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Light and M/s

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Light and M/s
Final Exam Review Questions

Multiple Choice
Identify the choice that best completes the statement or answers the question.

____ 1. In the diagram, the amplitude of the wave is shown by:

a. c c. b b. d d. a ____ 2. In the diagram, the crest of the wave is shown by:

a. c c. b b. d d. a ____ 3. In the diagram, the trough of the wave is shown by:

a. c c. b b. d d. a ____ 4. In the diagram, the wavelength is shown by:

a. c c. b b. d d. a ____ 5. If wavelength is kept constant, which increases as the speed of a wave increases?
a.
period
c.
amplitude
b.
frequency
d.
phase

____ 6. Which shows the correct angle of reflection given the incident ray shown?
a.

c.

b.

d.

____ 7. If the period of a certain wave (wavelength = 4.5 m) is 2 seconds, what is the speed of the wave?
a.
0.44 m/s
c.
9.0 m/s
b.
1.1 m/s
d.
2.3 m/s

____ 8. Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the speed of the waves?
a.
7.9 m/s
c.
39 m/s
b.
790 m/s
d.
2.6 m/s

____ 9. Which characteristic of sound is associated with frequency?
a.
pitch
c.
pressure
b.
loudness
d.
timbre

____ 10. Which type of resonator has harmonics in whole-number multiples of the fundamental?
a.
closed-pipe resonators
c.
strings
b.
open-pipe resonators
d.
both strings and open-pipe resonators

____ 11. Which frequency would be the third harmonic in a series for an open-pipe resonator if the fundamental is 440Hz?
a.
3080 Hz
c.
1320 Hz
b.
2200 Hz
d.
660 Hz

____ 12. Thirty beats are heard in one minute when two notes are played together. If the higher note has a frequency of 740 Hz, what is the frequency of the lower note?
a.
742 Hz
c.
739.5 Hz
b.
738 Hz
d.
740.5 Hz

____ 13. What is the formula to find speed of light?
a.
c = /f
c.
c = f/
b.
c = f
d.
2/f

____ 14. Which term describes a material that does not allow light to be transmitted?
a.
translucent
c.
polarized
b.
transparent
d.
opaque

____ 15. The reason lines on the bottom of a swimming pool can look wavy when seen from above is that
a.
water is denser in some places than in others.
b.
different layers of water refract the light at different angles.
c.
the lines actually are wavy.
d.
the uneven surface of the water refracts light at different angles.
e.
none of the above

____ 16. Light waves are
a.
longitudinal waves.
b.
transverse waves.

____ 17. The law of reflection says that
a.
the angle of reflection from a mirror equals the angle of incidence.
b.
waves incident on a mirror are partially reflected.
c.
all waves incident on a mirror are reflected.
d.
the angle a ray is reflected from a mirror is random.

____ 18. The critical angle for a light from the bottom of a swimming pool shining upward toward the pool's surface is the angle
a.
where light is refracted so it just skims the pool surface.
b.
43 degrees.
c.
at which all light is refracted out of the pool.
d.
42 degrees.
e.
at which some light is reflected from the surface.

____ 19. A beam of light emerges from water into air at an angle. The beam is bent
a.
away from the normal.
b.
not at all.
c.
48 degrees upward.
d.
96 degrees upward.
e.
towards the normal.

____ 20. When a light beam emerges from water into air, the average light speed
a.
remains the same.
b.
decreases.
c.
increases.

____ 21. Sound waves are produced by
a.
radio stations.
b.
vibrating objects.
c.
soft objects.
d.
objects under pressure.
e.
none of the above

____ 22. A sound wave is a
a.
standing wave.
b.
longitudinal wave.
c.
transverse wave.
d.
shock wave.
e.
none of the above

____ 23. A singer shattering crystal glass with her voice is a demonstration of
a.
beats.
b.
sound refraction.
c.
an echo.
d.
interference.
e.
resonance.

____ 24. Noise-canceling earphones are an example of
a.
constructive interference.
b.
destructive interference.
c.
beats.
d.
resonance.

____ 25. The time needed for a wave to make one complete cycle is its
a.
frequency.
b.
velocity.
c.
amplitude.
d.
period.
e.
wavelength.

____ 26. The distance between successive identical parts of a wave is called its
a.
frequency.
b.
period.
c.
velocity.
d.
amplitude.
e.
wavelength.

____ 27. The Hertz is a
a.
special radio wave.
b.
type of car.
c.
unit of period.
d.
unit of wavelength.
e.
unit of frequency.

____ 28. A wave created by shaking a rope up and down is called a
a.
Doppler wave.
b.
standing wave.
c.
longitudinal wave.
d.
constructive wave.
e.
transverse wave.

____ 29. Where can you touch a standing wave on a rope without disturbing the wave?
a.
At a node
b.
At any place along the wave
c.
At an antinode

____ 30. A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car, which is initially at rest. If the speed of the cars after they couple together is 36 km/h, what is the mass of the second car?
a.
40 metric tons
c.
10 metric tons
b.
20 metric tons
d.
5 metric tons

____ 31. Power can be described as
a.
the rate of energy transfer.
c.
the force exerted over a given distance.
b.
the change in kinetic energy.
d.
the ratio of work output to work input.

____ 32. What is the magnitude of the force required to accelerate an electron of mass 9.110–31 kg from rest to a speed of 2.0107 m/s for a distance of 0.50 cm?
a.
1.810–21 N
c.
1.810–16 N
b.
9.010–19 N
d.
3.610–14 N

____ 33. A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. How fast is the diver moving while falling past a diving board that is 3.0 m above the surface of the pool?
a.
12 m/s
c.
137 m/s
b.
14 m/s
d.
140 m/s

____ 34. A circus performer juggles four identical balls. The performer throws each ball upward with the same initial velocity and releases each ball at the same height. The ball with the greatest kinetic energy will be the ball that the juggler
a.
throws upward, just after it leaves her hand.
b.
is watching, just as it reaches its highest point.
c.
is about to catch, just before it lands in her hand.
d.
does not catch, just before it lands on the floor.

____ 35. A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s. What is the average friction force opposing its motion?
a.
16 N
c.
19 N
b.
18 N
d.
28 N

____ 36. The work done to move a spring away from its equilibrium position is equal to
a.
the ratio of force to mass.
c.
the ratio of force to displacement.
b.
the kinetic energy of the spring.
d.
the potential energy of the spring.

The graph below is a plot of displacement versus time of a mass oscillating on a spring.

____ 37. At which point on the graph is the acceleration of the mass zero?
a.
A
c.
C
b.
B
d.
D

____ 38. At which point on the graph is the net force on the mass at a maximum?
a.
A
c.
C
b.
B
d.
D

____ 39. At which point on the graph is the velocity of the mass zero?
a.
A
c.
C
b.
B
d.
D

____ 40. Consider a 20.0-kg pendulum clock that keeps good time. If the clock is moved to a location where it weighs 74 N, how many minutes will the minute hand move in 1 h?
a.
98
c.
38
b.
60
d.
23

____ 41. Which of these best explains how chlorophyll causes the leaves of a tree to appear green?
a.
Chlorophyll bends blue and red light without affecting green light.
b.
Chlorophyll absorbs blue and red light while reflecting green light.
c.
Chlorophyll destroys white and red light while storing green light to use.
d.
Chlorophyll combines sunlight with blue light from the sky to form green light.

____ 42. When magenta is taken out of white light, the color that remains is
a.
yellow.
c.
blue.
b.
cyan.
d.
green.

____ 43. When a light wave travels from one medium into another, which property or properties of the wave change?
a.
its speed only
c.
its speed and wavelength only
b.
its speed and frequency only
d.
its speed, wavelength, and frequency

____ 44. When a light wave not perpendicular to the surface travels from a mineral crystal into the air,
a.
v increases and  increases
c.
v decreases and  decreases
b.
v increases and f decreases
d.
v decreases and f increases

____ 45. The figure shows the path of a ray of light passing through three layers. Use information in the table to answer the question.

Which of these list, from 1 to 3, the materials that could make up each layer?
a.
air, ethanol, air
c.
ethanol, water, diamond
b.
diamond, air, crown glass
d.
water, crown glass, ethanol

____ 46. A submerged fish looks up and sees a fisherman seated at the edge of the water on the distant shore of the pond. At which of the following angles to the vertical must the fish be looking?
a.
36.9°
c.
48.8°
b.
41.2°
d.
90.0°

Completion
Complete each statement.

47. The change in direction of waves as they pass from one medium to another is called ____________________.

48. A sound wave results from variations in ____________________ transmitted through matter.

49. Sound travels ____________________ in a liquid than in a gas.

50. The lowest frequency that is resonant is called the ____________________.

Short Answer

51. An incident wave is propagated down a spring. When it meets another spring, some of the energy continues down the spring, while some is reflected back on the first spring, but inverted. Predict how the second spring compares to the first in terms of stiffness or heaviness.

Problems

52. The graph below displays how displacement varies with time when a wave passes a fixed point at a speed of 12.0 m/s. Calculate the frequency and wavelength of the wave.

53. A closed-pipe resonator has a length of 1.73 m. Calculate the frequency of its third harmonic if the velocity of sound is 343 m/s.

54. An open-pipe resonator has a length of 2.39 m. Calculate the frequency of its third harmonic if the velocity of sound is 343 m/s.

55. The harmonic series from a long tube is given below. Is this tube acting as an open-pipe resonator or a closed-pipe resonator? Explain your answer.
203 Hz, 609 Hz, 1015 Hz, 1421 Hz

56. A tuning fork (440 Hz) is struck over a closed-pipe resonator. If the speed of sound is 343 m/s, what is the shortest the column could be when it makes a loud tone?

57. A golfer uses a club to hit a 45-g golf ball resting on an elevated tee, so that the golf ball leaves the tee at a horizontal speed of +38 m/s.
a. What is the impulse on the golf ball?
b. What is the average force that the club exerts on the golf ball if they are in contact for 2.010–3 s?
c. What average force does the golf ball exert on the club during this time interval?

58. A tennis player receives a shot with the 60.0-g ball traveling horizontally at –50.0 m/s, and returns the shot with the ball traveling horizontally. The tennis ball and the tennis racket are in contact for 1.0010–3 s. The average force exerted on the ball by the tennis racket is 5.70103 N. Find the speed of the tennis ball after it leaves the racket.

59. A 180-kg crate is sitting on the flatbed of a moving truck. The coefficient of sliding friction between the crate and the truck bed is 0.30. Two taut cables are attached to the crate, one on each side. Each cable can exert a maximum horizontal force of 650 N either forward or backward if the crate begins to slide. If the truck stops in 1.8 s, what is the maximum speed the truck could have been moving without breaking the cables?

60. A single uranium atom has a mass of 3.9710–25 kg. It decays into the nucleus of a thorium atom by emitting an alpha particle at a speed of 2.10107 m/s. The mass of an alpha particle is 6.6810–27 kg.
a)What is the recoil speed of the thorium nucleus?
b) Calculate and compare the KE before and after the event. Use this comparison to determine the type of collision (elastic, inelastic, or explosion).

61. A 10.0-g bullet is fired into a stationary 5.00-kg block of wood. The bullet lodges inside the block. The speed of the block-plus-bullet system immediately after the collision is measured as 0.600 m/s.
a) What was the original speed of the bullet?
b) Calculate and compare the KE before and after the event. Use this comparison to determine the type of collision (elastic, inelastic, or explosion).

62. Two cars enter an icy intersection. A blue car with a mass of 2.50103 kg is heading east at 20.0 m/s, and a red car with a mass of 1.45103 kg is going north at 30.0 m/s. The two vehicles collide and stick together. Determine the speed and direction of the cars as they skid away together just after colliding.

63. At 9.0 s after takeoff, a 250-kg rocket attains a vertical velocity of 120 m/s.
a. What is the impulse on the rocket?
b. What is the average net force on the rocket?
c. What is the acceleration of the rocket?
d. What is its altitude?

64. While sitting in a stationary wagon, a girl catches a 2.6-kg medicine ball moving at a speed of 2.7 m/s.
a. If the 11-kg wagon is on frictionless wheels, what is the velocity of the girl, wagon, and ball after she has caught the ball? The girl’s mass is 55 kg. Ignore friction between the wheels and the road.
b. The girl then throws the ball back at a speed of 2.7 m/s relative to the wagon. What is the velocity of the girl and the wagon right after she has thrown the ball?

65. A collision between two identical pucks, one moving and the other stationary, takes place on ice. The puck with an initial momentum of 4.0 kg?m/s is deflected 60.0° eastward from its original path.
a) Using the law of conservation of momentum, what is the momentum of the puck that was originally stationary after the collision?
b) Calculate and compare the KE before and after the event. Use this comparison to determine the type of collision (elastic, inelastic, or explosion).

66. The following collisions take place on a flat, horizontal tabletop with negligible friction.
a.

A 2.1-kg cart, A, with frictionless wheels is moving at a constant speed of 3.4 m/s to the right on the tabletop, as shown above, when it collides with a second cart, B, that is initially at rest. The force acting on cart A during the collision is shown as a function of time in the graph below, where t = 0 is the instant of initial contact. Assume that friction is negligible. Calculate the magnitude and direction of the velocity of cart A after the collision.

b. In another experiment on the same table, an incident ball, C, of mass 0.15 kg is rolling at 1.3 m/s to the right on the tabletop. It makes a head-on collision with a target ball, D, of mass 0.50 kg at rest on the table. As a result of the collision, the incident ball rebounds, rolling backward at 0.80 m/s immediately after the collision. Calculate the velocity of ball D immediately after the collision. (Ignore friction in your calculations.)
c. In a third experiment on the same table, an incident ball, E, of mass 0.250 kg rolls at 5.00 m/s toward a target ball, F, of mass 0.250 kg. The incident ball rolls to the right along the x-axis, and makes a glancing collision with a target ball, F, that is at rest on the table. The velocity of incident ball E immediately after the collision is 4.33 m/s at an angle of 30.0 above the x-axis. Calculate the magnitude and direction of target ball F’s velocity immediately after the collision. (Ignore friction.)

67. A player pushes a 250-g hockey puck over frictionless ice with a constant force, causing it to accelerate at 24 m/s2 over a distance of 50.0 cm.
a. Find the work done by the hockey player on the puck.
b. What is the change in the kinetic energy of the puck?

68. You exert a horizontal force of 4.6 N on a textbook as you slide it 0.60 m across a library table to a friend.
a. Calculate the work you do on the book.
b. Your friend returns the book by pushing it with a force of 6.2 N at an angle of 30.0° below the horizontal. What is the work done by your friend on the book?

69. Shown below is a graph of force versus displacement for an object being pulled. Determine the work done by the force in pulling the object 7.0 m.

70. A ski lift carries a 75.0-kg skier at 3.00 m/s for 1.50 min along a cable that is inclined at an angle of 40.0° above the horizontal.
a. How much work is done by the ski lift?
b. How much power is expended by the ski lift?

71. A 0.300-kg baseball is thrown at a speed of 6.5 m/s. The batter hits the ball and it flies into the outfield at a speed of 19.2 m/s. How much work is done in hitting the baseball?

72. A worker pushes a lawn mower with a force of 23.0 N, exerted along the direction of the handle and at a speed of 1.25 m/s across a lawn that is 18.5 m wide. The handle of the lawn mower makes an angle of 60.0° with the horizontal.
a. How much work is done by the worker?
b. If the worker is pushing as hard as possible, how else can the amount of work done be increased?
c. How much power is exerted by the worker?

73. Hoover Dam has a capacity for producing 2.0106 kW of power. How much work is done by the turbines each day?

74. A stuntwoman, working on the set of an action movie, performs the following stunt. She steps out of the window of a building and lands on a giant inflated pillow directly beneath the window. The woman’s mass is m. When she reaches the pillow, her speed is v. Air resistance can be neglected. Use work and energy methods to solve the following.
a. Find an expression for the height of the window above the pillow.
b. The stuntwoman comes to rest in the inflated pillow after falling a distance, s, into it. Find an expression for the average force exerted by the pillow on her body.
c. For her next stunt, the woman slides down a rubber chute of length, l. She starts by climbing up a ladder, then sits on the chute and starts sliding down at a height H. Along the chute, a constant frictional force of Ff is exerted on the woman. Find an expression for the stuntwoman’s speed when she reaches the bottom of the chute.
d. Draw two energy diagrams to compare the work done, the initial and final potential energies of the stuntwoman, and the initial and final kinetic energies of the stuntwoman for the situations in b. and c.

75. Zeke slides down a snow hill on a rubber mat. Zeke’s mass is 76.0 kg and the mass of the mat is 2.00 kg. Zeke starts from rest at the crest of the hill. Frictional forces may be disregarded.
a. What is the change in the gravitational potential energy of Zeke and the mat when they slide to 1.20 m below the crest?
b. What is the change in the kinetic energy of Zeke and the mat when they slide to 1.20 m below the crest?
c. How fast are Zeke and the mat moving when they are 1.20 m below the crest?

76. Meena releases her 10.5-kg toboggan from rest on a hill. The toboggan glides down #he frictionless slope of the hill, and at the bottom of the slope it moves along a rough horizontal surface, which exerts a constant frictional force on the toboggan.
a. When the toboggan is released from a height of 15.0 m, it travels 6.0 m along the horizontal surface before coming to rest. How much work does the frictional force do on the toboggan?
b. From what height should the toboggan be released so that it stops after traveling 10.0 m on the horizontal surface?

77. Kuan stands on the edge of a building’s flat roof, 12 m above the ground, and throws a 147.0-g baseball straight down. The ball hits the ground at a speed of 18 m/s. What was the initial speed of the ball?

78. As shown below, a 450-kg roller-coaster car starts from rest at point A at a height of 47 m, rolls down the track, reaching point B at a speed of 25 m/s, and then rolls up a second hill where it reaches a height of 23 m before coming to rest (at point C). What are the gravitational potential energy and kinetic energy of the car when it is at points A, B, and C?

79. A loaded freight car of mass 50.0103 kg, moving at 18.0 km/h along a straight, level track, collides with a stationary empty freight car of mass 15.0103 kg. At the collision, the two boxcars lock together.
a. What is the velocity of the moving pair of boxcars after the collision?
b. How much energy is lost during the collision?

80. An arrow with a mass of 320 g is shot straight up into the air and reaches a height of 150 m before stopping and falling back to the ground.
a. How much work is done by gravity as the arrow reaches its maximum height?
b. If the arrow’s kinetic energy is zero at its maximum height, calculate the initial velocity of this arrow as it is shot from the ground.

81. A single proton has a mass of mp = 1.671027 kg. Traveling in a cyclotron at 90 percent of the speed of light, it strikes an atom of gold, Au, at rest. The proton rebounds with a velocity of 1.40108 m/s while the gold atom recoils at a speed of 1.64107 m/s. What is the mass of the gold atom? Assume all collisions are elastic and all motion is along the same axis.

82. A spring stretches by 25.0 cm when a 0.500-kg mass is suspended from its end.
a. Determine the spring constant.
b. How much elastic potential energy is stored in the spring when it is stretched this far?

83. A spring has a spring constant of 135 N/m. How far must it be compressed so that 4.39 J of elastic potential energy is stored in the spring?

84. On a planet where the gravitational acceleration is five times g on Earth, a pendulum swings back and forth with a period of 1.22 s. What is the length of the pendulum?

85. Sonya hears water dripping from the eaves of the house onto a porch roof. She counts 30 drops in 1.0 min.
a. What is the frequency of the drops?
b. What is the period of the drops?

86. Hiroshi is generating waves on a rope by flipping the rope up and down. Each motion up or down lasts 0.20 s. The distance from a crest to a trough is 0.40 m.
a. What is the amplitude of the wave?
b. What is the frequency of the waves?

87. A pulse with an amplitude of 0.53 m travels to the right along a rope. Another pulse, with an amplitude of –0.24 m, travels to the left along the same rope. The two pulses approach each other. What is the amplitude of the rope at the point where the midpoints of the pulses pass each other?

88. A physics teacher attaches an electric oscillator to one end of a 2.0-m horizontal spring and attaches the other end of the spring to a stationary hook in the wall. She adjusts the frequency of the oscillator to produce a standing wave in the spring. Students observe that the standing wave has three nodes and two antinodes. She then doubles the frequency of the oscillations and produces another standing wave. How many nodes and antinodes do the students observe in the new standing wave?

89. Each back-and-forth movement of the bob in a small pendulum clock releases a cog on a wheel. As the cog is released, the wheel undergoes a slight rotation. If the release of three cogs moves the second hand of the clock forward 1.0 s, what is the length of the pendulum?

90. The distance between four consecutive antinodes of a standing wave in a spring is 42 cm. What is the wavelength of the standing wave? Hint: The distance between two consecutive antinodes in a standing wave represents 0.5 .

91. A simple pendulum has an arm that is 20.0 cm long with a 100.0-g mass suspended from the end. Pulling the mass 2.0 cm to one side and releasing it starts the pendulum in motion.
a. What is the amplitude of this pendulum?
b. What is the period of this pendulum?
c. If the arm is lengthened to 50.0 cm, what happens to the period?
d. If the mass is increased to 500.0 g, what happens to the period?
e. If the mass is pulled 4.5 cm to one side and released, what happens to the period?
f. A grandfather clock uses a pendulum with a period of 2.0 s to keep time. If the clock uses a 275-g mass as a counterweight, how long should the pendulum arm be?

Assume that the speed of sound in air is 343 m/s, at 20°C, unless otherwise noted.

92. The sound a mosquito makes is produced when it beats its wings at the average frequency of 620 wing beats per second. What is the wavelength of the sound waves produced by the mosquito?

93. The pulse-echo technique is used in diagnostic medical imaging. A short ultrasound pulse is emitted from the device, and echoes are produced when the pulse is reflected at a tissue interface. The echo signals are received back at the device and then analyzed to build up an image of the organ. The speed of sound in soft tissue is 1540 m/s. If an echo is received 58.210–6 s after the pulse was emitted, how far is the tissue interface from the ultrasound device?

94. While fishing from a boat anchored offshore, you see another fishing boat between your boat and the shore. The other boat sounds a 510-Hz horn as it heads toward the shore at a speed of 18 m/s.
a. If your fishing boat is stationary, what is the frequency of the sound waves from the horn that reach you?
b. If your fishing boat now heads out to sea at a speed of 15 m/s, what is the frequency of the sound waves from the horn that reach you?

95. A species of bat navigates by emitting short bursts of sound waves that have a frequency range that peaks at 58.0 kHz.
a. If a bat is flying at 4.0 m/s toward a stationary object, what is the frequency of the sound waves reaching the object?
b. What is the frequency of the reflected sound waves detected by the bat?
c. What is the difference between the frequency of the sound waves emitted by the bat and the frequency of the sound waves detected by the bat if the bat is flying at 4.0 m/s and the object is a moth approaching at 1.0 m/s?

96. Hannah places an open, vertical glass tube into a container of water so that the lower end of the tube is submerged. She holds a vibrating tuning fork over the top of the tube while varying the water level in the tube. Hannah notices that the loudest sound is heard when the distance from the water to the top of the tube is 32.7 cm, and again when the distance is 98.2 cm. What is the frequency of the tuning fork?

97. The six strings of a standard guitar are tuned to the following frequencies: 165, 220, 294, 392, 494, and 659 Hz.
a. Find the lengths of the shortest open-ended organ pipes that would produce the same frequencies.
b. Sketch the pipes, showing their lengths to scale.

98. The fundamental tone of an open-pipe resonator with a length of 48 cm is the same as the second harmonic tone of a closed-pipe resonator. What is the length of the closed-pipe resonator?

99. You hear the siren of a fire engine as you stand on the side of the road. As it approaches, the siren which broadcasts at a frequency of 645 Hz is heard by you as being 660 Hz. How fast is the fire engine traveling?

100. An open tube is filled with water which is slowly drained as a tuning fork of frequency f = 1.00103 Hz is held over the open end. As the water drains, the level of the water is marked as a maximum of sound is heard in the tube. These maxima are detected at distances of 16.7 cm, 33.4 cm and 50.1 cm, measured from the open end of the tube. What is the speed of sound in the air within the tube?

101. Visible light has a frequency that ranges from about 4.01014 Hz to about 7.51014 Hz. What is the range of wavelengths for visible light?

102. A concave mirror with a focal length of 25.0 cm produces an image at 75.0 cm. What is the object distance for the object that produced the image?

103. A beam of light traveling through air strikes flint glass at an angle of 31.0° to the normal. At what angle does the beam enter the flint glass?

104. Light has a speed of 2.00108 m/s in clear acrylic.
a. What is the index of refraction of clear acrylic?
b. What is the wavelength of yellow light ( = 589 nm) in clear acrylic?
c. A beam of light traveling through air strikes a block of clear acrylic at an angle of 29° to the normal. At what angle does the light enter the block of clear acrylic?

105. A beam of light traveling through water is incident upon an unknown type of glass at an angle of 45.0° to the normal and is refracted at an angle of 33.6° to the normal.
a. What is the index of refraction for the unknown type of glass?
b. What is the speed of light in the unknown type of glass?

106. An even layer of oil floats on top of water (noil = 1.15). A beam of light strikes the oil at an angle of 32.1° to the normal. What is the beam’s angle of refraction in the water?

107. Calculate the critical angles for each of the following situations.
a. a beam of light passing from crown glass into air
b. a beam of light passing from diamond into flint glass
c. a beam of light passing from quartz into water
d. a beam of light passing from water into air

108. A beam of light has a wavelength of 550 nm. While traveling through an unknown substance, the light has a wavelength of 5.0010–7 m. What is the index of refraction of the unknown substance?

109. A beam of light traveling through water strikes the boundary between the water and air at an angle of 25° to the normal. At what angle does the light enter the air?

110. A ray of light from a region containing air (index of refraction n1 = 1.00) enters one end of an optical fiber at angle of incidence i, as shown in the figure below. The index of refraction of the optical fiber is n2 = 1.48.

a. If the angle of incidence at the end of the fiber is i = 48.5°, what is the angle of reflection, , from the sidewall of the fiber?
b. Optical fibers are surrounded by a shield known as cladding. The cladding has a slightly lower index of refraction and is fused to the optical fiber making a completely reflective interface. Using the index of refraction of the cladding (n = 1.42), show that at the sidewall of the fiber, internal reflection is total.
c. Is there an angle of incidence, i  90°, for which internal reflection from the sidewall is not total?
Final Exam Review Questions
Answer Section

MULTIPLE CHOICE

1. ANS: D
The amplitude of a wave is half the height of the wave from crest to trough.

2. ANS: B
The crest of a wave is the top of the wave.

3. ANS: C
The trough of a wave is the bottom of the wave.

4. ANS: A
The wavelength of a wave is the distance between successive crests.

5. ANS: B wave speed =

6. ANS: C
The angle of incidence equals the angle of reflection.

7. ANS: D

8. ANS: A

9. ANS: A
The frequency of a sound determines the sound’s pitch.

10. ANS: D
Strings and open-pipe resonators have harmonics in whole-number multiples of the fundamental.

11. ANS: C
The third harmonic in this open-pipe resonator is produced at .

12. ANS: C
30 bpm = 0.5 bps or 0.5 Hz. The lower note then has a frequency of 740 Hz

13. ANS: B
In the ray model of light, light is represented as a ray that travels in a straight path.

NOT: /a/ The speed of light is directly proportional to the frequency of light. /b/ Correct! /c/ The speed of a light wave is directly proportional to its wavelength. /d/ The wavelength of a wave is a linear function of its speed.

14. ANS: D
Opaque materials do not allow light to be transmitted.

15. ANS: D

16. ANS: B

17. ANS: A

18. ANS: A

19. ANS: A

20. ANS: C

21. ANS: B

22. ANS: B

23. ANS: E

24. ANS: B

25. ANS: D

26. ANS: E

27. ANS: E

28. ANS: E

29. ANS: A

30. ANS: B
Rationale
a. uses equation
b. correct answer
c. same as ma
d. uses equation
Explanation
To find the mass, mb, of the second car, use the equation for the conservation of momentum, pi = pf, where p = mv. Since the two cars can be thought of as a system, use the following equation: mavai + mbvbi = mavaf + mbvbf

Given mass of first car ma = 10 metric tons initial velocity of firstcar vai = 108 km/h final velocity of first car vaf = 36 km/h initial velocity of second car vbi = 0 m/s final velocity of second car vbf = 36 km/h
Rearrange the equation to solve for mb

= 20 metric tons

31. ANS: A
Rationale
a. correct answer
b. work
c. work
d. efficiency

32. ANS: D
Rationale
a. did not square velocity
b. multiplied by 0.005, rather than dividing
c. KE
d. correct answer
Explanation
To find the force required to accelerate the electron, first find the amount of work that will be done on the electron by the force. The amount of work will be equal to the change in kinetic energy of the electron.
Given
mass of electron = 9.110–31 kg velocity of electron = 2.0107 m/s

First calculate the kinetic energy using the equation KEelectron 5 , where m = mass v = velocity

Knowing the distance and the energy of the electron, the work equation can be used to find the force required.
Given
distance = 0.50 cm 5 0.0050 m
KEelectron = 1.810–16 kg·m2/s2
W = F  D
Rearrange the work equation to solve for force

33. ANS: A
Rationale
a. correct answer
b. ignored initial velocity
c. did not take square root when solving or add initial velocity
d. did not take square root when solving

Explanation
Given
Initial upward velocity = 2.5 m/s
Distance between platforms = 7.0 m
The diver has an initial upward velocity of 2.5 m/s. This decreases to zero at the top of the diver’s leap and changes to –2.5 m/s as the diver falls past the 10-m-high diving platform. Ignore air resistance. As the diver falls, the change in potential energy is equal to the change in kinetic energy. The total mechanical energy, PE + KE, is constant, therefore PE = –KE.
Since you are solving for velocity at the 3.0-m-high diving platform, let 3.0 m be your reference point for zero potential energy. Therefore, the change in height, h, of the 10.0-m-high diving platform in relation to the 3.0-m-high diving platform is –7.0 m.

Rearrange to solve for the increase in velocity as a result of falling 7.0 m.

34. ANS: D
Rationale
a. This is generally when a ball would have the greatest KE.
b. greatest PE
c. This is generally when a ball would have the greatest KE.
d. correct answer
Explanation
Because each ball is thrown from the same height with the same velocity, the mechanical energy of each ball is the same. The total mechanical energy of each ball is equal to the sum of its potential energy and kinetic energy. This means that the ball with the least potential energy (the ball that is closest to the floor) will have the most kinetic energy.

35. ANS: A
Rationale
a. correct answer
b. divides change in PE by 5 m
c. added PE and KE instead of finding difference
d. solves using 5 m as the height and length of the incline
Explanation
Given mass of block, m = 3.0 kg incline,  = 37° sliding distance, L = 5.0 m final speed, v = 2.0 m/s
Find the height, h, through which the mass falls. The equation that relates the height, length, and incline of the ramp is

Rearrange and solve for height h = L sin 
= (5.0 m) (sin 37°)
= 3.0 m
Find the expected speed of a block that falls 3.0 m.
The final energy of the block is less than its initial energy. The difference is energy lost to friction.
PEinitial + KEinitial = PEfinal + KEfinal + Wfriction
Because Wfriction = (Ffriction)(distance), then

Rearrange to solve for the force of friction.

36. ANS: D
Explanation
Work done on an object is equivalent to the change in its potential energy. As the spring has zero potential energy when it is in equilibrium, the work done on the spring to move it away from equilibrium is equal to its potential energy.

37. ANS: C
Explanation
As the object moves closer to the equilibrium position (position C), the speed increases and the force decreases. The speed is its maximum and the force is zero at the equilibrium position. In the equilibrium position, the upward force balances the object’s weight, therefore, there is no acceleration.

38. ANS: A
Explanation
With the block in position A, the spring is stretched to the maximum displacement or the amplitude of the object on the spring. At this point (maximum displacement or amplitude of the oscillation), the speed of the block is zero (v = 0 m/s) and the spring’s force is at its maximum.

39. ANS: A
Explanation
At the object’s maximum displacement (A), the net force is at its maximum and the motion of the object has stopped (v = 0 m/s) for an instant before moving in the opposite direction. As the block moves closer to the equilibrium position, the speed increases and the force decreases. The speed is its maximum and the restoring force is zero at the equilibrium position (C). As the block continues to move past its equilibrium position, the force acting on it tends to slow it down.

40. ANS: C
Rationale
a. 60  1.63
b. moves 60 minutes per hour on Earth, with g = 9.80 m/s2
c. correct answer
d. forgot to take square root; or calculated 60 – 37
Explanation
The period of the pendulum is given by

where g = gravitational acceleration and l = length of the pendulum.
On Earth, g = 9.80 m/s2, the period of the pendulum clock is 1 s, and the minute hand moves 60 min in 1 h.
First find g for this new location, using
Fg = mg
Given
Fg = 74 N m = 20.0 kg
Rearrange to solve for g:

We can determine the period for this new location by comparing it with Earth.

Because we know that the period of a pendulum clock on Earth is 1 s,
Tnew = 1.6 s
So T at the new location is 1.6 times T on Earth. This means that the minute hand takes 1.6 times longer to move the same distance. Or, during a given interval, the minute hand moves a distance that is 1.6 times smaller than on Earth. So the minute hand moves only 1/1.6 times the distance it would move on Earth.

On Earth, the minute hand would move 60 minutes in one hour. dEarth = 60 min
At the new location, the distance moved is

So at the new location the minute hand will move 38 min in 1 h.

41. ANS: B
Rationale
a. refraction
b. correct answer
c. Something that stored a green substance might appear green.
d. Yellow and blue are mixed to make green.
Explanation
Tree leaves absorb light energy that plants use to make food. Light has three primary colors: red, blue, and green. If they absorb red and blue light while reflecting green light, the leaves will appear green.

42. ANS: D
Explanation
White light is the three primary colors combined:
Red + Blue + Green = White
When two of the three primary colors are combined the results are:
Red + Blue = Magenta
Red + Green = Yellow
Blue + Green = Cyan
If you subtract magenta from white light (take away red and blue light), you are left with green light.

43. ANS: C
Explanation
Light travels at different speeds in different materials. The speed of light is lower in water and other transparent materials than through a vacuum. As the speed of light is reduced in the slower medium, the wavelength is shortened proportionately. The frequency is unchanged; it is a characteristic of the source of the light and unaffected by medium changes.

44. ANS: A
Explanation
Light travels at different speeds in different materials. The speed of light is lower in water and other transparent materials than through a vacuum. Therefore, when the light travels from the mineral to the air its speed increases. The refraction of light when it passes from a slow medium to a fast medium bends the light ray away from the normal to the boundary between the two media. Therefore,  decreases.

45. ANS: B
Rationale
a. pattern would need to be of bending towards then away from normal; plus a and c should be equal
b. correct answer
c. fits pattern of bending away from then towards normal, however c is greater than b
d. reverses the relationship between index and refraction—assumes a lower index brings the ray toward normal.
Explanation
When a light ray passes from one material into another, its angle of refraction can be predicted using Snell’s Law of Refraction: n1 sin 1 = n2 sin 2 n1 = refractive index of the first material
1 = incidence angle (angle that the ray hits the boundary between the first and second surface) n2 = refractive index of the second material
2 = angle of ray in the second material
The equation shows that when light passes from a substance with a lower index of refraction, n1, to a substance with a higher index, n2, the angle 1 must be greater than 2 in order to preserve the equality.
The figure shows that the angle of the light ray in material can be ranked as follows:
Angle of light ray in Material 2  Material 3  Material 1
Based on this, it is possible to compare the indices of refraction of each material:
Indices of refraction: Material 1  Material 3  Material 2
Option b is the only choice in which the materials follow this pattern.

46. ANS: C
Rationale
a. tan–1
b. cos–1
c. correct answer
d. takes index of refraction of water as 1.00
Explanation
To see the man on the distant shore, the light from the man must travel along the boundary between the water and the air. This light then will be transmitted through the water at the critical angle. (Light rays are reversible, so the ray that travels along the boundary and to the fish follows the same path that the ray from the fish along the boundary to the man follows.)
The angle at which the fish sees the man is the critical angle for total internal reflection. To calculate the critical angle c, use the equation

n1 = index of refraction of air = 1.00 n2 = index of refraction of water = 1.33
Thus,

COMPLETION

47. ANS: refraction A wave bends or refracts as it passes from one medium to another.

48. ANS: pressure Sound waves create slight pressure variations as they travel through matter.

49. ANS: faster Sound travels faster through substances where the molecules are tightly packed together.

50. ANS: fundamental The fundamental frequency is the lowest sound that resonates in a pipe or on a string.

SHORT ANSWER

51. ANS:
Because the reflected wave is inverted, the second spring is probably heavier and/or stiffer.

PROBLEM

52. ANS:
Frequency = 0.400 Hz
Wavelength = 30.0 m

NOT: The graph can be used to find the time period of the wave. The reciprocal of time period is the frequency of the wave. The wavelength is equal to the velocity divided by the frequency.

53. ANS:
545 Hz

NOT: The frequency of the third harmonic in a closed-pipe resonator is equal to the velocity of sound divided by four times the length of the resonator.

54. ANS:
144 Hz

NOT: The frequency of the third harmonic in an open-pipe resonator is equal to the velocity of sound divided by two times the length of the resonator.

55. ANS:
This is acting as a closed-pipe resonator. The fundamental frequency is 203 Hz. The relative frequencies of the remaining frequencies in the series are 3x wavelength, 5x wavelength, and 7x wavelength respectively. These are in odd-number ratios, which is descriptive of a closed-pipe resonator.

56. ANS:
The wavelength of a 440 Hz note traveling at 343 m/s is given by v = f.
343 m/s = Hz)
 = 0.780 m
The relationship between length and wavelength in a closed-pipe resonator is that the length is equal to half the wavelength. The length of the standing wave in this column would therefore be (1/2)(0.780 m) = 0.390 m

57. ANS:
a. Impulse = Ft = mv
Ft = mvf – mvi
= mvf – 0
= (0.045 kg)(+38 m/s)
= +1.7 kg·m/s or +1.7 N·s
b. Ft = Impulse = 1.7 N·s

= +8.5102 N
c. Fgolf ball on club = –Fclub on golf ball
Fclub on golf ball = 18.5102 N
Fgolf ball on club = –8.5102 N

58. ANS:

59. ANS:

60. ANS:
a)
b)

61. ANS:

62. ANS: p(B+R) f= pBi+pRi

63. ANS:
a.
= 250 kg(120 m/s 2 0.0 m/s)
= 3.03104 kg?m/s
b.
c.

64. ANS:
a.
b.

65. ANS: p1i + p2i = p1f + p2f p1i + 0 = p1f + p2f p1i = p1f + p2f p1f = p1i sin  = (4.0 kg·m/s)(sin 60.0°)
= 3.5 kg·m/s, 60.0° east of north p2f = p1i cos  = (4.0 kg·m/s)(cos 60.0°)
= 2.0 kg·m/s, 30.0° west of north

66. ANS:
a. vf = 0.54 m/s to the right
Explanation
Impulse = change in momentum for cart A
Ft = p = mv
The impulse can be determined from the area under the graph.
Ft = area under curve = 2 (area of triangle)
Ft = 2(1.510–3 s)(24.0103 N)
Ft = –6.0 N·s
The impulse for cart A is negative since cart A loses momentum.
Ft = mvf – mvi

vf = +0.54 m/s vf = 0.54 m/s to the right

b. vDf = 0.63 m/s to the right
Explanation
Use conservation of momentum. All of the motions take place along a straight line.

pf = pi pCf + pDf = pCi + pDi mCvCf + mDvDf = mCvCi + mDvDi
Given
mC = 0.15 kg mD = 0.50 kg vCi = 1.3 m/s vDi = 0.0 m/s (at rest) vCf = –0.80 m/s vDf = ? mCvCf + mDvDf = mCvCi + 0
Solving for vDf:

vDf = 0.63 m/s to the right
c. After the collision, the velocity of ball F is 2.50 m/s at 60.0 below the x-axis.
Explanation
Use conservation of momentum.
In the x-direction: pEf, x + pFf, x = pEi, x + pFi, x
In the y-direction: pEf, y + pFf, y = pEi, y + pFi,y
Given
mE = 0.250 kg mF = 0.250 kg

Since the masses are equal, let mE = mF = m vEi, x = 5.00 m/s vEi, y = 0 m/s vFi = 0 m/s (at rest) so vFi, x = 0 and vFi, y = 0 vEf = 4.33 m/s
E = 30.0

In the x-direction: pEf, x + pFf, x = pEi, x mEvEf, x + mFvFf, x = mEvEi, x mvFf, x = mvEi, x – mvEf, x vFf, x = vEi, x – vEf cos E vFf, x = 5.00 m/s – (4.33 m/s)(cos 30.0) vFf, x = 1.25 m/s
In the y-direction: pEf, y + pFf, y = 0 pEf, y = –ppEf, y mvFf, y = –mvEf, y vFf, y = –vEf sin E vFf, y = –(4.33 m/s)(sin 30.0) vFf, y = –2.17 m/s
Find the magnitude of the final velocity of ball F:

Find the direction of the final velocity of ball F:

After the collision, the velocity of ball F is 2.50 m/s at 60.0 below the x-axis.

67. ANS:
a. W = Fd
F = ma
W = ma  d
= (0.250 kg)(24 m/s2)(0.500 m)
= 3.0 kg·m2/s2
= 3.0 J
b. W = KE
KE = 3.0 J

68. ANS:
a. W = Fd
= (4.6 N)(0.60 m)
= 2.8 J
b. W = Fd cos 
= (6.2 N)(0.60 m)(cos(230.0°))
= 3.2 J

69. ANS:
W = area under the curve

W = W1 + W2 + W3
W1: 0.0 m  3.0 m (triangle)
W2: 3.0 m  7.0 m (rectangle)
W3: =.0 m  7.0 m (triangle)
W = (4.0 N)(3.0 m) + (4.0 N)(4.0 m) + (3.0 N)(2.0 m)
= 25 N·m
= 25 J

70. ANS:
a. Fy = Fg = mg t = 1.50 min = 90.0 s
Length of incline: L = vt dy = L sin 
= vt sin 
W = Fydy
= mgvt sin 
= (75.0 kg)(9.80 m/s2)(3.00 m/s)(90.0 s)(sin 40.0°)
= 1.28105 J
b.

71. ANS:

72. ANS:
a. W = Fd cos 
= (23.0 N)(18.5 m)(cos 60.0°)
= 213 J
b. Without increasing the force, more work can be done by reducing the angle  that the handle of the lawn mower makes with the ground.
c.

73. ANS:

74. ANS:
a. Use conservation of mechanical energy. initial = at the window final = at the pillow reference level for PE = the level of the pillow
Let h be the height of the window above the pillow.
Method 1:

Method 2:

b. Use work and energy. initial = at the top of the pillow, where she initially lands final = a distance, s, into the pillow, where she comes to rest reference level for PE = the level where she comes to rest

The negative sign indicates that the force is in a direction opposite to her direction of motion, i.e., upward.
c. Use work and energy. initial = at the top of the chute final = at the bottom of the chute reference level for PE = the level of the bottom of the chute
Let vbottom be the speed of the stuntwoman at the bottom of the chute.

Wfriction = Ffrictiond cos(180°) because friction is opposite the direction of motion d = l
Wfriction = –Ffrictionl
And the work done by friction is negative because it reduces the energy in the system.
So

d.
For the situation described in 5. b.:

For the situation described in 5. c.:

75. ANS:
a.
b.
c.

76. ANS:
a.
b.

77. ANS:

78. ANS:
Point A:
PEA = 2.1105 J
KEA = 0
Point B:
PEB = 0
KEB = 1.4105 J
Point C:
PEC = 1.0105 J
KEC = 0

79. ANS:
a.
b.

80. ANS:
a.
b.

81. ANS:
From the law of conservation of energy,
KEp1 + KEAu1 = KEp2 + KEAu2 mpvp12 1 mAuvAu12 5 mpvp22 1 mAuvAu22
Since the gold atom starts from rest, vAu1 = 0 m/s. Substituting and simplifying, mpvp12 = mpvp22 + mAuvAu2 mAuvAu2 = mp(vp12  vp22)

82. ANS:

83. ANS:

84. ANS:

85. ANS:

86. ANS:

87. ANS:
A = A1 + A2
= 0.53 m + (–0.24 m)
= 0.29 m

88. ANS:
For f1, there are three nodes and two antinodes, indicating that the standing wave has a wavelength of 2.0 m, the same as the length of the spring.

The new standing wave has a wavelength of half the length of the spring. The spring contains two full wavelengths; therefore, the students observe five nodes and four antinodes.

89. ANS:

90. ANS:
The distance between four antinodes represents 1.5 .

91. ANS:
a. The amplitude is the distance of the swing from the center line, A = 2.0 cm.
b.
c. Because the period and length of the pendulum arm are directly related, an increase in the arm length increases the period:

d. Nothing. The period, T, is not dependent upon the mass of the pendulum.

e. Nothing. The period, T, is not dependent upon the amplitude.

f.

92. ANS:

93. ANS:

94. ANS:

95. ANS:
a. The frequency of the sound waves reaching the stationary object is fd1.

b. The frequency of the reflected sound waves from the object is fd1 and the frequency of the sound waves detected by the bat is fd2.

96. ANS:

97. ANS:
a.
b.

98. ANS:
Second harmonic (closed pipe) = fundamental (open pipe)

99. ANS:

100. ANS:
With resonance points every 16.7 cm,

101. ANS:

102. ANS:

103. ANS:

104. ANS:

a.
b.
c.

105. ANS:

106. ANS:

107. ANS:

108. ANS:

109. ANS:

110. ANS:
a. 44.1°
Explanation
First, using Snell’s Law, compute the angle of refraction for the ray entering the fiber at the end.

The angle of reflection at the sidewall is  = 90.0° – 30.4° = 59.6°
b.
Explanation
The critical angle at the sidewall is

Because 59.6°  73.6°, the angle of incidence at the sidewall is less than the critical angle and internal reflection is total.
c. less than 24.7°
Explanation
The angle of refraction at the end of the fiber must not exceed
r = 90.0° – 73.6° = 16.4° for internal reflection to be total at the sidewall. Again using Snell’s law to attempt to compute the angle of incidence for this condition to be true:

Angles must be less than 24.7o.

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