PHYSICS 2010
Final Exam
Review Session
Note that this review is not covering everything.
Rather it is highlighting key points and concepts.
Information on Course Web Page
The final exam will cover material including Chapters 1-14; CAPA
homework assignments 1-15; All Lecture Material , and all Section labs
and assignments. Note that the exam is cumulative and there will not
be a disproportionate emphasis on the material since the 3rd midterm.
The exam will be 35-40 multiple choice questions. The exam is closed
book and closed notes. Calculators are allowed, but no sharing of
calculators. A formula sheet will be included with your exam, and a
preview of that identical sheet is linked here as a PDF file. There is no
need to print and bring this sheet as a copy will be included with the
exam. A practice exam and solutions are available on the course
CULearn page; you may ignore problem 37 on this exam.
The exam will be held on the west side of the Coors Event Center; your
seat assignments will be posted on CULearn.
Reminder of Materials Available
• Textbook and example problems in the text
• Professor Dubson’s Chapter Notes (see link on web page)
• Clicker questions as posted in lecture notes
• CAPA problems, solutions on CULearn
• Practice exam and solutions on CULearn
• Three midterm exams and solutions on CULearn
• Lab manuals and problems
CH. 2 Kinematics in 1D
2.1 Reference frames and displacement
2.2 Average velocity
2.3 Instantaneous velocity
2.4 Acceleration
2.5 Motion at constant acceleration
2.6 Solving problems
2.7 Falling objects
2.8 Graphical analysis of linear motion
Motion under constant acceleration
What are some example problems?
Motion under constant acceleration
Finally: algebra, trigonometry, unit conversion.
Clicker Question
Room Frequency BA
Clicker Question
Room Frequency BA
Clicker Question
Room Frequency BA
A sailboat is being blown across the sea at a
constant velocity.
What is the direction of the net force on the boat?
CH. 3 Kinematics in 2D: Vectors
3.1 Vectors and scalars
3.2 Addition of vectors graphically
3.3 Vector subtraction and multiplication by a scalar
3.4 Adding vectors by components
3.5 Projectile motion
3.6 Solving problems
Room Frequency BA
Clicker Question
The vector A has magnitude |A| = 6 cm and makes an angle of
q = 30 degrees with the positive y-axis as shown.
A
y
q
x
What is the x-component of the vector A?
A) +4.0 cm
B) -3.0 cm
C) +5.2 cm
D) - 5.2 cm
E) None of the above
Football Punter Physics
For a specific play, the punter wants
to kick the ball as far down the field
as possible (i.e. maximum range).
What is the optimal angle to
kick the ball at
assuming the same initial speed
when kicked regardless of the angle?
Football obeys the laws of physics.
Constant acceleration case (ignoring air resistance).
x  x0  v0 x t  1 a x t 2
2
x0  0
y0  0
v0 x
v0 y

| v0 | cosq
ax  0

x  (| v0 | cosq )t
q
y  y 0  v0 y t  1 a y t 2
2

| v0 | sin q
ay  g

y  (| v0 | sin q )t  1 gt 2
2
Approach
1. At what time (t) does
the ball hit the ground
(i.e. when is y=0)
2. Then evaluate the x
position at that time
(thus giving the Range)
Room Frequency BA
Clicker Question

x  (| v0 | cosq )t

y  (| v0 | sin q )t  1 gt 2
2
q
Clearly at t =0, x=0 and y=0
(our initial conditions).
At what later time does the football hit the ground?
A)
B)
C)
D)
E)
t = 10 seconds
t = |v0| tanq
t = Sqrt(2|v0|sinq/g)
t = 2|v0|sinq/g
None of the above

0  (| v0 | sin q )t  1 gt 2
2

0  t (| v0 | sin q )  1 gt
2

(| v0 | sin q )  1 gt  0
2

t  2(| v0 | sin q ) / g



x  (| v0 | cosq )t

y  (| v0 | sin q )t  1 gt 2
2
q
Time of football flight (i.e. hang time)

t  2(| v0 | sin q ) / g
Now plug into x-equation to find out position when it hits the ground.


x  (| v0 | cosq )2 | v0 | sin q / g 
 2
2 | v0 | cosq sin q
x
g
 2
| v0 | sin 2q

g
* Uses trigonometry identity 2cosqsinq=sin2q
Remember: A large part of what you’ll be expected to do is to apply
Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Linear Motion:
1. Draw a free-body diagram identifying and labeling all forces.
2. Choose a coordinate system – typically with the forces pointing in the coordinate
directions (x,y) as much as possible.
3. Write down Newton’s 2nd law in each coordinate direction (typically), summing
the forces. The equation perpendicular to the direction of motion often allows
you to find the Normal Force, which is needed to determine the force of friction.
Σ Fx = max
Σ Fy = may
(for rectilinear motion)
If a force is not in a coordinate direction, you must find its components in the
coordinate directions.
4. If there is more than one mass, then Newton’s 2nd law may be needed for each
mass.
Room Frequency BA
Clicker Question
A moving van collides with a sports car in a
high-speed head-on collision. Crash!
M
F truck
F car
m
During the impact, the truck exerts a force with magnitude Ftruck on
the car and the car exerts a force with magnitude Fcar on the truck.
Which of the following statements about these forces is true:
A) The magnitude of the force exerted by the truck on the car is
the same size as the magnitude of the force exerted by the car on
the truck: Ftruck = Fcar
B) Ftruck > Fcar
C) Ftruck < Fcar
Guaranteed by Newton’s Third Law
Room Frequency BA
Clicker Question
A moving van collides with a sports car in a
high-speed head-on collision. Crash!
M
F truck
F car
m
During the collision, the imposed forces cause the
truck and the car to undergo accelerations with
magnitudes atruck and acar. What is the relationship
between atruck and acar?
A) atruck > acar
B) acar > atruck
Newton’s Second Law: F = ma
C) atruck = acar
D) Indeterminate from information given.
Step 1: Draw a free-body diagram
Note that “Normal Force” is
always Perpendicular
(i.e. normal) to the surface.
Fg = mg
In this case, it is not in the
opposite direction to the
gravitational force.
Step 2: Choose a coordinate system
Fg = mg
y
x
We could choose our “usual” axes.
However, we know that we will have motion
in both the x and y directions.
If we pick these rotated axes, we know that
the acceleration along y must be zero.
Step 3: Write down the equations
Σ Fx = m ax , Σ Fy = m ay
Fg = mg
Problem:
Fg doesn’t point in a
coordinate direction.
Must break it into its
x- and y-components.
Room Frequency BA
Clicker Question
F(gravity)y
F(gravity)x
90 – q
a 90
|F(gravity)| = mg
What is the angle a labeled above?
A) a = 90 degrees
B) a = q
C) a = 90 – q
D) Cannot be determined
Step 3: Write down the equations
Σ Fx = m ax , Σ Fy = m ay
N
Fx = mg sin(q)
Fy = N – mg cos(q)
Fg = mg
We also know that ay = 0,
there is no motion in that
direction.
Fy =may=0= N – mg cos(q)
Step 4: Solve the Equations
Fx  ma x  mg sin(q )
a x  g sin(q )
Fy  may  N  mg cos(q )
N
a y   g cos(q )  0
m
N  mg cos(q )
As expected,
maximum
acceleration if
straight down
(q=90 degrees),
ax = g.
Recall x-y
definition
Room Frequency BA
Clicker Question
A mass m accelerates down a frictionless inclined plane.
+y
-mg cos θ
θ
+x
θ
-
Which statement is true?
A) N < mg
0 = Fy = N – mg cos θ
B) N > mg
N = mg cos θ < mg
C) N = mg
Room Frequency BA
Clicker Question
Frictionless table & pulley.
+x
M1
+T
1)Choose coordinates
2)Identify forces.
3) Write down F = ma
for each object
-T
M2
Fg = M2g
Mass 1: M1a = T
Mass 2: M2a = M2g - T
+x
What is F = ma for mass M2?
(A) M 2 a  T
(B) M 2 a  T  M 2 g
(C) M 2 a  M 2 g  T
(D)M 2 a  M 2 g
Room Frequency BA
Clicker Question
You are pushing horizontally with a force of 5000 Newtons on a
car that has a weight of 10,000 Newtons.
The car is not moving.
What can you say for certain about the coefficient of friction?
A)
B)
C)
D)
E)
ms = 0
ms = 0.1
ms = 0.5
mk = 0.5
None of the above
But actually, ms could be even larger
since we do not know if we have
reached the maximum.
Fnet = 0 = Fpush – Ffriction
0=Fpush – ms x Normal
ms x Mg = Fpush
ms = Fpush/Mg = 5000/10k
ms= 0.5
Incomplete Gallery of Problems Involving Newton’s Laws:
F
w & w/0 friction
m
w & w/0 friction
F
Box 1
M
Box 2
mm
fc
M1
w & w/0 friction
+x
M2
M
m
T
w & w/0 gravity
Practice setting up the free-body diagram and Fnet = ma
Chapter 5: Uniform Circular Motion with Gravity
Remember: A large part of what you’ll be expected to do is to apply
Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Uniform Circular Motion:
1. Draw a free-body diagram identifying and labeling all forces.
2. Choose a coordinate system – one of the directions will point in the
radial direction. Others directions: tangent direction (T) or vertical (y).
3. Write down Newton’s 2nd law in each coordinate direction (typically),
summing the forces.
Σ FR = maR = mv2/r
Σ FT = maT (for uniform circular motion)
4. If there is more than one mass, then Newton’s 2nd law may be
needed for each mass.
Room Frequency BA
Clicker Question
Wall-of-Death
Problem
Spinning
around with
back against the
wall.
What are the three forces #1, 2, 3?
1
2
3
A)
1 - gravity
2 - centrifugal force
3 – friction
B)
1 – friction
2 – normal force of the wall
3 – gravity
C)
1 - centripetal force
2 – normal force of the wall
3 – friction
D)
1 – friction
2 – centrifugal force
3 - gravity
For every case of uniform circular motion, there must
be a force directed towards the center.
We say there is a centripetal force. However, there is
always a specific force that is acting. There is no
“circle force”. Circular motion does not cause a force.
Ball circling
around tied to a Centripetal force  Tension Force
string.
Wall of Death
ride
Centripetal force  Normal Force
Race Car driving Centripetal force  Friction Force
in circle
Big G, Little g
Consider the force of gravity exerted by the Earth
with mass ME on a person of mass m on its surface?
RE

mM E
| F | gravity  G
2
RE


)

24

m
5
.
98

10
kg
11
2
2
| F | gravity  6.67 10 Nm / kg
2
6
6.37 10 m


2
| F | gravity  m  9.81m / s

| F | gravity  mg
)
)
Gravitational force on an object on the surface of the earth!
)
Room Frequency BA
Clicker Question
You are standing on the surface of
the earth.
The earth exerts a gravitational force
on you Fearth, and you exert a
gravitational force on the earth
Fperson.
Which of the following is correct:
A) Fearth > Fperson
B) Fearth < Fperson
C) Fearth = Fperson Newton’s Third Law
D) It’s not so simple, we need more
information.
Incomplete Gallery of Problems Involving
Newton’s Law of Universal Gravitation:
Room Frequency BA
Clicker Question
A rock of mass m is twirled on a
string in a horizontal plane.
The work done by the tension in the string on the rock is
T
A) Positive
B) Negative
C) Zero
The work done by the tension force
is zero, because the force of the
tension in the string is perpendicular
to the direction of the displacement:
W = F cos 90°= 0
Conservation of Mechanical Energy
Emechanical = KE + PE = constant
(isolated system, no dissipation)
Consider mass m swinging attached to a string of length L.
The swing is released from rest at a height h.
What is the speed v of the swing when it reaches height h/2?
MEi  ME f
KEi  PEi  KE f  PE f
0
KE = ½ mv2
PEgrav = mgy
1 2
1 2
h
mv i  mgh  mv f  mg
2
2
2
1 2 1
v f  gh
2
2
v f  gh
Clicker Question
Room Frequency BA
A block of mass m slides down a rough ramp of height h. Its initial
speed is zero. Its final speed at the bottom of the ramp is v.
m
h
Which is the amount of thermal energy, Ethermal, released from
the block’s motion down the ramp?
1
A) mgh  mv 2
2
1
C) mv 2  mgh
2
E) mgh
1
B) mgh  mv 2
2
1
D) mv 2
2
KEi  PEi  Eithermal  KE f  PE f  E thermal
f
1 2
0  mgh  0  mv  0  Ethermal
2
1 2
Ethermal  mgh  mv
2
A block of mass m is released from rest at height H
on a frictionless ramp. It strikes a spring with spring
constant k at the end of the ramp.
How far will the spring compress (i.e. x)?
0
0
KEi + PEi + Wfrict + Wexternal = KEf + PEf
0 + mgH + 0 +
x
2mgH
k
0
= 0 + (0 + ½ kx2)
gravitational
elastic
Clicker Question
Room Frequency BA
In which situation is the magnitude of the
total momentum the largest?
A) Situation I.
B) Situation II.
C) Same in both. Magnitudes are the same |ptotal|=mv
ptotal = mv + 0 = mv
ptotal = mv-2mv=-mv
Clicker Question
Room Frequency BA
A big ball of mass M = 10m and speed v strikes a small
ball of mass m at rest.
After the collision, could
the big ball come to a
complete stop and the
small ball take off with
speed 10 v?
A) Yes this can occur
B) No, because it violates
conservation of
momentum
C) No, because it violates
conservation of energy
pinitial  (10 m)v  10 mv
p final  m(10v)  10mv
1
KEinitial  (10 m)v 2  5mv 2
2
1
2
2
KE final  m(10 v)  50 mv
2
Rotational Kinematics
Describing rotational motion
angle of rotation (rads)

v 2

 2 f
r

angular velocity (rad/s)

atan
r
angular acceleration (rad/s2)
torque (N m)
moment of inertia (kg m2)
Newton’s 2nd Law
Kinetic energy (joules J)
KE  PE  constant
KEtrans  KErot  PE  constant
(Conservation of Mechanical
Energy)
IC
ptot   mi vi  constant
i
Tod
Ltot 
 Ii i  constant
i
angular momentum
(kg m2/s)
Clicker Question
Room Frequency BA
A small wheel and a large wheel are connected by a belt.
The small wheel turns at a constant angular velocity ωS.
There is a bug S on the rim of the small wheel and a bug
L on the rim of the big wheel? How do their speeds
compare?
A) vS = vL
B) vS > vL
C) vS < vL
Clicker Question
Room Frequency BA
  rF
Torque
Three forces labeled A, B, and C are applied to a rod which pivots
on an axis through its center.
Which force causes the largest size torque?
sin 45  cos 45  1 / 2  0.707
|  A | LF sin 45  0.707LF
L
|  B | F  0.5LF
2
L
|  C | 2F )  0.5LF
4
Clicker Question
Room Frequency BA
1
1 2
2
KEtot  Mv  I
2
2
2
I sphere  MR 2
5
I hoop  MR 2
M
Which object has the largest total
I disk
1
 MR 2
2
kinetic energy at the bottom of the ramp?
A) Sphere B) Disk
C) Hoop
D) All the same.
KEi  PEi  KE f  PE f
0  MgH  KE f  0
All have the same total KE.
KE f  MgH
Clicker Question
Room Frequency BA
H
1

2
2
 I hoop  MR , I disk  MR 
2
Which object will go furthest up the incline?
A) Puck
B) Disk
C) Hoop
KEi  PEi  KE f  PE f
2
1
1 v
2
Mv  I    MgH
2
2 r
D) Same height.
The hoop has the largest
moment of inertia, and
therefore the
highest total kinetic
energy.
Clicker Question
Room Frequency BA
Consider a solid disk of mass M and radius R with
an axis through its center.
An ant of mass m is placed on the rim of the disk.
I
1
MR 2  mR 2
2
The mass-disk system is rotating.
The ant walks toward the center of the disk.
The magnitude of the angular momentum L of
the system:
A) increases B) decreases C) remains constant
Unless an outside
torque is applied,
L = Iω = constant.
As ant moves inward, the kinetic energy of the Because I reduces, ω
system:
increases from the
A) increases B) decreases C) remains constant
ant’s motion.
Ch. 9 Statics and Static Equilibrium
Static Equilibrium: An object is
(1) not translating (not moving up, down, left, right)
(2) not rotating (not spinning CW or CCW).
Not translating:
F net  0   i Fi
F  F
x
Not rotating:
y
0
 net  0   i  i
(net force is zero)
(each component of the
net force is zero)
(net torque is zero)
A mass m is hanging (statically) from two strings.
The mass m, and the angles α and β are known.
What are the tensions T1 and T2?
Note: No lever arm. Thus no torques.
Two equations with two unknowns,
can solve for T and T after some algebra.
Ch. 10 Fluids
P   gh
Pascal’s
Principle
FOUT
Hydrostatic Pressure Equation
 AOUT 

FIN

 A 
IN
Fbuoy  m fluid g   fluidVg
Archimedes’
Principle
Vdisplaced
Floating Object
Vobject
object

 fluid
A1v1  A2 v2
Continuity Equation
1
2
The buoyant force equals the
weight of the fluid displaced.
or
1
2
1 A1v1  2 A2 v2
gy1  v12  P1   gy2  v22  P2
Bernoulli’s Equation
Clicker Question
Room Frequency BA
Cube A has edge length L and mass M. Cube B has edge
length 2L and mass 4M. Which has greater density?
A) A has larger density
B) B has larger density
C) A and B have the
same density.
Cube A has larger density. ρA = M/L3 , ρB = (4M)/(2L)3 = (1/2)M/L3 so
object A has twice the density of object B.
56
Clicker Question
Room Frequency BA
The air pressure inside the Space Station is P = 12 psi. There are
two square windows in the Space Station: a little one and a big
one. The big window is 30 cm on a side. The little window is 15
cm on a side. How does the pressure on the big window compare
to the pressure on the little window?
A) same pressure on both windows
B) 2 times more pressure on the big window
C) 4 times more pressure on the big window
D) 9 times more pressure on the big window
57
Clicker Question
Room Frequency BA
As shown, two containers are
connected by a hose and are filled
with water. Which picture correctly
depicts the water levels?
Different depths would give different
pressures! The net force on the fluid
at the connection tube would not be
zero and there would be flow!
P   gh
Pressure only depends on the depth; at
a given depth the pressure is the same.
58
Clicker Question
Room Frequency BA
Two identical bricks are held under water in a bucket. One of the
bricks is lower in the bucket than the other. The upward buoyant
force on the lower brick is…..
The weight of
displaced fluid does
not depend on depth
Fbuoy  m fluid g   fluidVg
59
Room Frequency BA
Clicker Question
“Incompressible” blood flows out of the heart via the aorta at a
speed vaorta. The radius of the aorta raorta = 1.2 cm. What is the speed
of the blood in a connecting artery whose radius is 0.6 cm?
A)
B)
C)
D)
E)
vaorta
2 vaorta
(2)1/2 vaorta
4 vaorta
8 vaorta
vartery
2
 Aaorta 
  raorta


vaorta   2  vaorta

 Aartery 
  rartery 
2
 raorta 

vaorta  4vaorta

 rartery 
60
CH. 11 &12 Vibrations, Waves, & Sound
Waves
Simple Harmonic Oscillator
Transverse & longitudinal.
F=-kx=ma

(Hooke’s Law)
v

T
 f
k
m
f 

1

2 2
T
1
m
 2
f
k
k
m
Standing waves on strings:
Etot = ½ mv2 + ½ kx2 = ½ kA2 (Cons of E)
Example: Pendulum

f 
g
L

1

2 2
g
L
1
L
T   2
f
g
Interpret plots of displacement, velocity,
and acceleration. Amplitude & period.
•
•
•
modes of oscillation
fundamental mode (n=1)
overtones (n=2,3,4,…)
fn = v/λn = nv/2L = nf1
Room Frequency BA
Clicker Question
The position of a mass on a spring as a function of time is shown
below. When the mass is at point P on the graph
A) The velocity v > 0
B) v < 0
C) v = 0
v is the slope at P
x
P
t
62
Wave Examples
Some common examples of waves are
• Sound waves (pressure waves) in air (or in any gas or solid or
liquid) – Longitudinal in gas/liquid, both transverse and
longitudinal in solid
• Waves on a stretched string – Transverse
• Waves on a slinky – Transverse and Longitudinal
• Waves on the surface of water - Transverse and Longitudinal
• "The Wave" at the ballpark stadium. The medium is the
people – Transverse
• Electromagnetic waves (light) – this is the only kind of wave
which does not require a medium! Their properties indicate
they are transverse, but you’ll learn more on EM waves next
semester.
63
Room Frequency BA
Clicker Question
The graph below shows a snapshot of a wave on a string which is
traveling to the right. There is a bit of paint on the string at point P.
y
vwave
P
x
At the instant shown, the velocity of paint point P has which
direction?
y
v
x
E) None of these
64
Room Frequency BA
Clicker Question
Three waves are traveling along identical strings (same mass per
length, same tension, same everything). Wave B has twice the
amplitude of the other two. Wave C has 1/2 the wavelength than A or
B. Which wave has the highest frequency? A B C D) All have same f
vwave
y
vf
A
x
y
B
x
No dependence on
amplitude
y
C
x
Smallest wavelength λ gives highest frequency f
65
Fundamentals and Overtones
When you pluck (or bow) the string it vibrates with many different
standing waves. These standing waves have to have nodes at the
fixed points at the ends.
λ = 2L and f1 = vstring /2L
This is the pitch of this note
λ = L and f2 = vstring /L = 2f1
λ = 2L/3 and f3 =3 vstring/2L = 3f1
In general overtones are
weaker, but not always!
66
CH. 13 & 14 Temperature & Heat
13.2
Temperature
13.4
14.1
14.3
14.4-5
Thermal expansion
Heat as energy transfer
Specific heat
Calorimetry & latent heat
EXTRAS
Clicker Question
Room Frequency BA
A projectile is fired straight up and then comes back
down to the ground. There is a force of air resistance.
During the entire flight, the total work done by the force
of gravity is:
A) Zero
B) Positive
C) Negative
During the entire flight, the total work done by the force
of air resistance is:
A) Zero
B) Positive
C) Negative
Clicker Question
Room Frequency BA
A pendulum is launched in two different
ways. During both launches, the bob is
given an initial speed 3 m/s and the same
initial angle from vertical.
Which launch will cause the pendulum to
swing the largest angle from the
equilibrium position to the left side?
A) Launch 1
B) Launch 2
C) Both are the same.
Clicker Question
Room Frequency BA
A block of mass m with initial speed v slides up a
frictionless ramp of height h inclined at angle q as
shown. Assume no friction.
Whether the block makes it to the top depend on the
angle q. What do you think about this statement?
A) True
B) False
Clicker Question
Room Frequency BA
A block of mass m slides down a rough ramp of height h. Its initial
speed is zero. Its final speed at the bottom of the ramp is v.
m
h
Which is the amount of thermal energy, Ethermal, released from
the block’s motion down the ramp?
1
A) mgh  mv 2
2
1
C) mv 2  mgh
2
E) mgh
1
B) mgh  mv 2
2
1
D) mv 2
2
KEi  PEi  Eithermal  KE f  PE f  E thermal
f
1 2
0  mgh  0  mv  0  Ethermal
2
1 2
Ethermal  mgh  mv
2
Clicker Question
Room Frequency BA
A pendulum consists of a mass m at the end of a string of
length L. When the string is vertical, the mass has speed
v0.
What is the maximum height h to which the mass swings?
Clicker Question
Room Frequency BA
Ball 1 of mass m moving right with speed v bounces off
ball 2 with mass M (M > m) and then moves left with
speed 2v.
What is the magnitude of the impulse of Ball #1
A) mv
B) 2mv C) 3mv D) ½ mv E) 0
Clicker Question
Room Frequency BA
A bullet of mass m traveling with initial speed v hits a
block of mass M on a frictionless table. The bullet
buries itself in the block, and the two together have a
final velocity vf.
The total kinetic energy of the bullet+block after the
collisions is _______ the total KE before the collision.
A) Greater than
B) Less than
C) Equal to
Clicker Question
Room Frequency BA
What is the torque about the origin?
A) r F sinq
B) r F cosq
C) zero
origin
Clicker Question
Room Frequency BA
Two light (massless) rods, labeled A and B,
each are connected to the ceiling by a
frictionless pivot as shown. Both rods are
released from a horizontal position.
Which one experiences the larger torque?
A) A
B) B
C) Same
I
2
m
r
 ii
i
θ
F  mg sin q
θ Fg = mg
 A  F L  Lmg sin q
L
 B  2mg sin q  Lmg sin q
2
Clicker Question
Room Frequency BA
Two light (massless) rods, labeled A and B,
each are connected to the ceiling by a
frictionless pivot as shown. Both rods are
released from a horizontal position.
Which one has the larger moment of inertia?
A) A
B) B
C) Same
2
I A  mL
2
1 2
 L
IB  (2m)    mL
 2
2
I
2
m
r
 ii
i
Clicker Question
Room Frequency BA
A small wheel and a large wheel are connected by a belt.
The small wheel turns at a constant angular velocity ωS.
There is a bug S on the rim of the small wheel and a bug
L on the rim of the big wheel? How do their speeds
compare?
A) vS = vL
B) vS > vL
C) vS < vL
Room Frequency BA
Clicker Question
A rock of mass m is twirled on a
string in a horizontal plane.
The work done by the tension in the string on the rock is
T
A) Positive
B) Negative
C) Zero
The work done by the tension force
is zero, because the force of the
tension in the string is perpendicular
to the direction of the displacement:
W = F cos 90°= 0
Clicker Question
Room Frequency BA
Roller Coaster Problem
N Fnet,y = N-mg = may = mv2/r
v
mg
Approximately circular arc
If the car moving rightward at the top of this hill:
A) The net force on the car is upward.
B) The net force on the car is downward
C) The net force on the car is zero
Clicker Question
Room Frequency BA
Two paths lead to the top of a big hill.
Path #1 is steep and direct and Path #2 is twice as long but less steep.
Both are rough paths and you push a box up each.
How much more potential energy is gained if you take the longer path?
A) none
B) twice as much
C) four times as much
D) half as much
d
h
#1
2d
h
θ
d 2  h2
#2
φ
4d 2  h 2
PE  mgh in both cases.
Clicker Question
Room Frequency BA
A big ball of mass M = 10m and speed v strikes a small
ball of mass m at rest.
After the collision, could
the big ball come to a
complete stop and the
small ball take off with
speed 10 v?
A) Yes this can occur
B) No, because it violates
conservation of
momentum
C) No, because it violates
conservation of energy
pinitial  (10 m)v  10 mv
p final  m(10v)  10mv
1
KEinitial  (10 m)v 2  5mv 2
2
1
2
2
KE final  m(10 v)  50 mv
2