1. Physics Final Review Dan C. and Ashwin M. – Exam 1
Allen W. + Vincent L. – Exam 2
Collin W. – Final [new material]
We will go through HW questions and Old exams Questions
Ask questions as they come up

2. Final Format Your exam will cover everything – 2/3 of the exam will be new material and 1/3 of the final will be old material [exam 1+2 stuff]. But it is important to know Exam 1+2 stuff really well because the professors will build that material into the new material Questions.

3. Physics 121 Midterm 1 Review
Ashwin Malhotra
and
Dan Carrero

4. Vectors Key points Know all your prefixes: Know basic trig
Micro 10^-6
Nano 10^-9
Pico 10^-12
Etc
Know basic trig
Sin(x)=O/H
Cos(x)=A/H
Tan(x)=O/A
Understand that X can change and that can change the relationships of the sides

5. Vectors

6. Solution
Explanation: You are give two sides of a right triangle, you can easily find the third side with a2+b2=c2. To find the angle you can use “inverse cosine” and use what they gave you or do “inverse sine” or “inverse tan” involving the y-component: Cos-1(2.2m/s/2.8m/s) = angle The units m/s cancel and you are left with the answer for the angle in degrees
2.8 m/s
2.2 m/s
The Angle in Question
Cos-1(X/Hyp) = angle
Sin-1(Y/Hyp) = angle
Tan-1(X/Y) = angle
X2 + Y2 = C2
Hyp aka C = sqt(X2 + Y2)

7. Vectors Key Theory What is a vector? How to add/substract Vectors?
A quantity which has magnitude and direction
How to add/substract Vectors?
Two ways  graphically versus components
Graphical lets you quickly see what going to happen
Components helps to get exact values i.e. previous problem
How to multiple/divide Vectors by scalars?
Scalars only alter the magnitude
The direction of the vector remains the same i.e. Arrow * 4 = 4Arrow
Do Not need to know Cross Product or Dot Product

8. Kinematics Difference between displacement and distance
Difference between velocity and speed
Difference between instantaneous velocity/acceleration and average velocity/acceleration
Motion key equations:
V=Vo+at
deltaX=Vot+.5at2
V2=Vo2+2a[deltaX]

9. Kinematics
You can find out that the CS is 5m/s easily. Then you can also find out that in 1.1seconds the car travels 5.5m. But that is not the answer to part A. You have to add 16m to that and get 21.5m ~22m to be correct. Why? Because they ask for the position at 1.1 not the change in position. Similar method for part B.

10. Kinematics

11. The amount of time it takes to get to max speed must be the same as the amount of time it takes to get to rest from max speed and the same amount of distance must be covered in both.
They gives us the Max Speed. We can calculate how much distance is covered and how long it takes to do so.
Using the max speed/max acceleration we can figure out how long it takes to accelerate and how much distance is covered
Decelerating to rest
Constant speed  its max velocity
Accelerating to max speed
This must also be 13.8m
This leaves us with this middle distance – we have it is being travelled at 6m/s. To find out how long, we take the distance and divide by speed:
[100 – 2(13.8)] / 6 = 21seconds
V=Vo+at
6=0+1.3t
T=6/1.3
deltaX=Vot+.5at2
deltaX=0+.5(1.3)(6/1.3)2
deltax=13.8m

12. Kinematics

13. Earth Moon
.5 = v^2 / (2*9.8) get v then use h = v^2 / (2*(9.8/6))
Earth Moon
1m = v^2 / (2*9.8) get V=4.42 then use that to get h = 4.42^2/ (2*(9.8/6))
Vf=V0+at
The Vf at the max point is zero.
0= V0 + (9.8*1.5); get V0=14.7
Same V0 on moon
0 = ((9.8/6)*(t)); get t  9seconds. This T is only half; have to double!

16. Relative Velocity
There two important cases to think about. The dumb boater and the smart boater.
One gets where he needs to be faster than the other. Why?
And the other gets exactly where he needs to be but slower. Why?
Make a good free body diagram for these and read carefully about directions of things travelling.
All kinematics equations are still valid here

17. Relative Velocity

18. The boat travels in one direct faster than the other  why
The boat travels in one direct faster than the other  why? the current must be helping. So we can figure out the direction of the water flowing. We can also figure out the speed:
29/4=7.25
29/6=4.83
7.25 – x = x
x = 1.2m/s
the x is helping one and slowing down the other; so we say if there was no x there would be an equal speed for traveling both ways – we can set up the above expression and find what x is and that equal speed

19. The way he will travel
The way he travels

20. Relative Velocity

21. This is how they should travel to get where they want to be: 17m/s
This how they will travel
West
South
West wind:
13m/s
With the picture above it should be easy to figure out what the angle [red star] of travel should be  Sin(angle)=13/17
Angle = ~ 50
to be exact the angle is 50 degrees Southwest

22. How he will travel
How he travels
X velocity comes from the current

23. Motion There is the simple linear motion.
Then there is projectile motion.
There are horizontal projectiles and there are vertical projectiles
Treat X and Y directions separately
Key thing to keep in mind is that all projectile motion problems have time in common between the X and Y direction.
Basic question would be to give enough things to get time from one direction and to find something using that time in the other direction

24. Motion

25. Sin(A) = Opp/Hyp
29{Sin(31)} = opposite = Velocity in the Y
Vf – Vi = a*t  0 – Vy = (9.8/6) * (t)
t = [this the time required to reach the max height, half total time. Have to double it to get answer~18s]
Cos(A) = Adj/Hyp
29{Cos(31)} = Adjacent = Velocity in X
Vx * t = Distance
Vx * t = distance [earth] 29{Cos(31)} t = x plug in for t, x= 75.7m
Vx * t = distance [moon] 29{Cos(31)} * 18s = 450m
Divide 450/75 ~ 6
Time earth comes from Vf – Vi = a * t  0 – 29{Cos(31)} = 9.8 * t
t = 3.045…

26. Motion

28. Force
F=ma
The ma is the cause and the F is effect; not the other way around. Think about this:
The mass does not change but acceleration does, together they create a force which can do work
You will have to put together force with work for the final
Drawing a free body diagram is the only way to solve force problems
Best thing to do is to draw sample free body diagrams on your cheat sheet so you do not miss any forces during the exam

29. Force

30. Force

32. Force

33. The Elevator Problem

39. Work, Force, and Energy WARM UPS!!
1st: Convert the velocity to meters/second
100 km/hr = 27.8 m/s
2nd: Once everything is in SI Units, we can start thinking about the problem!!
3rd: There are two ways to think about this problem KINEMATICS or WORK/ENERGY
KINEMATICS:
Step 1: Vf2 = Vi2 + 2ad Solve for a.
Step 2: Since F = ma and friction stops the car
F = ma = Ff = ukFn = ukmg
Therefore
uk = a/g
Work – Energy:
Step 1: Since the car has speed it has a kinetic energy (KE). Since the car stopped work was done (by friction) to change the kinetic energy KEf – KEi = work = Ff d
Step2: /2mvf2 – 1/2mvi2 = ukmg d
Therefore
uk = v2/(gd)

40. Newton’s 2nd law is: The SUM of all the on an object forces is equal to that objects mass multiplied by its acceleration.
F1 + F2 + F3 + ….. = ma
With forces you need to visualize what’s doing what! This can only be learned through practice.
1. Draw the direction of
forces acting on all of
the objects in the
problem.
2. The direction of your
arrows only gets
meaning when YOU decide which way is +
No force is ever negative
Unless YOU decide it is.
Your decision is based on
The direction you set your
Coordinates.
If a force arrow you draw
Is opposite your
Coordinates, then and only
then is it negative. T Ty
Fwall Tx +y
mb g +x

41. Normal Force: FN=ukFf M mg

43. Given:
mC = 1200 kg
v = 15 m/s
h = 5 m
Want:
v’ = ?
Initial
Final

44. Allen Wu and Vincent Lin
Midterm 2 review

45. KE + PE = KE’ + PE’ ½mv2 + mgh = ½mv’2 ½v2 + gh = ½v’2
Work/energy is scalar meaning that it does not matter on the path it takes, as long as the surface is frictionless. Only focus on the beginning and the end.
Initial: Car is moving (kinetic energy) and is 5 m above the “final” ground (potential energy)
Final: Car is moving (kinetic energy) on “final” ground (no potential energy)
W = W’
KE + PE = KE’ + PE’
½mv2 + mgh = ½mv’2
½v2 + gh = ½v’2
½(15 m/s) m/s2(5 m) = ½(v’)2
17.97 m/s = v’

47. vy
vx’
vy' vx v'

48. Energy is NOT conserved (inelastic)
Momentum is conserved
Given
mx = 1000 kg
my = 800 kg
vx = 15 m/s
vy = 20 m/s
Want
v‘ = ?
m1v1i + m2v2i = m1v1f + m2v2f
x: (1000 kg)(15 m/s) + (800 kg)(0 m/s) = (1000 kg kg)(v’x)
: kg m/s = (1800 kg)(v’x)
15000 kg * m/s = 8.33 m/s = v’x
1800 kg
y: (1000 kg)(0 m/s) + (800 kg)(20 m/s) = (1000 kg kg)(v’y)
: kg * m/s = (1800 kg)(v’y)
16000 kg * m/s = 8.89 m/s = v’y
1800 kg
a2 + b2 = c2
= c2
12.18 = c

49. Convert to SI units! Want Given Angular velocity (ω) = ?
Radius of disc = 2 m
Initial tangential velocity = 3 m/s
Angular acceleration (α) = 1 rad/s2
Angular position (θ) = 2 revolution
Convert to SI units!
2 revolutions * 2π radians = 4π radians
1 revolution

50. Look at your angular mechanics formulas, their format is identical to linear mechanics
θ = θ0 + ω0t + ½αt2
ω = ω0 + αt
ω2 = ω02 + 2α(θ – θ0)
Conversions from linear to angular mechanics
ω = v/r
α = a/r
Since we are working with angular stuff, convert tangential velocity to angular velocity
ω0 = 3 m/s = 1.5 rad/s
2 m
We are looking for angular velocity after the disc accelerated for after a measured distance.
Also, notice that there is no time mentioned in the question, so if would be harder to use the first 2 equations.
ω2 = ω02 + 2α(θ – θ0)
ω2 = (1.5 rad/s)2 + 2(1 rad/s2)(4π rad)
ω2 = rad2/s2
ω = 5.23 rad/s

52. fmdr2ω = (fmdr2 + fmpr2) (ω’)
Look at the statements being made, all of them are comparing moment of inertia to angular velocity.
So, look for an equation that compares moment of inertia to angular velocity
Equation Iω = I’ ω’ = L
L is angular momentum, and angular momentum is always conserved
We know I is defined as f*m*r2, where f is a variable defined by the object
So we can rewrite equation 8.14 as fmr2ω = fmr’2ω’
In the problem, the person is initially standing at the axis of rotation. Therefore, that person has no angular momentum.
Initial: Only the disk has an angular momentum fmdr2ω
After the person reaches the edge of the disk, that person is now not at the axis of rotation. This means there is a distance from the person to the axis, and that person now has angular momentum.
Final: Angular momentum of the disk fmdr2ω’
angular momentum of the person fmpr2ω’
fmdr2ω = fmdr2ω’ + fmpr2ω’
fmdr2ω = (fmdr2 + fmpr2) (ω’)
I ω = I’ ω’
This term is larger (moment of inertia of the system)
Therefore, ω’ must be smaller (in order to keep the left side constant)

54. The system is at equilibrium, which means Σ τ = 0
There are 4 torques on the system, the torque on mass 1, mass 2, left side of the ruler and right side of the ruler.
τ = F*d
τ3 (+)
τ4 (-)
I labeled the torques 1-4 as followed
τ2 (-)
τ1 (+)
Remember, counterclockwise torque is positive, clockwise torque is negative

55. τ3 = F * d = m3gd3 = .3mR * 9.81 * .5(.3 m – 0 m) = .44*mR
Given:
m1 = 3 kg
m2 = 1 kg
d1 = .3 m
d2 = .7 m
LR = 1 m
Want:
mR = ?
τ1 = F * d = m1gd1 = 3 * 9.81 * .3 m = 8.83
τ2 = -F * d = -m2gd2 = -1 * 9.81 * .7 m = -6.87
τ3 = F * d = m3gd3 = .3mR * 9.81 * .5(.3 m – 0 m) = .44*mR
τ4 = -F * d = -m4gd4 = -.7mR * 9.81 * .5(.7 m – 0 m) = -2.4*mR
Σ τ = 0 = mR mR
0 = 1.96 – 1.96mR
-1.96 = -1.96mR
1 = mR

57. What to Do Here:
Draw Free Body Diagrams!
Manipulate the equations you already have T1 T2 m3 m1 m2 T1 T2
F = Iα/r = T2 – T1
½m3r2α/r = T2 – T1
½m3r2a/r2 = T2 – T1
m3a = 2(T2 – T1) W1 W2
F1 = m1a = T1 – W1
F2 = m2a = W2 – T2
m2a = m2g – T2

59. Remember in springs
At maximum displacement (pulled or stretched)  maximum potential energy
At 0 displacement  maximum kinetic energy
Anything in between will have some kinetic energy and some potential energy
Total energy of the system is E. Total energy of the system is also at maximum potential energy (has 0 kinetic energy) or maximum kinetic energy (has 0 potential energy).
E = PEmax = KEmax
Since the question is asking for potential energy, we will use PE instead of KE
PEmax = PE + KE
PEmax = PE + ¾E = PE + ¾PEmax
PEmax – ¾PEmax = PE
½kA2 – ¾*½kA2 = PE
½kA2 – 3/8 kA2 = PE
4/8kA2 – 3/8kA2 = PE
1/8 kA2 = PE
Question is asking for the system’s potential energy, PE

60. When working with energy, just look at beginning and final parts of the question
We know at the end, the block has reached a maximum height and not moving, giving it PEmax = mgh.
At the initial step when the block is compressed, there is work being acted on the block. W = Fx. We need to find this net force.
What to Do Here
Draw Free Body Diagrams!

61. Initial N
ΣFy: N – mgcos(θ) = 0 Fs
ΣFx: Fs – mgsin(θ) = 0
mgcos(θ)
What we want
Initial θ θ
W = F * d
W = (Fs – mgsin(θ)) * d
W = Fsd – mgdsin(θ)
W = ½kd2 – mgdsin(θ)
Final
W = mgh mg
mgsin(θ)
½kd2 – mgdsin(θ) = mgh
½kd2 – mgdsin(θ) = h mg
kd2 – dsin(θ) = h
2mg

62. Free Body Diagram! N T Φ θ
Given
m = 10 kg
φ = 20°
Want
T = ? mg
Remember, normal force is the force exert by a surface onto the object. Imagine if there was no wall there to stop the beam from moving. The beam would move left.
But since the wire is not fixed at the center of the beam, the beam will tilt downwards.
N = normal force
T = tension

63. Using head to tail rule, we can make a vector
Using head to tail rule, we can make a vector. The bolded arrow is the direction the beam wants to go. Since normal force (dashed) is the force that opposes the beam from moving, it would point in the opposite direction.
Next, break up your T and N into their components
Tx = Tcos(Φ) Ty = Tsin(Φ)
Nx = Ncos(θ) Ny = Nsin(θ) N
This system is at equilibrium
ΣFx = ΣFy = Στ = 0
2 unknown variables
2 unknown variables
2 unknown variables, but 1 cancels out, use this.
ΣFx = Nx – Tx = 0
ΣFy = Ty + Ny – mg = 0
Στ = Ty¾L – mg½L = 0
Use trig here
sin(Φ) = Ty/T
sin(20) = 65.4/T
191.2 N = T T
Ty¾L = mg½L
¾Ty = ½mg
Ty = (2/3)*10 kg * 9.81 m/s2
Ty = 65.4 N Ty Φ Tx

64. Collin West
New Material

65. PHY 121 Review—Material after Midterm 2
Waves and Sounds
Doppler and Beats
Buoyancy
Fluids and Pressure
Gas laws
Heat Transfer
Laws of Thermodynamics*
Heat Engines*

66. Wave Velocities
A wave on a string has velocity of 200 m/s. What is the velocity of a wave on a string with the same tension and made of the same material, but with twice the radius?
A) 50 m/s
B) 100 m/s
C) 200 m/s
D) 400 m/s

67. Wave Velocities
A wave on a string has velocity of 200 m/s. What is the velocity of a wave on a string with the same tension and made of the same material, but with twice the radius?

68. Wave Velocities So doubling the radius halves the speed.
A wave on a string has velocity of 200 m/s. What is the velocity of a wave on a string with the same tension and made of the same material, but with twice the radius?
So doubling the radius halves the speed.
The answer is B, 100 m/s

69. Wave Velocities Other possible questions:
Give you the speed and the tension, find the mass
What tension would it take to produce the desired velocity, given the mass
What’s the density, given the radius, tension and speed?
Given the Tension, linear density and length, how long does it take a wave pulse to travel the length of the string?

70. Doppler Shifts
A submarine sends out a 10MHz sonar pulse, which reflects off a blue whale swimming the other direction at 1.0 m/s. What frequency difference is observed when ? The speed of sound in seawater is 1560 m/s.
A) 0 kHz
B) 6.4 kHz
C) 13 kHz
D) 1 MHz

71. Doppler Shifts Two steps:
A submarine sends out a 10MHz sonar pulse, which reflects off a blue whale swimming the other direction at 1.0 m/s. What frequency difference is observed when ? The speed of sound in seawater is 1560 m/s.
Two steps:

72. Doppler Shifts Getting the Equations Right:
A submarine sends out a 10MHz sonar pulse, which reflects off a blue whale swimming the other direction at 1.0 m/s. What frequency difference is observed when ? The speed of sound in seawater is 1560 m/s.
Getting the Equations Right:
Velocities are positive if the receiver and source are moving TOWARDS each other.
Be careful—other definitions exist. Just pick one that works for you and make sure
it’s well-labeled on your formula sheet.

73. Doppler Shifts
A submarine sends out a 10MHz sonar pulse, which reflects off a blue whale swimming the other direction at 1.0 m/s. What frequency difference is observed when ? The speed of sound in seawater is 1560 m/s.

74. Doppler Shifts
A submarine sends out a 10MHz sonar pulse, which reflects off a blue whale swimming the other direction at 1.0 m/s. What frequency difference is observed when ? The speed of sound in seawater is 1560 m/s.
(Answer C)

75. Wave Velocities Other possible questions:
The source might also be in motion
They could give you an observed frequency and ask you to find the speed of the source (one-step problem only!)
They could give a source , a frequency shift, and ask you to find the speed in the medium
They could ask about beats between transmitted and received waves.
Know how to convert between frequencies, wavelengths, and periods.

76. Buoyancy
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
A) Only the copper rises
B) Only the aluminum rises
C) Both blocks rise
D) Neither block moves

77. Buoyancy
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
The key idea: The buoyant force on an object in a fluid is the weight of the fluid it displaces
Contrary to your intuition, the buoyant force depends ONLY on the volume of the object and the density of the fluid it’s immersed in. It always points upwards. If the buoyant force is larger than the force of gravity, the object will float. Otherwise it will not.
More familiar result: an object floats if it is less dense than the fluid it is in.

78. Buoyancy The diagram in this problem is trying to trick you!
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
The diagram in this problem is trying to trick you!
Remember, it’s always assumed it is not to scale.

79. Buoyancy
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
Copper is more dense, so it takes a smaller block to make 1kg. Smaller block means less water displaced, which means less buoyant force.

80. Buoyancy
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
Copper is more dense, so it takes a smaller block to make 1kg. Smaller block means less water displaced, which means less buoyant force.

81. Buoyancy
1 kg of copper (8900 kg/m3) and 1 kg of Aluminum (2700 kg/m3) are suspended on opposite sides of an Atwood machine. Initially, both are at the same height. The entire system is immersed in water. What happens?
Either way, the answer is B: The aluminum rises because the copper feels a larger net force downward. Always draw free-body diagrams!

82. Buoyancy Other possible questions:
Given an object of a certain size, how heavy can it be before it sinks?
Given the buoyant force on an object of a certain volume, what is the density of the fluid it is immersed in?
Given a rectangular block with a known size and density which floats on water, how high will the water line be?

83. Fluids: Hydrostatics
What is the pressure somewhere, how high does the fluid rise, what’s the density of the fluid
Excellent review of concepts on Mastering Physics, chapter 11 practice problems: Tactics Box 13.2

84. Fluids: Ideal Hydrodynamics
An oil with density 900 kg/m3 flows upward through a pipe, gaining 10 m elevation and increasing it’s speed from 2 m/s to 3 m/s. If the gauge pressure beforehand was 200 kPa, what is the pressure afterwards?
A) 110 kPa
B) 190 kPa
C) 250 kPa
D) 300 kPa

85. Fluids: Ideal Hydrodynamics
An oil with negligible viscosity flows upward through a pipe, gaining 10 m elevation and increasing it’s speed from 2 m/s to 3 m/s. If the gauge pressure beforehand was 200 kPa, what is the pressure afterwards?
Bernoulli: So
(Answer A)
Beware units conversions. It’s absolute pressure, so if you get a negative answer, it’s not right!

86. Fluids: Ideal Hydrodynamics
Other possible questions:
Might not give you both velocities, but give you the radius of the pipe:
Might have multiple segments, in which case, just repeat the procedure multiple times
Might ask you to find the density of the fluid by giving you the pressures at two points
Might ask you about a trough or a tub; these are just differently shaped pipes!

87. Fluids: Viscous Hydrodynamics
Water (with viscosity coefficient 0.001) leaves the end of the syringe indicated at right with a velocity of 10 m/s. What is the gauge pressure at point P?
A) 3.2 kPa
B) kPa
C) kPa
D) kPa

88. Fluids: Viscous Hydrodynamics
Water (with viscosity coefficient 0.001) leaves the end of the syringe indicated at right with a velocity of 10 m/s. What is the pressure at point P?
Poiseuille Equation:
The secret:
So And
Finally,
(answer C)

89. Ideal Gasses: The Basics
A small, familiar toolbox:
P should be absolute pressure
n = number of moles
Might need to get this from the mass and molar mass, or the volume of a gas, or perhaps even from the number of atoms!
T = temperature in KELVIN
R = gas constant
R = 8.31 J/ mol K -- must use Pressure in Pa, volume in m3

90. Heat: The Basics Possible Questions…
How much work does it take to create a given amount of heat?
Find the amount of thermal expansion for a given material when heat is added (or vice-versa)
Find the amount that temperature increases for a given material when heat is added (or vice-versa)
What if there is more than one type of material?

91. Heat Transfer (Calorimetry)
A 40g block of Copper is heated to 400 degrees C, then dropped in an insulated bath containing 1 kg of water at 20 degrees C. What is the temperature of the water at some time much later? Use the information given at right.
Specific Heats (in J/gK)
Copper = 0.385
Water = 4.187
A) 21 degrees C
B) 44 degrees C
C) 294 degrees C
D) 420 degrees C

92. Heat Transfer (Calorimetry)
A 40g block of Copper is heated to 400 degrees C, then dropped in an insulated bath containing 1 kg of water at 20 degrees C. What is the temperature of the water at some time much later? Use the information given at right.
Specific Heats (in J/gK)
Copper = 0.385
Water = 4.187
Three Easy(ish) Steps:
1) Heat lost by one equals heat gained by the other:
2) , and each is the same:
Then simply (ish) solve for !
(In this case, answer is A)

93. Heat Transfer (Calorimetry)
Things to be careful of
Units! Especially for the specific heats.
If the specific heats are in grams, the masses must be in grams!
Temperatures must be converted into Kelvin, but the answer will be in Kelvin as well!
They might give you volumes instead of masses– you’ll have to convert to masses by using the densities
Be careful: one of the substances may change phase (e.g., melt)