PHYSICS 2010 Final Exam Review Session

PHYSICS 2010 Final Exam

Review SessionNote that this review is not covering everything. Rather it is highlighting key points and concepts.

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.

Information on Course Web Page

http://www.colorado.edu/physics/phys2010/phys2010_fa11/finalexam_2010_fa11_equations.pdf

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 displacement2.2 Average velocity2.3 Instantaneous velocity2.4 Acceleration2.5 Motion at constant acceleration2.6 Solving problems2.7 Falling objects2.8 Graphical analysis of linear motion

Motion under constant acceleration

What are some example problems?

Finally: algebra, trigonometry, unit conversion.

Motion under constant acceleration

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 scalars3.2 Addition of vectors graphically3.3 Vector subtraction and multiplication by a scalar3.4 Adding vectors by components3.5 Projectile motion3.6 Solving problems

The vector A has magnitude |A| = 6 cm and makes an angle ofq = 30 degrees with the positive y-axis as shown.

x

yA

q

What is the x-component of the vector A?A) +4.0 cmB) -3.0 cmC) +5.2 cmD) - 5.2 cmE) 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).

200 2

1 tatvxx xx 200 2

1 tatvyy yy

0cos||

0

00

0

x

x

avv

xq

ga

vvy

y

y

qsin||0

00

0

q

tvx )cos|(| 0 q 2

0 21)sin|(| gttvy q

Approach1. 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)

tvx )cos|(| 0 q

20 2

1)sin|(| gttvy q


q Clearly at t =0, x=0 and y=0 (our initial conditions).

At what later time does the football hit the ground?

A) t = 10 secondsB) t = |v0| tanqC) t = Sqrt(2|v0|sinq/g)D) t = 2|v0|sinq/gE) None of the above

20 2

1)sin|(|0 gttv q

gtvt 21)sin|(|0 0 q

021)sin|(| 0 gtv q

gvt /)sin|(|2 0 q

tvx )cos|(| 0 q

20 2

1)sin|(| gttvy q

qgvt /)sin|(|2 0 q

Time of football flight (i.e. hang time)

Now plug into x-equation to find out position when it hits the ground.

gvvx /sin||2)cos|(| 00 qq

gv

xqq sincos||2 2

0

g

v q2sin|| 20

* 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.

A moving van collides with a sports car in a high-speed head-on collision. Crash!

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

F

truck F

carM m

Guaranteed by Newton’s Third Law


A moving van collides with a sports car in a high-speed head-on collision. Crash!

F

truck F

car

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 C) atruck = acar

D) Indeterminate from information given.

M m


Newton’s Second Law: F = ma

Step 1: Draw a free-body diagram

Fg = mg

Note that “Normal Force” is always Perpendicular

(i.e. normal) to the surface.

In this case, it is not in the opposite direction to the

gravitational force.

Fg = mg

x

y We could choose our “usual” axes. However, we know that we will have motion

in both the x and y directions.

Step 2: Choose a coordinate system

x

yIf 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

Problem: Fg doesn’t point in a coordinate direction.Must break it into its x- and y-components.

Fg = mg

|F(gravity)| = mg

F(gravity)y

F(gravity)x


What is the angle a labeled above?A)a = 90 degrees B) a = qC) a = 90 – q D) Cannot be determined

a90 – q90

Fg = mg

Step 3: Write down the equations Σ Fx = m ax , Σ Fy = m ay

N Fx = mg sin(q)

Fy = N – mg cos(q)

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

)sin(qmgmaF xx

)cos(qmgNmaF yy

)sin(qgax

0)cos( qgmNa y

)cos(qmgN

As expected, maximum

acceleration if straight down

(q=90 degrees), ax = g.

Recall x-y definition

A mass m accelerates down a frictionless inclined plane.

Which statement is true?A) N < mgB) N > mgC) N = mg

+x

+y

θ

θ-mg cos θ

-

0 = Fy = N – mg cos θN = mg cos θ < mg


M1

M2

Frictionless table & pulley.

1)Choose coordinates2)Identify forces.3) Write down F = ma

for each object

+x

+x

+T-T

Fg = M2g

Mass 1: M1a = T

Mass 2: M2a = M2g - T(A) M 2a T (B) M 2a T M 2g (C) M 2a M 2g T(D)M 2a M 2g

What is F = ma for mass M2?


Clicker Question Room Frequency BAYou 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) ms = 0B) ms = 0.1C) ms = 0.5 D) mk = 0.5E) None of the above

Fnet = 0 = Fpush – Ffriction

0=Fpush – ms x Normalms x Mg = Fpush

ms = Fpush/Mg = 5000/10kms= 0.5

But actually, ms could be even larger since we do not know if we have

reached the maximum.

Incomplete Gallery of Problems Involving Newton’s Laws:

w & w/0 frictionw & w/0 friction

M m+x

F

fc

m

Box 1 Box 2

m

M

m

M1

M2

w & w/0 friction

F

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.

1

2

3

What are the three forces #1, 2, 3?

A) 1 - gravity2 - centrifugal force3 – friction

B) 1 – friction2 – normal force of the wall3 – gravity

C) 1 - centripetal force2 – normal force of the wall3 – friction

D) 1 – friction2 – centrifugal force3 - gravity


Wall-of-Death ProblemSpinning

around with back against the

wall.

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

string.Centripetal force Tension Force

Wall of Deathride Centripetal force Normal Force

Race Car driving in circle

Centripetal force Friction Force

Consider the force of gravity exerted by the Earth with mass ME on a person of mass m on its surface?

RE

Big G, Little g

2||E

Egravity

RmM

GF

) ) )26

242211

1037.6

1098.5/1067.6||m

kgmkgNmF gravity

)2/81.9|| smmF gravity

mgF gravity ||

Gravitational force on an object on the surface of the earth!

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

D) It’s not so simple, we need more information.


Newton’s Third Law

Incomplete Gallery of Problems Involving Newton’s Law of Universal Gravitation:

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

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

TA) Positive B) NegativeC) Zero


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?

KE = ½ mv2

PEgrav = mgy

Conservation of Mechanical Energy

fi MEME

ffii PEKEPEKE

221

21 22 hmgmvmghmv fi

0

ghv f 21

21 2

ghv f

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.

Which is the amount of thermal energy, Ethermal, released from the block’s motion down the ramp?

A) mgh 12

mv2 B) mgh 12

mv2

C) 12

mv2 mgh D) 12

mv2

E) mgh

h

m

KEi PEi Eithermal KE f PE f E f

thermal

0 mgh 0 12

mv2 0 Ethermal

Ethermal mgh 12

mv2


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)?

KEi + PEi + Wfrict + Wexternal = KEf + PEf

0 0

0 + mgH + 0 + 0 = 0 + (0 + ½ kx2)

kmgHx 2

gravitational

elastic

In which situation is the magnitude of the total momentum the largest?

A) Situation I.B) Situation II.C) Same in both.

ptotal = mv + 0 = mv


ptotal = mv-2mv=-mv

Magnitudes are the same |ptotal|=mv

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 occurB) No, because it violates conservation of momentumC) No, because it violates conservation of energy

mvvmpinitial 10)10( mvvmp final 10)10(

22 5)10(21 mvvmKEinitial

22 50)10(21 mvvmKE final

Rotational Kinematics Describing rotational motion

atan

r

vr

2

2 f

IC

angle of rotation (rads)

angular velocity (rad/s)

angular acceleration (rad/s2)

torque (N m)

moment of inertia (kg m2)

Newton’s 2nd Law

Last

Wee

kM

onda

y

Tod

KE PE constant KEtrans KErot PE constant(Conservation of Mechanical Energy)

ptot mii vi constant Ltot Ii

i i constant

Kinetic energy (joules J)

angular momentum (kg m2/s)

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


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

| B |L2

F 0.5LF

| C |L4

2F )0.5LF

rFTorque


KEtot 12

Mv2 12

I 2

I sphere 25

MR2

Ihoop MR2

Idisk 12

MR2

Which object has the largest total

kinetic energy at the bottom of the ramp?

A) Sphere B) Disk C) Hoop D) All the same.


ffii PEKEPEKE 00 fKEMgH

MgHKE f

M

All have the same total KE.

Which object will go furthest up the incline?

A) Puck B) Disk C) Hoop D) Same height.

Ihoop MR2 , Idisk 12

MR2

The hoop has the largest moment of inertia, and

therefore thehighest total kinetic

energy.

H


ffii PEKEPEKE

MgHrvIMv

22

21

21

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.

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

As ant moves inward, the kinetic energy of the system:

A) increases B) decreases C) remains constant

Unless an outside torque is applied,L = Iω = constant.

Because I reduces, ω increases from the

ant’s motion.

I 12

MR2 mR2


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 Fi

i

Fx Fy 0

(net force is zero)

(each component of the net force is zero)

net 0 ii (net torque is zero)Not rotating:

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 T1 and T2 after some algebra.

Ch. 10 Fluids

P gh

FOUT AOUT

AIN

FIN

Pascal’s Principle

Fbuoy m fluidg fluidVgArchimedes’ Principle

The buoyant force equals the weight of the fluid displaced.

Vdisplaced

Vobject

object

fluidFloating Object

A1v1 A2v2 or 1A1v1 2A2v2Continuity Equation

gy1 12v1

2 P1 gy2 12v2

2 P2 Bernoulli’s Equation

Hydrostatic Pressure Equation

56

Cube A has edge length L and mass M. Cube B has edge length 2L and mass 4M. Which has greater 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.

A) A has larger densityB) B has larger densityC) A and B have the same density.

57

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

58

P gh

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!

Pressure only depends on the depth; at a given depth the pressure is the same.

59

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…..

Fbuoy m fluidg fluidVg


The weight of displaced fluid does not depend on depth

60

“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) vaorta

B) 2 vaorta

C) (2)1/2 vaorta

D) 4 vaorta

E) 8 vaorta

vartery Aaorta

Aartery

vaorta

raorta2

rartery2

vaorta

raorta

rartery

2

vaorta 4vaorta


km

f 2

1

2km

T 1f

2mk

gL

f 2

1

2gL

T 1f

2Lg

v T

f

fn = v/λn = nv/2L = nf1

Simple Harmonic Oscillator

Example: Pendulum

Interpret plots of displacement, velocity, and acceleration. Amplitude & period.

Waves

Transverse & longitudinal.

Standing waves on strings:

• modes of oscillation• fundamental mode (n=1)• overtones (n=2,3,4,…)

CH. 11 &12 Vibrations, Waves, & Sound

Etot = ½ mv2 + ½ kx2 = ½ kA2 (Cons of E)

F=-kx=ma (Hooke’s Law)

62

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

t

P

63

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.

64

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.

At the instant shown, the velocity of paint point P has which direction?

E) None of these


y

x P

vwave

v

x

y

65

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


Smallest wavelength λ gives highest frequency f

y

x

vwave

y

x

y

x

A

C

B

v f

No dependence on amplitude

66

Fundamentals and OvertonesWhen 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/2LThis 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!

CH. 13 & 14 Temperature & Heat13.2 Temperature

13.4 Thermal expansion14.1 Heat as energy transfer14.3 Specific heat14.4-5 Calorimetry & latent heat

EXTRAS


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) ZeroB) PositiveC) Negative

During the entire flight, the total work done by the force of air resistance is:A) ZeroB) PositiveC) Negative


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 1B) Launch 2C) Both are the same.


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

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.

Which is the amount of thermal energy, Ethermal, released from the block’s motion down the ramp?

A) mgh 12

mv2 B) mgh 12

mv2

C) 12

mv2 mgh D) 12

mv2

E) mgh

h

m

KEi PEi Eithermal KE f PE f E f

thermal

0 mgh 0 12

mv2 0 Ethermal

Ethermal mgh 12

mv2



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?


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 #1A) mv B) 2mv C) 3mv D) ½ mv E) 0


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 thanB) Less thanC) Equal to

origin


What is the torque about the origin?

A) r F sinqB) r F cosqC) zero

I miri2

i

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

A FL Lmgsinq B L2

2mgsinq Lmgsinq

θ

Fg = mgθ

F mgsinq


I miri2

i

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

IA mL2 IB (2m)L2

2

12

mL2


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


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

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

TA) Positive B) NegativeC) Zero


Roller Coaster ProblemClicker Question Room Frequency BA

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 downwardC) The net force on the car is zero

mg

N Fnet,y = N-mg = may = mv2/r

v

Approximately circular arc

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

h hd 2d

d 2 h2 4d 2 h2

#1 #2θ φ

PE mgh in both cases.


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 occurB) No, because it violates conservation of momentumC) No, because it violates conservation of energy

mvvmpinitial 10)10( mvvmp final 10)10(

22 5)10(21 mvvmKEinitial

22 50)10(21 mvvmKE final

Documents

PHYSICS 2010 Final Exam Review Session