Transcript Physics 106P: Lecture 15 Notes

Physics 201: Review
Final Exam:
Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272
The exam will cover chapters 1 – 14
The exam will have about 30 multiple choice questions
Consultations hours the same as before.
Another review sessions will be held by your TA’s
at the discussion session
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Physics 201, Spring 2011
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Problem Solving

Read and understand the problem statement completely:
Often this is helped by a diagram showing the relationships of the
objects.
Be sure you understand what is wanted
Be sure you understand what information is available to you (or can
be found from the available information)

Translate the situation described to physics concepts.
Be alert for clues regarding the choice of relationships
(e.g. conservation of energy, conservation of momentum,
rotational or linear motion, …….)
Be alert for detail that would qualify the use of some concepts
(e.g. friction affecting conservation of energy.)
If there are other “unknowns” involved how can you find them
(or eliminate them).
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Problem solving…..


After choosing the appropriate relationship between the
concepts (equation), find the target quantities -- at first
algebraically, then substitute numbers at the end.
Check
Does the answer make sense?
Are the units consistent?
Often with multiple choice questions you can round the
numerical quantities and check the final choice without the
calculator.

Techniques and Hints
Be clear and organized -- neat in your solution.You must be
able to read and understand your own notes!
Go through the exam completely at first and complete those
questions that you are confident in solving. Then return to the
others.
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Kinematics:
Chapters 2, 3 (linear)
9 (rotational)
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Dynamics:
Chapters 4, 5, (the 2nd law)
6, 7, (energy and work)
10, 11 (rotational, gravity)
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No net force, No net torque:
Chapters 8 (linear: consequence of the 2nd law)
10 (rotational)
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Statics: Stationary balance
Chapters 12 (static equilibrium, elasticity)
Fluids
Archimedes’ Principle :
A body wholly or partially submerged in a fluid is buoyed
up by a force equal to the weight of the displaced fluid.
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Oscillations:
Resonance frequency:
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Question (Chapt 2)
An European sports car dealer claims that his product will
accelerate at a constant rate from rest to a speed of 100 km/hr in
8s. What is the speed after first 5 s of acceleration?
17.4 m/s
53.2 m/s
44.4 m/s
34.7 m/s
28.7 m/s
v  v 0  at
a 
v  v0
t
(fo r constant acceleration)

10 0 km / hr  0

100000 / 36 00
8s
A fter 5 secon ds: v  0 
m /s
2
8
10000 0 / 3 600
m / s  5 s  17.4 m / s
2
8
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Question (Chapt 4)
Two gliders of unequal mass mA<mB are placed on a frictionless air
track. Glider A is pushed horizontally as shown so that the gliders
accelerate together to the right.
Let FhA represent the magnitude of the force of the hand on the glider A.
Let FBA represent the magnitude of the force exerted by the glider A on
the glider B.
Which one of the following is
true?
FhA < FBA
Net external, FhA-FBA, is
causing block A to
accelerate to the right.
FhA = FBA
FhA > FBA
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Newton’s Second Law:
FBA = FAB < FhA
Physics 201, Spring 2011
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Question , Continued
Two gliders of unequal mass mA<mB are placed on a frictionless air
track. Glider A is pushed horizontally as shown so that the gliders
accelerate together to the right.
Let FhA represent the magnitude of the force of the hand on the glider A.
Let FBA represent the magnitude of the force exerted by the glider A on
the glider B.
Which one of the following is
true?
FBA < FAB
FBA = FAB
FBA > FAB
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Newton’s Third Law
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Question, Continued
Two gliders of unequal mass mA<mB are placed on a frictionless air
track. Glider A is pushed horizontally as shown so that the gliders
accelerate together to the right.
How does the net force on glider B (FB) compare to the magnitude of
the net force on glider A (FA)?
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FB < FA
Given : m A  m B and a A  a B
FB = FA
Second Law
FB > FA

FB
FA

mB
mA
: F A  m A a A and F B  m B a B
 FB  FA
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Question (Chapt. 6)
How much power is needed to lift a 75-kg student vertically
upward at a constant speed of 0.33 m/s?
25 W
12.5 W
243 W
115 W
230 W
Power =
Work
Time

Force  Distance
Time
Power = mgv  75 kg  9 .8 m/s
12/10/07
 Force  Speed
2
Physics 103, Fall 2007, U. Wisconsin
 0 .33 m/s  243 W
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Question (Chapt 8)
A moving object collides with an object initially at rest.
Is it possible for both objects to be at rest after the collision?
Yes
R ealize that m om entum is conserved
No
m 1 v1i  m 1 v1 f  m 2 v 2 f
For this to be true both speeds cannot be zero
Can one of them be at rest after the collision?
Yes
No
If the objects stick together after the collision, is the kinetic
m
m v  ( m  m )v  v 
energy conserved?
m m
Yes
1
KE  m v
2
No
1
1
m
1
1 1i
1
2
f
1
2
2
1 1i
i
KE f 
v1i
f
( m 1  m 2 )v f 
2
2
2
1
2 m1  m 2
v1i 
2
m1
m1  m 2
KEi
K inetic energy is reduced - an inelastic collision
E nergy is lost as heat
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Elastic collision (Q4)
v
m

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M
A block of mass m moving at to the right with speed v hits a
block of mass M that is at rest. If the surface is frictionless
and the collision is elastic, what are the final velocities of
the two blocks?
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Elastic collision
v
M
m

A block of mass m moving at to the right with speed v hits a
block of mass M that is at rest. If the surface is frictionless
and the collision is elastic, what are the final velocities of
the two blocks?
In the center of mass frame, velocities reverse after an elastic
collision
vCM = mv/(m+M)
v-vCM
m
-(vvCM)
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m
Physics 201, Spring 2011
-vCM
M
M
vCM
17
Elastic collision
v
m
M
Now find velocity of each block in lab frame:
Velocity of m = vCM - (v-vCM) = 2vCM – v = (m-M)v/(m+M)
Velocity of M = 2vCM = 2mv/(m+M)
vCM = mv/(m+M)
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Question (Chapt 9)
A boy is whirling a stone around his head by means of a string.
The string makes one complete revolution every second. The
boy then speeds up the stone, keeping the radius of the circle
unchanged, so that the string makes two complete revolutions
every second. What happens to the tension in the string?
The tension increases to four times its original value.
The tension increases to two times its original value.
The tension is unchanged.
The tension reduces to one half its original value.
The tension reduces to one fourth its original value.
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Concepts
Is there a net force acting on the system?
Yes
No
Yes, the direction of velocity is changing.
Centripetal acceleration is provided by the tension in the
string.
The centripetal acceleration is different in the two cases
presented, therefore, the tension will be different
Note that the radius has not changed in the two
conditions
Note also that angular velocity is given (in words).
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Solution
2
F1  m
v1
F2  m
v2
Centripetal force for the two situations:
Need to write in terms of change to angular
velocity because that is what is specified
Some of the quantities are not given but we
are comparing situations, I.e., take ratios
and cancel common factors!
r
2
r
v1  r  1
F2
F1
2

v2
v
2
1
2
2


2
1

2
2
1
2
 4
Tension goes up by a factor of 4!
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Question (Chapt. 9)
The picture below shows three different ways of using a wrench to
loosen a stuck nut. Assume the applied force F is the same in each
case.
In which case is the torque on the
nut the biggest?
Case 1
Case 2
Case 3
12/10/07
= F d sin 
Longest lever arm, d
at 90o angle
Physics 103, Fall 2007, U. Wisconsin
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Question (Chapt 12)
A sign of mass M is hung 1 m from the end of a 4 m long uniform beam
of mass m, as shown in the diagram. The beam is hinged at the wall.
What is the tension in the guy wire? Determine the tension T, and the
contact force F at the hinge.
wire
 o
1m
3m
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SIGN
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What are the concepts involved?
Is there a net force acting on the system?
Yes
No
wire
 o
1m
Is there a net torque acting on the system?
Yes
No
SIGN
Draw the free body diagram.
How many forces are acting on the system?
2
3
mg, Mg, tension, force from hinge
4
What is the direction of the contact force at the
5
hinge between the wall and the beam ?
Vertical
Horizontal
It has both vertical and horizontal components
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Solution
3m
Fy
T
2m
300
Fx
mg
Forces

F  0 
Mg
Fx  T cos 30  0
0
Fy  T sin 30  m g  M g
0
Torques
Hint: Choose axis of rotation at support because Fx & Fy are not known
   0  2 m g  3 M g  4 T sin 30
 T  g 2 m  3 M
0
/ 2
(T can now be com puted)
Substitute T in Force equations to get F x and F y
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Motion in Gravity (Chapt. 11)
A rock is thrown straight up from the Earth’s surface. Which one of the following
statements concerning the net force acting on the rock at the top of its path is
true?
It is equal to zero for an instant.
It is equal to the force used to throw it up but in opposite direction
It is equal to the weight of the rock
Its direction changes from up to down
Its magnitude is equal to the sum of the force used to throw it up
and its weight
12/10/07
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Question (Chapt 12)
A mass of 100 tons (105 kg) is lifted on a steel rod two cm
in diameter and 10 m in length.
(Young’s modulus of steel is 210 109 N/m2)
(a) How long does the rod stretch?
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A mass of 100 tons (105 kg) is lifted on a steel rod two cm
in diameter and 10 m in length.
(Young’s modulus of steel is 210 109 N/m2)
(a) How long does the rod stretch?
Y 
stress

F /A
strain
L / L
F
m gL
L 
L 
 0 .15 m
2
Y A
Yr
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F = force
A = area of rod
L = length of rod
ΔL = change of
length of the rod
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Question (Chapt 13)
1) The pressure on the roof of a tall building is 0.985 × 105
Pa and the pressure on the ground is 1.000 × 105 Pa. The
density of air is 1.29 kg/m3. What is the height of the
building?
A. 100 m
B. 118 m
C. 135 m
D. 114 m
E. None of the above
P1   h1 g; P2   h 2 g  h  h 2  h1 
0.015  10 Pa
P2  P1
g
5
h 
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1 .29 kg / m  9 .8 m / s
3
2
 118 m
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Question (Chapt 13)
A venturi tube may be used as the inlet to an automobile
carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm
diameter, what is the pressure drop in the constricted
section for an airflow of 3.0 m/s in the 2.0-cm section?
(fuel density = 1.2 kg/m3.)?
Velocity is faster in constricted section because mass flow is conserved
(mass that flows into constriction must also flow out).
Pressure drops because of Bernoulli principle:
(applies to incompressible, frictionless fluid)
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P
1
 v   gh  constant
2
2
30
Calculate velocity in constriction
Fluid flow without friction
• Volume flow rate: V/t = A x/t = Av (m3/s)
• Continuity: A1 v1 = A2 v2
i.e., mass that flows in must then flow out
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Question, continued
A venturi tube may be used as the inlet to an automobile
carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm
diameter, what is the pressure drop in the constricted
section for an airflow of 3.0 m/s in the 2.0-cm section?
(fuel density = 1.2 kg/m3.)?
70 Pa
C onstant of volum e flow rate resulted
continuity equation:
85 Pa
100 Pa
A1 v1  A 2 v 2  v 2 
A1
A2
 r1
2
v1 
 r2
2
2
v1 
115 Pa
B ern ou lli E qu atio n (sam e h eigh t):
81 Pa
P1 
1
2
 v1  P2 
2
1
2
 v2   P 
2
1
2
d1
v1
2
d2
 v 2  v1
2
2

 d 14
 2
C o m bin in g w ith ab ov e:  P    4  1  v1
2  d2

1
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Question (Chapt 13)
The water level in identical bowls, A and B, is exactly the
same. A contains only water; B contains ice as well as
water. When we weigh the bowls, we find that
WA < WB
WA = WB
WA > WB
WA < WB if the volume of the ice cubes is greater than
one-ninths the volume of the water.
WA < WB if the volume of the ice cubes is greater than
one-ninths the volume of the water.
Eureka! Archimedes Principle.
Weight of the water displaced = Bouyant Force
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Question (Chapt 13)
1) A block of aluminum (density 3041 kg/m3) is lifted very slowly
but at constant speed from the bottom of a tank filled with water. If
it is a cube 20 cm on each side, the tension in the cord is:
A. 160 N
T
B.
4N
Fb
C. 80 N
D.
8N
E. None of the above
W

V olum e of fluid displaced: V  V A l  20  10
B uoyant force: FB   w V g  1000  8  10
W eight: W  M
3
g   A l V g  3041  8  10
Al
2

3
 8  10
3
m
3
 9.8  78 N
3
 9.8  238 N
T ension: T  W  FB  160 N
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Question (Chapt 13)
A wind with velocity 10 m/s is blowing through a wind
generator with blade radius 5.0 meters. What is the
maximum power output if 30% of the wind’s energy can be
extracted? (air density = 1.25 kg/m3.)
7.2 kW
14.7 kW
B ernoulli E quation (sam e height):
21.3 kW
P1 
1
 v1  P2 
2
1
 v2   P 
2
 v 2  v1
2
2

29.4 kW
2
2
2
Pressure difference results in net force on the blades
39.6 kW
M agnitude of the force = Pressure x B lade A rea
Pow er =
W ork

Fo rce x D istan ce
T im e

Physics 201, Spring 2011
 Fo rce x V elo city
T im e
1

2
2
Pow er    v i 3 0%   R
2

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v 
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Question (Chapt 13)
Firemen connect a hose (8 cm in diameter) to a fire hydrant. When the nozzle is open, the
pressure in the hose is 2.35 atm. (1 atm. = 105 Pa). The firemen hold the nozzle at the same
height of the hydrant and at 45o to the horizontal. The stream of water just barely reaches a
window 10 m above them. The diameter of the nozzle is about:
A. 8 cm
B. 6 cm
C. 4 cm
D. 2 cm
E. None of the above Flow is constant in hose  A v  A v
1 1
v1
  r1 v1   r2 v 2  r2  r1
2
2
P
Point 3
1
2
Point 2
Point 1
1
v2
 v 2 x are co n stan t, so
2
2 gh3 
196  14 m / s
45 angle  v 2 
o
P1 
1
2
1
2
v2 x  v2 y 
2
 v1  P2 
2
1
2
2
2 v 2 y  19.799 m / s
 v 2  v1 
2
P

2
2
 P1 
1
2
 v2 
2
 v 2  196000 P a
v1 
2
2
1000
100000  235000  196000   11.045 m
d2  8
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v1
  g h 3  1 00 0  9.8  10  9 8 0 0 0 P a 
2
2y
v2 y 
10m
 d 2  d1
2
2
v
2
 v 2   g h  co n stan t (B ern o u lli's E q uatio n).
B u t, P an d
1
2
v2
2
Physics 201, Spring 2011
11.045
/s
 5.975  6 cm
19.799
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Question (Chapt 14)
At t=0, a 795-g mass at rest on the end of a horizontal spring
(k=127 N/m) is struck by a hammer, giving it an initial
speed of 2.76 m/s. The position of the mass is described
by x (t )  A co s   t    , with   k / m  12 .6 4 rad /s.

2
What is period of the motion? period = 2π/ω
0.497 s
What is the frequency of the motion?
2.01 Hz
What is the maximum acceleration?
34.9 m/s2
What is the total energy?
3.03 J
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Question (Chapt 14)
The amplitude of a system moving with simple harmonic
motion is doubled. The total energy will then be
4 times larger
2 times larger
the same as it was
half as much
quarter as much
U 
1
kx 
2
1
mv
2
2
2
at x  A , v  0
U 
1
kA
2
2
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