Exam 1: Tomorrow 8:20-10:10pm
• Room Assignments:
• Breakdown of the 20 Problems
Last Name
Room
Material
# of Problems
A-K
CAR 100
Chapter 1
1
L-R
NPB 1001
Chapter 2
5
S-Z
TUR L007
Chapter 3
3
Chapter 4
6
Chapter 5
3
Chapter 6
2
• Crib Sheet:
• Calculator:
You may bring a single hand written formula sheet on 8½ x 11
inch paper (both sides).
You should bring a calculator (any type).
• Scratch Paper:
R. Field 10/01/2013
University of Florida
We will provide scratch paper.
PHY 2053
Page 1
Exam 1: Tomorrow 8:20-10:10pm
• Room Assignments:
• Breakdown of the 20 Problems
Last Name
Room
Material
# of Problems
A-K
CAR 100
Chapter 1
1
L-R
NPB 1001
Chapter 2
5
S-Z
TUR L007
Chapter 3
3
The exams are difficult!
Chapter 4
6
We have to find out who is good
Chapter 5
3
at problem solving and who is not.
2
Expect an average of Chapter
~12/206
with high = 20 and low = 2!
You may bring a single 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 10/01/2013
University of Florida
We will provide scratch paper.
PHY 2053
Page 2
Exam 1 Fall 2010: Problem 4
• A motorist drives along a straight road at a constant
speed of 80 m/s. Just as she passes a parked motorcycle
police officer, the officer takes off after her at a constant
acceleration. If the officer maintains this constant value
of acceleration, what is the speed of the police officer
when he reaches the motorist?
Answer: 160 m/s
% Right: 18%
acar  0
vcar (t )  v0
xcar (t )  v0t
R. Field 10/01/2013
University of Florida
Let tc be the time it takes for officer
to reach the motorist.
xcar (tc )  xcop (tc )
acop  a
vcop (t )  at
v0tc  at
1
2
xcop (t )  12 at 2
2
c
2 v0
tc 
a
 2v0 
vcop (tc )  atc  a
  2v0
 a 
 2(80m / s)  160m / s
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 10/01/2013
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 Spring 2012: Problem 12
• A beanbag is thrown horizontally from a dorm room
window a height h above the ground. It hits the ground a
horizontal distance d = h/2 from the dorm directly below
the window from which it was thrown. Ignoring air
resistance, find the direction of the beanbag's velocity
just before impact.
2h
y-axis
2
Answer: 76.0° below the horizontal
th 
% Right: 22%
g
h
v y (t )   gt
d
vx (t )  v0
d
x-axis
v

0
2
1

th
y(t )  h  2 gt
x(t )  v t
0
Let th be the time the beanbag hits the ground.
y(th )  0  h  gt
x(th )  d  v0th
1
2
R. Field 10/01/2013
University of Florida
2
h
v y (t h )
gth gth2 2h
tan 



4
v x (t h )
v0
d
d
  76
PHY 2053
Page 5
Exam 1 Fall 2010: Problem 8
A
B
C
F
• Three blocks (A,B,C), each having mass MA = M, MB= 2M,
MC = M are connected by strings on a horizontal
frictionless surface as shown in the figure. Block C is
pulled to the right by a horizontal force of magnitude F
that causes the entire system to accelerate. What is the
magnitude of the net horizontal force acting on block B
due to the strings?
F
F
F  ( M A  M B  M C )a
Answer: F/2
A
B
F
C
% Right: 41%
FB  ( M A  M B )a
FA  M Aa
A
B
MB
2M
F  FB  FA  M B a 
F
F  12 F
M A  M B  MC
M  2M  M
B
net
R. Field 10/01/2013
University of Florida
PHY 2053
Page 6
Exam 1 Fall 2010: Problem 9
• The figure shows two blocks with masses m1 and
m2 connected by a cord (of negligible mass) that
passes over a frictionless pulley (also of negligible
mass). If m2 = 5m1 what is the magnitude of the
acceleration of block 2 when released from rest?
Answer: 2g/3
% Right: 85%
m2 g  FT  m2 ax
FT  m1 g  m1ax
(m2  m1 ) g  (m1  m2 )ax
x-axis
FT
ax 
FT
m2  m1
5m  m1
g 1
g  23 g
m1  m2
m1  5m1
m1g
R. Field 10/01/2013
University of Florida
PHY 2053
x-axis
m2g Page 7
Exam 1 Spring 2012: Problem 20
• A conical pendulum is constructed from a stone
of mass M connected to a cord with length L and L
negligible mass. The stone is undergoing
uniform circular motion in the horizontal plane
M
as shown in the figure. If the cord makes an
angle  = 30o with the vertical direction and the
period of the circular motion is 4 s, what is the y-axis
2
L
length L of the cord (in meters)? t an  4 R
Answer: 4.59
% Right: 34%
FT cos  Mg  0
v2
FT sin   Max  M
R
2
v
tan 
Rg
R. Field 10/01/2013
University of Florida
2R
T
v
4 2 R 2
2
v 
T2
R
sin  
L
gT 2
gT 2 t an
R
4 2
R
L
sin 



FT
x-axis
R
Mg
gT 2
(9.8m / s 2 )(4s) 2
L 2

 4.59m
2

4 cos
4 cos(30 )
PHY 2053
Page 8
Exam 1 Spring 2012: Problem 37

a
• Consider a mass M = 6 kg suspended by a very light string
from the ceiling of a railway car near the surface of the
Earth. The car has a constant acceleration as shown in the
figure, causing the mass to hang at an angle  with the
vertical. If the acceleration of the railway car is a = 6 m/s2,
y-axis
what is the tension in the string (in N)?

F
Answer: 68.9
F
cos


Mg
Mg  FT cos  0
T

x-axis
% Right: 63%
FT sin   Max
Mg
FT2  M 2 ( g 2  ax2 )
T
FT  M g 2  a x2  (6kg ) (9.8m / s 2 ) 2  (6m / s 2 ) 2  68.9 N
R. Field 10/01/2013
University of Florida
PHY 2053
Page 9
Exam 1 Fall 2010: Problem 11
Fext
M
y-axis

M
Fext

x-axis
• Near the surface of the Earth, a block of mass M = 2 kg
slides along the floor while an external force Fext is
applied at an upward angle  = 26o. If the coefficient of
kinetic friction between the block and the floor is 0.488,
and the magnitude of the acceleration of the block is 1.89
m/s2, what is the magnitude of the external force?
Answer: 12 N
% Right: 59%
Fext cos  f k  Max
Fext sin   FN  Mg  0
f k  k FN
R. Field 10/01/2013
University of Florida
M (a x   k g )
Fext 
cos  k sin 
(2kg)(1.89m / s 2  (0.488)9.8m / s 2 )

 12.0 N


cos(26 )  (0.488) sin(26 )
PHY 2053
Page 10
Exam 2 Spring 2011: Problem 2
• A race car accelerates uniformly from a speed of 40 m/s to
a speed of 58 m/s in 6 seconds while traveling around a
circular track of radius 625 m. When the car reaches a
speed of 50 m/s what is the magnitude of its total
acceleration (in m/s2)?
Answer: 5
% Right: 49%
R. Field 10/01/2013
University of Florida
v2  v1 (58m / s )  (40m / s )
at 

 3m / s
t 2  t1
6s
v 2 (50m / s ) 2
ar 

 4m / s
R
625m
atot  at2  ar2  5m / s
PHY 2053
Page 11
Exam 2 Spring 2011: Problem 4
• Near the surface of the Earth, a car is traveling at a
constant speed v around a flat circular race track with a
radius of 50 m. If the coefficients of kinetic and static
friction between the car’s tires and the road are k = 0.1,
s = 0.4, respectively, what is the maximum speed the car
can travel without slipping?
2
y-axis
Answer: 14 m/s
% Right: 74%
f s  Max  Maradial
v
M
R
FN  Mg  May  0
FN
R
fs  s FN
x-axis
R
R
R
2
vmax 
( f s ) max 
(  s FN ) 
(  s Mg )   s gR
M
M
M
fs
vmax   s gR  (0.4)( 9.8m / s 2 )( 50 m)  14 m / s
R. Field 10/01/2013
University of Florida
PHY 2053
Page 12
Exam 1 Fall 2010: Problem 20
• A horizontal force of magnitude 35 N pushes a block of
mass 4 kg across a floor where the coefficient of kinetic
friction is 0.6. What is the increase in the kinetic energy
of the block when the block slides through a displacement
of 5 m across the floor?
Answer: 57.4 J
% Right: 52%
E f  Ei  W
E  E f  Ei  W
W  ( F  f k )d  ( F  k Mg )d
F
fk
M
KE  E  ( F  k Mg)d  (35N  23.52N )(5m)  57.4J
R. Field 10/01/2013
University of Florida
PHY 2053
Page 13
Exam 1 Fall 2010: Problem 17
y-axis
h
M
v
M
fk
45o
x=0
x-axis
45o
x-axis
d
h
x=0
• Near the surface of the Earth a block of mass M and initial
velocity 16.97 m/s is sliding to the right along the
(negative) x-axis as shown. The surface is frictionless for
x < 0. At x = 0 the block encounters a 45o incline ramp. If
the block stops at a height h = 9.8 m, what is the kinetic
coefficient of friction of the ramp? d  h / sin  Answer: 0.5
2
1
% Right: 41%
Mgh

Mv

f
d
E f  Ei  W
k
2
 2

Ei  12 Mv
E f  Mgh
W   fk d
2
R. Field 10/01/2013
University of Florida
FN  Mg cos  0
f k  k FN
f k  k Mg cos
v
 1
 2 gh 
 k  tan 
2


(
16
.
97
m
/
s
)

 tan(45 )
 1  0.5
 2(9.8m / s 2)(9.8m) 
PHY 2053
Page 14