Physics 100: Conceptual Problemksmcf/Aran/concep.pdfPhysics 100: Conceptual Problem Where is Venus (#1) The planet Venus is referred to as the morning star or the evening star because

Physics 100: Conceptual Problem

Warm-Up Problem (#0)

We have asserted that the statement,

“Michael Jordan was the greatest basketball player to have ever played”,

is not scientifically verifiable.

Which of the following modifications of this statement and proposed test is not apairing of a scientific hypothesis and an observation to definitively confirm or refute thehypothesis.

(A). Michael Jordan scored more points than any other basketball player. Confirm bylooking at NBA Statistics.

(B). Michael Jordan is the most popular baseketball player to have ever played. Confirmby putting Michael in an arena full of basketball fans and listening to how loud theapplause is.

(C). If one asks a group of people to name the greatest basketball player to have everplayed, Michael Jordan will be the most common answer. Confirm by doing asurvey of many thousands of people.

(D). Michael Jordan won more NBA MVP titles than any other player. Confirm bylooking a list of MVPs on the NBA web site.

Physics 100: Conceptual Answer

Warm-Up Problem (#0)

(B).All of the other answers propose a hypothesis, and give a direct test as to whether

that hypothesis is true or false. In this case, although the hypothesis is reasonable, thetest proposed won’t directly confirm or refute the hypothesis.


Where is Venus (#1)

The planet Venus is referred to as the morning star or the evening star because it alwaysappears in the sky only at one time of the day: either just before sunrise or just aftersunset.Which of the following hypotheses could explain this?

(A). Venus is orbiting the earth very rapidly

(B). Venus is very distant and is a fixed point on the sky, like a star

(C). You can draw a nearly straight line connecting, in order, Sun-Venus-Earth

(D). You can draw a nearly straight line connecting, in order, Sun-Earth-Venus

(E). None of the above explain the observation


Where is Venus (#1)

(C).If Venus is the morning or evening star, then it must be near the sun always for an

earth observer. That would happen if you could connect Sun-Venus-Earth with a straightline. In the first two answers, this would result in motion of Venus independent of thesun which would not explain why Venus should always be near the sun.


Phases of Venus (#2)

Ptolemeic Copernican

E

S

V

E

SV

Venus is visible from the earth because it relfects light from the sun, just as the moondoes. Like the moon, it also has phases, which were first observed with Galilean tele-scopes. Phases result from the illuminated side of a planet or satellite not fully facingthe earth.Which of the following statements is true of Venus’ phases?

(A). There is no “new” (dark) phase in the Ptolemeic picture

(B). There is no “full” (bright) phase in the Ptolemeic picture

(C). There is no “new” (dark) phase in the Copernican picture

(D). There is no “full” (bright) phase in the Copernican picture

(E). None of the above are true


Phases of Venus (#2)

(B).In the Copernican, heliocentric (sun-centered) picture, all of the phases are possi-

ble, depending on whether or not the sun is between Earth and Venus or whether Venusis between the Earth and the Sun. In the Ptolemic picture, though, only one of these ispossible since Venus and the sun both orbit the Earth at a definite radius. Ptolemy pickedVenus to orbit inside the sun’s orbit, so it would always have been between Earth and thesun; therefore, no full phase.


Concept Problem (#3): Aristotle vs. Galileo

Imagine the motion of a balloon filled with air as compared with an uninflated balloon.Here are some statements regarding the motion of the two balloons:

1. The inflated balloon falls more slowly than the deflated one because it containsmore air and air wants to rise.

2. The difference in motion is entirely due to a difference in the resistance of the twoballoons.

3. The two balloons will fall at the same speed under ideal conditions.

4. The two balloons have different composition and therefore have a different naturalmotion.

Associate these statements with either Aristotle or Galileo.

(A). Aristotle: 1, 2; Galileo: 3, 4

(B). Aristotle: 1, 2, 4; Galileo: 3

(C). Aristotle: 1, 2, 3; Galileo: 4

(D). Aristotle: 1, 4; Galileo: 2, 3

Whose statements give a better explanation of the motion of the balloon in your view,Aristotle’s or Galileo’s. Briefly, why?


Aristotle vs. Galileo (#3)

(D). You could make a good case for (B) also.Aristotle’s theories of motion had a sense of “natural order” behind them which

said air always wants to rise about earth and water, so 1 and 4 sound very Aristotilean.Galileo, on the other hand, would have emphasized that his law of falling would predictidentical results in the absence of resistance as implied by 3 and 2. Aristotle might alsohave come up with a statement along the lines of 2 since resistance to motion was soimportant in his theories, but it would be also important for Galileo to have this statementin his repetoire to explain the outcome of the falling experiment violating his law!


Concept Problem (#4): Speed and Acceleration

time

dis

tan

ce

A B CThere following graph shows how far an object has travelled as a function of time. Whichof the following statements describing this motion is false.

(A). The object has its largest positive acceleration (i.e., speed is increasing most rapidly)at time A

(B). The object is moving fastest at time B

(C). The object is not moving at time C

(D). None of the above


Speed and Acceleration (#4)

(D).The speed of motion is the slope of distance vs. time, i.e., how rapidly distance

changes with time. Acceleration is change of that slope (sometimes called “curvature”).At point A, the speed is increasing rapidly. At point B, the speed (slope) is large, and atpoint C, the distance doesn’t change with time (speed is zero).


Concept Problem (#5): Galilean Analysis of the Professor and theBubbles

Prof. McFarland has demonstrated the motion of soap bubbles. Which statement(s) de-scribe(s) the violation(s) of the Galilean laws of motion.

(A). The bubbles’ fall is not the same constant acceleration of a bowling ball because ofair resistance and air currents.

(B). The bubbles fall slowly because of their inertia

(C). The bubbles appear to have no inertia because of the resistance from the still air inthe room

(D). (A) and (B)

(E). (A) and (C)

(F). (B) and (C)

Imagine you were travling by plane and you blew a bubble while Galileo watched from ahot-air balloon from outside the plane. Watching the bubble cruising along at 500 mph,he would say,

(A). The bubble moves so fast because of the plane’s inertia

(B). The bubble moves so fast because of the bubble’s inertia

(C). The bubble is pulled by the plane

(D). All of the above


Galilean Analysis of the Professor and the Bubbles (#5)

(E) and (B), respectively.For the first question, the bubble doesn’t fall with constant acceleration because of

air resistance. And sometimes, with the very light bubbles, air currents from the roomventillation can even blow a bubble upward. Also, why don’t the bubbles follow thewalking professor as the law of inertia would predict? Same reason... the air causingthat resistance isn’t moving with the professor, so it opposes the inertial motion.

For the second question, the key point is to realize tha the plane and its inertia areirrelevant to why the bubble moves! The bubble moves forward with the plane becauseof its inertia, and it just so happens the the environment of the plane ensures there is noair resistance to stop it doing so! You might argue this is really a cheat since it’s all aboutair resistance (imagine for example that the bubble started at rest according to Galileo’sviewpoint! then the answer is very different!), and I’d have to admit you have a point.Nevertheless, only answer (B) is strictly correct in the Galilean picture.


Concept Problem (#6): Accelerating Observers

An bored astronaut is playing with a ball as his rocket is accelerated into space by itsengines. If he drops the ball (towards earth and against the direction of his rocket’sacceleration), the balls fall towards the floor of the rocket

(A). Faster than the ball would have before lift off

(B). At the same speed that the ball would have before lift off

(C). Slower than the ball would have before lift off

(D). The ball does not fall since the astronauts are going into space


Accelerating Observers (#6)

(A).In a non-accelerating frame, Galileo tells you the ball falls towards the floor, but

he also would say that the floor accelerates upward at the same time due to the thrust ofthe rocket engines. So relative to the floor, the downward acceleration is greater than justthe acceleration due to falling,


Concept Problem (#7): Rochester Piano Tuners

Using any set of reasonable assumptions and approximations, estimate the number ofpiano tuners in the Rochester Yellow Pages.

(A). Approximately 2

(B). Approximately 20

(C). Approximately 200

(D). Approximately 2000

Hint: break the problem down into small pieces. What limits the number of pianotuners? Do they have to have pianos to tune? How many can they tune? How many mustthey tune to stay in business? etc.


Rochester Piano Tuners (#7)

(B).There are lots of ways to analyze this. The point is that when you only need to get

something right to a factor of ten, you can make a lot of guesses along the way, and stillbe approximately correct!

Start with��

Million people in the Rochester metro area. Say they live in � �� Mil-lion households and further imagine that

� �� have pianos, for about� �� pianos. A

piano must be tuned twice a year, but lots of people don’t bother, so let’s imagine thatthere are only about

� �� tunings per year. It takes about an hour to tune a piano.Distance are relatively short here, so lets say that a single piano tuner can drive aroundand tune four pianos in a day. There are about

�� working days per year, so each tunertunes

� � �� pianos per year. Therefore, to satisfy the demand, there should be� � piano

tuners in Rochester.A quick glance at the Yellow pages shows you that the number listed there is � � ;

however,��

of these are duplicates (urban and suburban listings, for example). So� �

was not a bad guess at all!


Concept Problem (#8): Sample Sizes and Experimental Results

Two pollsters ask the same yes/no question of two large groups. Pollster A gathers� � �

responses; pollster B gathers� � �� responses. Which of the following statements are true

of their results?

(A). The two pollsters asked the same question and will measure the same fraction ofpositive answers

(B). Pollster A will have a smaller margin of error than Pollster B

(C). Pollster B’s fraction of positive answers will always be closer to the true fraction ofpositive answers than Pollster A.

(D). None of the above


Sample Sizes and Experimental Results (#8)

(D).As we saw, the first answer is not correct because there are statistical fluctuations

in the responses of individuals, even if you know the answer on average. The secondanswer is incorrect because asking more individuals results in a smaller margin of error.However, the third is also incorrect! Even if the pollster B’s answers have a smallermargin of error, it is still possible that she gets unlucky while pollster A gets lucky andends up with the right answer.


Concept Problem (#9): Probability and Experimental Results

Medical doctors perform a series of tests to correlate different eating particular foodswith cancer risk. Assume that their standard for finding a “cancer risk” is a � �� proba-bility that the measured cancer rate in a particular group is above normal.If the doctors test

� � � independent groups, each representing one food, how many “can-cer risks” would the doctors find on average if none of the foods tested caused cancer?

(A). None

(B). 1

(C). 5

(D). 10


Probability and Experimental Results (#9)

(C).If there are no true cancer causing agents, then only fakes will be found. If the

standard for a “cause” is a higher probability of cancer than would happen randomly99% of the time, then 1% of the time a random fluctuation will give a positive result forcausing cancer!

Therefore for� � � food samples,

�on average will pass this standard and “cause

cancer”.


Concept Problem (#10): Momentum

Jim drives a Mazda Miata that weighs� � �� pounds down the road at

� � � mph. Jackis driving a Mack Truck in the opposite direction that weighs

� � � �� pounds at� � mph.

They collide head on and their bumpers lock, causing the vehicles to stick together.Immediately after the collision, which of the following statements is true?

(A). Both vehicles are at rest

(B). Jim (Miata) will be more badly hurt

(C). Momentum is conserved in the collision

(D). All of the above

(E). None of the above


Momentum (#10)

(D).We can dispatch with the third one immediately. Momentum is always conserved!

(Unless you are forgetting to account for an outside force.)The momentum of each before the collision is

� � �� pounds � mph, but in op-posite directions. Therefore, the sum of the momenta of the two vehicles is... 0! So ifthey are stuck together afterward, their total momentum must be zero and therefore theirspeed is zero.

Finally, who will be more badly hurt? It’s intuitive, yes, but the reason that Jim isthe victim here is that Jim will decclerate from

� � � mph to � and Jack from only� � mph

to � . Therefore, Jim will have much more momentum to lose against the steering columnand windshield than Jack, and will likely be badly hurt.


Concept Problem (#11): Force

A constant force is exerted on a frictionless car initially at rest. The force acts for a shorttime interval and gives the cart a certain final speed.

Force

A force half as big is exerted for on another cart. For this other force to acceleratethe second cart to the same final speed as the first, this force will have to have a timeduration

(A). Half of the force on the first cart

(B). The same as the force on the first cart

(C). Twice the force on the first cart

The same initial force is exerted for the same length of time on a third cart withtwice as much mass. The speed of the third cart is

(A). Half of the speed of the first cart

(B). The same as the speed of the first cart

(C). Twice the speed of the first cart

The same initial force is exerted for the same period of time on a fourth cart, whichhas a mass identical to the first. However. it did not start at rest but was moving at thesame speed as the first cart was after receiving the initial force. The increase in speed ofthe fourth cart after the force is

(A). Half of the increase of speed of the first cart

(B). The same as the increase of speed of the first cart

(C). Twice the increase of speed of the first cart


Force (#11)

(C), (A) and (B), respectively.In the first problem, two identical carts receive a force that differs by a factor of

two. � �� , so the accelerations will also differ by a factor of two. Acceleration is therate of change of speed, so in order to change speed the same amount, the cart receivingless force will have to receive that force for twice as long!

In the second problem, � � �� with an identical force, so the doubled mass ofthe third cart means it has less acceleration, and therefore less speed.

In the third problem, Newton’s force laws make no reference to whether or not theobject started in motion. So although the total speed might be greater, the increase inspeed (which comes from acceleration) will be the same.


Concept Problem (#12): Recoilless Rifle

Many gun manufacturers claim to sell “recoilless rifles” which can shoot a bullet withoutcausing the gun to recoil. What law(s) of motion might you think a recoilless rifle wouldviolate?

(A). Galileo’s Law of Inertia

(B). Conservation of Momentum

(C). Newton’s Third Law

(D). (A) and (B)

(E). (A) and (C)

(F). (B) and (C)

Bonus: any ideas about how a “recoilless rifle” must operate?


Recoilless Rifle (#12)

(F).Newton’s third law says there must be an equal and opposite force on the gun

from the bullet to counter the force of the bullet on the gun. As we explained, that comesdirectly from conservation of momentum. Galileo’s law of inertia says that the bullet,once in motion would tend to stay in motion and that it would require something tochange its momentum to start in motion, but it is not vioated per se by a recoilless rifle.

Recoilless rifles work by expelling gasses out the back of the rifle against thedirection of force on the bullet. They also have spring cushioning systems which spreadthe recoil out over a longer time to make it feel like less of a “kick”.


Concept Problem (#13): Supernova Shockwave

Professor McFarland has illustrated the generation of a shock wave in a supernova. Thetop ball can soar to new heights because of

(A). The Law of Inertia (Newton’s First Law)

(B). Conservation of Momentum

(C). Conservation of Energy

(D). Balancing of Forces


Supernova Shockwave (#13)

(C).It’s the fact that much of the energy from the big falling mass gets transferred into

the little mass at the top that causes it to zoom off after the impact.


Concept Problem (#14): Energy and Force

A moving person is lifting a � � � lb piano with a system of pulleys. If the rope is pulleda length � , the piano is raised a length �� . Using what you know about conservation ofenergy, determine the average force must the moving person exert to raise the piano. Isit

Piano

(A). 160 lbs?

(B). 320 lbs?

(C). 640 lbs?


Energy and Force (#14)

(A).The amount of energy put into lifting this piano against the force of gravity is that

force ( � � � lb here) times the distance lifted ( �� ). The amount of work done pulling theother end of the rope is also force ( � ) times the distance over which the force is exerted( � ). Equating those two, the lifting force exerted on the rope must be half that of theforce of gravity on the piano because the distance lifted on the rope is twice as long.


Concept Problem (#15): Energy and Friction

A stone is thrown straight up in the air from the ground and while moving it encountersfriction in the air. Compared to its initial speed while leaving the ground, its final speedis

(A). Larger

(B). The same

(C). Smaller


Energy and Friction (#15)

(C).Friction results from collisions with air molecules that, on average, take energy

away from the moving body and transfer it into the air (into heat). So the stone losesenergy on its flight. Therefore, when it is the same distance off the ground as when itstarted, the lost energy means less kinetic energy, and therefore slower motion.


Concept Problem (#16): Sound Waves

Different musical notes sound different because they have different frequencies. Thehigher the “pitch” the higher the frequency.

Imagine that you are listening to two notes played on a piano in this room: note Aand note B. Note B has the higher pitch. Which note is carried by a wave with a longerwavelength?

(A). Note A

(B). Note B

(C). They are the same

(D). Not enough information to tell


Sound Waves (#16)

(A).Wavelength times frequency is the wave speed, which is a constant of the medium

(air in this case). So the higher pitch (freqency) must have the shorter wavelength.


Concept Problem (#17): InterferenceTwo speedboats, travelling in opposite directions on a lake, each produce a wake ofwaves that are � inches in height from crest to trough. You were calmly sitting in yourinner tube when the speedboats pass you in either direction.

1. When the crests from waves from each of the boat reach you simultaneously, howhigh is your boat lifted from its original height?

2. When the crest from one boat and the trough from the other boat reach you simul-taneously, how high is your boat lifted from its original height?

You

6"

(A). (1) 3”, (2) 0”

(B). (1) 3”, (2) 3”

(C). (1) 6”, (2) 0”

(D). (1) 6”, (2) 3”

(E). None of the above are correct


Interference (#17)

(C).The distance from crest to trough is � inches, so one wave can lift the boat � inches

at its crest.If two waves constructively interfere (crests reach you at the same time) the total

height is � �!�"�#� inches.If two waves destructively interfere (crest of one and trough of other reach you at

the same time) the total height is �%$&�'�#� inches.


Concept Problem (#18): Electrostatic Forces and Atoms

So, here’s your chance to figure out the big picture on your own. Imagine a hydrogenatom, made up of a heavy nucleus and a light electron which have opposite charges andthus attract each other. Since we’ve said that the electrostatic (Coulomb) force is likegravity, it shouldn’t shock you if I told you that the light electron orbits the nucleus andlike the earth orbits the sun.

One difference, however, is that gravity is a much weaker force than electromag-netism.

SO, which of the following statements about the orbit of the electron around thenucleus should be correct.

1. The acceleration of the electron will depend on its distance from the nucleus andthe charge of the nucleus.

2. The electron goes one heck of a lot faster around the nucleus than the earth doesaround the sun

3. If somehow there were two electrons stuck together that were suddenly orbiting thenucleus, the acceleration of the two electrons around the nucleus would be half aslarge as the one electron

(A). 1 and 2

(B). 1 and 3

(C). 2 and 3

(D). All three statements are correct


Electrostatic Forces and Atoms (#18)

(A).From Coulomb’s Force law, you know that the electrostatic force is an inverse

square law (inversely proportional to the square of the distance) and that it depends onthe charge of the two objects, so clearly 1 is correct. 2 makes sense also, since we learnedhow much stronger electromagnetism is than gravity and since the distances are muchsmaller.

3 might seem tempting, but it is wrong. � � �� and twice as much mass wouldhalve the acceleration, but the force on the two electrons is twice the force on a singleelectron. So � would be the same for the two as for each individually.


Concept Problem (#19): Loops of Current

We just argued that a loop of current makes a magnet with a north pole on one side ofthe loop and a south pole on the other. Using what you know about forces between barmagnets, what should happen if these two loops of current are brought near each other?

(A). There will be no force between the two loops

(B). There will be an attractive force between the two loops

(C). There will be a repulsive force between the two loops


Loops of Current (#19)

(C).Remember that the “bar magnet” made by the loop of current was perpendicular

to the loop. So each of these loops is like a bar magnet pointed up and down with thenorth pole up. But we remember that N repels N and S repels S, so the two will pusheach other apart in this configuration.


Concept Problem (#19): Interference Patterns in Sound

You are at a rock concert with two speakers�

meters apart playing the same music.Because you were too cheap to get good seats, you are stading against the back wall ofthe stadium

� � � meters away. As you walk along the back wall, you hear

(A). Maxima and minima of sound interference

(B). No interference because sound doesn’t interfere

(C). No interference for another reason

(D). Need more information

(Hint: remember that the speed of sound is (*) � � � m/s, that a typical frequencyfor audible sound is +,)-� �� Hz, and that ( �-+/. . This allows you to get the wavelength)


Interference Patterns in Sound (#19)

(A).The separation between two maxima is .102��3 , where 3 is

�meters and 0 is� � � meters in this problem. So the separation is

� �. .A typical . for sound, given the parameters above, is

�meter.

So the separation between maxima will be about� � meters. If you walk along the

back wall, � � � Hz sound will go through a maximum then a minimum and back to amaximum after

� � meters. Of course that distance is proportional to . so will vary fordifferent frequency sounds, so sounds will disappear (complete destructive interference)at different locations for different frequencies!


Concept Problem (#20): Electromagnetic Waves

Which of the following should be true of electromagnetic waves?

1. Two electromagnetic waves can interfere with each other

2. Electromagnetic waves cannot travel instantaneously

3. Electromagnetic waves must carry energy

(A). 1 and 2

(B). 1 and 3

(C). 2 and 3

(D). All three statements are true


Electromagnetic Waves (#20)

(D).All of these are properties of all waves and should be shared by electromagnetic

waves as well.


Concept Problem (#21): Dopper Shift of Light

In 1927, an astronomer named Edwin Hubble showed that light from distant galaxieswas “red-shifted” as seen from earth. That is, if light was emitted from a distant source,it had a longer period (lower frequency) than was expected. Furthermore, the furtheraway the source was, the bigger the shift. Based on this data, Hubble concluded that

(A). Distant galaxies are not moving relative to the earth (static universe)

(B). Distant galaxies are moving closer to the earth (collapsing universe)

(C). Distant galaxies are moving farther away from the earth (expanding universe)

Hubble got this problem right only 4 � years ago, and has lots of cool stuff namedafter him (Hubble shift, Hubble constant, Hubble Telescope)! There is hope for us all!


Dopper Shift of Light (#21)

(C).Think of a police car with a siren. When it goes away from you, the siren is lower-

pitched, or lower frequency. Therefore, the galaxies are moving away and the Universeappears to be expanding!


Concept Problem (#22): Swimmer in a Current

A swimmer who swims at a speed ( is swimming in a stream in the direction of thecurrent which moves at velocity 5 . What is the apparent speed of the swimmer as viewedfrom shore?

swimmer

current v

c

shore

(A). 5(B). (6$75(C). ((D). (��&5What is the swimmer’s apparent speed as view from shore if she swims against thedirection of the current?

(A). 5(B). (6$75(C). ((D). (��&5Is it possible that the swimmer is unable to swim upstream?

(A). Yes

(B). No


Swimmer in a Current (#22)

(D), (B) and (A), respectively.If the swimmer moves with the current, the swimmer and stream speeds add. If

the swimmer is swimming upstream, the apparent speed is the difference of the two. If aswimmer is swimming upstream more slowly the the current, then the current will makethe swimmer appear to be heading downstream from the viewpoint of shore. (Think ofevery bad adventure movie you’ve seen where the hero tries to swim away from the edgeof a waterfall but is washed over by the current!)


Concept Problem (#23): Null result of Michelson-Morley

The 1887 Michelson-Morley experiment (and many subsequent experiments) did notobserve the change the expected change in the propogation of light due to the aether.Which of the following could be an explanation for this?

(A). The aether moves along with the earth

(B). Light moves at the same speed, regardless of motion through the aether

(C). The period (time for one complete wave) of a lightwave depends on its direction oftravel through the aether

(D). (A) and (C)

(E). All of the above


Null result of Michelson-Morley (#23)

(E).If the earth dragged the aether along, the experiment wouldn’t show anything since

it depends on earth moving through the aether. If light speed isn’t affected by the aether,that would destroy the effect. We assumed also that the period of light didn’t depend onits direction, so that also could affect the result.


Concept Problem (#24): The Light Clock

We have shown that from the point of view of an observer (call him Ishmael) watching atrain-based “light clock”, that the train-based light clock ticks more slowly than his ownlight clock by a Lorentz factor, 8 .

Casey

Ishmael

v

Now imagine that train-based observer (Casey Jones) ignored for a moment the troubleahead and trouble behind and was watching Ishmael’s light clock. Casey Jones wouldobserve that

(A). Ishmael’s light clock ticks more quickly than Casey’s

(B). Ishmael’s light clock ticks at the same rate as Casey’s

(C). Ishmael’s light clock ticks more slowly than Casey’s


The Light Clock (#24)

(C).The same analysis that showed that Ishmael should observe a slow clock moving

with Casey can be reversed (since there is no preferred viewpoint for the laws of physics!)and Casey will also conclude that Ishmael has a slow clock!


Concept Problem (#25): Separated at Birth

Two twins are born, and as the result of a really unusual custody settlement when theparents divorce immediately after the birth, one twin (Astra) is put into a spaceshiptraveling at almost the speed of light (say � � � � ��( ) while the other (Terra) remains onearth. The spaceship travels in a path that goes by a nearby star

� � light years (thedistance light travels in 20 years) away. No attempt is made to to communicate back andforth between the ship and earth until the ship arrives at the distant star. Which of thefollowing statements is/are true?

(A). From Astra’s point of view, Astra should be older than Terra

(B). From Terra’s point of view, Terra should be older than Astra

(C). Only Terra can correctly assert that she should be older since she is stationary

(D). If Astra sends a message as she goes by the star predicting Terra’s age from Astra’spoint of view at the time Terra receives the message, Terra will agree with Astra’sprediction


Separated at Birth (#25)

All but (C) are true.From each twin’s point of view, the other should be younger because her clock will

be apparently slow. We’ve already said that “stationary” doesn’t mean anything special,so you know the third statement is wrong. We also have argued that simultaneous eventsat the same point in space (Terra receiving Astra’s message and Terra evaluating her ownage) must be simultaneous to all observers, so it makes sense that the prediction shouldagree.


Concept Problem (#26): Relativistic Garage

Imagine a car with a length � and a garage with a length � �:9 � and a door on either end.Dudley Driver desides that he will attempt to make the car fit into the garage by goingnear the speed of light and contracting the length of his car.Dudley enlists his friend, Percepta, to assist in his effort. From her viewpoint, Perceptawill close both the front and back doors when the middle of Dudley’s car is exactly inthe middle of the garage. There will be a (brief) moment when both doors are closed(before Dudley smashes through the back), thus proving the the car is smaller than thegarage.Dudley, bright guy that he is, realizes that this shouldn’t work since the garage will belength contracted from his point of view, and therefore smaller than the car.If Dudley and Percepta would do the experiment, which of the following statementswould be a correct statement regarding the outcome.

(A). Percepta would claim the experiment worked as designed and the car fit in thegarage.

(B). Dudley would see that Percepta is right, and there is a moment where the doors areboth closed and his car is inside.

(C). Percepta would not be able to close the car in the garage because the car doesn’t fitinto the garage if both are at rest.

(D). Dudley would conclude that the attempt to close the garage doors failed becausethey both hit his car (longer than the garage) at the same time.

(E). None of the above are true statements.


Relativistic Garage (#26)

(A).This is a tough one! The key is to realize that both observers have to agree on

things like “the door gets smashed”, but they don’t have to agree on the order of the twoevents because of relativity of simultaneity!

So if the outside observer believes the front door is left intact and the back issmashed, then both observers have to see that. That alone determines the answer!

You also know which one will be observed first by Dudley according to the outsideobserver. Since Dudley is rushing towards the back door, the light would meet him onthe way, according to Percepta. However, the light from the front door will get there latersince Dudley is running away from it! Therefore, Percepta would predict that Dudleysees the back door light while the front door light is still en route. Here again, the twoobservers must agree since we are talking about two measurements at the same time(when Dudley sees the back door light) and same place (Dudley’s head). From Dudley’sviewpoint, that means that the back door closed first.

Weird stuff, huh? But you didn’t have to do any math – it was all just understand-ing relativity of simultaneity that gives you the answer.


Concept Problem (#27): Trying to Break the Light Barrier

Donald Rumsfeld says, “The US Military broke the sound barrier, and I’ll be damned ifwe can’t break the light barrier! Besides, my boss needs some help with the vision thing,and let’s face it, sending someone to Mars is just crazy talk. Too dangerous”

The US Congress passes new tax increases on mass-produced domestic beer, bot-tles of wine under $10, currency exchanges, rent-to-own furniture stores, increases theSocial Security payroll tax on the first $50,000 of income and vows to cut down onearned income tax credit fraud to raise $10 ;=< “from Americans least likely to createjobs”, and the mega defense contractor Lockheed-Martin-Boeing-Haliburton-Grumund-Enron2TheRevenge builds a plane which is capable of delivering “nearly infinite” thruston demand. Chuck Yeager comes out of retirement for the flight.

The Joint Chiefs’ winning strategy is to “sneak up” on the light barrier by gettingvery close to a speed of ( and pushing the “nearly infinite thrust button”. Which of thefollowing will be true when the button is pushed?

(A). Yeager will experience no significant acceleration (from his viewpoint).

(B). An earth-based observer will see the mass of the ship increase dramatically

(C). Yeager will observe the mass of the ship increasing

(D). The ship will break the light barrier according to an observer (not based at thePentagon) on earth


Trying to Break the Light Barrier (#27)

(B).The last answer is false, since this can’t happen.The first answer and third answers are also wrong. From Yaeger’s point of view,

the ship is still before pressing the button, so why would the answer be any differencedepending on how fast the earth happens to be going by him?

(B) however, is right. From the earth point of view, the energy released by theinfinte thrust button will go into increasing Yaeger’s ship’s mass although it will continueto travel at just short of the light barrier.


Concept Problem (#28): Inertial Forces and “Weightlessness”

In the ISS (“International Space Station”, also known as the last rest stop leaving earth)which orbits the earth, Ludmila tells you that she feels “weightless”. Is this because

(A). There is no gravitational force on her?

(B). The gravitational force on her is balanced by her inertial force?

(C). The gravitational force on her is balanced by a force from the ship?

(D). The ship is falling as fast as she is in the earth’s gravitational field?


Inertial Forces and “Weightlessness” (#28)

(B) or (D).Both of these statements are ways of saying that there would be no net force ex-

erted on her. In the first answer the “inertial force” balances gravity; in the secondanswer, the ship falls with her and so doesn’t exert a force on her to keep her on her path.

By the way, why doesn’t the ship crash if (D) is true? Well, as the ship falls, it isalso whizzing around the earth, so that by the time the ship “falls” enough to reach thesurface, it has gone some distance around the earth. Since the earth is curved, it never“catches” up with the surface (if it is going fast enough around the earth, of course!)


Concept Problem (#29): Spinning Space Station

In order to provide “gravity” in a space station, it has been proposed that humans mightlive in big spinning rings where the inertial forces might approximate gravity.

FI

Light Beam

spacewalker

If a beam of light passes by the space station, explain what the path looks like to anoutside observer (a space walker, for example). Explain what it looks like to a residentof the spinning space station.


Spinning Space Station (#29)

A space walker outside the station will just observe the light to go straight. BUT,the resident of the space station, who is moving in a curved path, will see the light bendfrom her point of view because she will be unable of her own path bending. (She willjust think it is gravity!)


Concept Problem (#30): How Big is an Atom?

One of Einstein’s other amazing contributions to science in 1905 (the year he proposedthe theory of special relativity) was to explain how Brownian motion could be used todiscover the masses of atoms.Temperature is related to kinetic energy, and kinetic energy of a mass is given by �>5@?A� � ,so B�CED ) �� >5 ?If you mistakenly think that the mass of an atom is

� � � times bigger than it really is, thenyou will predict the velocity to be

(A).� � � times bigger than if you had the correct mass

(B).� � times bigger than if you had the correct mass

(C). The same as if you had the correct mass

(D).� � times smaller than if you had the correct mass

(E).� � � times smaller than if you had the correct mass

Hint: you can solve this using the formula above, but just use your intuition todecide if it is bigger or smaller!


How Big is an Atom? (#30)

(D). B C Dis a number that just depends on temperature, and you are trying to figure out

the velocity from the mass. If you think the mass is� � � times bigger than it really is,

then to keep

BFCGDthe same, 5 ? would have to be

� � � times smaller. Therefore 5 wouldbe� � times smaller.


Concept Problem (#31): Quanta of Light

Imagine that the same amount of energy, H , is about to be released in one of two colorsof light, either orange (

� � � � ;JI Hz) or violet ( 4K� � � ;JI Hz). How do the number of lightquanta (photons) compare?

(A). More violet photons would be released than would be released in orange

(B). The same number of photons would be released in both colors

(C). More orange photons would be released than would be released in violet


Quanta of Light (#31)

(C).Each orange photon has less energy than each violet photon because H � L@+ . So

for a fixed amount of energy, more orange photons can be created.


Concept Problem (#32): Photoelectron energy

The photoelectric effect has a threshold. Imagine that this threshold is right at the fre-quency of orange light (

� � � � ;JI Hz). Using energy conservation arguements, whatshould the kinetic energy of the ejected electron be if it completely absorbs violet lightquanta ( 4M� � �N;=I Hz).

(A). Zero kinetic energy

(B). L,�PO � � � �N;=IRQ Hz

(C). L,�PO � � � � ;=I Q Hz

(D). L,�POS4M� � �N;=IRQ Hz


Half-Life (#32)

(B).If the threshold is at

� � � �T;JI Hz, then LU� � � � �N;=I Hz is the energy required toremove the electron from the metal. A photon with the violet frequency will give energyLV�W42� � �;=I Hz to an electron, so after removal from the metal takes its energy away,conservation of energy will leave kinetic energy of L,�PO � � � � ;JI Q Hz.


Concept Problem (#33): deBroglie Interference

Imagine you wanted to find deBroglie interference. The deBroglie wavelength is

.>� L�>5 and the double slit experiment gives interference maxima separated by

X � 02.3 where 3 is the slit separation and 0 is the distance from the slits.

As an experimentalist, what would you do to try to see deBroglie wave interferenceof atom-sized objects? Which option do you think would be the most helpful?

(A). Speed up the atoms

(B). Slow down the atoms

(C). Decrease the slit separation

(D). Use heavier atoms

(E). Use lighter atoms


deBroglie Interference (#33)

(B), (C) and (E).(B) and (E) would increase . , and therefore increase

X. (C) would increase

Xas

well.


Concept Problem (#34): deBroglie Interference Again

You have a two slit experiment and label your slits A and B. You know that if you blockslit B (or slit A), you won’t see interference. Imagine instead of block the slit, you haveIgor, your assistant, set up a detector which will identify which slit a given particle passesthrough. Igor will then keep track of the position of each photon and which slit it wentthrough, but he won’t tell you which photons are which.

What will Igor see if he just looks at the photons which went through slit A?

(A). Igor will see interference

(B). Igor will see just what he would have seen if he blocked slit B

What will you see if you don’t know which photons are which?

(C). You will see the interference

(D). You will see no interference


deBroglie Interference Again (#34)

(B) and (D).In the first question, if Igor knows that all photons go through slit A that is abso-

lutely equivalent to doing the experiment with the slit B blocked. In both cases, photonscan only go through slit A.

In the second question, think about this from Igor’s point of view. From his pointof view, your answer will be the same as if Igor added the results of two experiments,one where he know all photons go through slit A and one where they all go through slitB. So if you add those two, each without interference, you wouldn’t expect interference.Igor and you have to agree on the result, so you see no interference either. That is tosay that the fact that ANYONE (e.g., Igor) knows which photons are which destroysquantum interference for all. In other words, the measurement, even if you don’t knowthe result, changes the answer.


Concept Problem (#35): Spin Selectomatic

You have three spin selectors. You know that if you take a beam of atoms and orient thespin selectors in exactly opposite directions, you will get nothing out. Also, if you taketwo spin selectors that aren’t completely aligned or misaligned, you’ll get some of whatyou started with out.Now you arrange your spin selectors as shown below. What will you see?

(A). You will see some atoms

(B). You will see no atoms

What would you see if you removed the middle selector?

(C). You will see some atoms

(D). You will see no atoms


Spin Selectomatic (#35)

(A) and (D)Because the middle spin selector will redefine the state with its measurement, it

is possible to have atoms pass through all three selectors. However, if the middle one isremoved, then you are asking essentially, if the spin is up, “is it now down”? That can’thappen without the intermediate measurement to redefine the state.


Concept Problem (#36): Uncertainty in Energy

Imagine that you have an object which can be at two different energies: a “ground state”(lowest energy) and an “excited state” (higher energy). The higher energy state is usuallynot stable because the system will want to go to a lower energy, and so these excited statesusually have a “lifetime”. Let’s imagine that the way a system goes from the excited toground state is by emitted a photon. We will see more about this when we talk about theBohr model of the atom.

For the energy of a state, the Heisenberg Uncertainty relation says that

O X H"QYO X[Z QK\) L �The lifetime as a state can be thought of as the uncertainty in time,

X[Z.

In a spectrum split by a prism, a “narrow line” indicates a smaller range of frequen-cies and therefore smaller uncertainty in energy (remember H �]L@+ ). The narrowest linewill correspond to an excited state which is:

(A). “Meta-stable” (lives for a very long time) ( ) �sec)

(B). “Meso-stable” (lives for� �T^_ sec)

(C). “Short-lived” (lives for� �T^`;a? sec)


Uncertainty in Energy (#36)

(A).The uncertainty in energy

X H can be small only ifXMZ

is big to pick up the slackso that the product can be bigger than L .

XMZis biggest, as the question suggests, if the

state lives for a long time.


Concept Problem (#37): Thomson Deflection of Cathode Rays

Thomson knew (’cause Maxwell said so) that cathode rays would be affected by electric( H ) and magnetic ( b ) fields according to the following formulas

�]�#cd5�b� �ec�H

If the magnetic forces appear very large compared to electric forces, that means that

(A). The cathode rays aren’t moving at all

(B). The cathode rays move at a very low velocity

(C). the cathode rays are moving at a very high velocity


Thomson Deflection of Cathode Rays (#37)

(C).The force increases as the velocity increases; therefore a large deflction which

implies a large force would only occur if the particle were moving very fast.


Concept Problem (#38): Rutherford Experiment with Cathode Rays

Imagine that Rutherford had sugested that his student Marsden investigate the scatteringof cathode rays in atoms. (Cathode rays consist of electrons which are ) 9 � � � timeslighter than f particles.) What would have been expected in the planetary and plumpudding models?

(A). Planetary: back scattering, Plum Pudding: no back scattering

(B). Planetary: no back scattering, Plum Pudding: no back scattering

(C). Planetary: back scattering, Plum Pudding: back scattering

(D). Planetary: no back scattering, Plum Pudding: back scattering


Rutherford Experiment with Cathode Rays(#38)

(C).Thinking back to our analogy with the heavy alpha particle barrelling through

the pudding but bouncing off the nucleus, what made the difference was that the alphaparticle was much more dense than anything thought to be inside the atom. But in thecase of cathode rays (electrons) we know that similar (identical!) elements are to befound inside the atom. Therefore, in either picture, one might expect back-scattering.


Concept Problem (#39): Bohr Atom as gih j

We have said in this class that as Llk � , quantum effects turn off. (Light quanta havesmaller and smaller energies, deBroglie wavelengths go to zero, etc.)What happens if Lmk � in a Bohr atom?

(A). The electron no longer sticks to the nucleus

(B). The electron spirals inward (size goes to 0)

(C). The electron gets bigger

Does this answer make sense based on the Maxwell idea of how a planetary atomshould behave?


Bohr Atom as geh j (#39)

(B).If Lmk � , then the size of the atom, set by L will also go to zero. This is consistent

with Maxwell’s idea since the atom would keep radiating energy as it spiraled inwards inthe Maxwell picture. Quantum effects, however, keep that from happening in real atoms.


Concept Problem (#40): Spectral Lines and Energy Levels

Which of the transitions shown with arrows will emit a photon with the longest wave-length? The shortest wavelength?

3

45

21 E

(A). 5 k 4

(B). 5 k 3

(C). 5 k 1

(D). 4 k 2

(E). 4 k 1


Spectral Lines and Energy Levels (#40)

(A) and (C), repsectively.Long wavelength means low frequency. H � L@+ , so low frequency means low

energy. Low energy means a small difference in energy levels (conservation of energy).So (A) fills the bill for the smallest difference in energy.

Conversely short wavelength will mean a big difference in energy levels, so (C).


Concept Problem (#41): X-rays

X-rays are photons produced in atomic transitions which have energies of ) � �`^`;on Joules,or about

� �� times more energy than the lowest energy state of the Bohr hydrogen atom.For atoms other than hydrogram, the Bohr model predicts that

H6pK�rq ? H6st ? where q is the charge of the nucleus in units of electron charges.

In which of the atoms below would transitions be most likely to produce X-rays?

(A). Hydrogen atoms ( qe� � )(B). Helium atoms ( qe� � )(C). Oxygen atoms ( qi� 9 )(D). Iron atoms ( q#� � � )(E). Silver atoms ( qe� � 4 )


X-rays (#41)

(E).The higher the q , the bigger the energy differences between energy levels. The

biggest those differences, the higher energy the photons emitted by the atom can be.To get a change of energy

� � � times greater than any possible in the Hydrogenatom, you’d need an element with q ?,u � � � . Iron barely gets you there, but Silver( q ? =2209) is a better bet.


Concept Problem (#42): What holds it together?

Identify which of the forces below hold together each of the following objects.

1. A galaxy

2. The solar system

3. The earth

4. The atmosphere and the earth

5. Atoms in a crystal

6. Atoms in molecule

7. Electrons and nuclei in an atom

8. Protons and neutrons in a nucleus

(A). Electromagnetic Forces

(B). Strong Froce

(C). Gravitational Force


What Holds It Together? (#43)

1–4 are held together by gravity; 5–7 are held together by electromagnetism; 8 is heldtogether by the strong force.


Concept Problem (#43): Half-Life

In radioactive decay, half of the nuclei in a sample decay in a period called a “half-life”.Imagine you start with

� � pennies and you played a game where you toss all yourcoins and remove (decay) all those that come up tails. You repeat this procedure.

About how many times would you have to toss these coins to get down to onepenny?

(A). 2 times

(B). 4 times

(C). 8 times

(D). 16 times

What is the “half-life” of a penny in this game?


Half-Life (#43)

(B).The “average” chain of the game would be:� �"k 9 k � k � k ��

Therefore it takes four coin tosses (on average) to get down to one coin. The half-life isone toss!


Concept Problem (#44): Fusion in the Sun

The sun emits� � � � ?v_ Watts of power from fusion processes. How much mass does the

sun lose per second? ( H ��( ? �/w �kg is ) � � ;ax Joules)

(A). 4 kg

(B). 4 million kg

(C). 4 billion kg

(D). 4 trillion kg


Fusion in the Sun (#44)

(C).Each kilogram lost per second results in

� �y;z4 Joules or a power of� �T;z4 Watts in

that second. (�

Watt is�

Joule per second.)To get

� � � � ?{_ Watts, a mass loss of

� � � �F|}� � � � � ?{_� � ;axis required.


Concept Problem (#45): Fision of ~��A� U

Chemists write an element as ��`�� H�where H is the symbol for the element, q is the number of protons in the nucleus and �is the number of neutrons. ( q�� e� , the atomic mass number.)In fision of ?v<zn U, the following reaction is observed

��y�� ?{<�n| ? � k ;=Iz?n{_ b*�y� | ;�K� �� Y�y�R�� `�Find � and � .

(A). X=91, Y=2

(B). X=36, Y=3

(C). X=56, Y=0

(D). X=92, Y=1


Fission of U-235 (#45)

(B).In the reaction shown, the total atomic number and the number of protons must

balance on each side of the equation. Balancing protons,

� � � � � �P��so � �#� � .

Balancing atomic number,

� � � � � � ��N� �!� � ��where we use the fact that the atomic number of a neutron is

�. So � �e� .


Concept Problem (#46): Quantum Tesla Coil

Remember the Tesla coil that we used to light up a fluorescent tube at a distance?Formulate the correct description of this demonstration in terms of quantum fields. Whichof the following does your formulation not include?

(A). Electrons in Tesla coil emit photons as they are accelerated

(B). Photons travel between electrons in the Tesla coil and electrons in the

(C). An electromagnetic wave causes the atoms in the tube to vibrate and emit light

(D). Photons exite atomic electrons in the tube, causing them to eventually drop to theirground state and radiate light


Concept Problem (#47): Interpreting a Cloud Chamber Picture

This is an actual picture from a modern version of a cloud chamber called a “bubblechamber”. Only particles with charge show up in the photo, and because of the magneticfield, particles with positive charge bend one way and particles with negative chargebend another way? Which of the following could describe the process which is pointedto by the arrow? Which of these is physically allowed?

(A). An electron suddenly changes into a positron

(B). An electron suddenly changes its direction

(C). An uncharged particle (photon?) interacts to produce an electron and a positron

(D). An electron and positron are created together travelling in the same direction


Concept Problem (#48): Antimatter and You

One of the strange but true facts about our Universe is that matter and antimatter can becreated together but if we look around us, we only see matter.

What would happen if a� � kg (critical mass of ?{<�n � ) meteor of antimatter struck

the earth?

(A). Nothing

(B). An explosion much smaller than a fission bomb

(C). An explosion the same size as a fission bomb

(D). An explosion much larger than a fission bomb

Which of the following might an astronomer look for if he or she was searchingfor an antimatter galaxy in space?

(A). A really dark (anti-light) spot in the sky

(B). Photons resulting from intergalactic dust or anti-dust anihilating with antigalaxiesor galaxies, respectively

(C). Huge explosions when they collide with matter galaxies

(D). Reverse gravitational lensing (light bends away from the matter)


Concept Problem (#49): Energy of Particle Accelerators

The deBroglie wavelength tells us that

�!� Ly(H where E was the energy of the probe used to look for structure of length � . This saysthat we needed a probe with energy

� � ^� Joules to see distances of� � ^`;=_ meters.

Let’s say that we want to try� �T^`;o� meters. What energy probe would we need?

(A).� � ^`; s Joules

(B).� �N^� Joules

(C).� � ^_ Joules

(D).� �N^�I Joules


Concept Problem (#50): Building a Proton

Protons and neutrons are made of a combination of three quarks total, either of up ordown quarks. The proton has electric charge +1, the neutron has charge 0; the up quarkhas charge +2/3 and the down has charge -1/3. Which of the following combinations isthe proton; which is the neutron?

(A). � �¡�(B). � � ¢(C). �1¢¢(D). ¢N¢¢

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