#$&* course Phy 231 1/19 6pm 006. Physics
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first time, the car will gain 2mph each second, passing the lamppost 10 seconds later at a speed of 30mph (because it started at 10mph and gained 2mph*10sec or 20mph). The second time, assuming that the car increases speed at the same rate, it will still gain 2mph each second, which means after 10 seconds it will be travelling 40mph. Since the average speed during those 10 seconds is higher the second time, that means it covered more distance during the same amount of time. So it must have passed the lamppost before 10 seconds had past. Again, if we assume it is gaining speed at the same rate after it begins to coast, then yes, the speed at the lamppost will be 10mph greater than the first time. The problem doesn't specify this, though. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, I see my error. I worked through the logic by comparing how far it would travel during 10 seconds, which led me to the right answer for the first part (because I reasoned that if it covered more distance in the same amount of time, it passed the lamppost earlier). But this paved the way for me to mess up the second part, because I was still focused on what happens during the 10-sec interval vs. what happens during the interval it takes to reach the lamppost. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first car speeds up by 10mph in five seconds, for an average of 10/5 or 2mph/sec. The second car speeds up by 50mph in 20 seconds, for an average of 50/20 or 2.5mph/sec. The second car speeds up at the greater rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first team is exerting 3000 Newtons per 1500 kg, or 2 Newtons/kg. The second team is exerting 5000 Newtons per 2000 kg, or 2.5 Newtons/kg. The second team will win because their car will accelerate faster as a result of the great .5 Newtons/kg of force. If someone pulled the second team's car in the opposite direction with 500 Newtons of force, it would reduce the net force to 4500 Newtons in the right direction. For a 2000 kg car, that would result in 4500 Newton/2000 kg, or 2.25 Newtons/kg. The would still win, because their Newton/kg force(rate?) still exceeds the 2 Newtons/kg the first team is exerting. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???Is Newtons/kg a rate? I'm not sure what terminology to use to describe this situation.??? ------------------------------------------------ Self-critique Rating:
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was pretty sure momentum was the term for this relationship, but I didn't remember that it's a simple multiplication relationship. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let's assume an ounce of Cheerios gives one the energy to pull 50 pounds of body weight 100 feet. That would mean that the first climber has eaten enough Cheerios to pull 50 pounds of body weight 12*100 feet, or 1200 feet. But he weighs 200 pounds, so he'll use up his energy four times as fast and only get 300 feet. The second climber has eaten enough Cheerios to pull 50 pounds of body weight 10*100 feet, or 1000 feet. But he weighs 150 lbs, so he'll use up his energy 3 times as fast and only get 333.3 feet. Still, he'll go further than the first climber. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Similarly to the sample student comment you include, I was completely confident that I figure it out with my made-up numbers. But it feels all wrong to use made-up numbers, especially when I know we'll be looking at calculus-based concepts that involve curves, not lines (so the relationship between concepts might not always be obvious when looking at just one or a few examples). I do realize now that I could have taken my reasoning one step further and gotten the precise oz/lb values. That would have felt a less arbitrary and more correct way of finding the answer. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the cars are taken out of gear, their speed should decrease at the same rate, since the forces acting on them should be exactly the same from that moment on. So, say the speeds start out at x mph and 2x mph. They will both eventually reach 0 mph. When the slower car reaches 0 mph, its speed will have decreased by x mph (x- x=0). During that same amount of time, the faster car should have ALSO decreased by the same amount, or x mph. But since it started out at 2x mph, it will be traveling at 2x-2 or x mph when the slower car stops. So, if it continues to decrease at the same rate, it is halfway to stopping, and should take twice as long to stop. To find the average coasting velocity we add initial velocity and final velocity and divde by 2. For the first car, this is (x-0)/2, or one-half x. For the second car, this is (2x-0)/2, or x. The average coasting velocity of the second car is twice that of the first car. The distance traveled by the faster car will be more than twice that of the slower. If we imagine a halfway point on its path, the second half of the path is equal to the entire path traveled by the first car, because the average velocities of each of those situations is the same. But, the first half of the path traveled by the faster car involved an average velocity that is greater than the second half. So, it travelled much further during the first half of its path, meaning the total path is more than double the length of the slower car's path. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like I understand this pretty well, but I didn't get so far in my logic as to realize that the distance would actually be 4 times greater. I like to think that if pressed, I would have figured it out, but it's helpful to notice that I actually had more information than I realized. More conclusions could have been drawn with no specific numbers other than ""twice the velocity."" In this case I found it very helpful to actually write this down as math. It helps me organize my thinking, but I wonder if I'm using that x as a bit of a crutch to avoid looking at it from a more purely logical standpoint. ------------------------------------------------ Self-critique Rating: 2
.............................................
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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I nailed this one too. But, I again wonder if referring to ""my graph"" means I'm not quite there on the larger concept at play. I can see clearly how it works in this specific example that I've put to paper, but I feel less sure on how to explain why this is so. ------------------------------------------------ Self-critique Rating: 3
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I actually think I'm wrong about this part, or at the very least I am less sure of myself when I said: ""If the first foot of pullback causes x feet of coasting, the second foot will cause 2x feet of coasting. This means that a 2-ft pullback will actually cause x feet (from the first foot of pullback) PLUS 2x feet (from the second foot of the pullback), for a total of 3x feet of coasting."" Is this essentially the error made by the student who said ""She should travel three times as far,"" above? As I did this, I was thinking only in terms of the 1st and 2nd feet of pullback (and didn't necessarily think she would travel three times as far, in total). The numbers are all swimming at this point... I think I am clear on the general concept, but if asked to justify it with exact numbers, I don't know that I'd be able to. Perhaps it is because I'm still not sure I am clear on what force is, mathematically. Let me try without the x, and just make up different numbers even though that bothers me a little. The first foot of pullback causes a 2-foot coast, for example. The second foot of pullback causes a 4-foot coast, because we've doubled the force over the same distance. (Right?) Doubling again, the third and fourth feet of pullback, together, would cause an 8-foot coast. Doubling again, the fifth-through-eighth feet of pullback cause a 16-foot coast. That means a four-foot pullback should cause a 2+4+8 or 14 ft. coast An eight-foot pullback should cause a 2+4+8+16 or a 30 ft. coast. This is definitely not right, since you said the answer is that she actually goes four times as far. Where is my logic off? Is it that, looking at my third step, the third and fourth feet of pullback should actually cause an 8-foot coast per foot of pullback within that range? (so, 8*2) And the fifth-through-eighth should cause a 16-foot pullback per foot of pullback within that range? (so 16*4) That would add up to 2 feet + 4 feet + 2(8) = 16 feet + 4(16) = 64 feet for a total of 86 feet. But maybe I'm not supposed to be adding at all here? When you say ""twice the pullback equals twice the force,"" do you mean twice the average force, spread out over the entire pull? Or twice the force at that exact moment? I'm afraid I've gotten myself very lost here. Self-critique 1 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The light emitted from the light bulbs will have the same energy, which should mean the size of the spheres should not matter from half a mile away. The moth will see the same amount of light in both cases, so the total amount of brightness, as perceived from a great distance, should be equivalent regardless of size. On the other hand, I believe that on the surface of the sphere, the brightness from each bulb will be ""spread out"" different amounts. More so on the larger sphere. It seems that the energy from the bulb, if divided into 1-sq.-inch pieces, will have more such pieces to divide among on the larger sphere, giving less energy and therefore less brightness to each. To calculate how much less, we'd need to find the surface area of each sphere. I can't remember how to do this, but I am fairly certain that doubling the diameter more than doubles the surface area of a sphere. So a 1-sq-in segment of the larger sphere will have less than one-half the brightness of the smaller one. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got to the right conclusion (though without the specificity of a four-times greater area). But I think my reasoning was faulty. I reasoned that the light has to be ""shared"" by different amounts of surface area in each case. But you mention energy per second, leading me to believe that this is more about the time it takes the light to get to the surface of the sphere, rather than how much it has to spread out. Or is it both? ------------------------------------------------ Self-critique Rating: 2
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Hmm. So, I see now that I did not read carefully enough. The third thing to look for was what it takes to raise the temperature after ALL the ice has melted. But that is not something we can calculate from the information given, because the final minute, during which the temperature went up to 40, did not actually start with all of the ice melted. I was assuming I had more of a basis for calculating this specifically than I really did. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I'm picturing this as a series of triangles. Each wave is essentially an area above the natural surface of the water, filled in with ""extra"" water that is borrowed from an adjacent valley. Using the x-axis of a graph as the normal surface of the water when still, and the y-axis as depth in feet, I imagine the top of the first wave coming towards me from the right at (0,.5) since the wave is 6 inches high. The next wave is 6 feet away, at (6, .5). The bottom of the valley in between should occur at (3, -.5). This is where the wave ""borrowing"" water comes in-- in order for the water level to rise in certain points, it must drop in equal measure somewhere else. So the first wave looks like a triangle with three points at (-1.5,0), (0,.5), and (1.5,0). The area beneath it along that exact line-- the amount of ""extra water"" coming up over the natural surface- is (1/2)(3*.5), or .75 sq feet. But, if the waves reach me at the same time, there will be twice as much water, or 1.5 sq. feet. The waves will still be the same distance apart, so the base of the triangle will still be three. Now we need to find the height of the triangle. 1.5=(1/2)(3h) 3=3h h=1. The wave will be 1 foot tall when they meet at the middle. I know it's really a curve, not triangles with sharp points, but this is the best way I can reason through it right now. Since each peak is 3 ft away from the closest valley, I would have to move 3 feet in one direction for peaks to meet valleys at the spot where I'm standing. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well I probably made this way more complicated than I needed to, but I felt like the only way I could be sure the height would double would be to figure out how much water was involved. I see now how unnecessary that is; two waves occupying the same space, essentially, will always contain twice the water and need to reach twice as high. My error in the second part came from forgetting that getting closer to one set gets me farther from the other. I was still thinking of the behavior of the farther wave at the midpoint, rather than at the point where I'm standing. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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, between release and coming to rest at maximum height, 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The final speed at the end of the ascent is zero in both cases. We can assume that the initial speed at the beginning of the ascent was the same, because the problem specificies that without air resistance, their paths would be the same. We know that the steel ball takes longer to reach a speed of zero. That means it is moving upward longer, which means it gets to a greater height. This tells us that it will have to fall a greater distance than the wood ball. We don't know for sure if air resistance acts the same on the way down as it did on the way up. But I don't think that matters for this question. Either air resistance on the way down is not very significant here, and we can assume (since both start their descent at a speed of zero), the steel ball will take longer to reach the ground, OR if air resistance is significant, it will serve only to make the wood ball fail to accelerate towards the ground as quickly as it otherwise would. Since we know the steel ball has longer to fall, which means greater distance to build up speed, AND that the steel ball will encounter little air resistance to interfere with the natural acceleration of gravity, we can assume that its final speed at the point of landing will be greater than the wood ball, which means it will have a greater average velocity over the course of the fall. We don't know for sure which will spend longer falling. That depends on how air resistance behaves on the way down. I suppose it is possible that, even though the wood ball has less distance to fall, air resistance might slow it down enough make that actually take longer than the steel ball falling from a greater height. I don't think it's likely, but I do believe it is possible, so we can't be certain. My guess is that if we can assume that air resistance affects things exactly the same way in each direction, then the steel ball will spend longer falling by the same ratio (compared to the wood ball) as it rose. If we can assume air resistance is equivalent going both directions, then we know the wood ball will hit the ground first, because it started falling sooner and we are assuming the steel ball's descent is larger than the wood ball's by the same ratio as the two ascents (the point being, we assume it takes longer and it starts earlier). confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q014. If you double the voltage across a certain circuit you double the current passing through it. The power required to maintain the circuit is equal to the product of the current and the voltage. How many times as much power is required if the voltage is doubled? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It helps me to write this out with made-up coefficients. 1 power = 1 current * 1 voltage X power = 2 current * 2 voltage X power = 4 (current * voltage) The power needs to be quadrupled if the voltage is doubled. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:"