Final Exam: 4/26/14 8:00pm-10:00pm
• Room Assignments:
Last Name
Room
A-G
• Breakdown of the 20 Problems
Material
# of Problems
CLB C130
Your Exam 1
4
H-R
NRN 137
Your Exam 2
4
S-U
FLG 280
Chapter 11.1-12.8
4
V-Z
MAT 18
Chapter 1.1-10.8
8
• 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 4/20/2014
University of Florida
We will provide scratch paper.
PHY 2053
Page 1
Make-Up Exam: 4/23/14 5:10-7:00pm
For anyone who had to miss an exam during the semester.
• Room: MAEA 327.
• 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 4/20/2014
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 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 gt h2
y-axis
v0
h + 12 gt h2
vo =
th
2vo 2(h + 12 gt h2 ) 2h
2(0.98m)
tf =
=
=
+ th =
+ (0.5s ) = (0.4 s ) + (0.5s ) = 0.9 s
2
g
gt h
gt h
(9.8m / s )(0.5s )
R. Field 4/20/2014
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 + vo tup − 12 gtup
votup = 12 gtup2 − H
vot down = − gt
1
2
v0 = g
1
2
2
down
(t − t
2
up
+H
2
down
)
(tup + t down )
R. Field 4/20/2014
University of Florida
2
ydown (t down ) = 0 = H − vot down − 12 gt down
v0 (tup + t down ) = 12 g (t − t
2
up
y-axis
2
down
)
H
x-axis
= 12 g (tup − t down ) = 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 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 = Max = 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 4/20/2014
University of Florida
PHY 2053
Page 5
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%
M B g − FT = M B a
FT = ( M A + M C )a
fs = M Ca
M C amax = ( f s ) max = μ s FN = μ s M C g
amax = μ s g
μs =
MB =
MB
M A + M B + MC
M B g = ( M A + M B + M C )a
FN
MB
a=
g
M A + M B + MC
FT
FT
μs (M A + M B + M C ) = M B
0.6
μs
(M A + M C ) =
(30kg + 10kg ) = 32 (40kg ) = 60kg
1 − μs
1 − 0.6
R. Field 4/20/2014
University of Florida
Fs
PHY 2053
MBg
Page 6
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?
Answer: 50.8
% Right: 12%
α (t ) = α
ω (t ) = αt
θ (t ) = 12 αt 2
aradial (t ) = Rω 2 (t ) = Rα 2t 2
1⎛ α
N (t1 ) = ⎜
2 ⎝ 2π
α=
⎞2
⎟t1
⎠
Let t1 be the time for the
first revolution, N(t1) = 1.
4πN (t1 )
t12
at (t ) = Rα
aradial (t ) = Rα 2t 2 = at = Rα
t=
R. Field 4/20/2014
University of Florida
1
α
=
t1
(3 min)(60 s / min)
=
≈ 50.8s
4πN (t1 )
4π (1)
PHY 2053
Page 7
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
=
1
+
f
=
(
2
/
s
)
≈ 3.0m
=
ω
⎜1 +
⎟
⎜
⎟
⎜
⎟
2
2
2
2 g ⎝ MR ⎠
g ⎝ MR ⎠
9.8m / s ⎝ 2 ⎠
H=
R. Field 4/20/2014
University of Florida
PHY 2053
Page 8
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)?
Hoop or Ball
d
θ
v = Rω
α = Rax
Answer: 2.39
% Right: 53%
Mg sin θ − f = Ma x
I
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 + 52 ) 7
=
=
= 10 = 0.7
I
a x (ball ) ⎛
⎞ (1 + 1)
⎜⎜1 + hoop2 ⎟⎟
⎝ MR ⎠
Mgsinθ
R
θ
x-axis
d = axt
1
2
2
t=
2d
ax
1
t (hoop )
a x (ball )
=
=
= 1.1952
t (ball )
a x (hoop )
0. 7
t (hoop ) = 1.1952 × t (ball ) = 1.1952 × (2 s ) ≈ 2.39 s
R. Field 4/20/2014
University of Florida
PHY 2053
Page 9
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 4/20/2014
University of Florida
PHY 2053
Mg
Page 10
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?
g
1
0.60MR2
Answer:
% Right: 20%
yhoop = 12 ahoop t 2
ycylinder = 12 acylinder t 2
ahoop =
MR 2
R. Field 4/20/2014
University of Florida
4Δy
4(2m)
=
≈ 0.25
gt 2 (9.8m / s 2 )(1.81s ) 2
4Δy
1− 2
1 − 0.25
gt
=
≈
≈ 0.6
4Δy 1 + 0.25
1+ 2
gt
PHY 2053
R
=2g
(1 + I hoop / MR )
g
acylinder =
(1 + I cylinder / MR 2 )
g
g
Δa = acylinder − ahoop =
−
(1 + I cylinder / MR 2 ) 2
2
2Δy
Δa = 2
t
Δy = ycylinder − yhoop = 12 Δat 2
I cylinder
R
R
FT
y-axis
Mg
Mg − FT = Ma y
τ = RFT = Iα =
Ia y
R
g
ay =
(1 + I / MR 2 )
Page 11
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%
cos θ =
( px )i = ( px ) f
( M car + M Ball )Vi = M Ball (Vi + Vmuzzle cos θ )
M carVi
(5000kg )(2m / s )
=
= 0.6667
M BallVmuzzle (100kg )(150m / s )
R. Field 4/20/2014
University of Florida
PHY 2053
θ ≈ 48.2o
Page 12
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?
132.8o
Answer:
% Right: 16%
y-axis
r
r
r
r
r
r
p A + pB + pC = Mv A + MvB + MvC = 0
r
r
vC = v0 xˆ
vB = v0 cos θxˆ + v0 sin θyˆ
r
r r
v A = −(vB + vC )
B
v0
vA
A
θ
v0
C
x-axis
r
v A = −v0 (1 + cos θ ) xˆ − v0 sin θyˆ
r 2
2
v A = v A = v02 (1 + cos θ ) 2 + sin 2 θ = v02 [2 + 2 cos θ ]
[
]
v A2
( 4m / s ) 2
cos θ = 2 − 1 =
− 1 = −0.68
2
2v0
2(5m / s )
θ ≈ 132.8o
R. Field 4/20/2014
University of Florida
PHY 2053
Page 13
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/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
ωf = 2
ω= 2
ω = 54 ω
2
2
R1 + R2
R + ( R / 2)
R. Field 4/20/2014
University of Florida
PHY 2053
Page 14
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.
2
2
24
−11
Answer: 8.67
% Right: 18%
g=
GM E (6.674 ×10 Nm / kg )(5.974 ×10 kg )
=
≈ 9.823m / s 2
2
6
2
RE
(6.471× 10 m)
E f = Ei
⎡ GM E GM E ⎤ 2GM E
v 2f = 2⎢
−
⎥=
h + RE ⎦
RE
⎣ RE
GmM E
h + RE
GmM E
E f = 12 mv 2f −
RE
Ei = −
h
ME
RE
⎡
RE ⎤ 2 gRE h
−
1
⎥=
⎢
⎣ h + RE ⎦ h + RE
2 gRE h
2 g (1.5RE )
2(9.823m / s 2 )(1.5)(6.371×106 m)
vf =
⎯h⎯
⎯
⎯→
=
≈ 8.67km / s
=1.5 RE
h + RE
2.5
2.5
R. Field 4/20/2014
University of Florida
PHY 2053
Page 15
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%
Li = I iωi = ( MR 2 + 12 MR)ωi = L f = ( Mr 2 + 12 MR)ω f
r=R
R. Field 4/20/2014
University of Florida
3ωi 1
3 1 R
− =R
− =
2ω f 2
4 2 2
PHY 2053
Page 16
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 = 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
F
ball? Neglect hydrodynamic drag forces.
d = 12 a y t 2
ρ V g − ρ waterVball g = ρ downVball adown
Mg
Answer: 3ρwater/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
⎞
− 1⎟ g = adown = ⎜⎜1 − water ⎟⎟ g
aup = ⎜
⎜ ρ
⎟
water
⎝ ρ down ⎠
⎝ up
⎠
ρ waterVball g − ρ upVball g = ρ upVball aup d
y-axis
ρ water
ρ water 3
F
ρ up =
=
= 5 ρ water
⎛ ρ water
⎞
ρ water − ρ up
1
2 − ρ water / ρ down 2 − 3
− 1⎟ g
aup =
g =⎜
⎜
⎟
ρ up
Mg
⎝ ρ up
⎠
buoyancy
buoyancy
R. Field 4/20/2014
University of Florida
PHY 2053
Page 17
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 4/20/2014
University of Florida
Fbuoyancy
PHY 2053
(
)
1
3
≈ 1.33m
Page 18
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?
Answer: 93.3
% Right: 18%
Pout
Pin
L
L
Pout A − Pin A − M tot g = 0
APout
L
APin Mtotg
y-axis
M tot g
5Mg
5(25kg )(9.8m / s 2 )
Pin = Pout −
= Patm − 2 = Patm −
= 101kPa − 7.656kPa ≈ 93.3kPa
A
L
(0.4m) 2
R. Field 4/20/2014
University of Florida
PHY 2053
Page 19
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
2E k 2 2E
2 E 4π 2 2
2 2
− x =
−ω x =
− 2 x
v =
m T
m
m m
2
x
k
ω=
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 4/20/2014
University of Florida
PHY 2053
Page 20
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 4/20/2014
University of Florida
PHY 2053
Page 21