final exam: 4/26/14 8:00pm-10:00pm - university of florida · your exam 2 4 your exam 1 4 ... for...

R. Field 4/20/2014 University of Florida PHY 2053

Final Exam: 4/26/14 8:00pm-10:00pm

MAT 18V-ZFLG 280S-UNRN 137H-RCLB C130A-GRoomLast Name

You may bring two hand written formula sheet on 8½ x 11 inch paper (both sides).

• Room Assignments:

• Crib Sheet:

• Breakdown of the 20 Problems

• Calculator:

8Chapter 1.1-10.84Chapter 11.1-12.84Your Exam 24Your Exam 1

# of ProblemsMaterial

You should bring a calculator (any type).

• Scratch Paper: We will provide scratch paper.


Make-Up Exam: 4/23/14 5:10-7:00pm

You may bring two hand written formula sheet on 8½ x 11 inch paper (both sides).

• Room:

• Crib Sheet:

• Breakdown of the 20 Problems:

• Calculator: You should bring a calculator (any type).

• Scratch Paper: We will provide scratch paper.

Similar to the Final Exam.

For anyone who had to miss an exam during the semester.

MAEA 327.


Final Exam Fall 2010: Problem 9• Near the surface of the Earth a startled armadillo leaps vertically

upward at time t = 0, at time t = 0.5 s it is a height of 0.98 m above the ground. At what time does it land back on the ground?

Answer: 0.9 s% Right: 47%

v0

y-axis )()( 2

1221 gtvtgttvty oo −=−=

)(0)( 21

foff gtvtty −==

gvt o

f2

=

221)( hhoh gttvhty −==

h

ho t

gthv2

21+

=

ssssssm

mtgt

hgt

gthgvt h

hh

hof 9.0)5.0()4.0()5.0(

)5.0)(/8.9()98.0(22)(22

2

221

=+=+=+=+

==


Final Exam Fall 2011: Problem 7• Near the surface of the Earth a suspension bridge is a height H

above the level base of a gorge. Two identical stones are simultaneously thrown from the bridge. One stone is thrown straight down with speed v0 and the other is thrown straight up a the same speed v0. Ignore air resistance. If one of the stones lands at the bottom of the gorge 2 seconds before the other, what is the speed v0(in m/s)?

Answer: 9.8% Right: 36%

)()( 2221

0 downupdownup ttgttv −=+

221

221

0)()(

upupoupup

oup

gttvHtygttvHty−+==

−+=

HgttvHgttv

downdowno

upupo

+−=−=

221

221 y-axis

x-axis

H

221

221

0)()(

downdownodowndown

odown

gttvHtygttvHty−−==

−−=

smssmttgtttt

gv downupdownup

downup /8.9)2)(/8.9()()()( 2

21

21

22

21

0 ==−=+

−=


Exam 1 Fall 2010: Problem 12

θ

Mg x-axis

y-axis

fs FN

• Near the surface of the Earth, a block of mass M is at rest on a plane inclined at angle θ to the horizontal. If the coefficient of static friction between the block and the surface of the plane is 0.7, what is the largest angle θ without the block sliding?

Answer: 35o

% Right: 83%

θ

M

0sin ==− xs MafMg θ

0cos ==− yN MaMgF θ

o35≈θ

θμμθ cossin MgFfMg sNss ===

7.0tan == sμθ


Exam 1 Spring 2014: Problem 22• In the figure, blocks A and C have a mass of 30 kg and 10 kg,

respectively. If the surface of the table is frictionless and the static coefficient of friction, μs, between block A and block C is 0.60, what is the maximum mass of block B (in kg) such that, when the system is released from rest, block C will not slip off block A?

Answer: 60% Right: 19%

aMf Cs =

gMFfaM CsNssC μμ === maxmax )(

ga sμ=max

aMFgM BTB =−

aMMF CAT )( +=

aMMMgM CBAB )( ++=

gMMM

MaCBA

B

++=

CBA

Bs MMM

M++

=μ

BCBAs MMMM =++ )(μ

kgkgkgkgMMM CAs

sB 60)40()1030(

6.016.0)(

1 23 ==+

−=+

−=

μμ MBg

Fs

FT

FT

FN


Exam 1 Spring 2014: Problem 55• A race car starts from rest at t = 0 and travels around a circular track of radius R with a

constant angular acceleration. If the race car completes its first revolution around the track in 3 minutes, at what time t (in s) is the magnitude of the tangential acceleration of the car equal to the magnitude of the radial acceleration (i.e. centripetal acceleration) of the car?


221)(

)()(

tttt

t

αθαωαα

===

222 )()( tRtRtaradial αω ==

αRtat =)(

αα RatRta tradial === 22)(

sstN

tt 8.50)1(4

min)/60min)(3()(4

1

1

1 ≈===ππα

211 22

1)( ttN ⎟⎠⎞

⎜⎝⎛=

πα

21

1)(4t

tNπα =

Let t1 be the time for the first revolution, N(t1) = 1.


Final Exam Spring 2011: Problem 12• A uniform disk with mass M, radius R = 0.50 m, and moment of

inertia I = MR2/2 rolls without slipping along the floor at 2 revolutions per second when it encounters a long ramp angled upwards at 45o

with respect to the horizontal. How high above its original level will the center of the disk get (in meters)?


x-axis

v v

θ = 45ο H

R

)(2212

21 RHMgEMgRIMvE fi +==++= ω

mssmmf

MRI

gR

MRI

gR

MIR

gMIv

gH

0.3)/2(23

/8.9)5.0(2121

2

21

21

22

222

2

222

2

2

22222

≈⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

ππω

ωωωf

Rvπωω

2==


Final Exam Spring 2012: Problem 40• A thin hoop (I = MR2) and a solid spherical ball (I =

2MR2/5) start from rest and roll without slipping a distance d down an incline ramp as shown in the figure. If it takes 2 s for the ball to roll down the incline, how long does it take for the hoop (in s)?

xMafMg =−θsinxRa

Rv==

αω

θ

Hoop or Ball

d


θ

x-axis

f R Mgsinθ

RfaRII x === ατ

xx MaaRIMg =− 2sinθ

221 tad x=⎟

⎠⎞

⎜⎝⎛ +

=

21

sin

MRI

gaxθ

ssballthoopt 39.2)2(1952.1)(1952.1)( ≈×=×=

( )( ) 7.0

111

1

1

)()(

1075

2

2

2

==++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎠⎞

⎜⎝⎛ +

=

MRIMRI

ballahoopa

hoop

ball

x

x

xadt 2

=

1952.17.0

1)()(

)()(

===hoopaballa

ballthoopt

x

x



R FT

Mg

y-axis

• A cloth tape is wound around the outside of a non-uniform solid cylinder (mass M, radius R) and fastened to the ceiling as shown in the figure. The cylinder is held with the tape vertical and then release from rest. If the acceleration of the center-of-mass of the cylinder is 3g/5, what is its moment of inertia about its symmetry axis?

Answer: % Right: 48%

232 MR

R

yT MaFMg =−

RIa

IRF yT === ατ

)( yT agMF −=

2322

53

53

22

MRMRg

ggMRa

agaFRI

y

y

y

T =⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛ −==


Exam 2 Spring 2014: Problem 30• Near the surface of the Earth a cloth tape is wound around the

outside of a thin uniform hoop with mass M, radius R, and moment of inertia Ihoop = MR2. A cloth tape is also wound around the outside of a non-uniform cylinder with the same mass M and the same radius R, but with moment of inertia Icylinder. Both the hoop and the cylinder are fastened to the ceiling as shown in the figure. They are both released from rest at t = 0 at the same distance from the ceiling. If at t = 1.81 s the two objects are a vertical distance of 2 meters apart, what is the moment of inertia of the cylinder Icylinder?

R R

Answer: 0.60MR2

% Right: 20% R FT

Mg

y-axis

yT MaFMg =−

RIa

IRF yT === ατ

221

221

taytay

cylindercylinder

hoophoop

==

)/1( 2MRIgay +

=6.025.0125.01

41

41

2

2

2 ≈+−

≈Δ

+

Δ−

=

gty

gty

MRIcylinder

gMRI

gahoop

hoop 21

2 )/1(=

+=

221 atyyy hoopcylinder Δ=−=Δ

)/1( 2MRIga

cylindercylinder +

=

2)/1( 2

gMRI

gaaacylinder

hoopcylinder −+

=−=Δ

2

2t

ya Δ=Δ

25.0)81.1)(/8.9(

)2(44222 ≈=

Δssm

mgt

y


Exam 2 Spring 2012: Problem 10

• A cannon on a railroad car is facing in a direction parallel to the tracks as shown in the figure. The cannon can fires a 100-kg cannon ball at a muzzle speed of 150 m/s at an angle of θ above the horizontal as shown in the figure. The cannon plus railway car have a mass of 5,000 kg. If the cannon and one cannon ball are travelling to the right on the railway car a speed of v = 2 m/s, at what angle θ must the cannon be fired in order to bring the railway car to rest? Assume that the track is horizontal and there is no friction.

Answer: 48.2o

% Right: 57%iBallcarix VMMp )()( +=

θ v

x-axis

)cos()( θmuzzleiBallballBallfx VVMVMp +==

o2.48≈θ

fxix pp )()( = )cos()( θmuzzleiBalliBallcar VVMVMM +=+

6667.0)/150)(100()/2)(5000(cos ===smkgsmkg

VMVM

muzzleBall

icarθ


Exam 2 Spring 2014: Problem 16• A firecracker is tossed straight upward into the air. It explodes into three pieces

(A, B, and C) of equal mass just as it reaches the highest point. If piece A moves off at a speed of 4 m/s and piece B and C move off with equal speeds of 5 m/s, what is the angle between piece B and C?

Answer: 132.8o

% Right: 16% v0

A vA

y-axis

x-axis v0

B

C θ

0=++=++ CBACBA vMvMvMppp rrrrrr

)( CBA vvv rrr+−=

xvvC ˆ0=r

yvxvvA ˆsinˆ)cos1( 00 θθ −+−=r

yvxvvB ˆsinˆcos 00 θθ +=r

[ ] [ ]θθθ cos22sin)cos1( 20

2220

22 +=++== vvvv AAr

o8.132≈θ

68.01)/5(2)/4(1

2cos 2

2

20

2

−=−=−=smsm

vvAθ



• A uniform solid disk with mass M and radius R is mounted on a vertical shaft with negligible rotational inertia and is initially rotating with angular speed ω. A non-rotating uniform solid disk with mass M and radius R/2 is suddenly dropped onto the same shaft as shown in the figure. The two disks stick together and rotate at the same angluar speed. What is the new angular speed of the two disk system? Answer: 4ω/5% Right: 47%

R M

M

ffffiii MRMRILMRIL ωωωω )( 222

1212

1212

1 +=====

ωωωω 54

22

2

22

21

21

)2/(=

+=

+=

RRR

RRR

f


Exam 1 Spring 2014: Problem 36• A stone of mass m is released from rest a height h = 1.5RE above the surface of the

Earth, where RE is the radius of the Earth. What is the speed of the stone (in km/s) when it hits the Earth? Ignore air resistance and take RE = 6.371×106 m, ME = 5.974×1024 kg, and G = 6.674×10-11 N·m2/kg2, where ME is the mass of the Earth and G is Newton’s constant.


E

Ei Rh

GmME+

−=if EE =

E

Eff R

GmMmvE −= 221

226

242211

2 /823.9)10471.6(

)10974.5)(/10674.6( smm

kgkgNmR

GMgE

E ≈×

××==

−

E

E

E

E

E

E

E

E

E

Ef Rh

hgRRh

RR

GMRh

GMR

GMv+

=⎥⎦

⎤⎢⎣

⎡+

−=⎥⎦

⎤⎢⎣

⎡+

−=21222

skmmsmRgRh

hgRv ERh

E

Ef E

/67.85.2

)10371.6)(5.1)(/823.9(25.2

)5.1(22 62

5.1 ≈×

=⎯⎯⎯ →⎯+

= =

h

RE

ME


Final Exam Spring 2011: Problem 5• A toy merry-go-round consists of a uniform disk of mass M and

radius R that rotates freely in a horizontal plane about its center. A mouse of the same mass, M, as the disk starts at the rim of the disk. Initially the mouse and disk rotate together with an angular velocity of ω. If the mouse walks to a new position a distance r from the center of the disk the new angular velocity of the mouse-disk system is 2ω. What is the new distance r?

Answer: R/2% Right: 65% ffiiii MRMrLMRMRIL ωωω )()( 2

12212 +==+==

221

43

21

23 RRRr

f

i =−=−=ωω



d water y-axis

Mg

Fbuoyancy

• Two small balls are simultaneously released from rest in a deep pool of water (with density ρwater). The first ball is reseased from rest at the surface of the pool and has a density three times the density of water (i.e. ρ1 = 3ρwater). A second ball of unknown density is released from rest at the bottom of the pool. If it takes the second ball the same amount of time to reach the surface of the pool as it takes for the first ball to reach the bottom of the pool, what is the density of the second ball? Neglect hydrodynamic drag forces. Answer: 3ρwater/5% Right: 22%

221 tad y=

upballupballupballwater aVgVgV ρρρ =−

yadt 2

=

ggaup

water

up

upwaterup ⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

−= 1

ρρ

ρρρ

Mg

d

water

Fbuoyancy

y-axis

downballdownballwaterballdown aVgVgV ρρρ =−

ggadown

water

down

waterdowndown ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−=

ρρ

ρρρ 1

gagadown

waterdown

up

waterup ⎟⎟

⎠

⎞⎜⎜⎝

⎛−==⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

ρρ

ρρ 11

waterwater

downwater

waterup ρρ

ρρρρ 5

3

312/2

=−

=−

=


Exam 2 Spring 12: Problem 46• What is the minimum radius (in m) that a

spherical helium balloon must have in order to lift a total mass of m = 10-kg (including the mass of the empty balloon) off the ground? The density of helium and the air are, ρHE = 0.18 kg/m3

and ρair = 1.2 kg/m3, respectively. Answer: 1.33% Right: 37%

0)( =+− gmVgV ballonHEballonair ρρ

0)( =+− gmMF HEbuoyancy (MHE+m)g

Fbuoyancy

y-axis r

m

HEairballon

mrVρρ

π−

== 334

( ) mmmkg

kgmrHEair

33.134.2)/02.1(4

)10(3)(4

3 31

31

31

33 ≈=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=πρρπ


Exam 2 Spring 2014: Problem 43• A cubical metal box with sides of mass M and length L has a square lid

also with mass M and length L. The lid is not attached to the box, however, the lid and the box form an airtight seal. Near the surface of the Earth, the lid is held at rest by a steel cable, as shown in the figure. The pressure outside the box is the atmospheric pressure, Pout = Patm = 101 kPa. The box is partially evacuated to an inside pressure Pin so that the box remains at rest. If M = 25 kg and L = 0.4 m, what is the maximum inside pressure Pin(in kPA) such that the box will not fall?

Pout

Pin L

L


APin

L

y-axis

APout

Mtotg

0=−− gMAPAP totinout

kPakPakPam

smkgPLMgP

AgMPP atmatm

totoutin 3.93656.7101

)4.0()/8.9)(25(55

2

2

2 ≈−=−=−=−=


Exam 2 Fall 2011: Problem 50• An ideal spring-and-mass system is undergoing simple harmonic

motion (SHM) with period T = 3.14 s. If the mass m = 0.5 kg and the total energy of the spring-and-mass system is 5 J, what is the speed (in m/s) of the mass when the displacement is 1.0 m?

Answer: 4.0% Right: 34% 2

212

212

21 kxmvkAE x +==

22

22222 4222 x

TmEx

mEx

mk

mEvx

πω −=−=−= T

mk

πω

ω

2=

=

( )smsmsm

skgJx

TmEvx /4)/()/20()1(

14.34

5.0)5(242 22222

2

22

2

2

=−=−=−=ππ


Final Exam Spring 2011: Problem 20

Ambuance Car VA Vcar

vsound

Ambuance Car Vcar VA

• An ambulance moving at a constant speed of 20 m/s with its siren on overtakes a car moving at a constant speed of 10 m/s in the same direction as the ambulance. As the ambulance approaches the car the driver of the car perceives the siren to have a frequency of 1200 Hz. What frequency does the driver of the car perceive after the ambulance has passed? This takes place on a cold night in Alaska where the speed of sound is 320 m/s.

Answer: 1127 Hz% Right: 53%

vsound

0

0

fvvvvf

fvvvvf

Asound

carsoundaway

Asound

carsoundtoward

++

=

−−

=

))(())((

carsoundAsound

Asoundcarsound

toward

away

vvvvvvvv

ff

−+−+

=

HzHzfvvvvvvvvf towardcarsoundAsound

Asoundcarsoundaway 1127)1200(

)10320)(20320()20320)(10320(

))(())((

≈−+−+

=−+−+

=

final exam: 4/26/14 8:00pm-10:00pm - university of florida · your exam 2 4 your exam 1 4 ... for...

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