#$&* course PHY 122 6/16/13 11:30am If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time: • Will the vehicle require more or less than 10 seconds to reach the lamppost? • Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because the speed is double the first time, at 20mph, and still accelerating, but the distance to the lamppost is still the same, in theory it should get to the lamppost quicker. So it will require less than 10 seconds to reach the lamppost. Since it is traveling down a hill, and the speed is increasing 2mph every second and it is traveling faster than the initial run, there is less time to increase the speed before reaching the lamppost. As a result the speed will increase less than 10mph between the post and the lamppost. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: First vehicle: Speeds up from 20-30mph change of 10mph This happens in 5 seconds so the rate of change is 10mph/5seconds This means that the rate of change is 2mph/second Second vehicle: Speeds up from 40-90mph change of 50mph This happens in 20 seconds so the rate of change is 50mph/20seconds This means the rate of change is 2.5mph/second This shows that the second vehicle is speeding up at a greater rate by 0.5mph per 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 ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Part 1: which team would win Team 1: 3000 Newtons on 1500kg Newtons/kg = 3000/1500 = 2Newtons/kg Team 2: 5000 Newtons on 2000kg Newtons/kg = 5000/2000 = 2.5 Newtons/kg Team 2 would win, even from the beginning of the race Part 2: opposing force added, who would win? New team 2: Because 500N of force is being added in the opposite direction, this force should be subtracted from the initial force being applied (reference frame specifies is negative) So team two is now exerting 5000-500 Newtons on a 2000kg car So Newtons/kg = (5000-500)/2000 = 4500/2000 = 2.25 Newtons/kg Team 2 would still beat out team 1 by 0.25 Newtons/kg even from the beginning of the race. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First player: 250lb; 10ft/s Since both mass and velocity are significant to effectiveness of collision, To figure out his effectiveness in the collision his mass and velocity should be multiplied together: 250lb*10ft/s = 2500 lb*ft/s Second player: 200lb; 20ft/s following the same procedure 200lb*20ft/s = 4000 lb*ft/s 4000 is far greater than 2500, almost double it! Therefore the second player would very easily be able to push the first player backwards immediately after the collision. Poor 1st player, he will be moving backwards very quickly! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: They key here is to see how much cheerios they ate per pound of body weight, this will help to tell who has more energy to use from their breakfast, and thus would be able to climb farther. To do this, the ounces of cheerios must be divided by their body weight to determine the ounces per pound. The first climber: 12 ounces Cheerios, 200lb body weight 12 ounces/ 200lb 0.06 oz/lb The second climber: 10 ounces Cheerios, 150lb body weight 10 ounces/150lbs 0.0667 oz/lb From this calculation you can see that the second climber has a slightly greater amount of ounces per pound of body weight. This gives him a slight advantage, so he will be able to climb for a slightly longer period of time. Let me go off on a tangent for a second. While from common sense, and reading the problem, it is easy to see the value meant to be calculated here. In reality, from what I have learned in my major, I can clearly state that the situation set forth in this problem would not work the way we have just calculated it. This is because various other outside variables would affect metabolism and give one climber the advantage over another. The amount of food consumed does not determine energy availability, though it is nice to simplify it for the purpose of this problem! 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Lets assume air resistance is negligible because it will throw off the answers slightly. From common sense it is easy to tell that the automobile going twice as fast will require longer to stop his vehicle. It also is easy to see that it should take exactly twice as long to stop without air resistance. If air resistance is included, the faster automobile will experience more of it, but will still take almost 2x as long to stop. The automobile will travel longer with a velocity. As a result he has a greater velocity for a greater time. He will therefore have a greater average coasting velocity. Neglecting air resistance, it again should be exactly double the slower car. A similar, but slightly different situation holds true for distance traveled. The automobile going twice as fast should travel exactly double the time of that of the slower car neglecting air resistance. This is easy to see because velocity is defined as change in position over time and the faster car is traveling twice as much distance per second as the slower car, and time is also doubled. Because time is doubled and velocity is doubled, neglecting air resistance the faster car should travel exactly 4x the distance of the slower car. With air resistance it should be just slightly under this value. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I probably should have thought before formulating this one, and thought after. My words are kind of jumbled. I need to approach each part by itself, and remember in the final part that both time and velocity are doubled. This means that displacement must be 4x that of the slower car. Also, nowhere in the problem was I given the permission to neglect air resistance. I need to be aware of this when reading! ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To figure out this problem it would be important to see how the stretch changes with weight, because just from looking at the problem it easy to see that the change in stretch as weight increases is not a linear relationship From 100-150lbs 50 lbs is added The cord stretches from 5-9feet so a change of 4 feet Then from 150lbs-200lbs another 50 lbs is added The cord stretches from 9-12feet so a change of 3 feet This shows that the rate of stretch is decreasing as the weight increases, so a value closer to 100 lbs would stretch the cord at a faster rate than a weight closer to 150lbs 125lbs is exactly in between 100 and 150lbs Because more stretch occurs closer to 100lbs, the cord should stretch more than half of the difference of stretch between 100 and 150 lbs (5-9 feet stretch) by the time it reaches 125lbs. Because exactly half way in between the stretch is 7 feet, the 125lbs should be able to stretch the cord past 7 feet if it follows the above stated pattern. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should have graphed this probably to make my solution clearer. It is hard to describe this in clear wording! ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are two things to consider in this situation. When she is pulled back to 8 feet, not only is the distance to release double, but the length of the cord is doubled. As stated above the cord will exert a greater force when it is pulled back further. Because 8 feet is double of 4, not only does the force exerted on the skater double, but the distance over which that force can be exerted on the skater is also doubled. As a result the skater should go 4x as far when the cord is stretched to 8 feet, assuming the relationship stated above remains true. This indicates that she easily will travel greater than twice as far! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I’m okay on this one, but all of the student-teacher comments are throwing me off a bit. By doubling the distance the cord is stretched, the force applied to the skater is doubled. By doubling the distance the cord is stretched the distance to release, where the cord will apply force to the skater, is also doubled. So you must double the distance traveled twice because there are 2 factors to get 4x the original distance……am I on the right track here? ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: Part 1: For the first part of this problem, the key is to look at the lightbulbs. They are both the same bulb, giving off the same amount of energy. The globe around them should distribute the same amount of light from each bulb at some point regardless of size. Because the distance is so far, and the bulbs are the same, the same amount of energy must be distributed through each globe. From so far away, the size of the globes doesn’t matter, just the equality of the bulbs. Thus both lights would look identical to the moth far away. Part 2: When on the surface of the globe the major difference is seen. The surface area of a sphere is 4πr^2. In the larger of the globes the radius is doubled, meaning the r^2 quantity is doubled. Since 2^2 is 4, the larger of the globes should have 4x as much surface area as the smaller globe. 4x as much surface area means that the lightbulb must distribute the same amount of energy over 4x as much area before it can pass through the frosted globe. This means, while the net sum of all energy passing through each globe is the same, at any one point on the 2x2 globe the light will appear dimmer than on the 1x1 globe. As a result, the moth walking on the larger sphere will only be able to see ¼ the brightness of a moth walking on the smaller sphere when only looking at 1 square inch of sphere. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I believe I neglected part 1 of this problem slightly. Because ultimately the same amount of energy is being given off by the bulbs, and the same amount of energy will reach the eye, From far away both bulbs appear identical. Yet, close up, because of the discrepancy in surface area the larger globe will only appear ¼ as bright. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Least Energy: Increasing the temperature of ice by 20 degrees to its melting point. - From what I learned in physics 1 I know that less energy is required to raise the temperature of ice 20 degrees than water. The ice also takes less time to reach its melting point at 0 Celsius as stated in the above problem. It only takes 20 seconds to reach 0 Celsius, as opposed to taking 1 minute to raise the water temperature 20 degrees Celsius. Increasing the temperature of water by 20 degrees above its melting point - Because the ice uses less energy it is obvious that the water will have to be the middle value, because raising its temperature 20 Celsius only took 1 minute. Most energy: Melting the ice at 0 degrees C will take the most energy, which is obvious to see from the problem because it takes 2:10 to melt the ice at 0 Celsius. Also, from physics 1 we learned about the behavior of fluids. It takes more energy to change states than to change temperature at the same state. The melting/freezing point of ice/water is obviously 0 degrees Celsius. This could be obtained from common knowledge, or from looking at the problem. The ice does not change states until it reaches 0 Celsius, and it remains at this temperature consistently throughout its melting. Only when all the ice has melted can the temperature of the water be increased further. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I actually think I’m okay here, I was a little hesitant because my memory of the behavior of fluids was a little bit fuzzy. I remember drawing a graph where we showed the amount of energy required for each change of state and in between states, and I used this knowledge to apply to this problem. ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: If the peaks are 6 inches high each, and two reach you at the same time in the center point, they will combine together to form one complete peak. The same holds true for the valleys. So you will rise 2x6 inches, and fall 2x6 inches below the starting point (assuming the valleys are the same height as the peaks as they cancel each other out as stated above). The total distance you travel from rest will be 12 inches up to the top of a peak and 12 inches to the bottom of a valley for a grand total of 24 inches bobbed from the absolute top of a peak to the absolute bottom of a valley. If the peaks are 6 feet apart, and they meet by you, moving 3 feet to either side will not make a difference, because if you move 3 feet one way, you are 3 feet closer to one end and 3 feet farther from another end. This means that there will still be that 6 foot discrepancy, and the water will be choppiest here. You need to find a place where the distance only changes by 3 feet for total so that you will be in a place where the valley of one wave can meet the peak of another. This will be achieved by moving 1.5 feet in one direction from the center. This is because this will place you 1.5 feet farther from one end and 1.5 feet closer to the other, for a grand total of 3 feet. This should put you in the calmest water with the littlest movement. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay, for the first part. I understand that the combined height of the peaks is 12 inches, but why can you neglect the combined height of the valleys? Because the peaks and the valleys end up combining wouldn’t the total distance you bob be 12 inches up from start and 12 inches down from the starting point, meaning you are bobbing 24 inches total from bottom to top? Displacement is ultimately 0 because you return to the starting point, but that is not what the problem is asking for. The problem merely asks how far you bob. Perhaps a definition of “bobbing” is in order, because I assumed that bobbing would take into account both the peaks and the valleys. While each is 12 inches net from both waves meeting, I don’t see why the total movement should only be 12 inches since we also move through the valleys. ------------------------------------------------ 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 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: When the balls are falling, because air resistance is taken into account here and we know that the wood ball is more affected by air resistance, it will not be able to move as fast as the steel ball. Also the steel ball very likely has a greater mass than the wood ball, and thus, because of gravitational acceleration (g=9.8 m/s^2 down), the steel ball will most likely be moving at a faster velocity as it falls. By similar comparison, because the wood ball is affected by air resistance, it is expected that it will spend more time falling than the steel ball. However, this questions really depends on the numerical details given. Depending on the discrepancy in the maximum height of the steel and wood balls, there could be significant difference in the distance each ball has to fall. This could also affect which ball spends greater time falling. Looking at what is said in the original problem, it is stated that both balls would come to rest at the same time if it were not for resistance. Since the wood ball is only caused to stop rising sooner because of air resistance, wouldn’t this same condition also be true in the falling condition. As a result, perhaps the steel ball will spend more time in the air. This really, again, depends on specific numbers given. For the final part of this question, which ball hits the ground first, again this is a really number dependent solution. The steel ball has farther to fall, and because of gravitational acceleration it will be hitting the ground with a higher velocity. However the wood ball has less of a distance to fall and is affected by air resistance. Depending on the discrepancy in the heights of the two balls, the magnitude of the effect of air resistance, and the mass of each ball respectively which one hits the ground first would change. In this case, because the wood ball stopped before the steel ball when rising it is implied that it would hit the ground first. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2 ********************************************* Question: `q014. If you double the voltage across a certain circuit you double the current passing through it. The power required to maintain the circuit is equal to the product of the current and the voltage. How many times as much power is required if the voltage is doubled? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P= VI where P=power, V=voltage, I=Current The easiest way to do this is to assume that there is some base power P, some base voltage V, and some base current I. To figure out the new P, 2 should be substituted for V and I in the original equation because each of these values are doubled. This is because of the relationship between voltage current and power stated above. This gives you: ?P= 2V*2I where the original V, I and P are constant so can be removed to get ?=2*2 ?=4 Therefore the original P must be multiplied by 4 to keep the relationship the same. As a result 2x the power is required if the voltage is doubled confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course PHY 122 6/16/13 11:30am If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time: • Will the vehicle require more or less than 10 seconds to reach the lamppost? • Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because the speed is double the first time, at 20mph, and still accelerating, but the distance to the lamppost is still the same, in theory it should get to the lamppost quicker. So it will require less than 10 seconds to reach the lamppost. Since it is traveling down a hill, and the speed is increasing 2mph every second and it is traveling faster than the initial run, there is less time to increase the speed before reaching the lamppost. As a result the speed will increase less than 10mph between the post and the lamppost. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK