Physics

#$&*

course PHY 121

5/30 8

Question: `q001. There are two parts to this problem. Reason them out using common sense.

If the speed of an automobile changes by 2 mph every second, then how long will it take

the speedometer to move from the 20 mph mark to the 30 mph mark?

Given the same rate of change of speed, if the speedometer initially reads 10 mph, what

will it read 7 seconds later?

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Your solution:

To get from 20 mph to 30 mph, increasing the speed steadily at a rate of 2 mph, it will

take 5 seconds. That is 30-20 = 10; 10/2-5.

If the speedometer initially reads 10 mph, then the speed will be 24 mph after 7 seconds

(7*2)+10=24.

confidence rating #$&*: 3

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Given Solution:

`aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2

mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph

Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and

the speedometer now reads 24 mph.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q002. An automobile traveling down a hill passes a certain milepost traveling

at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill,

with its speed increasing by 2 mph every second. The time required to reach the lamppost

is 10 seconds.

It then repeats the process, this time passing the milepost at a speed of 20 mph. This

time:

Will the vehicle require more or less than 10 seconds to reach the lamppost?

Since its initial speed was 10 mph greater than before, does it follow that its speed at

the lamppost will be 10 mph greater than before?

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Your solution:

The second time that the car goes downhill, it will take less than 10 seconds to cover

the area it covered in 10 seconds in the first run. This is because it is going faster.

I think that the speed at the lamppost may be a little more than 10 mph greater than

before because it will have a little more time to gather speed before it passes the

lamppost.

confidence rating #$&*: 2

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Given Solution:

`aIf it starts coasting down the same section of road at 20 mph, and if velocity changes

by the same amount every second, the automobile should always be traveling faster than

if it started at 10 mph, and would therefore take less than 10 seconds.

The conditions here specify equal distances, which implies less time on the second run.

The key is that, as observed above, the automobile has less than 10 seconds to increase

its speed. Since its speed is changing at the same rate as before and it has less time

to change it will therefore change by less.

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Self-critique (if necessary): I was thinking that the car would have more time to

gather speed, but it really would have less time since it is going faster. It makes

sense that the speed will change by less because it has less time to change.

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Self-critique Rating: 3

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Question: `q003. The following example shows how we can measure the rate at which an

automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second

hand of a watch moves from the 12-second position to the 16-second position, and its

speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change

equal to 20 mph / 4 seconds = 5 mph / second.

We wish to compare the rates at which two different automobiles increase their speed:

Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30

mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?

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Your solution:

The second car speeds up at the greater rate. The first car changes by 10 mph (30-20)

in 5 seconds, or 2 mph/sec. The second car changes by 50 mph (90-40) in 20 seconds,

which is 5/2 or 2.5 mph/second.

confidence rating #$&*: 3

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Given Solution:

The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which

occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of

change of 2 mph per second.

The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20

seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second.

Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second

greater than that of the first.

Self-critique: OK

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Self-critique Rating: 3

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Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800

Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an

automobile speeds up is determined by the net number of Newtons per kg. Two teams

pulling on ropes are competing to see which can most quickly accelerate their initially

stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg

automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile.

Which team will win and why?

If someone pulled with a force of 500 Newtons in the opposite direction on the

automobile predicted to win, would the other team then win?

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Your solution:

The second team will win. For the first car, 3000/1500 = 2. For the second car,

5000/2000=5/2 or 2.5.

If someone pulled with a force of 500 Newtons in the opposite direction of the second

car, you would get 4500/2000, which is 2.25, so it would still win.

confidence rating #$&*: 2

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Given Solution:

`aThe first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while

the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The

second team therefore increases velocity more quickly. Since both start at the same

velocity, zero, the second team will immediately go ahead and will stay ahead.

The second team would still win even if the first team was hampered by the 500 Newton

resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000

kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first

team

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Self-critique (if necessary): OK but I should have said ""Newtons per kg"" after my

answers.

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Self-critique Rating: 3

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Question: `q005. Both the mass and velocity of an object contribute to its effectiveness

in a collision. If a 250-lb football player moving at 10 feet per second collides head-

on with a 200-lb player moving at 20 feet per second in the opposite direction, which

player do you precidt will be moving backward immediately after the collision, and why?

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Your solution:

I would expect that the 200-lb player moving at 20 feet per second would have more

momentum than the other player, so the 250-lb player would be the one moving backward

after the collision.

confidence rating #$&*: 2

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Given Solution:

`aGreater speed and greater mass both provide advantages. In this case the player with

the greater mass has less speed, so we have to use some combination of speed and mass to

arrive at a conclusion.

It turns out that if we multiply speed by mass we get the determining quantity, which is

called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb

ft / sec, so the second player will dominate the collision.

In this course we won't use pounds as units, and in a sense that will become apparent

later on pounds aren't even valid units to use here. However that's a distinction we'll

worry about when we come to it.

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Self-critique (if necessary): OK--I know know that speed times mass equals momentum.

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Self-critique Rating: 3

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Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep

mountain as far as they can until they use up all their energy from the meal. All other

things being equal, who should be able to climb further up the mountain, the 200-lb

climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10

ounces of Cheerios?

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Your solution:

I would expect the 150-lb climber who ate 10 ounces of Cheerios to be able to climb

further up the mountain. This is because it will take more energy for the 200-lb

climber to take himself up the mountain, so the extra 2 ounces of Cheerios won't give

him so much extra energy that he would be the one to go farther.

confidence rating #$&*: 3

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Given Solution:

`aThe comparison we make here is the number of ounces of Cheerios per pound of body

weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight,

while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has

more energy per pound of body weight.

It's the ounces of Cheerios that supply energy to lift the pounds of climber. The

climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will

climb further.

STUDENT COMMENT

I am satisfied with how I worked out the problem, though it would be nice to know what

formulas to use in case my instinct is wrong. I should have got the energy used per

pound by rereading the question.

INSTRUCTOR RESPONSE

There are two points to these problems:

1. You can go a long ways with common sense, intuition or instinct, and you often don't

need formulas.

2. Common sense, intuition and instinct aren't the easiest things to apply correctly,

and it's really easy to get things turned around.

A corollary: When we do use formulas it will be important to understand them, as best we

can, in terms of common sense and experience.

Either way, practice makes the process easier, and one of the great benefits of studying

physics is that we get the opportunity to apply common sense in situations where we can

get feedback by experimentally testing our thinking.

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Self-critique (if necessary): OK--now I have a formula for it

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Self-critique Rating: 3

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Question: `q007. Two automobiles are traveling up a long hill with an steepness that

doesn't change until the top, which is very far away, is reached. One automobile is

moving twice as fast as the other. At the instant the faster automobile overtakes the

slower their drivers both take them out of gear and they coast until they stop.

Which automobile will take longer to come to a stop? Will that automobile require about

twice as long to stop, more than twice as long or less than twice as long?

Which automobile will have the greater average coasting velocity? Will its average

coasting velocity by twice as great as the other, more than twice as great or less than

twice as great?

Will the distance traveled by the faster automobile be equal to that of the slower,

twice that of the slower or more than twice that of the slower?

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Your solution:

The car that is going faster will take longer to come to a stop. Because it is going

twice as fast, it should take twice as long to stop.

The faster car will have the greater average coasting velocity. It's average coasting

velocity should be twice as great at the other.

I am thinking that the distance traveled by the faster automobile will be twice that of

the slower car.

confidence rating #$&*: 2

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Given Solution:

`aIt turns out that, neglecting air resistance, since the slope is the same for both,

both automobiles will change velocity at the same rate. So in this case the second would

require exactly twice as long.

If you include air resistance the faster car experiences more so it actually takes a bit

less than twice as long as the slower.

For the same reasons as before, and because velocity would change at a constant rate

(neglecting air resistance) it would be exactly twice as great if air resistance is

neglected.

Interestingly if it takes twice as much time and the average velocity is twice as great

the faster car travels four times as far.

If there is air resistance then it slows the faster car down more at the beginning than

at the end and the average velocity will be a bit less than twice as great and the

coasting distance less than four times as far.

STUDENT COMMENT: I do not understand why the car would go four times as far as the

slower car.

INSTRUCTOR RESPONSE: The faster car takes twice as long to come to rest, and have twice

the average velocity.

If the car traveled at the same average velocity for twice as long it would go twice as

far.

If it traveled at twice the average velocity for the same length of time it would go

twice as far.

However it travels at twice the average velocity for twice as long, so it goes four

times as far.

STUDENT COMMENT:

it’s hard to know this stuff without having first discussed it in notes or read it in

the book, or

have an equation handy. I guess this will all come with the class.

INSTRUCTOR RESPONSE

One purpose of this and similar exercises is to get students into the habit of thinking

for themselves, as opposed to imitating what they see done in a textbook. You're doing

some good thinking. When you get to the text and other materials, ideally you'll be

better prepared to understand them as a result of this process.

This works better for some students than others, but it's beneficial to just about

everyone.

STUDENT COMMENT

I understand, it seems as though it would be easier if there were formulas to apply. I

used a little common sense on all but

the last one. Reading the responses I somewhat understand the last one. ?????The problem

doesn’t indicate the vehicle

travels twice the average velocity for twice as long. Should I have known that by

reading the problem or should that have

become clear to me after working it some?????

INSTRUCTOR RESPONSE

You did know these things when you thought about the problem.

You concluded that the automobile would take twice as long to come to rest, and that it

would have twice the coasting velocity. You just didn't put the two conclusions together

(don't feel badly; very few students do, and most don't get as close as you did).

You should now see how your two correct conclusions, when put together using common

sense, lead to the final conclusion that the second automobile travels four times as

far.

No formula is necessary to do this. In fact if students are given a formula, nearly all

will go ahead and use it without ever thinking about or understanding what is going on.

In this course we tend to develop an idea first, and then summarize the idea with one or

more formulas. Once we've formulated a concept, the formula gives us a condensed

expression of our understanding. The formula then becomes a means of remembering the

ideas it represents, and gives us a tool to probe even more deeply into the

relationships it embodies.

There are exceptions in which we start with a formula, but usually by the time we get to

the formula we will understand, at least to some extent, what it's about.

I suppose this could be put succinctly as 'think before formulating'.

STUDENT COMMENT

I feel that I did decent on the problem, but I am the student that likes to have

formulas. Your insight has opened my eyes to a different way of looking at this problem.

I like the comment “Think before Formulate”

INSTRUCTOR RESPONSE

Your solution was indeed well thought out.

I should probably add another comment:

'Think after formulating.'

Formulas are essential, but can't be applied reliably without the thinking, which should

come first and last.

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Self-critique (if necessary): I can see that 2 times as long coasting coupled with 2

times the average speed will cause the faster car to go 4 times as far. I just need

more practice with this to make it gel in my brain.

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Self-critique Rating: 2

@&

This one confuses almost everyone.

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Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord

stretches by 5 feet beyond its initial unstretched length. When a person weighing 150

lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial

unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is

stretched by 12 feet beyond its initial unstretched length.

Based on these figures, would you expect that a person of weight 125 lbs would stretch

the cord more or less than 7 feet beyond its initial unstretched length?

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Your solution:

The change in weight between the 100 pound person and the 150 pound person is 50 pounds.

The change in the length of the stretch of the bungee cord is 9-5, or 4 feet. However,

when the change in weight remains 50 pounds but the people weigh 150 and 200 pounds, the

cord change is 12-9, which is 3. This makes me think that, even though 7 is halfway

between 5 and 9, that the cord will not go exactly 7 for the 125 pound person. I drew a

little picture showing the 4 and the 3. My hunch is that, because the stretch got

smaller for the heavier people, the 125 person would stretch the cord a little more than

7 feet beyond its initial unstretched length.

confidence rating #$&*: 2

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Given Solution:

`aFrom 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the

increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands

each additional pound results in less increase in length than the last and that there

would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This

leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft

to 9 ft, or more than 7 ft.

A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of

this graph and would also show that the stretch at 125 lb must be more than 7 feet (the

graph would be concave downward, or increasing at a decreasing rate, so the midway

stretch would be higher than expected by a linear approximation).

STUDENT COMMENT

I feel like I nailed this one. Probably just didn’t state things very clearly.

INSTRUCTOR RESPONSE

You explanation was very good.

Remember that I get to refine my statements, semester after semester, year after year.

You get one shot and you don't have time to hone it to perfection (not to say that my

explanations ever achieve that level).

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q009. When given a push of 10 pounds, with the push maintained through a

distance of 4 feet, a certain ice skater can coast without further effort across level

ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push)

through the same distance, the skater will be able to coast twice as far, a distance of

60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous

distance) the skater will again coast a distance of 60 feet.

The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type

cord slung between two posts in the ice. The cord, as one might expect, exerts greater

and greater force as it is pulled back further and further. Assume that the force

increases in direct proportion to pullback (ie.g., twice the pullback implies twice the

force).

When the skater is pulled back 4 feet and released, she travels 20 feet. When she is

pulled back 8 feet and released, will she be expected to travel twice as far, more than

twice as far or less than twice as far as when she was pulled back 4 feet?

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Your solution:

When I look at the skater and the pushing, I notice that, if you double the force of the

push and keep the length of the push the same, you also double the distance that the

skater coasts. If you keep the force of the push the same and double the length of the

push, then the distance doubles from the original. For this reason, I believe that

doubling the pull on the slingship will double the distance. The skater will be

expected to travel twice as far as the first time, with the 4-foot pull.

confidence rating #$&*: 2

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Given Solution:

`aThe distance through which the force acts will be twice as great, which alone would

double the distance; because of the doubled pullback and the linear proportionality

relationship for the force the average force is also twice as great, which alone would

double the distance. So we have to double the doubling; she will go 4 times as far

STUDENT COMMENT: I do not understand the linear proportionality relationship for the

force.

If the skater is pulled back an extra four feet, does that mean that the amount of

pounds propelling her is also doubled?

INSTRUCTOR COMMENT: That is so. However the force propelling her isn't the only thing

that influences how far she slides. The distance through which the force is applied is

also a factor.

Doubling the force alone would double the sliding distance.

Doubling the distance through which the force is applied would double the sliding

distane.

Doubling both the applied force and the distance through which it is applied quadruples

the sliding distance.

STUDENT SOLUTION AND QUESTION

She should travel three times as far. The first four feet pulled back yield 20 feet of

travel. The second four feet (i.e., feet 5 through 8) will propel her with twice the

force as the first four feet. So this interval, by itself, would propel her 40 feet. The

20 feet of the first four-foot interval plus the 40 feet of the second four-foot

interval is 60 feet total.

But wouldn’t it be the case that by the time the slingshot reaches the four-foot

position, the force exerted on the skater would only be half of that exerted when she

was eight feet out? I understand why it would be a multiplier of four if the force were

the same throughout, but I’m assuming that the force will decrease as the slingshot is

contracts.

I would appreciate help with this question. Thanks.

INSTRUCTOR RESPONSE

The average force for the entire 8-foot pull would be double the average force for the

4-foot pull. At this point we don't want to get too mathematical so we'll stick to a

numerical plausibility argument. This argument could be made rigorous using calculus

(just integrate the force function with respect to position), but the numerical argument

should be compelling:

Compare the two pulls at the halfway point of each. For a convenient number assume that

the 4-foot pull results in a force of 100 lb. Then the 8-foot pull will therefore exert

a force of 200 lb.

When released at the 4-foot mark, the skater will be halfway back at the 2-foot mark,

where she will experience a 50-lb force.

When released at the 8-foot mark, the skater will be halfway back at the 4-foot mark,

where she will experience a 100-lb force.

Since the force is proportional to pullback, the halfway force is in fact the average

force.

Note that during the second 4 ft of the 8 ft pull the force goes from 100 lb to 200 lb,

so the average force for the second 4 ft is 150 lb, three times as great as the average

force for the first 4 ft. The max force for the second 4 ft is double that of the first

4 ft, but the second 4 ft starts out with 100 lbs of force, while the first 4 ft starts

out with 0 lbs.

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Self-critique (if necessary): So, for the second example, doubling the pull on the

slingshot has the effect of doubling both the size (length) of the push as well as

doubling the force. This means that it's going to double-double (or multiply by 4).

Again, still need more work on this to get into this mode of thinking.

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Self-critique Rating: 2

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Question: `q010. Two identical light bulbs are placed at the centers of large and

identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2

feet.

To a moth seeking light from half a mile away, unable to distinguish the difference in

size between the spheres, will the larger sphere appear brighter, dimmer or of the same

brightness as the first?

To a small moth walking on the surface of the spheres, able to detect from there only

the light coming from 1 square inch of the sphere, will the second sphere appear to have

the same brightness as the first, twice the brightness of the first, half the brightness

of the first, more than twice the brightness of the first, or less than half the

brightness of the first?

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Your solution:

I believe that the larger sphere will appear dimmer because the lightbulb is farther

from the sphere, so the light is more diffuse.

I believe that the second sphere will appear to have half the light for the same reason

as above.

confidence rating #$&*: 1

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Given Solution:

`aBoth bulbs send out the same energy per second. The surface of the second bulb will

indeed be dimmer than the first, as we will see below. However the same total energy per

second reaches the eye (identically frosted bulbs will dissipate the same percent of the

bulb energy) and from a great distance you can't tell the difference in size, so both

will appear the same. The second sphere, while not as bright at its surface because it

has proportionally more area, does have the extra area, and that exactly compensates for

the difference in brightness. Specifically the brightness at the surface will be 1/4 as

great (twice the radius implies 4 times the area which results in 1/4 the illumination

at the surface) but there will be 4 times the surface area.

Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with

twice the diameter will have four times the surface area and will appear 1 / 4 as bright

at its surface. Putting it another way, the second sphere distributes the intensity over

four times the area, so the light on 1 square inch has only 1 / 4 the illumination.

STUDENT COMMENT: I understand the first part of the problem about the distances. But

the second part really confuses me. Looking straight down from the top of the spheres,

the bulb is the same intensity and the frosted glass is exactly the same, so why would

it seem dimmer? I would think that if a person was standing in front of the spheres,

that person would be able to tell a difference, but not extremely close.

INSTRUCTOR RESPONSE: Imagine a light bulb inside a frosted glass lamp of typical size.

Imagine it outside on a dark night. If you put your eye next to the glass, the light

will be bright. Not as bright as if you put your eye right next to the bulb, but

certainly bright. The power of the bulb is spread out over the lamp, but the lamp

doesn't have that large an area so you detect quite a bit of light.

If you put the same bulb inside a stadium with a frosted glass dome over it, and put

your eye next to the glass on a dark night, with just the bulb lit, you won't detect

much illumination. The power of the bulb is distributed over a much greater area than

that of the lamp, and you detect much less light.

STUDENT COMMENT:

I also didn’t get the second part of the question. I still don’t really see where the ¼

comes from.

INSTRUCTOR RESPONSE:

First you should address the explanation given in the problem:

'Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with

twice the diameter will have four times the surface area and will appear 1 / 4 as bright

at its surface. Putting it another way, the second sphere distributes the intensity over

four times the area, so the light on 1 square inch has only 1 / 4 the illumination. '

Do you understand this explanation?

If not, what do you understand about it and what don't you understand?

This simple image of a 2x2 square being covered by four 1x1 squares is the most basic

reason the larger sphere has four time the area of the smaller.

There is, however, an alternative explanation in terms of formulas:

The surface area of a sphere is 4 pi r^2.

If r is doubled, r^2 increases by factor 2^2 = 4.

So a sphere with double the radius has four time the area.

If the same quantity is spread out over the larger sphere, it will be 1/4 as dense on

the surface.

STUDENT COMMENT:

I also have no clue why the extra area doesn’t take away some brightness.

INSTRUCTOR RESPONSE:

All the light produced by the bulb is passing through either of the spheres. From a

distance you see all the light, whichever sphere you're looking at; you see just as much

light when looking at one as when looking at the other.

From a distance you can't tell whether you're looking at the sphere with larger area but

less intensity at its surface, or the sphere with lesser area and greater intensity at

its surface.

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Self-critique (if necessary):

I have read and re-read these explanations. My understanding is that, from a distance,

the fact that the light bulbs inside the spheres are the same means that the size of the

spheres don't affect the light that is given off and seen from a distance. Up close,

doubling the diameter results in 4 times the surface on the sphere, so the brightness

would be 1/4 of the other sphere. I still need to continue to work on these types of

problems to increase my understanding.

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Self-critique Rating: 2

@&

Again, some of these questions, including this one, are intended to be quite challenging. Not many students get this one, and a lot of the ones who do are probably faking it.

*@

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Question: `q011. The water in a small container is frozen in a freezer until its

temperature reaches -20 Celsius. The container is then placed in a microwave oven, which

proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds

the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a

little bit of the cube is melted and the temperature is 0 Celsius. After another minute

most of the ice is melted but there is still a good bit of ice left, and the ice and

water combination is still at 0 Celsius. After another minute all the ice is melted and

the temperature of the water has risen to 40 degrees Celsius.

Place the following in order, from the one requiring the least energy to the one

requiring the most:

Increasing the temperature of the ice by 20 degrees to reach its melting point.

Melting the ice at its melting point.

Increasing the temperature of the water by 20 degrees after all the ice melted.

At what temperature does it appear ice melts, and what is the evidence for your

conclusion?

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Your solution:

Increasing the temperature by 20 degrees after the ice is melted (it took 600

Joules/second times 60 seconds, or 36,000 to get the ice to 40 degrees, so I cut that

number in half and got 18,000)

Increasing the temperature by 20 degrees to reach melting (it took 10 seconds times 600

Joules per second to get the ice to -1 and 10 more seconds at 600 Joules per second to

bring it up to zero, for a total of 24,000 Joules)

Melting the ice at its melting point took 60 seconds at 600 Joules per second, or 36,000

Joules.

Ice starts to melt at 0 degrees Celsius. The ice was still solid at -1, but at 0 a

little bit of the cube was melted and at zero most of the ice was melted but the

ice/water combination was still 0 degrees Celsius.

confidence rating #$&*: 2

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Given Solution:

Since the temperature is the same when a little of the ice is melted as when most of it

is melted, melting takes place at this temperature, which is 0 Celsius.

The time required to melt the ice is greater than any of the other times so melting at 0

C takes the most energy. Since we don't know how much ice remains unmelted before the

final minute, it is impossible to distinguish between the other two quantities, but it

turns out that it takes less energy to increase the temperature of ice than of liquid

water.

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Self-critique (if necessary): I put way too much into it and don't know if the math was

even necessary. I got he part correct about the time required to melt the ice being

greater than any of the other times. I was hesitant to make the statement about

increasing the temperature by 20 degrees being least because I could see that it

probably didn't make sense to just cut it in half. If I were presented with another

problem like this, I don't know if I'd try to just make educated guesses or if I'd try

the Math again.

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Self-critique Rating: 2

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Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a

lap pool). Two friends with kickboards, one at either end of the pool, are using them to

push waves in your direction. Their pushes are synchronized, and the crests of the waves

are six feet apart as they travel toward you, with a 'valley' between each pair of

crests. Since your friends are at equal distances from you the crests from both

directions always reach you at the same instant, so every time the crests reach you the

waves combine to create a larger crest. Similarly when the valleys meet you experience a

larger valley, and as a result you bob up and down further than you would if just one

person was pushing waves at you.

Now if you move a bit closer to one end of the pool the peak from that end will reach

you a bit earlier, and the peak from the other end will reach you a little later. So the

peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't

bob up and down as much. If you move far enough, in fact, the peak from one end will

reach you at the same time as the valley from the other end and the peak will 'fll in'

the valley, with the result that you won't bob up and down very much.

If the peaks of the approaching waves are each 6 inches high, how far would you expect

to bob up and down when you are at the center point?

How far would you have to move toward one end or the other in order for peaks to meet

valleys, placing you in relatively calm water?

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Your solution:

The combined peaks would be 12 inches, 6"" from each side.

You would have to move halfway between the middle and one end (1/4 of the way into the

pool) in order for the peaks to meet valleys and place yourself in calm water.

confidence rating #$&*: 2

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Given Solution:

`aIf the two 6-inch peaks meet and reinforce one another completely, the height of the

'combined' peak will be 6 in + 6 in = 12 in.

If for example you move 3 ft closer to one end you move 3 ft further from the other and

peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you

move 3 ft closer to one end you move 3 ft further from the other. This shifts your

relative position to the two waves by 6 feet (3 feet closer to the one you're moving

toward, 3 feet further from the other). So if you were meeting peaks at the original

position, someone at your new position would at the same time be meeting valleys, with

two peaks closing in from opposite directions. A short time later the two peaks would

meet at that point. ]

However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting

valleys so you will be in the calmest water.

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Self-critique (if necessary): I did not expect to get the 12 inches correct, but I did.

I am having trouble visualizing the question, I guess. I have drawn a rectangle with 2

kickboard people on either end, a person in the middle, and a peak in the middle, where

the water meets. I was thinking that moving 3 feet in either directions would cause you

to be standing in a place where the water was starting to come up on one side and was

going down on the other side. Moving 1.5 feet in one direction or the other would put

you in a spot where the water was rising. Moving would not change where the waves would

peak. I guess, for this one, I am still confused.

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Self-critique Rating: 2

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Question: `q013. This problem includes some questions that are fairly straightfoward,

some that involve more complicated considerations, and possibly some that can't be

answered without additional information.

We're hoping for some correct answers, but we expect that few students coming into this

course will be able to think correctly through every nuance of the more complex

situations. On these questions we are hoping for your best thinking without being

particularly concerned with the final answer.

A steel ball and a wood ball are both thrown upward and both rise with the same average

speed. If not for air resistance they would both come to rest at the same time, at the

same height. However air resistance causes the wood ball to stop rising more quickly

than the steel ball.

Each ball, having risen to its maximum height, then falls back to the ground.

Which ball would you expect to have the greater average velocity as it falls?

Which ball would you expect to spend the greater time falling?

Which ball would you expect to hit the ground first?

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Your solution:

The ball having the greater velocity would be the steel ball because it has more time to pick up speed since it went higher.

The ball that would have the greater time falling would be the steel ball because it went up higher.

I would expect the wood ball to hit the ground first because it stopped rising first.

confidence rating #$&*: 1

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Question: `q014. If you double the voltage across a certain circuit you double the

current passing through it. The power required to maintain the circuit is equal to the

product of the current and the voltage. How many times as much power is required if the

voltage is doubled?

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Your solution:

If the formula is power = current x voltage, and you double the voltage, you would double the power as well.

confidence rating #$&*: 1

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Self-critique Rating: I don't see answers, so I don't know."

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Very good thinking throughout.

Some questions are intentionally challenging so errors are almost inevitable.

*@

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

Very good thinking throughout.

Some questions are intentionally challenging so errors are almost inevitable.

*@

#*&!