physics 220 - final exam - summer 2014 version 31 - answer key

Physics 220 - Final Exam - Summer 2014Version 31 - Answer Key

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1. A pendulum on the Earth has a period T. The acceleration due to gravity on Mars isless than that on the Earth, and the acceleration due to gravity on the Moon is evenless than on Mars. Identical pendulums have the same mass and length. Where wouldthe period of an identical pendulum be the smallest?

(a)4 The period is smallest on the Earth.

(b) The period is the same on the Earth, the Moon, and Mars.

(c) The period is smallest on Mars.

(d) The period is smallest on the Moon.

(e) The period is smallest on both Mars and the Moon.

Solution:The period of a pendulum is given by:

T = 2π

√L

g

The length of the pendulum does not change. The acceleration due to gravity (g) changeson the Earth, the Moon, and Mars. The smallest period is when the acceleration due togravity (g) is the largest. This occurs on the Earth.

Version 31 - Answer Key Physics 220 - Final Exam - Summer 2014 of 20

2. Joule’s machine is shown on the right. A block with amass of 6.1 kg is dropped some distance (z). A ropeis attached both to the block and to a pulley whichis connected to a rotating paddle. When the blockdrops, the paddle turns causing the temperature of thewater to increase. If the mass of the water is 0.3 kg,what distance (z) must the mass drop to increase thetemperature of the water by 0.5 ◦C? (Assume thatthere is no friction.)

(a) 50.00 m

(b) 8.35 m

(c)4 10.50 m

(d) 20.07 m

(e) 2.46 m

Solution:Energy is conserved. The potential energy of the block is converted into the heat energyof the water. This gives the following:

Einitial = Efinal

PEinitial = Qfinal

mblockgz = mwatercwater∆T

Solving for z:

z =mwatercwater∆T

mblockg=

(0.3)(4186)(0.5)

(6.1)(9.8)= 10.50 m


3. Which of the following changes the speed of sound in air?

(a) frequency

(b)4 temperature

(c) period

(d) amplitude

(e) wavelength

Solution:The speed of sound depends in a medium only depends on the properties of the medium.The temperature of the air changes the speed of sound in air.


4. A treasure chest filled with gold has a mass of 930 kg and is sitting at the bottom of theocean in a place that is 2.6 m deep. A chain is attached to the treasure chest so thatthe treasure chest can be pulled up to the surface. Since the treasure chest is filled withgold, assume that the chest has the density of gold. Assume the ocean water has thedensity of regular water. What force is required to lift the treasure chest?

(a) 4,722 N

(b) 3,274 N

(c) 15,205 N

(d)4 8,642 N

(e) 9,978 N

Solution:There are three forces acting on the treasure chest: the force of gravity, tension in thechain, and the force of buoyancy. The sum of the forces for the treasure chest gives thefollowing:

ΣFy : T + FB − Fg = ma

T + FB − Fg = 0

Solving for tension:

T = Fg − FB

T = mchest g − ρwater Vchest g

The volume of the chest can be found using the density and mass:

ρ =m

V→ Vchest =

mchest

ρgold

Substituting this into the tension equation gives:

T = mchest g − ρwater

(mchest

ρgold

)g

T = (930)(9.8)− (1000)

(930

19, 300

)(9.8)

T = 8,642 N


5. A certain string has a mass of 0.15 kg and a length of 0.64 m. A tension of 39 N isapplied to the string. What is the velocity of a wave on the string?

(a) 0.69 m/s

(b) 166.40 m/s

(c)4 12.90 m/s

(d) 15.05 m/s

(e) 3.02 m/s

Solution:The velocity of a wave on a string is given by:

v =

√T

µ

The mass density is the mass divided by the length. (µ = M/L) Substituting this intothe previous equation gives the following:

v =

√TL

M=

√(39)(0.64)

0.15= 12.90 m/s


6. An air bubble ascends from the bottom of a lake that is 22.3 m deep. The temperatureat the bottom of the lake is 263 K, and near the surface the temperature is 289 K. Ifthe volume of the bubble is 83.6 m3 at top of the lake, what is the volume of the bubbleat the bottom of the lake?

(a) 29.10 m3

(b)4 24.10 m3

(c) 41.88 m3

(d) 35.26 m3

(e) 17.84 m3

Solution:

PV = nRT

The number of moles is constant. From the Ideal Gas Law, this gives the following ratio:

PbotV

bot

Tbot

=PtopVtop

Ttop

The pressure at the top of the lake is atmospheric pressure. Ptop = 1.013× 105 Pa. Thepressure at the bottom of the lake depends on the depth, as shown:

Pbot = Ptop + ρgh

Pbot = 1.013× 105 + (1000)(9.8)(22.3)

Pbot = 319.84× 103 Pa

Solving for the volume at the bottom of the lake:

Vbot

=PtopVtopTbot

PbotTtop

Vbot

=(1.013× 105)(83.6)(263)

(319.84× 103)(289)

Vbot

= 24.10 m3


7. Two metal blocks are placed in a room and have the same initial temperature. Bothblocks are made of steel, and both blocks are heated by the same amount (Q). Block Twohas twice the mass of Block One. How does the change in temperature of Block Onecompare to the change in temperature of the Block Two?

(a) ∆Tone = ∆Ttwo/2

(b) ∆Tone = ∆Ttwo/4

(c) ∆Tone = 4 ∆Ttwo

(d) ∆Tone = ∆Ttwo

(e)4 ∆Tone = 2 ∆Ttwo

Solution:Both blocks are heated by the same amount, which gives the following:

Qone = Qtwo

monecsteel∆Tone = mtwocsteel∆Ttwo

mone∆Tone = mtwo∆Ttwo

Block Two has twice the mass of Block One, which means mtwo = 2mone. This givesthe following:

mone∆Tone = (2mone)∆Ttwo

∆Tone = 2 ∆Ttwo

The temperature change in Block One is twice the temperature change in Block Two.


8. A speaker with a power of 34 W is playing music. What is the sound intensity level ofthe music at a distance of 22 m? Assume that the sound can only be heard in front ofthe speaker and not behind it.

(a) 36.4 dB

(b)4 100.5 dB

(c) 180.4 dB

(d) 78.5 dB

(e) 11.2 dB

Solution:In order to find the sound intensity level, first the intensity must be calculated. Sincethe sound only goes in front of the speaker, the area is half of the surface area of asphere, which is 2πr2. The intensity is found by:

I =Power

Area=

Power

2πr2=

34

2π(22)2= 1.118× 10−2 W/m2

The sound intensity level is given by:

β = (10 dB) log10

I

I0

= (10 dB) log10

1.118× 10−2

1× 10−12

= (10 dB) log10(1.118× 1010)

= (10 dB)(10.05)

= 100.5 dB


9. A 2.9 kg mass is attached to a spring on a table. Assume that there is no friction. Themass oscillates with a period of 9.5 s. If the maximum amplitude of the mass is 5.3 m,what is the maximum velocity of the mass on the spring?

(a) 1.92 m/s

(b) 1.44 m/s

(c) 0.62 m/s

(d) 6.37 m/s

(e)4 3.51 m/s

Solution:The angular velocity can be found from the period as shown:

ω =2π

T=

2π

9.5= 0.66 rad/s

The max velocity is given by:

vmax = A ω = (5.3 m)(0.66 rad/s) = 3.51 m/s

‘


10. A bank vault is opened by applying a force of 102 N perpendicular to the plane of thedoor, 1.5 m from the hinges. Find the torque due to this force about an axis throughthe hinges.

(a) 68.0 Nm

(b)4 153.0 Nm

(c) 0.0 Nm

(d) 280.4 Nm

(e) 137.3 Nm

Solution:Torque about an axis is equal to the radius times the force times the sine of the anglebetween the force and distance.

τ = rF sin(θ) = (1.5 m)(102 N)(sin(90)) = 153.0 Nm


11. A suction cup has an air pressure of 20.0 kPa inside the suction cup when it is attachedto a ceiling. The suction cup covers an area of 2.04 cm2, and a mass is hanging fromthe suction cup. What is the largest mass that the suction cup can hold before falling?

(a) 1.18 kg

(b) 0.42 kg

(c) 2.81 kg

(d)4 1.69 kg

(e) 16.59 kg

Solution:The suction cup is held up because of the force due to the difference of pressure insideand outside of the suction cup. The pressure outside is atmospheric pressure. Sum ofthe forces for the suction cup:

ΣFy : Fout − Fin − T = ma = 0

PoutA− PinA− T = 0

(Pout − Pin)A = T

Sum of the forces for the hanging mass

ΣFy : T − Fg = ma = 0

T = Fg = mg

Substituting the second equation into the first and solving for mass gives:

m =(Pout − Pin)A

g=

(101.3× 103 − 20.0× 103)(2.04× 10−4)

(9.8)= 1.69 kg


12. A heat engine has two reservoirs, one at a temperture of 350 K, and the other at 900 K.The heat engine takes 8,400 J of heat from the hot reservoir, expels 7,400 J of heat tothe cold reservoir, and does 1,000 J of work. Which of the following is true about thisheat engine?

(a) This heat engine is possible since the entropy of the system decreases.

(b) This heat engine is possible but is an ideal reversible heat engine.

(c)4 This heat engine is possible and is a regular irreversible heat engine.

(d) This heat engine is not allowed by the second law of thermodynamics.

(e) This heat engine is not allowed by the first law of thermodynamics.

Solution:The highest possible efficiency for a heat engine is for an ideal or irreversible heat engineand is given by:

eideal = 1− TcTh

eideal = 1− 350

900eideal = 1− 0.39

eideal = 0.61

The efficiency for a real heat engine is given by:

e =W

QH

e =1,000

8,400

e = 0.12

This heat engine is possible since the calculated efficiency is less than the ideal efficiency.


13. A container has a volume of 5.76 m3 and holds 15.8 moles of an ideal gas. The pressurein the container is 83.5 kPa. Wat is the temperature of the gas?

(a) 6,713.40 K

(b) 90.57 K

(c)4 3,663.12 K

(d) 580.08 K

(e) 1,104.10 K

Solution:

PV = nRT

From the Ideal Gas Law, solving for temperature gives:

T =PV

nR=

(83.5× 103 Pa)(5.76 m3)

(15.8 mol)(8.31 J/(K mol))= 3,663.12 K


14. A cart with a mass of 3.3 kg is pushed to the right with a velocity of 8.8 m/s. A secondcart with a mass of 7.5 kg is in front of the first cart and is moving to the right with avelocity of 3.2 m/s. The first cart collides with the back of the second cart in an ines-lastic collision. After the collision, the first cart moves with a final velocity of 0.5 m/s.What is the final velocity of the second cart?

(a) 7.77 m/s

(b) 1.85 m/s

(c) 16.57 m/s

(d)4 6.85 m/s

(e) 8.57 m/s

Solution:Momentum is converserved for all collisions. Conservation of Momentum gives the fol-lowing:

~p1,i + ~p2,i = ~p1,f + ~p2,f

m1~v1,i +m2~v2,i = m1~v1,f +m2~v2,f

This problem only has motion in one direction (the x-direction) so only the x-componentsof the momentum are needed, and the notation can be simplified.

m1v1,i +m2v2,i = m1v1,f +m2v2,f

Solving for v2,f :

v2,f =m1v1,i +m2v2,i −m1v1,f

m2

=(3.3 kg)(8.8 m/s) + (7.5 kg)(3.2 m/s)− (3.3 kg)(0.5 m/s)

7.5 kg

= 6.85 m/s


15. A figure skater initially has her arms extended. She starts spinning on the ice at 3 rad/s.She then pulls her arms in close to her body. Which of the following happens?

(a)4 She spins at a higher angular velocity.

(b) She spins at the same angular velocity.

(c) Her angular momentum increases.

(d) Her angular momentum decreases.

(e) She spins at a lower angular velocity.

Solution:Angular momentum of the system is convserved.

~Lskater,initial = ~Lskater,final

Iskater,initial ~ωskater,initial = Iskater,final ~ωskater,final

mskaterr2skater,initial ~ωskater,initial = mskaterr

2skater,final ~ωskater,final

As the figure skater pulls in her arms, her radius decreases. In order to conserve angularmomentum, her angular speed (ω) increases.


16. The equation for a traveling wave is:

y(x, t) = (3.7 m) sin(83 t− 7.7x)

I) What is the amplitude of the wave?II) What is the frequency of the wave?III) What is the wavelength of the wave?

(a)4 I) 3.7 m II) 13.21 Hz III) 0.82 m

(b) I) 3.7 m II) 6.17 Hz III) 1.94 m

(c) I) 7.7 m II) 0.82 Hz III) 13.21 m

(d) I) 7.7 m II) 13.21 Hz III) 1.94 m

(e) I) 3.7 m II) 0.82 Hz III) 13.21 m

Solution:The equation for a traveling wave is given by:

y(x, t) = A sin(ω t− k x)

y(x, t) = A sin(2πft− 2π

λx)

From the equation given in the problem, the amplitude (A) is 3.7 m. The angular fre-quency (ω) is 83 rad/s. and the wave number (k) is 7.7 m−1.

The frequency (f) is related to the angular frequency (ω) and can be found by thefollowing:

ω = 2πf → f =ω

2π=

83

2π= 13.21 Hz

The wavelength (λ) is related to the wave number (k) and can be found by the following:

k =2π

λ→ λ =

2π

k=

2π

7.7= 0.82 m


17. A Pitot-Static tube is used to measure pressure inorder to find an airplane’s velocity. The pressure at 1,where the velocity of air is zero, is 5.33 ×104 Pa, andthe pressure at 2 is 2.53 ×104 Pa. The airplane is flyingat an altitude where the density of air is 0.77 kg/m3.What is the velocity of the airplane?

(a) 229.38 m/s

(b) 200.33 m/s

(c) 499.50 m/s

(d) 372.09 m/s

(e)4 269.69 m/s

Solution:The height of point 1 and point 2 are similar and aproximately zero. The velocity ofpoint 1 is zero. This gives the following:

P1 +1

2ρv2

1 + ρgh1 = P2 +1

2ρv2

2 + ρgh2

P1 +1

2ρ(0)2 + ρg(0) = P2 +

1

2ρv2

2 + ρg(0)

P1 = P2 +1

2ρv2

2

Solving for v2:

v2 =

√2(P1 − P2)

ρ=

√2(5.33× 104 − 2.53× 104)

0.77= 269.69 m/s


18. A ball with a mass of 2.3 kg is thrown up at a 67.6◦ angle above the horizontal. Theball initially has a velocity of 54.9 m/s and starts approximately on the ground. Howlong is the ball in the air before it hits the ground?

(a) 12.30 s

(b) 5.18 s

(c) 2.13 s

(d)4 10.36 s

(e) 2.36 s

Solution:The initial velocity in the x-direction and y-direction are given by:

v0x = v0 cos θ v0y = v0 sin θ

The ball is being thrown straight up. At the very top, the velocity in the y-directionis zero (vfy = 0). The acceleration in the y-direction is -9.8 m/s2. Use the kinematicequation with velocity, time, and acceleration.

vfy = v0y + ayt

Solving for time gives:

t =vfy − v0y

a=

0− v0 sin θ

a=−(54.9 m/s) sin 67.6◦

−9.8 m/s2 = 5.18 s

The total time the ball in in the air is twice the time it takes for the ball to travel tothe top of its path.

ttotal = 2t = 2(5.18 s) = 10.36 s


Refer to the following P-V diagram for Problem 19 and Problem 20.

In the figure above, P1 = 30.2 kPa, P2 = 62.1 kPa, P3 = 88.1 kPa, V1 = 2.24 m3, andV2 = 7.01 m3.

19. A gas undergoes a two part process from A to C as shown in the P-V diagram above.The first part is from A to B and the second from B to C. What is the work done bythe gas during this process (all parts from A to C)?

(a) −420.24 kJ

(b)4 +358.23 kJ

(c) +741.87 kJ

(d) −296.22 kJ

(e) +144.57 kJ

Solution:The work for A→ B is the area under the curve which can be broken up into a rectangeand a triangle. This gives the following:

WAB = P2(V2 − V1) + (1/2)(P3 − P2)(V2 − V1)

= (62.1 kPa)(7.01 m3 − 2.24 m3) + (1/2)(88.1 kPa− 62.1 kPa)(7.01 m3 − 2.24 m3)

= 296.22 kJ + 62.01 kJ

= 358.23 kJ

The work done on a PV-diagram is the area under the curve. Now calculate the totalwork:

WA→C = WAB +WBC

WA→C = WAB + PBC∆VBC

WA→C = WAB + PBC(0)

WA→C = WAB + 0 = 358.23 kJ

Physics 220/Final Exam – of 20 – SOLUTIONS

20. After going through the two part process from A to C as explained in Problem 19, thegas undergoes another two part process. Part one is from C to D, and part two is from Dto A. The gas has now undergone one complete cycle and is back at A (where it started).What is the work done by the gas during the entire cycle (all four parts A-B-C-D-A)?

(a) +406.09 kJ

(b) +33.26 kJ

(c) −296.22 kJ

(d) −220.14 kJ

(e)4 +214.17 kJ

Solution:The work done on a PV-diagram is the area under the curve.

WA→A = WAB +WBC +WCD +WDA

WA→A = WAB + PBC∆VBC +WCD + PDA∆VDA

WA→A = WAB + PBC(0) +WCD + PDA(0)

WA→A = WAB +WCD

From Problem 19:

WAB = P2(V2 − V1) + (1/2)(P3 − P2)(V2 − V1)

= (62.1 kPa)(7.01 m3 − 2.24 m3) + (1/2)(88.1 kPa− 62.1 kPa)(7.01 m3 − 2.24 m3)

= 296.22 kJ + 62.01 kJ

= 358.23 kJ

The work for D → A is found by the following:

WCD = P1(V1 − V2) = (30.2 kPa)(2.24 m3 − 7.01 m3) = −144.05 kJ

Now calculate the total work:

WA→A = WAB +WCD = 358.23 kJ +−144.05 kJ = 214.17 kJ

physics 220 - final exam - summer 2014 version 31 - answer key

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