2/18/2014
Exam 1: Thursday, 2/20, 8:20-10:10pm
• Room Assignments:
• Breakdown of the 20 Problems
Exam 1: Thursday, 2/20, 8:20-10:10pm
• Room Assignments:
• Breakdown of the 20 Problems
Last Name
Room
Material
# of Problems
Last Name
Room
Material
# of Problems
A-G
NRN 137
Chapter 1
0, but?
A-G
NRN 137
Chapter 1
0, but?
H-O
TUR L007
Chapter 2
4
H-O
Tur L007
Chapter 2
4
P-Z
CLB C130
Chapter 3
4
P-Z
CLB C130
Chapter 3
4
Chapter 4
4
Chapter 5
4
Chapter 6
4
You may bring a single hand written formula sheet on 8½ x 11
inch paper (both sides).
• Crib Sheet:
• Calculator:
• Scratch Paper:
PHY 2053
Page 1
Exam 1 Fall 2010: Problem 4
acar = 0
Let tc be the time it takes for officer
to reach the motorist.
vcar (t ) = v0
acop = a
vcop (t ) = at
xcar (t ) = v0t
xcop (t ) = 12 at 2
University of Florida
xcar (tc ) = xcop (tc )
v0 tc = 12 atc2
tc =
Page 3
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
Answer: 76.0°below the horizontal
t h2 =
% Right: 22%
g
h
v y (t ) = − gt
v x (t ) = v0
d
d
x-axis
v0 =
2
1
θ
y
(
t
)
=
h
−
gt
t
x(t ) = v0t
h
2
Let th be the time the beanbag hits the ground.
y (t h ) = 0 = h − 12 gth2
x(t h ) = d = v0th
University of Florida
tan θ =
PHY 2053
v y (t h )
v x (t h )
University of Florida
PHY 2053
Page 2
• 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
2
vot down = − 12 gtdown
+H
2v0
a
 2v 
vcop (tc ) = atc = a 0  = 2v0
 a 
= 2(80m / s ) = 160m / s
PHY 2053
We will provide scratch paper.
Final Exam Fall 2011: Problem 7
• 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%
You should bring a calculator (any type,
no wifi).
• Scratch Paper:
We will provide scratch paper.
University of Florida
• Crib Sheet:
• Calculator:
You should bring a calculator (any type,
no wifi).
The exams are difficult!
Chapter 4
4
We have to find out who is good
Chapter 5
4
at problem solving and who
is not.
4
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).
gt h gt h2 2h
=
=
=4
v0
d
d
o
θ ≈ 76
=
Page 5
v0 = 12 g
(t − t
2
up
2
down
)
(tup + t down )
2
ydown (t down ) = 0 = H − vot down − 12 gtdown
y-axis
2
v0 (tup + t down ) = 12 g (tup2 − t down
)
H
x-axis
= 12 g (tup − t down ) = 12 (9.8m / s 2 )( 2s ) = 9.8m / s
University of Florida
PHY 2053
Page 4
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
Answer: F/2
F = ( M A + M B + M C )a
A
B
F
C
% Right: 41%
FB = ( M A + M B ) a
FA = M Aa
A
FnetB = FB − FA = M B a =
University of Florida
B
MB
2M
F=
F = 12 F
M A + M B + MC
M + 2M + M
PHY 2053
Page 6
1
2/18/2014
Exam 1 Fall 2010: Problem 9
Exam 1 Spring 2012: Problem 20
• 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%
• 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 θ = 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)? tan φ = 4π R
m2 g − FT = m2 a x
FT − m1 g = m1ax
( m2 − m1 ) g = ( m1 + m2 ) a x
FT
5m − m1
m − m1
ax = 2
g= 1
g = 23 g
m1 + m2
m1 + 5m1
m1g
PHY 2053
x-axis
m2g Page 7
v2
FT sin φ = Ma x = M
R
v2
tan φ =
Rg
University of Florida
Exam 1 Spring 2012: Problem 37
T
FT = M g 2 + a x2 = (6kg ) (9.8m / s 2 ) 2 + (6m / s 2 ) 2 ≈ 68.9 N
PHY 2053
Page 9
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)?
at =
v2 − v1 (58m / s ) − (40m / s )
=
= 3m / s
t2 − t1
6s
2
2
v
(50m / s )
ar =
=
= 4m / s
R
625m
2
2
atot = at + ar = 5m / s
FT
x-axis
R
Mg
gT 2
(9.8m / s 2 )( 4s ) 2
=
≈ 4.59m
4π 2 cos φ
4π 2 cos(30o )
Page 8
Exam 1 Fall 2010: Problem 11
Fext
y-axis
θ
M
Answer: 5
% Right: 49%
L=
φ
PHY 2053
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 θ 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)?
θ
F
Answer: 68.9
FT cos θ = Mg
Mg − FT cos θ = 0
θ
% Right: 63%
x-axis
FT sin θ = Ma x
Mg
FT2 = M 2 ( g 2 + a x2 )
φ
R=
Fext
θ
University of Florida
gT 2
gT 2 tan φ
4π 2
R
L=
sin φ
T=
FT cos φ − Mg = 0
FT
University of Florida
2πR
v
4π 2 R 2
v2 =
T2
R
sin φ =
L
Answer: 4.59
% Right: 34%
x-axis
φ
θ
M
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 θ = 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 θ − f k = Ma x
Fext sin θ + FN − Mg = 0
f k = µ k FN
University of Florida
Fext =
=
M (a x + µ k g )
cos θ + µ k sin θ
(2kg )(1.89m / s 2 + (0.488)9.8m / s 2 )
≈ 12.0 N
cos(26o ) + (0.488) sin(26o )
PHY 2053
Page 10
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 µk = 0.1,
µs = 0.4, respectively, what is the maximum speed the car
can travel without slipping?
2
y-axis
Answer: 14 m/s
% Right: 74%
v
2
max
f s = Ma x = Maradial = M
FN − Mg = Ma y = 0
v
R
fs ≤ µ s FN
x-axis
R
R
R
=
( f s ) max =
( µ s FN ) =
( µ s Mg ) = µ s gR
M
M
M
FN
R
fs
vmax = µ s gR = (0.4)(9.8m / s 2 )(50m) = 14m / s
University of Florida
PHY 2053
Page 11
University of Florida
PHY 2053
Page 12
2
2/18/2014
Exam 1 Fall 2010: Problem 20
Exam 1 Fall 2010: Problem 17
• 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
F
fk
M
∆E = E f − Ei = W
W = ( F − f k )d = ( F − µ k Mg )d
∆KE = ∆E = ( F − µ k Mg )d = (35 N − 23.52 N )(5m) ≈ 57.4 J
University of Florida
PHY 2053
Page 13
y-axis
M
h
v
45
x=0
x-axis
M
fk
o
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 θ Answer: 0.5
2
% Right: 41%
E f = Ei + W Mgh = 12 Mv − f k d
2
Ei = 12 Mv 2
E f = Mgh
W = − fk d
University of Florida
FN − Mg cos θ = 0
f k = µ k FN
f k = µ k Mg cos θ
 v

− 1
 2 gh 
µ k = tan θ 
 (16.97m / s) 2

= tan(45o )
− 1 ≈ 0.5
 2(9.8m / s 2)(9.8m) 
PHY 2053
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