Final Exam: 4/27/13 12:30pm-2:30pm
• Room Assignments:
Last Name
Room
A-G
• Breakdown of the 20 Problems
Material
# of Problems
TUR L007
Your Exam 1
4
H-K
TUR L005
Your Exam 2
4
L-M
TUR L011
Chapter 11.1-12.8
4
N-R
WEIL 270
Chapter 1.1-10.8
8
S-Z
WM 100
• Crib Sheet:
• Calculator:
You may bring two hand written formula sheet on 8½ x 11 inch
paper (both sides).
You should bring a calculator (any type).
• Scratch Paper:
R. Field 4/21/2013
University of Florida
We will provide scratch paper.
PHY 2053
Page 1
Make-Up Exam: 4/24/13 5:10-7:00pm
For anyone who had to miss an exam during the semester.
• Room: 1220 NPB.
• Breakdown of the 20 Problems:
Similar to the Final Exam.
• Crib Sheet:
You may bring two hand written formula sheet on 8½ x 11 inch
paper (both sides).
• Calculator:
You should bring a calculator (any type).
• Scratch Paper:
R. Field 4/21/2013
University of Florida
We will provide scratch paper.
PHY 2053
Page 2
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%
y(t )  vot  12 gt2  t (vo  12 gt)
y(t f )  0  t f (vo  gt f )
1
2
tf 
2vo
g
y(th )  h  voth  12 gth2
y-axis
v0
h  12 gth2
vo 
th
2vo 2(h  12 gth2 ) 2h
2(0.98m)
tf 


 th 
 (0.5s)  (0.4s)  (0.5s)  0.9s
2
g
gth
gth
(9.8m / s )(0.5s)
R. Field 4/21/2013
University of Florida
PHY 2053
Page 3
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)?
ydown (t )  H  vot  12 gt 2
yup (t )  H  vot  12 gt 2
Answer: 9.8
2
% Right: 36% yup (tup )  0  H  votup  12 gtup
votup  12 gtup2  H
votdown   gt
1
2
v0  g
1
2
2
down
(t  t
2
up
H
2
down
)
(tup  tdown )
R. Field 4/21/2013
University of Florida
2
ydown (tdown )  0  H  votdown  12 gtdown
v0 (tup  tdown )  12 g(t  t
2
up
y-axis
2
down
)
H
x-axis
 12 g (tup  tdown )  12 (9.8m / s 2 )(2s)  9.8m / s
PHY 2053
Page 4
Exam 1 Fall 2010: Problem 12
• Near the surface of the Earth, a block of mass M is at
rest on a plane inclined at angle  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
 without the block sliding?
Answer: 35o
% Right: 83%
M

y-axis
Mg sin   f s  Max  0
FN
fs
FN  Mg cos  May  0
Mg sin   f s  s FN  s Mg cos
tan  s  0.7

x-axis
Mg
  35
R. Field 4/21/2013
University of Florida
PHY 2053
Page 5
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)?
Answer: 3.0
% Right: 39%
v
2
H

R
Ei  Mv  I  MgR  E f  Mg(H  R)
1
2
v
1
2
2
1  2 I 2 1  2 2 I 2
H
v    
R    
2g 
M
M
 2g 

x-axis
v  R
  2f
R2 
I  2 2 2 R 2 
I  2 2 2 (0.5m) 2  3 
2



1

f

(
2
/
s
)
 3.0m
1 





2
2
2
2 g  MR 
g  MR 
9.8m / s  2 
R. Field 4/21/2013
University of Florida
PHY 2053
Page 6
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)?
v  R
Answer: 2.39
% Right: 53%
Hoop or Ball
Mg sin   f  Max
I
  Rax
  I  a x  Rf
R
x-axis
g sin 
I
a

x
Mg sin   2 a x  Ma x
I 

R
d  12 axt 2
1 
2 
 MR 
I ball 

1 

a x (hoop)  MR 2  1  52  7


 10  0.7
I
a x (ball) 
 1  1
1  hoop2 
 MR 
d

Mgsin
R

t
2d
ax
ax (ball)
t (hoop)
1


 1.1952
t (ball)
ax (hoop)
0.7
t (hoop)  1.1952 t (ball)  1.1952 (2s)  2.39s
R. Field 4/21/2013
University of Florida
PHY 2053
Page 7
f
Exam 2 Fall 2011: Problem 17
• A cloth tape is wound around the outside of a nonuniform 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?
R
2
Answer: 23 MR
% Right: 48%
Mg  FT  May
FT  M ( g  a y )
  RFT  I 
Iay
R
R
FT
y-axis
 g  53 g 
R 2 FT  g  a y 
2
 MR 2  23 MR 2
I

MR   3

 a 
ay
g
y
5




R. Field 4/21/2013
University of Florida
PHY 2053
Mg
Page 8
Exam 2 Spring 2012: Problem 10

v
x-axis
• 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  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  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 ( px )i  (M car  M Ball )Vi
( px ) f  M BallVball  M Ball (Vi  Vmuzzle cos )
% Right: 57%
cos 
( px )i  ( px ) f
(M car  M Ball )Vi  M Ball (Vi  Vmuzzle cos )
M carVi
(5000kg)(2m / s)

 0.6667
M BallVmuzzle (100kg)(150m / s)
R. Field 4/21/2013
University of Florida
PHY 2053
  48.2
Page 9
Exam 2 Fall 2011: Problem 31
• 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 . 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: 4/5
% Right: 47%
M
R
M
Li  Iii  12 MR12  Lf  I f  f  ( 12 MR12  12 MR22 ) f
R12
R2
f  2
 2
  54 
2
2
R1  R2
R  ( R / 2)
R. Field 4/21/2013
University of Florida
PHY 2053
Page 10
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 . 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 2. What is the new distance r?
Answer: R/2
% Right: 65%
Li  Iii  (MR2  12 MR)i  Lf  (Mr 2  12 MR) f
rR
R. Field 4/21/2013
University of Florida
3i 1
3 1 R
 R
 
2 f 2
4 2 2
PHY 2053
Page 11
Exam 2 Fall 2011: Problem 36
• Two small balls are simultaneously released from rest in a deep pool of
water (with density rwater). 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. r1 = 3rwater). 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
F
ball? Neglect hydrodynamic drag forces.
d  12 ayt 2
r V g  rwaterVball g  rdownVball adown
Mg
Answer: 3rwater/5 down ball
2d
 r

r  r water
% Right: 22%
d water
t
adown  down
g  1  water  g
y-axis
r down
ay
 r down 
 r water 
 r

aup  
 1 g  adown  1  water  g
 r

water
 r down 
 up

rwaterVball g  rupVball g  rupVball aup d
y-axis
r water
r water 3
F
rup 

 5 r water
 r water 
r water  rup
1
2  r water / r down 2  3
aup 
g 
 1 g


rup
Mg
 rup

buoyancy
buoyancy
R. Field 4/21/2013
University of Florida
PHY 2053
Page 12
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, rHE = 0.18 kg/m3
and rair = 1.2 kg/m3, respectively.
Answer: 1.33
% Right: 37%
r
y-axis
m
Fbuoyancy  (M HE  m) g  0
(MHE+m)g
rairVballon g  ( rHEVballon  m) g  0
Vballon  43 r 3 
m
r air  r HE
1
3
1
3

 

3m
3(10kg )
3
  

r  

2
.
34
m
3 
 4 ( r air  r HE )   4 (1.02kg / m ) 
R. Field 4/21/2013
University of Florida
Fbuoyancy
PHY 2053


1
3
 1.33m
Page 13
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%
E  12 kA2  12 mvx2  12 kx2
2E k 2 2E
2 E 4 2 2
2 2
v 
 x 
 x 
 2 x
m m
m
m T
2
x
vx 
2 E 4 2 2
 2 x 
m
T
R. Field 4/21/2013
University of Florida
k

m
2

T
2(5 J )
4 2
2
2
2
2
2

(
1
)

(
20
m
/
s
)

(
m
/
s
)  4m / s
2
0.5kg 3.14 s 
PHY 2053
Page 14
Exam 2 Fall 2011: Problem 52
• In the figure, two blocks (m = 5 kg and M = 15 kg) and
a spring (k = 196 N/m) are arranged on a horizontal
frictionless surface. If the smaller block begins to slip
when the amplitude of the simple harmonic motion is
greater than 0.5 m, what is the coefficient of static
friction between the two blocks? (Assume that the
system is near the surface of the Earth.)
spring
Answer: 0.5
% Right: 26%
f s  m ax  s FN
FN  m g
s 
R. Field 4/21/2013
University of Florida
amax  s g
2
amax  spring
A
k

(M  m)
k
A
( M  m)
k
196N / m
A
(0.5m)  0.5
( M  m) g
(20kg )(9.8m / s)
PHY 2053
Page 15
Final Exam Spring 2011: Problem 20
• 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
vsound
speed of sound is 320 m/s.
Answer: 1127 Hz
% Right: 53%
f away
f away
f toward
f away 

vsound  vcar
f0
vsound  v A
vsound  vcar

f0
vsound  v A
f toward 
VA
Ambuance
Vcar
Car
vsound
VA
Vcar
Car
Ambuance
(vsound  vcar )(vsound  v A )
(vsound  v A )(vsound  vcar )
(vsound  vcar )(vsound  v A )
(320 10)(320 20)
f toward 
(1200Hz)  1127Hz
(vsound  v A )(vsound  vcar )
(320 20)(320 10)
R. Field 4/21/2013
University of Florida
PHY 2053
Page 16