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PHY 222 University Physics II

8-10:30 pm Dec 6, 2019

Practice Final Exam Your Name _________________________________________________________________ Your C-number ______________________________________________________________ Your Signature ______________________________________________________________ Your Discussion (circle one) 5O (9:30 am – 10:20 am)

5Q (12:30 pm – 1:20 pm)

5R ( 2:00 pm – 2:50 pm)

Problems 1-20 have multiple choice answers and are worth 5 points each. Please copy your answers to the table on the front page. Problem 21 (25 points) and 22 (25 points) are the essay problems. To receive full credit you need to write down the complete solution and to circle your answers. Multiple-choice answers 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Exam grade MC

21

22

Total

The honor code is enforced. Format (FM) 1

1" "

Equation sheet Pressure & depth: dP = ρ gdy ; if ρ is constant: P = P0 + ρgh 1 2

FT

Speed of propagation of a wave on a string: v = Waves: D( x, t) = Asin(kx − ωt ) or 2π

ω=

λ

area

I∝

v=

1 r2

λ

=

T

ω k

I ∝ A2

⎛ ⎞ Sound level: β (in dB) = 10 log10 ⎜ I ⎟ I ⎝

Doppler effect: f ' = f

µ

D( x, t) = Asin(kx + ωt ) dependent on wave travelling direction

2π = 2π f T

Intensity: I = Power

1 2

Bernoulli: P1 + ρ gh1 + ρ v12 = P2 + ρ gh2 + ρ v22

Continuity: A1v1 = A2v 2

k=

Buoyant force: FB = ρgV

vsnd ± vobs vsnd ∓ vsource

0

Beat frequency: f = f1 − f2

⎠

First sign: move approach each other; second sign: move apart. volume expansion ΔV = β V ΔT

Thermal expansion: linear expansion ΔL = α LΔT Specific heat: Q = cmΔT

Latent heat: Q = mL

Ideal gas law: PV = nRT = NkT

k=

1 2

R NA

3 2

rms speed vrms = v2 = 3kT

kinetic energy of gas molecule: K = mv 2 = kT

m

1st law of thermodynamics: ΔEint = Q −W Internal energy:

1 nRT ×number of degrees of freedom 2

Internal energy for monatomic molecule: 3 nRT ; Internal energy for diatomic molecule: 5 nRT 2

2

Work: dW = PdV Adiabatic process: PV γ =constant γ=

Cp CV

=

C 5 7 for monatomic molecule; γ = p = for diatomic molecule; Cp = CV + R 5 CV 3

2" "

Heat transfer:

T −T ΔQ = kA H L Δt L

dQ Entropy: dS = T

Engine efficiency: e = W

QH

Carnot engine: e = 1− TL

TH

Coefficient of performance (refrigerator): COP =

Ray Model of Light: Mirror equations

Refraction v =

QL W

Carnot refrigerator: COP =

d 1 1 1 R + = (where f = ) and m = − i 2 do d o di f

c n

Thin lens equations

λn =

λ n

Snell’s Law n1 sinθ 1 = n2 sinθ 2

1 1 1 + = d o di f

and m = −

di do 1 dark spots d sinθ = (m + )λ 2

Double slit: bright spots d sinθ = mλ

Diffraction:

Single slit: D sinθ = mλ dark: m= ±1, ±2… bright: m= 0, ±

Rayleigh criterion θ ≈

"

Δt = γΔt0 =

"

1.22λ D

"

l0 v 2 Δt0 " l = = l0 1− ( ) γ v c " 1− ( )2 c 2 K = (γ −1)mc

p = γmv "

"

Relativistic addition of velocities: u =

"

E = K + mc2 ""

"

u'+ v v 1+ 2 u' c "

Doppler shift for light: S, D move towards λ = λ0

3" "

5 3 ,± … 2 2

Iθ = I 0 cos2 θ

Polarization: Relativity:""

TL TH − TL

c−v c+v

S, D move away λ = λ0

c+v c−v

1. In its own reference frame, a box has the shape of a cube 2.0 m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship then flies past us with a horizontal speed of 0.80c. What is the volume of the box as we observe it? Please write down the equation and put the numerical values in. You do not need to solve the maths. [A] 8m 3 [B] 4.8m 3 [C] 6.4m 3 [D] 2.88m 3 [E] 1.728m 3 Solution:"The"dimension"along"the"direction"of"motion"is"contracted,"and"the"other"two"dimensions"are" 2

( ) ( )

unchanged l = l 0 1− v 2 c 2 ; V = l l 0 = l 0

3

(

1− v 2 c 2 = 2.0 m

)

3

(

)

2

1− 0.80 = 4.8m3 "

2. What is the speed of a pion if its average lifetime is measured to be 4.40×10-8 s? At rest, its average lifetime is 2.60×10-8 s. Please write down the equation and put the numerical values in. You do not need to solve the maths. The"speed"is"determined"from"the"time"dilation"relationship."

Δt0 = Δt 1 − v 2 c 2 "

"

→

2

⎛ 2.60 × 10− 8 s ⎞ ⎛ Δt ⎞ v = c 1−⎜ 0 ⎟ = c 1− ⎜ −8 ⎟ ⎝ Δt ⎠ ⎝ 4.40 × 10 s ⎠

2

"

3. A spaceship moving toward Earth at 0.8c transmits radio signals at 95 MHz. At what frequency should earth receivers be tuned? [A] 95 MHz [B] 190 MHz [C] 285 MHz [D] 380 MHz [E] 47.5 MHz

f = f0

c+v 1+ v c = f0 = 95.0 MHz c−v 1− v c

(

)

1+ 0.80 = 285MHz " 1− 0.80

"

Use the following condition to solve problems 4-5. A large, sealed container with area A and height L is filled with water (density ρ) up to level y. The remainder of the container is filled with n moles of air at temperature T. A small hole with area σ is punched at the bottom of the container to let water come out. 4. Calculate the pressure of the air in the container when the water level is y. nRT A(L − y) nRT nRT [A] [B] [C] [D] A(L − y) AL nRT Ay

PA(L − y) = nRT ⇒ P =

[E]

Ay nRT

nRT " A(L − y)

5. Calculate the speed at which water comes out of the container. Assume A>> σ. The atmosphere is P0. 4" "

2( [A]

[D]

nRT − P0 + ρgy) A(L − y) ρ

nRT − P 0 + ρ gy A(L − y) ρ

2( [B]

2( [E]

nRT + ρgy) A(L − y) ρ

2( [C]

nRT − P0 ) A(L − y) ρ

nRT − P0 + ρ gy) Ay ρ

Bernoulli’s equation: ρv 2 " P + ρ gy = P0 + 2 2 nRT 2 v= ( + ρgy − P0 ) " (P + ρ gy − P0 ) = ρ ρ A(L − y) "

6. A heat engine receives 8000 J of heat from the burning fuel and releases 5000 J of heat to the environment in each cycle. If the cycle is a Carnot cycle and the hottest temperature in the cycle is 800K, find the coldest temperature in the cycle. (A) 400K (B) 308K (C) 2133K (D) 300K (E) 500K QH=8000 J, QL=5000 J, W=QH-QL=8000-5000=3000J

T W 3 = =1− L " QH 8 TH TL 3 5 =1− = " 8 8 TH 5 TL = TH = 500K 8 e=

7. A refrigerator uses an ideal gas with constant volume molar heat capacity cv as working substance. The gas is initially at temperature T0, Volume V0, and pressure p0. The gas first expands at constant pressure (isobaric) two write the volume, then it is compressed adiabatically to the initial volume, finally it returns to the original state through an isochoric process. Write your results ONLY in terms of cv, T0, V0, and p0. Calculate the temperature of the gas after the adiabatic compression.

5" "

R +1 cV

[A] T0 (2)

R cV

[B] T0 (2)

[C]

3 T0 2

[D] 2T0

[E] 3T0

For ideal gas law: P0V0 = nRT0 γ For adiabatic process PV is constant P0 (2V0 )γ = PcV0γ γ

Pc = P0 2 = P0 2

Cp CV

= P0 2 R +1 CV

PV PV 2 Tc = c c = 0 0 nR nR

CV +R CV

= P0 2

R +1 CV

= T0 2

R +1 CV

"

"

Use the following condition to solve problems 8-10. A heat engine consists of n moles of diatomic ideal gas taken through the rectangular, reversible cycle shown.

6" "

8. Find the total heat entering the system and the work done per cycle in terms of p0 and V0. 31 31 p0V0 ; 4 p0V0 p0V0 ; 4 p0V0 [B] [C] 0;2 p0V0 [A] 2 4 31 31 p0V0 ;2 p0V0 [D] [E] p0V0 ;−2 p0V0 2 2

9. What is the efficiency of the engine? 2 4 [B] [C] 0 [A] 31 31

[D]

4 15

[E]

27 31

10. What is the entropy change in going from point c to point a in the diagram? [express your answer in terms of n and R (the gas constant)] nR nR [B] [A] [5ln(3) + 7 ln(2)] [7 ln(3) + 5 ln(2)] 2 2 nR nR [C] − [5ln(3) + 7 ln(2)] [D] − [7 ln(3) + 5 ln(2)] 2 2 [E] −nR[5ln(3) + 7 ln(2)]

7" "

11. A visible light with wavelength λ is most strongly reflected by a soap bubble (index of refraction is n). λ is the wavelength in vacuum or air. What is the minimum thickness of the soap bubble? (Hint: light reflected by a material with index of refraction greater than that of the material in which it is traveling, changes phase by 180°.) [A]

λ 4

[B]

λ 2

[C]

λ n

[D]

λ 4n

[E]

λ 2n

Front surface reflection phase shift 180°; back surface reflection phase shift 0, so front and back surface reflection has a total phase shift 180°. 2t=(m+1/2)λn==(m+1/2)(λ/n) when m=0, tmin=

λ 4n

Use the following condition to solve problems 12 and 13. A single slit of width 100 µm is illuminated (at normal incidence) by light of wavelength λ=600 nm and the diffraction pattern viewed on a screen 1 m away. 12 What is the distance on the screen between the first and second diffraction minima? (A 6mm

(B) 4 mm

(C) 8 mm

(D) 12 mm

8" "

(E) 3mm

13. Suppose in addition to λ=600 nm light, monochromatic light of unknown wavelength λ’= also illuminates the slit. If the distance on the screen between the second diffraction minima for λ and λ’ is 1 mm, what is λ’? (A) 550 nm

(B) 650 nm

(C) 700 nm

(D) 550 nm or 650 nm

(E) 600 or 700 nm

14. A particular organ pipe can resonate at 264 Hz, 440 Hz, and 616 Hz, but not at any other frequencies in between. Is this an open or a closed pipe? What is the fundamental frequency of this pipe? (A) Open, 176 Hz (D) Closed, 88Hz

(B) Open, 88 Hz (E) None of the above

(C) Closed, 176 Hz

15. A guitar string is 100 cm long and has a mass of 1 g. From the bridge to the support post is 50 cm and the string is under a tension of 90N. What is the fundamental frequency and second harmonic frequency? (A) 300Hz, 600Hz (B) 300 Hz, 900 Hz (C) 600Hz, 1200Hz (D) 450Hz, 900Hz 9" "

(E) 150Hz, 300Hz

16. Two submarines are traveling with speeds v1 and v2 relative to the water as shown in the figure. The sound generator in submarine #1 is set to emit with a frequency f0. Say the speed of sound waves in the water is vw. Find the frequency f1 received by submarine #1 after the waves are reflected from submarine #2.

(A) f1 = f0 (

vw − v2 ) vw + v1

(D) f1 = f0 (

vw − v2 vw − v2 )( ) vw − v1 v w − v1

(B) f1 = f0 (

vw + v1 v w + v2 )( ) vw − v2 vw − v1

(E) f1 = f0 (

vw − v2 vw + v1 )( ) vw − v1 v w + v2

10" "

(C) f1 = f0 (

vw + v2 vw − v1 )( ) vw + v1 v w − v2

Use the following condition to solve problems 17 and 18. Two stationary submarines communicate using sound waves of wavelength λ. They are both at depth h=λ below the ocean surface and separated by a horizontal distance d. Sub 1 sends a sonar signal o sub 2. Sub 2 receives a superposition of the signal that travels directly to it and a reflection from the ocean surface (assume no phase shift upon reflection).

17. If the interfering signals produce a minimum intensity, what is the possible distance between the subs d? (A)

15 λ 4

(B)

7 λ 12

(C)

3 λ 2

(D)

11" "

15 λ 4

and

7 λ (E) 12

15 λ, 4

3 λ 2

and

7 λ 12

When m=0, d=

7 15 λ When m=1, d= λ "Both"are"possible." 12 4

18. If the interfering signals produce a maximum intensity, what is the distance between the subs d? (A)

15 λ 4

(B)

7 λ 12

(C)

3 λ 2

(D)

12" "

15 λ 4

and

7 λ (E) 12

15 λ, 4

3 λ 2

and

7 λ 12

3 2

When m=1, d= λ "

Use the following conditions to solve problems 19 and 20. Two rods with same length d and same sectional area A are made of copper (thermal conductivity kc) and aluminum (thermal conductivity ka). The two rods are connected in series between two systems. One consisting of a mass m of ice at T1=0°C (latent heat L), the other maintained at a constant temperature T2>T1 (see figure). Assume that there is no dissipation of heat to the surroundings.

19. Calculate the temperature of the joint between the two rods.

(A)"

kcT1 + kaT2 kc + k a

(B)

k aT1 + k cT 2 kc + ka

(D)

kcT1 −kaT2 kc + ka

(E)

k aT1 − k cT 2 kc + ka

(C)

T1 + T2 2

ΔQ ka A(T2 − T ) kc A(T − T1 ) = = " d d t ka (T2 − T ) = kc (T − T1 ) "

T (ka + kc ) = kaT2 + kcT1 " T=

kaT2 + kcT1 ka + kc

20. Calculate the heat transferred per unit time between the two systems. (A)" k c k a

d T2 −T1 A kc + ka

(B) k c k a

A T2 − T1 d kc + ka

(D) k c k a

d T2 + T1 A kc + ka

(E) k c k a

A T2 + T1 d kc + ka

(C) k c k a

13" "

T2 − T1 kc + ka

ΔQ ka A(T2 − T ) = = t d

ka A(T2 −

kcT1 + kaT2 ) kc + ka A(T2 − T1 ) = ka kc d(kc + ka ) d

14" "

21. [25 points]

a) bob’s exam time is dilated in Alice’s reference frame.

TB =

T 'B 1−

u2 c2

b) Alice’s exam time is dilated in Bob’s reference frame. TA = T ' A 1−

u2 c2

c)"bob’s"exam"time"is"TB’"in"his"reference"frame"and"the"earth"in"his"reference"frame"is"moving"away"with" a"speed"of"u.""

T 'B u " d)"Bob’s"exam"time"in"Alice"reference"frame"is"TB"and"the"Bob"in"her"reference"frame"is"moving"away" with"a"speed"of"u."" "

TB u =

T 'B u2 1− 2 c

u"

15" "

22. [25 points]

"""""""""""e)""""Find"the"total"magnification.""""""""""""" """""""""""f)"""""Is"the"image"real"or"virtual?" "

" Solution:""

16" "

" a)"and"b)"Rays"and"images"are"plotted"" c) do,a = s "

1 1 1 = " + do, a di,a f

di,a = (

1 1 −1 fs " − ) = f s s− f

You"will"find"that" di,a < 0 "so"the"image"is"on"the"left"of"lens"a."" distance"between"lens"a"and"image"is" di,a =

d)" m = ma mb =

do,b =

fs " f −s

f f f −s = " × f −s 2f −s 2f −s

f2 fs " +f= f −s f −s

1 1 1 + =− " do,b di,b f

17" "

di,b = (−

1 f

−

1 f −s 1 )−1 = (− 1 1 −1 2 f − s −1 f2 " − f 2 ) = (− − 2 )−1 = (− ) = − 2f −s d o, b f f f f2 f −s

You"will"find"that" di,b < 0 "so"the"image"is"on"the"left"of"lens"b."" distance"between"lens"b"and"image"is" di,b =

f2 " 2f −s

sf d i,a f s− f " e)" ma = − =− = s d o, a f −s

f2 d 2f −s = f −s " mb = − i,b = − f2 d o, b 2f −s f −s −

m = ma mb =

f −s f f × = " f −s 2f −s 2f −s

f)"Image"is"virtual.""

18" "...