph below.

What is the coefficient of linear thermal expansion of the rod’s material? A. 7.4×10−8 K−1 B. 2.0×10−5 K−1 C. 0.00020 K−1 D. 0.016 mm K−1 E. 0.50 mm K−1

Solution

ΔL =αLΔT ⇒α = ΔLLΔT

= 1L

ΔLΔT

⎛⎝⎜

⎞⎠⎟ =

10.8 m( ) 0.0161×10−3 m/K( ) = 2.0 ×10−5 K−1

92. In one of the labs, a microphone was used to capture the sound produced when we

pluck a steel wire under tension. The signal was then fed onto an oscilloscope.

The image below shows a screenshot of one of the oscilloscope displays obtained in the experiment. The horizontal scale is 2 milliseconds per tickmark. What is the fundamental frequency of this wave?

A. 100 Hz B. 200 Hz C. 300 Hz D. 400 Hz E. 500 Hz Solution

T = 2.5 tickmarks = 5 ×10−3 s ⇒ f = 1T= 200 Hz

2ms

93. A small aluminum cylinder of volume 30 cm3 is tied to the end

of a string and lowered onto a container with water, as shown in the figure to the right. The tension in the string is measured with a force probe. As the cylinder becomes slowly submerged, we take simultaneous measurements of the tension in the string (T) and of the volume of the cylinder that is submerged (Vsub). Which of the following plots shows the relationship we expect to find between the two quantities?

Solution T + FB = mcylg⇒ T = mcylg − FB; FB = msubmg = ρcylVsubmg

T = mcylg − ρcylVsubmg = ρcylVcylg − ρcylVsubmg = ρcylg Vcyl −Vsubm( )

The tension decreases proportional with the submerged volume.

Physics 111 Final Exam - KEY

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