physics semester 2 review and tutorial
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SEMESTER 2 REVIEW
&TUTORIAL
Slides with red backgrounds involve word problems.
Slides with tan backgrounds involve matching concepts.
Slides with olive backgrounds involve reading data tables.
Slides with green backgrounds involve graphing.
The slides used in this tutorial are color coded. If you are experiencing difficulty
with one aspect of your understanding than another you might find this coding useful.
Make sure you know how to use the following formulas.These will be given to you on a separate card in finals day.
Displacement, Time, Velocity, Acceleration, Mass, Force, Weight, Tension, Impulse, Momentum, Work, Energy, Power, Spring Constant, Radius, Tangential Velocity, Centripetal Acceleration, Centripetal Force, Angular Momentum, Wavelength, Frequency, Time Period, Mach Number, Index of Refraction
MatchSelect the units that match the physics quantity.
Hz or
cycles/s
J or
kgm2/s2
J/s kg kgm/s
kgm2/s
m m/s m/s2 N or
kgm/s2
N/m None Ns s s/cycle
Displacement- mTime- sVelocity- m/sAcceleration- m/s2
Mass- kg Force- N Weight- N Tension- NImpulse- NsMomentum- kgm/sWork- JEnergy- J
Matched
Power- J/sSpring Constant- N/mRadius- mTangential Velocity- m/sCentripetal Acceleration- m/s2
Centripetal Force- NAngular Momentum- kgm2/sWavelength- mFrequency- Hz (cycles/s)Time Period- s/cycle or just s Mach Number- NoneIndex of Refraction- None
Analyze the Energy Bar Graph for a 1000 kg rollercoaster. Calculate the height of each hill.
Work Ug KE Ug KE Ug KE Ug KE0
50000
100000
150000
200000
250000
Energy Bar Graph
Analyzed Graph
20.41 m
15.31 m
5.1 m
17.86 m
Work Ug KE Ug KE Ug KE Ug KE0
50000
100000
150000
200000
250000
Energy Bar Graph
Ug = mgΔy
Analyze the Energy Bar Graph for a 1000 kg rollercoaster. Calculate the velocity on each hill.
Work Ug KE Ug KE Ug KE Ug KE0
50000
100000
150000
200000
250000
Energy Bar Graph
Analyzed Graph
10.00 m/s
17.32 m/s
7.08 m/s
Work Ug KE Ug KE Ug KE Ug KE0
50000
100000
150000
200000
250000
Energy Bar Graph
0 m/s
KE = ½ mv2
Solve for Force
What is the net forcerequired to get therollercoaster to the topof the first hill according to theprevious energy bar graph?
SOLUTION:
K U E
Find the force.
W = 200,000 J FΔy = 20.41 m
F = 9799.12 N
Solve for kinetic energy.
What is the kinetic enerycontained in a 35 kg object
thathas been displaced 7 metershorizontally by a force of 6 N?
SOLUTION:
K U E
Find the kinetic energy.
F = 6 N KEΔx = 7 m
KE = 42 J
Solve for units.
If that same 35 kg object wasdisplaced 7 meters horizontally
bya force of 6 N in 2 seconds,
whatwould be the units of your
answer?
W = J Unitst = s
Units = J/s
Find units.
P = W/t
EnergyImpulseMomentumPowerWork
Match
• The product of force and the time interval during which the force acts.
• The product of the constant force on an object and the straight line distance through which the object is moved.
• Rate at which work is done.• Inertia in motion.• The ability to do work.
Energy- The ability to do work.Impulse- The product of force and the time interval during which the force acts.Momentum- Inertia in motion.Power- Rate at which work is done.Work- The product of the constant force on an object and the straight line distance through which the object is moved.
Matched
Analyze the graph. Determine the magnitude and units of the slope.
0 5 10 15 20 25 30 35 40 450
5
10
15
20
25
30
Force vs Change of Length
Δ l (m)
F (
N)
Analyzed graph.
0 5 10 15 20 25 30 35 40 450
5
10
15
20
25
30
Force vs Change of Length
Δ l (m)
F (
N)
Rise = 15 NRun = 25 mSlope = Rise/RunSlope = .6 N/m
Analyze the graph. Determine the units of the area.
0 5 10 15 20 25 30 35 40 450
5
10
15
20
25
30
Force vs Change of Length
Δ l (m)
F (
N)
Analyzed graph.
0 5 10 15 20 25 30 35 40 450
5
10
15
20
25
30
Force vs Change of Length
Δ l (m)
F (
N)
Area =( ½) Rise x RunArea = N x mArea = kgm/s2 x mArea = kgm2/s2
Area = J
Determine where the yintercept on a force vs
changeof length graph would be if
thespring contains no energy.
Determine y-intercept
SOLUTION:
K U E
Determine the y-intercept.
Us = 0 J y-intercept Δl = 0 m
y-intercept = 0,0
Two identical 924 kg cars begin breaking
at exactly the same time with the same
constant force of 1250 N on a level road.
Car “A” comes to a stop in 50 meters. Car
“B” comes to a stop in 100 meters.Determine the velocity of each
vehicle.
Solve for velocity.
SOLUTION:
K U E
To solve for velocity remember the Work-Kinetic Energy Theorem
F = 1250 N Vf”A”
Xi = 0 m Vf”B”
Xf”A” = 50 mXf”A” = 100 mm = 924 kg Vf”A”= 11.63 m/s Vf”B”= 16.45 m/s
Work Us KE Us KE Us KE0
2,000
4,000
6,000
8,000
10,000
12,000
Energy Bar Graph
Analyze the Energy Bar Graph. Determine what is happening to this energy transfer over time.
Analyzed graph. Energy is being transferred from a spring to a moving object.
Work Us KE Us KE Us KE0
2,000
4,000
6,000
8,000
10,000
12,000
Energy Bar Graph
Solve for velocity
A stopped car was leftunattended on a 17 meterhill. It rolls down hill for 5meters. Determine thevelocity of the car.
SOLUTION:
K U E
Determine the velocity.
Yi = 17 m V Yf = 12 m Δy = 5 m
V = 9.9 m/s
Analyze the graph. Choose the part of the graph best represents the picture.
Work KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug0
10000
20000
30000
40000
50000
60000
Energy Bar Graph of a Pole Vaulter
Analyzed graph.
Work KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug0
10000
20000
30000
40000
50000
60000
Energy Bar Graph of a Pole Vaulter
Analyze the graph. Choose the part of the graph best represents the picture.
Work KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug0
10000
20000
30000
40000
50000
60000
Energy Bar Graph of a Pole Vaulter
Analyzed graph.
Work KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug0
10000
20000
30000
40000
50000
60000
Energy Bar Graph of a Pole Vaulter
Solve for frequency
Determine the frequency that anobject is moving around a 30meter diameter circle if it istraveling at 5 m/s.
SOLUTION:
K U E
Calculate the frequency.
r = 15 m TV = 5 m/s f
f = .053 Hz
You could also use 2πr as λ!
Radius (m)
Time Period
(s)
Centripetal
Acceleration (m/s2 )
Centripetal
Force (N )
Frequency (Hz)
1 .898
4 12.25
7 14
13 .086
Complete the data table of a 2 kg object moving at a constant tangential velocity of 7 m/s.
Completed data table.Radius
(m)Time
Period(s)
Centripetal Acceleratio
n (m/s2 )
Centripetal
Force (N )
Frequency (Hz)
1 .898 49 98 1.11
4 3.59 12.25 24.5 .279
7 6.28 7 14 .159
13 11.63 3.77 7.54 .086
Frequency (Hz)
TimePeriod
(s)
Centripetal Acceleratio
n (m/s2 )
Centripetal Force (N )
.5 2
.2 37.9
.111 35.1
.077
Complete the data table of a 3 kg object moving around a constant radius of 24 meters.
Completed data table.Frequency
(Hz)Time
Period(s)
Centripetal Acceleration
(m/s2 )
Centripetal Force (N )
.5 2 236.87 710.61
.2 5 37.9 113.7
.111 9 11.7 35.1
.077 13 5.61 16.83
Match
Select the graph that best represents the inverse square relationship between centripetal acceleration and radius if ac
(m/s2) is plotted on the y-axis and r (m) is plotted on the x-axis.
Graph Options
Matched Graph Options
Select the graph that best represents the inverse square relationship between centripetal acceleration and radius if ac
(m/s2) is plotted on the y-axis and r (m) is plotted on the x-axis.
Caramel, a 1.9 × 1030 kg star was just discovered inside the Milky Way. Determine the force of attraction between the star and its orbiting planets.
Completed data table. Fg= G (m1m2)
r2
Longitudinal WaveTransverse WaveReflected WaveRefracted Wave
Match
• Bending of an oblique ray of light when it changes velocity due to a change in the medium in which it is traveling.
• Wave in which the individual particles of a medium vibrate back and forth in the direction in which the wave travels.
• Return of light from a surface in such a way that the angle at which the ray is retruned is equal to the angle at which it strikes the surface.
• Wave in which the individual particles of a medium vibrate back and forth perpendicular to the direction in which the wave travels.
Matched
Longitudinal Wave- Wave in which the individual particles of a medium vibrate back and forth in the direction in which the wave travels.Transverse Wave- Wave in which the individual particles of a medium vibrate back and forth perpendicular to the direction in which the wave travels.Reflected Wave- Return of light from a surface in such a way that the angle at which the ray is retruned is equal to the angle at which it strikes the surface.Refracted Wave- Bending of an oblique ray of light when it changes velocity due to a change in the medium in which it is traveling.
Analyze each of these sound waves. Which of these is longitudinal? Which is quieter. Which is lower pitched? Which contains more amplitude (energy)? Which has the longest wavelength?
“A” “B”
“A” “B”
Analyzed graph.
Longitudinal? “A” and “B” because they are both sound waves!Quieter? “A”Lower Pitched? “B”More Energy? “B”Longest Wavelength? “B”
Wavelength
Amplitude
Frequency
Solve for speed of sound
Determine the speed of sound at a
temperature of 23.25° C.
SOLUTION:
K U E
Calculate speed of sound.
T = 23.5° C Vsos
Vsos = 343.95 m/s
Solve for frequency when you’re standing still and the temperature is zero.
Determine the frequency of sound
you will hear when the ice creamtruck is coming toward you and
thenwhen it is traveling away from
you ifthe truck emits a bell at a
frequency of244 Hz and is traveling at 2 m/s.
SOLUTION:
K U E
Calculate frequency.
fi = 244 Hz ff
V = 2 m/s Towards ff = 245.48 Hz
Away ff = 242.53 Hz
Who buys ice cream in that kind of weather???
Solve for frequency
Compare the frequencies of a sound
that has a .75 meter wavelength when
traveling in a temperature of -10° C
and the same sound wave when the
temperature is 10° C.
SOLUTION:
K U E
Calculate frequencies.
T = -10° C ff
T = 10° C -10 C f = 432 Hz
10 C f = 448 Hz
Match Graph Options
Select the graph that best represents the position of a sound wave that echoes off the ocean floor and returns to the source at the surface of the ocean. The ship creating the sound is the reference point.
Graph Options
Select the graph that best represents the position of a sound wave that echoes off the ocean floor and returns to the source at the surface of the ocean.
Matched
Point at which sound wave hits ocean floor.
Solve for velocity.
It is a beautiful 28° C day without
a cloud in the sky and you arewatching the air show on
LakeMichigan. A plane is traveling
atMach 2.5. Determine the
plane’svelocity.
SOLUTION:
K U E
Calculate velocity.
T = 28° C VMach = 2.5
V = 867 m/s
Find the velocity of sound for that day and then multiply that number by 2.5!
Solve for time.
If the same plane is flying 300 meters
off the ground, determine the time it
takes after the plane passes directly
overhead for you to hear the sonic
boom, and how far the plane would
travel in that same amount of time.
SOLUTION:
K U E
Calculate time and displacement.
T = 28° C tMach = 2.5 Xf
Vp = 867 m/sVs = 346.8 m/s t = .865 s
Xf = 749.96 m
Solve for wavelength
Determine the wavelength of asound wave that has a frequencyof 128.75 Hz and is traveling at 330 m/s.
SOLUTION:
K U E
Calculate wavelength.
f = 300 Hz λV = 769 m/s
λ = 2.56 m
UmbraPenumbraAdditive Primary ColorSubtractive Primary ColorComplementary Color
Match
• Three colors of light absorbing pigments that when mixed in certain proportions will reflect any color of the spectrum.
• A partial shadow that appears where some of the light is blocked and other light can fall.
• Any two colors of light that when added together produce white light.
• Darker part of a shadow where all light is blocked.
• Three colors of light that when added together in certain proportions will produce any color of the spectrum.
Matched
Umbra- Darker part of a shadow where all light is blocked.Penumbra- A partial shadow that appears where some of the light is blocked and other light can fall.Additive Primary Color- Three colors of light that when added together in certain proportions will produce any color of the spectrum.Subtractive Primary Color- Three colors of light absorbing pigments that when mixed in certain proportions will reflect any color of the spectrum.Complementary Color- Any two colors of light that when added together produce white light.
Solve for velocity
Compare the velocities of a radio
wave and a sound wave when the
temperature is 17° C.
SOLUTION:
K U E
Calculate and compare velocities.
T = 17° C Vs
c = 3 x 108 m/s Vr = 3 x 108 m/s
Vs = 340.2 m/s
Radio waves are part of the electromagnetic spectrum!
Solve for velocity
Compare the change in velocities
of a sound wave and a radio wave
after hitting glass if the index of
refraction is 1.43 when thetemperature is 17° C .
SOLUTION:
K U E
Calculate and compare velocities.
T = 17° C Vs
c = 3 x 108 m/s Vr = 2.09 x 108 m/s
Vs = 340.2 m/s
Reflective velocity will equalIncidental velocity for sound.
Solve for frequency
Determine the frequency of aradio wave that has awavelength of 300 meters.
SOLUTION:
K U E
Calculate frequency.
c = 3 x 108 m/s fλ = 300 m
f = 1,000,000 Hz or 1,000 kHz or 100 MHz Did you remember radio waves are part of the electromagnetic spectrum?
Match Polarizer Options
Select the set of polarizers that would allow the most light to pass through.
Polarizer Options
Select the set of polarizers that would allow the most light to pass through.
Matched
Match Polarizer Options
Select the set of polarizers that would allow the most light to pass through.
Polarizer Options
Matched
Select the set of polarizers that would allow the most light to pass through.
Solve for index of refraction
Mr. Floyd is trying to separate the color
pink from white light by shining white
light through a diamond prism. If the
angle of incidence is 35° and the angle of
refraction is 13.7°, determine the index
of refraction for a diamond.
SOLUTION:
K U E
Calculate the index of refraction.
nair = 1 ndiamond Θi = 35°Θr = 13.7°
n = 2.42
Solve for image size
A person is 1.5 meters tall andis standing 5 meters in front ofa pinhole camera. The camerascreen is .1 meters from thepinhole. Determine the size ofthe image.
SOLUTION:
K U E
Calculate the size of the image.
So = 1.5 m Si p = 5 mq = .1 m
Si = .03 m
Solve for focal point.
A picture of a 1.5 meter objectproduces an image of 1.5 cm when the object is 4 metersfrom the camera. Determine the focal point of the camera.
SOLUTION:
K U E
Calculate the focal point.
So = 1.5 m f Si = .015 m qp = 4 m
f = .0396 m
Match
Select the statements that match each type of mirror. Concave Plane
Upright Inverted Real Virtual Magnified Reduced Same size Reversed True (Not Reversed)
Concave- Upright or Inverted, Real or Virtual, Magnified, Reduced, or the same size, Reversed or True.
Plane- Upright, Virtual, Same size, and Reversed.
Matched
Match
Select the statements that match a concave mirror.
The object is outside the center point.The object is at the center point.The object is at the focal point.The object is inside the focal point.
Upright Inverted Real Virtual Magnified Reduced Same size Reversed True (Not Reversed)
The object is outside the center
point.
Inverted, Real, Reduced, and
Reversed.
The object is at the center point.
Inverted, Real, Same size, and
Reversed.
The object is at the focal point.
No image is produced.
The object is inside the focal
point.
Upright, Virtual, Magnified, and
True.
Matched
Solve for image size, image distance and type of image.
A 7.5 cm object is placed 10 cm infront of a concave mirror that hasfocal point of 20 cm. Determinethe image size and distance. Then,determine if the image is real orvirtual.
SOLUTION:
K U E
Calculate the image size and distance and type.
So = 7.5 cm Si p = 10 cm qf = 20 cm Real or Virtual
Si = -15 cm (meaning the image has flipped)
q = -20 cm (meaning the image is in the mirror)
Virtual
v
v
Match
Black White Blue Cyan Green Magenta Red Yellow
Select the color that each object would appear if only red light was incident upon the objects.
Matched
Black White Blue Cyan Green Magenta Red Yellow
Select the color that each object would appear if only red light was incident upon the objects.
Match
Black White Blue Cyan Green Magenta Red Yellow
Select the color that each object would appear if only cyan light was incident upon the objects.
Matched
Black White Blue Cyan Green Magenta Red Yellow
Select the color that each object would appear if only cyan light was incident upon the objects.
Match
Select the most likely reason that each cloud would appear the color illustrated.
-The cloud is made up of small sized particles that reflect high frequency waves.-The cloud is made up of medium sized particles that reflect medium frequency waves.-The cloud is made up of large sized particles that reflect low frequency waves.
The cloud is made up of small sized particles that reflect high frequency waves.
Matched
The cloud is made up of large sized particles that reflect low frequency waves.
The cloud is made up of medium sized particles that reflect medium frequency waves.
Solve for velocity
While playing the “milk bottle” game at the
amusement park, a .448 kg ball is thrown
at a constant horizontal velocity of 10.4
m/s and collides with a stationary .577 kg
milk bottle. If the two objects then stick
together, determine the velocity at which
they would continue to travel.
SOLUTION:
K U E
Find the velocity.
m1 = .448 kg Vf
m2 = .577 kg pg = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s
p = 4.66 kgm/s2
Vf = 4.55 m/s
Solve for impulse
Dr. Fiala, who has a
mass of 100 kg is traveling at a
constant velocity of 1.5 m/s.Determine the impulse felt by
theunfortunate freshman sittingstationary in their bumper car.
SOLUTION:
K U E
Find the impulse.
m = 100 kg pVi = 1.5 m/s I
I = 150 Ns
Solve for force
If the unfortunate freshman sitting
in the bumper car experienced the
impact for .03 seconds, determine
the force that Dr. Fiala applied to
their bumper car.
Ha, Ha, Ha, Ha
SOLUTION:
K U E
Find force.
m = 100 kg FVi = 1.5 m/sp = 150 kgm/s2
t = .03 sI = 150 Ns F = 5,000 N
Solve for time.
If it takes 85,000 W of power to
raise Sky Trek Tower requiring2,000,000 J of energy,
determine thetime required to lift the ride to
thetop.
SOLUTION:
K U E
Find time.
P = 85,000 W tET = 2,000,000 J t = 23.53 s
Solve for height.
Fully loaded the Sky Trek Towerhas a mass of 2349.28 kg. Determine the maximum
height ofthe ride.
SOLUTION:
K U E
Find height.
P = 85,000 W ΔyET = 2,000,000 Jt = 23.53 sm = 2349.28 kgg = 9.8 m/s2
Δy = 86.87 m
Solve for velocity.
Sky Trek Tower is fully enclosed to prevent objects
from falling out. If you did drop
your accelerometer out of thewindow by accident, determine
itsvelocity just before reaching
theground.
SOLUTION:
K U E
Find velocity.
P = 85,000 W VET = 2,000,000 Jt = 23.53 sm = 2349.28 kgg = -9.8 m/s2
Δy = 86.87 m V = -41.26 m/s
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the acceleration it would be experiencing on the fall from the Sky Trek Tower.
Match
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the acceleration it would be experiencing on the fall from the Sky Trek Tower.
Matched
Force Diagra
ms
Select the force diagram that best matches the reading on the previous vertical accelerometer.
Force Diagra
ms
Matched
Select the force diagram that best matches the reading on the previous vertical accelerometer.
Solve for height.
To confirm the results from the
height slide, you decide to
triangulate the height of Sky TrekTower. Using a baseline of 20
meters,and a sightline height of 1.5
meters, you findθ1 (22°) andθ2 (20°).
Determine the triangulated height.
SOLUTION:
K U E
Find height.
b = 20 m hSLH = 1.5 mΘ1 = 22°Θ2 = 20°
h = 74.93 m
Solve for angle.
At a certain point in your ride on
your roller coaster your horizontal
acceleration is 14.53 m/s2. Determine the angle at which
yourhorizontal accelerometer
would beindicating.
SOLUTION:
K U E
Find angle.
a = 14.53 m/s2 Θg = 9.8 m/s2
Θ = 56°
Solve for g’s.
At a certain point in your ride on your roller coaster your
horizontal accelerometer has adeflection of 76°. Determine
thenumber of g’s being produced
atthat point.
SOLUTION:
K U E
Find g’s.
g = 9.8 m/s2 gΘ = 76° g = 4 g’s
Solve for acceleration.
At a certain point in your ride on your roller coaster your
vertical accelerometer is halfway
between the second and third line.
Determine your acceleration at that
point.
SOLUTION:
K U E
Find acceleration.
g = 9.8 m/s2 ag’s = 1.5 a = 4.9 m/s2
Force Diagra
msSelect the force diagram that best matches the acceleration you calculated for the previous problem.
Force Diagra
ms
Matched
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the most force of support.
Match
Vertical Accelerometer Readings
Matched
Select the vertical accelerometer reading that best matches the most force of support.
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the roller coaster traveling at constant velocity to the top of the first hill.
Match
Vertical Accelerometer Readings
Matched
Select the vertical accelerometer reading that best matches the roller coaster traveling at constant velocity to the top of the first hill.
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the roller coaster traveling down the first hill.
Match
Vertical Accelerometer Readings
Matched
Select the vertical accelerometer reading that best matches the roller coaster traveling down the first hill.
Vertical Accelerometer Readings
Select the vertical accelerometer reading that best matches the roller coaster actually accelerating at 4.9 m/s2.
Match
Vertical Accelerometer Readings
Matched
Select the vertical accelerometer reading that best matches the roller coaster actually accelerating at 4.9 m/s2.
It has been my pleasure to spend the last 10 months investigating physics with you. I hope you have enjoyed the journey as much as I have. Have a great summer and if I don’t see you around the halls of GBN again, have a great life!
With Kindest
Regards,
Doc
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