Final Exam: 04/28/12 10:00am-noon
• Room Assignments:
Last Name
• Breakdown of the 20 Problems
Room
Material
# of Problems
A-F
TUR L007
Your Exam 1
4
G-L
BRY 130
Your Exam 2
4
M-Q
LIT 109
Chapter 11.1-12.8
4
R-S
TUR L005
Chapter 1.1-10.8
8
T-Z
FLG 280
You may bring two hand written formula sheet on 8½ x 11 inch
paper (both sides).
• Crib Sheet:
• Calculator:
You should bring a calculator (any type).
• Scratch Paper:
R. Field 04/22/2012
University of Florida
We will provide scratch paper.
PHY 2053
Page 1
Make-Up Exam: 04/23/12 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 04/22/2012
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 ) = vo t − 12 gt 2 = t (vo − 12 gt )
y (t f ) = 0 = t f (vo − gt f )
1
2
tf =
2vo
g
y (t h ) = h = vot h − 12 gth2
y-axis
v0
h + 12 gt h2
vo =
th
2vo 2(h + 12 gt h2 ) 2h
2(0.98m)
tf =
=
=
+ th =
+ (0.5s ) = (0.4s ) + (0.5s ) = 0.9 s
2
g
gt h
gt h
(9.8m / s )(0.5s )
R. Field 04/22/2012
University of Florida
PHY 2053
Page 3
Final Exam Spring 2011: Problem 14
• A projectile hits the ground at an angle of 30o with respect to the
vertical at a speed of 30 m/s. The horizontal component of its
velocity at launch (in m/s) was:
y-axis
Answer: 15.0
% Right: 53%
vxf
x-axis
φ
vf
v x (t ) = v xo = vxf = v f sin φ = (30m / s ) sin(30o ) = 15m / s
R. Field 04/22/2012
University of Florida
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 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%
M
θ
y-axis
Mg sin θ − f s = Ma x = 0
FN
fs
FN − Mg cos θ = Ma y = 0
Mg sin θ = f s = µ s FN = µ s Mg cos θ
tan θ = µ s = 0.7
θ
x-axis
Mg
θ ≈ 35o
R. Field 04/22/2012
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
θ = 45ο
R
Ei = Mv + Iω + MgR = E f = Mg ( H + R)
1
2
v
1
2
2
x-axis
v = Rω
ω = 2πf
1  2 I 2 1  2 2 I 2
v + ω  =
R ω + ω 
2g 
M
M
 2g 

R2 
I  2 2π 2 R 2 
I  2 2π 2 (0.5m) 2  3 
2
ω
=
=
1
+
f
=
(
2
/
s
)
≈ 3.0m
1 +





2
2
2
2 g  MR 
g  MR 
9.8m / s  2 
H=
R. Field 04/22/2012
University of Florida
PHY 2053
Page 6
Example: Ball & Hoop
• 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 1 s for the ball to roll down the
incline, how long does it take for the hoop (in s)?
Hoop or Ball
d
θ
v = Rω
Answer: 1.2
Mg sin θ − f = Ma x
I
α = Rax
τ = Iα = a x = Rf
R
g sin θ
I
a
=
x
Mg sin θ − 2 a x = Ma x
I 

R
1 +
2 
 MR 
I ball 

1 +

a x (hoop)  MR 2  (1 + 25 ) 7
=
= = 0 .7
=
I hoop  (1 + 1) 10
a x (ball ) 
1 +

2 
 MR 
Mgsinθ
θ
R
θ
x-axis
d = axt
1
2
2
t=
2d
ax
t (hoop )
a x (ball )
1
=
=
= 1.1952
t (ball )
a x (hoop )
0.7
t (hoop ) = 1.1952 × t (ball ) ≈ 1.2 s
R. Field 04/22/2012
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 = Ma y
FT = M ( g − a y )
τ = RFT = Iα =
Ia y
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 04/22/2012
University of Florida
PHY 2053
Mg
Page 8
Example Problem: Pulley
• The ideal mechanical advantage is defined to be the
ratio of the weight W to the force of the pull FP for
equilibrium (i.e. W/FP in equilibrium). Assuming that
the pulleys rotate without friction and without the
rope slipping, what is the ideal mechanical advantage
of the combination of pulleys shown in the figure?
Answer: 7
T1 = FP
T2 = 2T1
T3 = 2T2
T1 + T2 + T3 = W
T3
T2
T2
W T1 + T2 + T3 T1 + 2T1 + 4T1
=
=
=7
FP
T1
T1
T3
R. Field 04/22/2012
University of Florida
PHY 2053
T1
Page 9
Exam 2 Spring 2012: Problem 10
θ
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 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 ( p x ) i = ( M car + M Ball )Vi
( p x ) f = M BallVball = M Ball (Vi + Vmuzzle cos θ )
% Right: 57%
( px )i = ( px ) f
( M car + M Ball )Vi = M Ball (Vi + Vmuzzle cos θ )
M carVi
(5000kg )(2m / s )
cos θ =
=
= 0.6667
M BallVmuzzle (100kg )(150m / s)
R. Field 04/22/2012
University of Florida
PHY 2053
θ ≈ 48.2o
Page 10
Previous Problem: Equivalent
At rest
θ
v
x-axis
θ
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 θ above the horizontal as shown
in the figure. The cannon plus railway car have a mass of 5,000 kg. If the
cannon, one cannon ball, and the railway car are initially at rest, at what
angle θ must the cannon be fired so that the recoil speed of the railway
car plus cannon is 2 m/s toward the left? Assume that the track is
horizontal and there is no friction.
( p x ) i = 0 ( p x ) f = M BallVball − M carV f = M BallVmuzzle cos θ − M carV f
Answer: 48.2o
( px )i = ( px ) f
cos θ =
− M carV f
M BallVmuzzle
R. Field 04/22/2012
University of Florida
=
M BallVmuzzle cos θ − M carV f = 0
− (5000kg )(−2m / s)
= 0.6667
(100kg )(150m / s )
PHY 2053
θ ≈ 48.2o
Page 11
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 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
ω/5
% Right: 47%
M
R
M
Li = I iωi = 12 MR12ω = L f = I f ω f = ( 12 MR12 + 12 MR22 )ω f
R12
R2
4
ωf = 2
ω
=
ω
=
5ω
2
2
2
R1 + R2
R + ( R / 2)
R. Field 04/22/2012
University of Florida
PHY 2053
Page 12
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
ω. What is the new distance r?
Answer: R/2
% Right: 65%
Li = I iωi = ( MR 2 + 12 MR)ωi = L f = ( Mr 2 + 12 MR)ω f
r=R
R. Field 04/22/2012
University of Florida
3ωi 1
3 1 R
− =R
− =
2ω f 2
4 2 2
PHY 2053
Page 13
Exam 2 Fall 2011: Problem 36
• 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 = 3ρ
ρwater). 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 a y t 2
ρ V g − ρ waterVball g = ρ downVball adown
Mg
Answer: 3ρ
ρwater/5 down ball
2d
 ρ

− ρ water
ρ
% Right: 22%
d water
adown = down
g = 1 − water  g
t=
y-axis
ρ down
ay
 ρ down 
 ρ water 
 ρ

aup = 
− 1 g = adown = 1 − water  g
 ρ

water
 ρ down 
 up

ρ waterVball g − ρ upVball g = ρ upVball aup d
y-axis
ρ water
ρ water 3
F
ρup =
=
= 5 ρ water
 ρ water

ρ water − ρ up
1
2 − ρ water / ρ down 2 − 3
aup =
g =
− 1 g


ρ up
Mg
 ρ up

buoyancy
buoyancy
R. Field 04/22/2012
University of Florida
PHY 2053
Page 14
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%
r
y-axis
m
Fbuoyancy − ( M HE + m) g = 0
(MHE +m)g
ρ airVballon g − ( ρ HEVballon + m) g = 0
Vballon = 43 πr 3 =
m
ρ air − ρ HE
1
3
1
3

 

3m
3(10kg )
3
 = 

r = 
=
2
.
34
m
3 
 4π ( ρ air − ρ HE )   4π (1.02kg / m ) 
R. Field 04/22/2012
University of Florida
Fbuoyancy
PHY 2053
(
)
1
3
≈ 1.33m
Page 15
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 kx 2
ω=
2E k 2 2E
2 E 4π 2
2 2
v =
− x =
−ω x =
− 2 x
m m
m
m T
2
2
x
k
m
2π
ω=
T
2 E 4π 2 2
2(5 J )
4π 2
2
2
2
2
2
vx =
− 2 x =
−
(
1
)
=
(
20
m
/
s
)
−
(
m
/
s
) = 4m / s
2
m T
0.5kg (3.14 s )
R. Field 04/22/2012
University of Florida
PHY 2053
Page 16
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.)
Answer: 0.5
% Right: 26%
f s = ma x ≤ µ s mg
µs =
R. Field 04/22/2012
University of Florida
amax = µ s g
amax = ω
2
spring
k
A=
A
( M + m)
k
196 N / m
A=
(0.5m) = 0.5
( M + m) g
(20kg )(9.8m / s )
PHY 2053
Page 17
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 =
Vcar
VA
Car
Ambuance
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 =
(1200 Hz ) ≈ 1127 Hz
(vsound + v A )(vsound − vcar )
(320 + 20)(320 − 10)
R. Field 04/22/2012
University of Florida
PHY 2053
Page 18