#$&* course Phy 231 6/9 around 10 Question: `q001. There are two parts to this problem. Reason them out using common sense.
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Given Solution: `aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time: Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:The vehicle will require less time, but no the vehicle would have already passed the lamppost by the time it reached a speed 10 mph faster than the original run. confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: automobile 1:10/5 2mph/second automobile 2:50/20 2.5mph/second the second car moves faster confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first. Self-critique: ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3000/1500=2 5000/2000=2.5 win, team would still win even if if there were 500 newtons pulling in the opposite direction. confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: 250*10=2500 200*20=4000 The 250 player will move backward immediately after the collision because the second player is bring more force. confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: 12/200=.06 ounce/lb 10/150=.067 ounce/lb the second climber has more ounce per pound so he has more energy
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Given Solution:OK `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): ------------------------------------------------ Self-critique Rating:OK ********************************************* 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: The skater is expected to move 4 times as far as being pulled back 4 because 8 feet is double 4 which gives the skater double the distance and the force pushing the skater is double the force it would be at 4 so added together would make 4 times the distance. confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating:OK ********************************************* 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:from far away the lights should appear the same but when the moth is walking on the light the brightness of the light will be less than half half the brightness of the first confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating:OK ********************************************* 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. Increasing the temperature of the ice by 20 degrees to reach its melting point. greatest. Melting the ice at its melting point. midway. Increasing the temperature of the water by 20 degrees after all the ice melted. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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. Increasing the temperature of the ice by 20 degrees to reach its melting point. greatest. Melting the ice at its melting point. midway. Increasing the temperature of the water by 20 degrees after all the ice melted. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!