Physics 1200 Final Exam Spring 2018 (downloaded from course website on Carmen) Spring 2018 Physics 1200 Final Exam Rooms

Section Day, Date, and Time of Final Exam12:40 Tue 5/01/2018 12:00 pm - 1:45 pm1:50 Tue 5/01/2018 4:00 pm - 5:45 pm3:00 Fri 4/27/2018 4:00 pm - 5:45 pm4:10 Mon 4/30/2018 4:00 pm - 5:45 pm

Identify your Final Exam location based on your Lecture Section (column) and Recitation Section (Row) Building names and addresses are given below the table.

LectureRecitation

Yang12:40

Yang1:50

Humanic3:00

Humanic4:10

Burrola Gabilondo SM 1153 SM 1153 SM 1153 `

Codecido SM 1153 SM 1153

Guo SM 1153 SM 1153

Humanic MP1000 SM 1153

Lake MP1000 SM 1153

Moore SM 1153 SM 1153

Nandyala SM 1153 SM 1153

Pfendner SM 1153 MP1000 EL 1008

Pollock EL 1008 MP1000

Suresh MP1000 EL 1008

Teng MP1000 EL 1008

Vasquez EL 1008 MP1000

Wei EL 1008 MP1000

EL = Evans Lab, 88 W 18th Ave.MP = 140 W 18th Ave.SM = Smith Lab, 174 W 18th Ave.

Physics 1200 Final Exam -- continued

•  105 minutes •  Comprehensive – covers all of the course material (more

or less) uniformly •  35 multiple-choice questions •  Equations and constants will be provided on the exam

Page 1 of equations given on the Final Exam

Final Exam -- Physics 1200 Spring 2014 -- 10:20 am lecture section --Humanic

Tuesday, April 29, 2014, 8:00 am - 9:45 am

Name _________________________________Rec. Instructor_____________________ * Exam is closed book and closed notes. * Write your name and the name of your recitation instructor on every page of the exam. * This exam consists of 35 multiple-choice questions (5 points each, 175 points total) * You have 1 hr 45 minutes to complete the exam.

Fx =max Fy =may F =G m1m2

r2∑ G = 6.67×10−11 Nm2 kg2 g(Earth)∑ = 9.80 m s2

vx = v0 x + axt x = 12 v0 x + vx( ) t x = v0 xt + 1

2 axt2 vx

2 = v0 x2 + 2axx

vy = v0 y + ayt y = 12 v0 y + vy( ) t y = v0 yt + 1

2 ayt2 vy

2 = v0 y2 + 2ayy W = F cosθ( )s

fk = µkFN h2 = ho2 + ha

2 sinθ = hoh

cosθ = hah

tanθ = hoha

F = −kx

!v = Δ!rΔt

!a = Δ

!vΔt

F =mg+ma aC =v2

rFC =

mv2

rfsMAX = µsFN

PEgravity =mgh PEspring = 12 kx

2 KE = 12mv

2 WNC = Ef −E0 E = KE +PE P =Wt

!Pf =

!P0

!P =!p1 +!p2

!p =m!v

!J =

!F∑( )Δt = !p f −

!p0

ω =ω0 +αt θ = 1

2 (ω0 +ω)t ω 2 =ω02 + 2αθ θ =ω0t + 1

2αt2 xCM =

m1x1 +m2x2

m1 +m2

vCM =m1v1 +m2v2

m1 +m2

m1v01 +m2v02 =m1vf 1 +m2vf 2 v01 − v02 = − vf 1 − vf 2( ) 12 m1v

201 + 1

2 m2v202 = 1

2 m1v2f 1 + 1

2 m2v2f 2

s = rθ ω =ΔθΔt

α =ΔωΔt

vT = rω aT = rα aC = rω2 1 rev = 2π rad = 360o ITHIN RING =MR

2

τ EXT = Iα τ = Fl I = mr2( )∑∑ IDISK = 1

2MR2L = I ω

Lf =

L0 τ EXT∑( )Δt = ΔL

ISOLID SPHERE = 25MR

2 WR = τθ KER = 12 Iω

2 WR = ΔKER E = 12 Iω

2 + 12Mv

2 +Mgh Ef = E0

Page 2 of equations given on the Final Exam Name ______________________________Rec. Instructor_____________________ Humanic Physics 1200 Final Exam April 29, 2014 8:00am-9:45am

ρ =mV

P = FA

P2 = P1 + ρgh FB =Wdisplaced fluid = ρVg

A1v1 = A2v2 Q = Av = ΔVΔt

ρwater =1000 kg/m3 Patm =1.013×105 Pa

P1 + 12 ρv1

2 + ρgy1 = P2 + 12 ρv2

2 + ρgy2 x = Acosωt vx = −Aω sinωt

ax = −Aω2 cosωt f = 1

Tω = 2π f ωspring-mass =

km

ωpendulum =gL

vsoundair = 343 m/s v = f λ I1

I2

=r2

2

r12 I = P

Aβ = 10 dB( ) log I

I0

!

"#

$

%&

I0 =1.00×10−12 W/m2 fn = nv

2L!

"#

$

%&, n =1, 2,3,.... v = F

m Lfn = n

v4L!

"#

$

%&, n =1,3, 5,.....

sinθbright =mλd

, m = 0,1, 2,3,.... sinθdark = m+ 12( ) λd

, m = 0,1, 2,3,.... fbeat = f1 − f2

sinθ1

v1

=sinθ2

v2

θ(radians) ≈ λD

sinθdarksingle-slit =m λ

W, m =1, 2,3,.... sinθdark

aperture =1.22 λD

source moving towards stationary observer: fo = fs1

1− vs v!

"#

$

%& , away: fo = fs

11+ vs v!

"#

$

%&

observer moving towards stationary source: fo = fs 1+ vov

!

"#

$

%& , away: fo = fs 1− vo

v!

"#

$

%&

M =vobjectvsound

sinθ = vsoundvobject

=1M

Chapter 12

Doppler effect and Shock waves

Page 1 of 1

Stickies 4/12/14, 10:25 PM

The Doppler Effect

Doppler effect for a moving source and fixed observer when vs ≥ v

fo = fs1

1− vs v"

#$

%

&' ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+=

vvff

sso 1

1moving toward observer moving away from observer

vs = ½ v

vs = v

vs = 2v

fo = 2 fs fo =23fs

fo =∞ fo =12fs

fo = − fs fo =13fs

not possible

not possible

observer only hears the sound after the source has passed by her

Page 1 of 1

Stickies 4/12/14, 10:25 PM

Page 1 of 1

Stickies 4/12/14, 10:25 PM

Shock waves Shock waves are produced by objects moving through a medium at a speed greater than the speed of sound.

Mach number =M =vobjectvsound

Mach angle: sinθ = vsoundtvobjectt

=vsoundvobject

=1M

vobject

vobjectt

Conical shock wave

vsoundt

θθ

Shock waves

t = time for object to go from point S0 to S5

Page 1 of 1

Stickies 4/12/14, 10:27 PM

Water vapor condensing into a cone due to a shock wave caused by the jet exceeding the speed of sound. Estimate M from the picture:

A sonic boom is heard on the ground whenever a plane exceeding the speed of sound passes an observer.

Shock waves

θ ≈ 70o sinθ = 1M

M ≈1

sin70o≈1.1

Example: The French/British supersonic passenger jet Concorde was in service from 1976 to 2003. It had a top speed of 1350 mph. Determine what Mach number this speed corresponds to and the angle of the shock cone produced.

The speed of sound in air = 343 m/s = 767 mph

MConconde =vobjectvsound

=1350767

=1.76

sinθ = 1M

=1

1.76= 0.568 ⇒ θ = 34.6o

Mach 1.76 θ = 34.6o (

Shock waves

Shock cone