{"id":27,"date":"2017-06-30T18:31:32","date_gmt":"2017-06-30T18:31:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/atd-monroecc-physics\/?post_type=chapter&p=27"},"modified":"2017-07-21T21:45:13","modified_gmt":"2017-07-21T21:45:13","slug":"conservation-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-monroecc-physics\/chapter\/conservation-laws\/","title":{"raw":"Conservation Laws","rendered":"Conservation Laws"},"content":{"raw":"Conservation Laws<\/h1>\r\nConcepts and Principles<\/h2>\r\nWhat is a Conservation Law?<\/h3>\r\nIn general, a conservation law is a statement that a certain quantity does not change over time. If you know how much of this quantity you have today, you can be assured that the exact same amount of the quantity will be available tomorrow. A famous (at least to physicists) explanation of the nature of a conservation law was given by Richard Feynman.\r\n
Imagine your child has a set of 20 wooden blocks. Every day before bedtime you gather up your child's blocks to put them away. As you gather up the blocks, you keep count in your head. Once you reach 20, you know you have found all of the blocks and it is unnecessary for you to search any longer. This is because the number of blocks is conserved. It is the same today as it was yesterday. <\/em><\/p>\r\nIf one day you only find 18 blocks, you know to keep looking until you find the missing 2 blocks. Also, with experience, you discover the typical hiding places for the blocks. You know to check under the sofa, or behind the curtains. <\/em><\/p>\r\nIf your child is rambunctious, you may even have to look outside of the room. Perhaps he threw a block or two out of the window. Even though blocks can disappear from inside of the room, and appear out in the yard, if you search everywhere you will always find the 20 blocks.<\/em><\/p>\r\nPhysicists have discovered a number of quantities that behave exactly like the number of wooden blocks. We will examine two of these quantities, energy and momentum, below.\r\n\r\n \r\nThe Impulse-Momentum Relation<\/h3>\r\nWhile Newton\u2019s Second Law directly relates the total force that acts on an object at a specific time to the object\u2019s acceleration at that exact same time, conservation laws relate the amount of a certain quantity present at one time to the amount present at a later time.\r\n\r\nThe first conserved quantity we will investigate is momentum<\/em>. Of course, just because momentum is conserved doesn't mean that the momentum of any particular object or system of objects is always constant. The momentum of a single object, like the number of blocks in the playroom, can change. Just as blocks can be thrown out of the window of the playroom, the momentum of a single object can be changed by applying impulse<\/em> to it. The relationship between impulse and momentum is, conceptually,\r\ninitial momentum + impulse = final momentum<\/em><\/p>\r\n \r\n\r\n \r\n\r\nPictorially, we can visualize this as\r\n\r\n \r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21193903\/Screen-Shot-2017-07-21-at-12.38.48-PM-300x127.png\")
\r\n\r\n \r\n\r\nIn practice, we will identify an object or collection of objects (a system<\/em>) and determine the amount of momentum the system contains at some initial time. This quantity cannot change unless impulse is done to the system. We call processes that bring momentum into the system as positive impulses, and processes that remove momentum from the system as negative impulses.\r\n\r\nMathematically this is written as\r\ninitial momentum + impulse = final momentum<\/em><\/p>\r\nPi<\/sub> + Jif<\/sub> = Pf<\/sub><\/em><\/p>\r\nwhere\r\n\r\n \t- momentum (P) is the product of an object\u2019s mass and velocity,<\/li>\r\n \t
- impulse (J) is the product of a force external to the system<\/em> and the time interval over which it acts,<\/li>\r\n \t
- and S indicates that you must sum the momentum of all of the objects in the system and all of the impulses acting on the system.<\/li>\r\n<\/ul>\r\n \r\n\r\nIn short, if no impulse is applied to a system, its momentum will remain constant. However, if an impulse is applied to the system, its momentum will change by an amount exactly equal to the impulse applied. This momentum does not appear or disappear without a trace. It is simply transferred to the object supplying<\/em> the impulse. In this sense, impulse is the transfer of momentum into or out of a system, analogous to tossing blocks into or out of a playroom.\r\n\r\n \r\n
The Work-Energy Relation<\/h3>\r\nThe second conserved quantity we will investigate is energy<\/em>. Just like momentum, or wooden blocks, the conservation of energy doesn't mean that the energy of any particular object is always constant. The energy of a single object or system of objects can be changed by doing work <\/em>to it. The relationship between work and energy is, conceptually,\r\ninitial energy + work = final energy<\/em><\/p>\r\nPictorially, we can visualize this as\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21195224\/Screen-Shot-2017-07-21-at-12.52.08-PM-300x121.png\")
\r\n\r\n \r\n\r\n \r\n\r\nThe similarity between momentum and energy is not complete, however. While there is only one form of momentum (i.e., one hiding place for momentum \u201cblocks\u201d) there are several forms of energy. These different forms of energy will be introduced as you progress through more and more complicated models of the physical world. For now, the only \u201chiding place\u201d I want to discuss is kinetic energy<\/em>. In terms of kinetic energy, the above conceptual relationship between work and energy becomes, expressed mathematically,\r\ninitial energy + work = final energy<\/em><\/p>\r\nKEi<\/sub> + Wif<\/sub> = KEf<\/sub><\/em><\/p>\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21195357\/Screen-Shot-2017-07-21-at-12.53.36-PM-300x67.png\")
<\/p>\r\nwhere\r\n
\r\n \t- kinetic energy (KE) is the product of one-half an object\u2019s mass and squared velocity,<\/li>\r\n \t
- work (W) is the product of a force (even an internal force) <\/em>and the displacement over which it acts (with more subtle details discussed below),<\/li>\r\n \t
- S indicates that you must sum the kinetic energy of all of the objects in the system and all of the work done to the system,<\/li>\r\n \t
- and we define a new unit, Joule (J), as J = kg (m\/s)2<\/sup> = N m.<\/li>\r\n<\/ul>\r\n \r\n
Unlike anything we\u2019ve studied up to this point, the work-energy relation is a scalar<\/em> equation. This will become especially important when we study objects moving in more than one dimension. For now, all this means is that in the expression for work,\u00a0![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21195551\/Screen-Shot-2017-07-21-at-12.55.42-PM.png\")
, we should use the magnitude<\/em> of the force and the magnitude<\/em> of the change in position. This product is then multiplied by cos f , where f is defined to be the angle between the applied force and the displacement of the object. If the force and displacement are in the same direction f = 0\u00b0, and the work is positive (the object gains energy). If the force and displacement are in the opposite direction f = 180\u00b0, and the work is negative (the object loses energy). Note that the actual directions of the force and the displacement are unimportant, only their directions relative to each other<\/em> affect the work.<\/p>\r\n \r\n\r\nIn general, if no work is done to a system, its kinetic energy will remain constant. However, if work is done to system, its total energy will change by an amount exactly equal to the work done. Work is the transfer of energy from one system to another, again analogous to tossing blocks from the playroom into the yard.\r\n\r\n \r\nAnalysis Tools<\/h2>\r\nApplying the Impulse-Momentum Relation to a Single Object<\/h3>\r\nLet\u2019s investigate the following scenario:\r\n\r\n
A 0.35 kg model rocket is fitted with an engine that produces a thrust of 11.8 N for 1.8 s. The rocket is launched vertically upward.<\/p>\r\n\r\n<\/div>\r\nTo apply the impulse-momentum relation, you must clearly specify the initial and final events at which you will tabulate the momentum. For example:\r\n
\r\n\r\n\r\nEvent 1: The instant the engine is ignited.\r\n\r\n \r\n\r\nP1<\/sub> = 0<\/td>\r\nEvent 2: The instant the rocket reaches maximum height.\r\n\r\nP2<\/sub> = (0.35) v2<\/sub><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nJ12<\/sub> = +Fexhaust<\/sub>(1.8)-Fgravity<\/sub>(1.8)\r\n\r\n\r\n\r\n\u00a0![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21195932\/Screen-Shot-2017-07-21-at-12.59.14-PM-205x300.png\") <\/td>\r\n | \r\n\r\nNote that each external force acting on the rocket is multiplied by the time interval over which it acts. (Also note that the rocket\u2019s engine does not produce a force on the rocket! The engine produces a downward force on the hot exhaust gases emitted from the engine and these hot gases exert an equal magnitude force back up on the rocket. That is why the force on the rocket is labeled as Fexhaust rather than Fengine.)\r\n\r\n <\/td>\r\n | <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nApplying impulse-momentum to the rocket during this time interval yields:\r\n P<\/em>1<\/sub> + J<\/em>12<\/sub> = P<\/em>2<\/sub><\/p>\r\n0 + Fexhaust<\/sub><\/em>(1.8) - Fgravity<\/sub><\/em>(1.8) = 0.35v<\/em>2<\/sub><\/p>\r\n0 + (11.8)(1.8) - (0.35)(9.8)(1.8) = 0.35v<\/em>2<\/sub><\/p>\r\nv<\/em>2<\/sub> = 43.0 m\/s<\/em><\/p>\r\nThus, the rocket is traveling at 43.0 m\/s at the instant the engine shuts off.\r\nOf course, there is no reason why we had to analyze the rocket\u2019s motion between the two instants of time we selected above. We could have selected the events:\r\n\r\n\r\n\r\nEvent 1: The instant the engine is ignited.\r\n\r\n \r\n\r\nP<\/em>1<\/sub> =<\/td>\r\nEvent 2: The instant the rocket reaches maximum height.\r\n\r\n \r\n\r\nP<\/em>2<\/sub> = 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21200738\/Screen-Shot-2017-07-21-at-1.05.46-PM-300x48.png\") \r\n\r\nDuring this time interval, the force of the exhaust gases only act on the rocket for a portion<\/em> of the entire time interval. Noting that the rocket\u2019s velocity when it reaches its maximum height is zero, impulse-momentum would look like this:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21200743\/Screen-Shot-2017-07-21-at-1.05.57-PM-300x123.png\") \r\n\r\nThus, the rocket is in the air for 6.19 s before reaching its maximum height.\r\n\r\n \r\nApplying the Work-Energy Relation to a Single Object<\/h3>\r\nThe work-energy relation also has many uses for investigating physical scenarios. For example, let\u2019s look again at our model rocket:\r\n\r\n A 0.35 kg model rocket is fitted with an engine that produces a thrust of 11.8 N for 1.8 s. The rocket is launched vertically upward.<\/em><\/p>\r\n\r\n<\/div>\r\nAssuming we\u2019ve already analyzed this scenario using impulse-momentum, what additional information can we extract using work-energy?\r\n\r\n\r\n\r\nEvent 1: The instant the engine is ignited.\r\n\r\n \r\n\r\nKE1<\/sub> = 0<\/td>\r\nEvent 2: The instant the engine shuts off.\r\n\r\nKE2<\/sub> = \u00bd (0.35) (43)2<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21201354\/Screen-Shot-2017-07-21-at-1.11.22-PM-300x29.png\") \r\n\r\nTherefore,\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21201357\/Screen-Shot-2017-07-21-at-1.11.33-PM-300x128.png\") \r\nWhat if we apply work-energy between the following two events:Thus, the rocket rises to a height of 39.1 m before the engines shuts off.\r\n\r\n\r\n\r\nEvent 1: The instant the engine is ignited.\r\n\r\n \r\n\r\nKE1<\/sub> = 0<\/td>\r\nEvent 2: The instant the rocket reaches its maximum height\r\n\r\n \r\n\r\nKE2<\/sub> = 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202005\/Screen-Shot-2017-07-21-at-1.15.49-PM-300x25.png\") \r\n\r\nDuring this time interval, the force of the exhaust gases only act on the rocket for a portion<\/em> of the entire displacement, namely 39 m, while the force of gravity acts over the entire displacement.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202008\/Screen-Shot-2017-07-21-at-1.16.06-PM-300x89.png\") \r\n\r\nThus, the maximum height reached by the rocket is 134 m.\r\n\r\n \r\n\r\n \r\nGravitational Potential Energy<\/h3>\r\nIn any situation in which an object changes its height above the surface of the earth, the force of gravity does work on the object. It is possible to calculate this work in general, and to rewrite the work-energy relation in such a way as to incorporate the effects of this work. This is referred to as constructing a potential energy function<\/em> for the work done by gravity.\r\n\r\nLet\u2019s imagine an object of mass, m, located an initial height, hi, above the zero of a vertical coordinate system, with the upward direction designated positive. It moves to a final height of hf.\r\n\r\n\r\n\r\n\u00a0![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202010\/Screen-Shot-2017-07-21-at-1.16.46-PM-157x300.png\") <\/td>\r\n | To calculate the work done by gravity on this object:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202012\/Screen-Shot-2017-07-21-at-1.16.57-PM-300x129.png\") <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe \u201cmgh\u201d terms are referred to as gravitational potential energy<\/em>. Thus, the work done by gravity can be thought of as changing the gravitational potential energy of the object. Let\u2019s plug the above result into the work-energy relation:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202015\/Screen-Shot-2017-07-21-at-1.17.38-PM-300x182.png\") \r\n\r\nTherefore, this final relation:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202017\/Screen-Shot-2017-07-21-at-1.17.48-PM-300x72.png\") \r\n\r\ncan (and will) be used in place of the standard work-energy relation provided:\r\n\r\n \t- You do not include the force of gravity a second time by calculating the work done by gravity. Basically, in this relationship gravity is no longer thought of as a force that does work on objects but rather as a source of potential energy.<\/li>\r\n \t
- You calculate the initial and final heights, hi and hf, using a coordinate system in which the upward direction is positive.<\/li>\r\n<\/ol>\r\n \r\n
Applying Work-Energy with Gravitational Potential Energy<\/h3>\r\nLet\u2019s use the work-energy relation, with gravitational potential energy terms, to re-analyze the previous scenario:\r\n\r\n A 0.35 kg model rocket is fitted with an engine that produces a thrust of 11.8 N for 1.8 s. The rocket is launched vertically upward.<\/em><\/p>\r\n\r\n<\/div>\r\nLet\u2019s apply work-energy between the following two events, setting the initial elevation of the rocket equal to zero:\r\n\r\n\r\n\r\nEvent 1: The instant the engine is ignited.\r\n\r\n \r\n\r\nKE1<\/sub> = 0\r\n\r\nGE1<\/sub> = 0<\/td>\r\n| Event 2: The instant the rocket reaches its maximum height\r\n\r\n \r\n\r\nKE2<\/sub> = 0\r\n\r\nGE2<\/sub> = (0.35)(9.8) h2<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n W<\/em>12<\/sub> = Fexhaust<\/sub><\/em>(39)cos0<\/p>\r\n \r\n\r\nDuring this time interval, the force of the exhaust gases only act on the rocket for a portion<\/em> of the entire displacement, namely 39 m. Remember, the force of gravity does not do work<\/em> in this way of modeling nature, rather the gravitational energy of the rocket changes as it changes its elevation.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202508\/Screen-Shot-2017-07-21-at-1.24.33-PM-300x119.png\") \r\n\r\nresults in, of course, the same maximum height reached by the rocket.\r\n\r\n \r\nApplying the Impulse-Momentum Relation to a Collision<\/h3>\r\nProbably the most useful application of the impulse-momentum relation is in the study of collisions. For example:\r\n\r\n Far from the earth, a 250 kg space probe, moving at 5 km\/s, collides head-on with a 60 kg piece of space debris initially at rest. The debris becomes entangled in the probe\u2019s solar collectors.<\/em><\/p>\r\n\r\n<\/div>\r\nLet\u2019s choose:\r\n\r\n\r\n\r\nEvent 1: The instant before the collision.\r\n\r\n \r\n\r\nObject: Space Probe\r\n\r\n \r\n\r\nP1 = (250)(5000)\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202749\/Screen-Shot-2017-07-21-at-1.27.24-PM-300x55.png\") \r\n\r\nObject: Debris\r\n\r\n \r\n\r\nP1 = 0\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202751\/Screen-Shot-2017-07-21-at-1.27.30-PM-300x55.png\") <\/td>\r\n | Event 2: The instant the debris and probe reach a common velocity.\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nP2 = 250 v2\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nP2 = 60 v2\r\n\r\n \r\n\r\n \r\n\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\nThe free-body diagrams for the two objects during this time interval are shown below.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21202924\/Screen-Shot-2017-07-21-at-1.29.00-PM-300x182.png\") \r\n\r\nApplying the impulse-momentum relation to each object separately yields:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21203037\/Screen-Shot-2017-07-21-at-1.30.15-PM-300x81.png\") \r\n\r\nNotice that the final velocities of the two objects are the same, because they remain joined together following the collision. Also, the Dt\u2019s are the same because the time interval over which the force of the debris acts on the probe must be the same as the time interval over which the force of the probe acts on the debris. In fact, these two forces must be equal to each other in magnitude by Newton\u2019s Third Law.\r\n\r\nThus, the impulses must cancel if the two equations are added together:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21203132\/Screen-Shot-2017-07-21-at-1.31.14-PM-300x145.png\") \r\n\r\nThe probe slows to a speed of 4032 m\/s (and the debris changes direction and accelerates to a speed of 4032 m\/s) via the collision. Thus, even though we do not know the magnitude of the force involved, or the duration of the collision, we can calculate the final velocities of the two objects colliding. This is because the forces involved comprise an interaction, and by Newton\u2019s Third Law forces that comprise an interaction are always equal in magnitude and opposite in direction.\r\n\r\nIn fact, in problems involving collisions (or explosions, which to physicists are merely collisions played backward in time!), you should almost always apply the impulse-momentum relation to the interacting objects because the forces involved comprise an interaction. Thus, by adding your equations together, these terms will always add to zero. This will often allow you to determine the final velocities of the colliding objects.\r\n\r\nIn conclusion, I should point out that the probe loses momentum during the collision and that the debris gains the exact same amount of momentum. (Check the numbers to verify this statement.) The momentum is transferred from the probe to the debris through the action of the impulse the probe and debris exert on each other. The momentum transfer from the probe to the debris is analogous to throwing a wooden block from the playroom into the yard: The playroom now has one less block and the yard has one more!\r\n\r\n \r\n\r\n \r\nApplying the Work-Energy Relation to the Same Collision<\/h3>\r\nLet\u2019s return to the collision scenario discussed above and attempt to investigate it using work-energy.\r\n\r\n Far from the earth, a 250 kg space probe, moving at 5 km\/s, collides head-on with a 60 kg piece of space debris initially at rest. The debris becomes entangled in the probe\u2019s solar collectors.<\/em><\/p>\r\n\r\n<\/div>\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21203310\/Screen-Shot-2017-07-21-at-1.32.36-PM-300x177.png\") \u00a0<\/em>\r\n\r\nApplying the work-energy relation to each object separately yields:\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21203421\/Screen-Shot-2017-07-21-at-1.33.44-PM-300x58.png\") \r\nThe final velocities of the two objects are the same, because they remain joined together following the collision, and the two forces are the same by Newton\u2019s Third Law. However, these two equations cannot be added together and solved because the two distances over which the forces act, <\/strong>Drprobe and <\/strong>Drdebris, are not necessarily equal.<\/strong> During the collision, the center of the probe will move a different distance that the center of the debris[footnote]If the two objects were actually particles, rather than being approximated as particles, then the two distances would have to be the same and the two works would cancel when the equation were added together. [\/footnote]. Since these two distances are different, the works will not<\/em> cancel as the impulses did, and the equations are not <\/em>solvable!\r\n\r\n \r\n\r\nIn fact, since we know v2 = 4032 m\/s from our momentum analysis,\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21203625\/Screen-Shot-2017-07-21-at-1.36.01-PM-300x67.png\") \r\nObviously, the two works do not cancel. In fact, the internal work<\/em>, or work done by the objects on each other, totals -0.61 x 109 J. This means that there is 0.61 x 109 J less<\/em> kinetic energy in the system of the probe and the debris after the collision than before the collision. This is sometimes referred to as the energy lost in the collision, although the energy is not lost but rather converted into other forms of energy (i.e., other hiding places for the wooden blocks that have yet to be discussed), such as thermal energy.\r\n\r\nIn short, the work-energy relation (as it now stands) cannot be used to effectively analyze collisions unless additional information regarding the internal energy is available. Occasionally, an approximation is made in which the total internal work is zero. When this approximation is made, the collision is referred to as an elastic<\/em> collision. Realistic collisions, in which the total internal energy is not zero and kinetic energy is \u201clost\u201d, are referred to as inelastic<\/em> collisions.\r\n\r\n \r\n\r\n \r\nActivities<\/h2>\r\nBelow are momentum vs. time graphs for six different objects.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210217\/aa-300x206.png\") \r\n\r\na. Rank these graphs on the basis of the change in momentum of the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\nb. Rank these graphs on the basis of the total impulse on the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nBelow are kinetic energy vs. time graphs for six different objects. All of the objects move horizontally.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210224\/bb-300x196.png\") \r\n\r\nRank these graphs on the basis of the change in kinetic energy of the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\nRank these graphs on the basis of the total work on the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nBelow are velocity vs. time graphs for six equal-mass objects.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210226\/cc-300x209.png\") \r\n\r\nRank these graphs on the basis of the change in momentum of the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nRank these graphs on the basis of the change in kinetic energy of the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nBelow are velocity vs. time graphs for six equal-mass objects. All of the objects move horizontally.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210229\/dd-300x212.png\") \r\n\r\nRank these graphs on the basis of the total impulse on the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\nRank these graphs on the basis of the total work on the object over the time interval indicated.\r\n\r\nLargest Positive\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Largest Negative\r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nBelow are six automobiles initially traveling at the indicated velocity. The automobiles have different masses and velocities.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210231\/ee-300x135.png\") \r\n\r\nAll automobiles will be stopped in the same amount of time. Rank these automobiles on the basis of the magnitude of the force needed to stop them.\r\n\r\n \r\n\r\nLargest\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest\r\n\r\n \r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\n \r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nAll automobiles will be stopped in the same amount of distance. Rank these automobiles on the basis of the magnitude of the force needed to stop them.\r\n\r\n \r\n\r\nLargest\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest\r\n\r\n \r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\n \r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nBelow are six automobiles initially traveling at the indicated velocity. The automobiles have different masses and velocities.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210233\/ff-300x129.png\") \r\nRank these automobiles on the basis of the magnitude of the force needed to stop them.\r\n\r\nLargest\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest\r\n\r\n \r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\n \r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nRank these automobiles on the basis of the magnitude of the work needed to stop them.\r\n\r\n \r\n\r\nLargest\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest\r\n\r\n \r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\n \r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nRank these automobiles on the basis of the magnitude of the impulse needed to stop them.\r\n\r\n \r\n\r\nLargest\u00a0 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest\r\n\r\n \r\n\r\n_____ The ranking cannot be determined based on the information provided.\r\n\r\n \r\n\r\nExplain the reason for your ranking:\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nFor each of the collisions illustrated below, sketch a graph of the momentum and velocity of asteroid A, the momentum and velocity of asteroid B, and the total momentum in the system of the two asteroids. Begin your graph before the collision takes place and continue it after the collision is over. Use a consistent scale on all graphs.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210236\/gg-300x142.png\") \r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210239\/hh-300x136.png\") \r\n\r\nFor each of the collisions illustrated below, sketch a graph of the momentum and velocity of asteroid A, the momentum and velocity of asteroid B, and the total momentum in the system of the two asteroids. Begin your graph before the collision takes place and continue it after the collision is over. Use a consistent scale on all graphs.\r\n\r\na. The two asteroids remain joined together after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210242\/ii-300x156.png\") b.\u00a0The two asteroids remain joined together after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210246\/kk-300x157.png\") \r\n\r\nFor each of the collisions illustrated below, sketch a graph of the momentum and velocity of asteroid A, the momentum and velocity of asteroid B, and the total momentum in the system of the two asteroids. Begin your graph before the collision takes place and continue it after the collision is over. Use a consistent scale on all graphs.\r\n\r\na.\u00a0The two asteroids remain joined together after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210248\/ll-300x163.png\") \r\n\r\nb.\u00a0The two asteroids remain joined together after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210251\/mm-300x159.png\") \r\n\r\nFor each of the collisions illustrated below, sketch a graph of the momentum and velocity of asteroid A, the momentum and velocity of asteroid B, and the total momentum in the system of the two asteroids. Begin your graph before the collision takes place and continue it after the collision is over. Use a consistent scale on all graphs.\r\n\r\na. Asteroid A rebounds at 5 m\/s after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210253\/nn-300x159.png\") \r\n\r\nb. Asteroid B moves at 20 m\/s after the collision.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210256\/oo-300x187.png\") \r\n\r\nFor each of the explosions illustrated below, sketch a graph of the momentum and velocity of fragment A, the momentum and velocity of fragment B, and the total momentum in the system of the two fragments. Begin your graph before the explosion takes place and continue it while the fragments travel away from the sight of the explosion. Use a consistent coordinate system and scale on all graphs.\r\n\r\na. The exploding egg is initially at rest.\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210259\/pp-300x151.png\") \r\n\r\nb. The exploding egg is initially at rest.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210303\/rr-300x156.png\") \r\n\r\n \r\n\r\nA 200 kg astronaut is initially at rest on the extreme edge of a 1000 kg space platform. She wears special magnetic shoes that allow her to walk along the metal platform. For each of the situations illustrated below, sketch a graph of the momentum and velocity of the astronaut, the momentum and velocity of the platform, and the total momentum in the system of the two objects. Begin your graph before the astronaut begins to walk and continue it while she walks along the platform. Use a consistent coordinate system and scale on all graphs.\r\n\r\na. The astronaut and platform are initially at rest.\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210308\/tt-300x157.png\") \r\n\r\nb. The astronaut and platform are initially drifting to the right.\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210305\/ss-300x155.png\") \r\n\r\nFor each of the scenarios described below, indicate the amount of kinetic energy and gravitational potential energy of the object at each of the events listed. Use a consistent scale throughout each motion. Set the lowest point of the motion as the zero-point of gravitational potential energy\r\n\r\n \r\n\r\na. A 4000 kg rocket\u2019s engine produces a thrust of 70,000 N for 15 s. The rocket is fired vertically upward.<\/em>\r\n<\/div>\r\n ![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210311\/uu-300x161.png\") \r\n\r\n \r\n\r\n \r\n\r\n b. To practice falling, a 55 kg pole-vaulter falls off of a wall 6.0 m above a 2.0 m thick foam cushion resting on the ground. However, he misses the cushion. The pole-vaulter sinks about 0.10 m into the ground before stopping.<\/em>\r\n\r\n \r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210315\/vv-300x155.png\") <\/div>\r\n \r\n\r\n \r\n\r\nFor each of the scenarios described below, indicate the amount of kinetic energy and gravitational potential energy of each object at each of the events listed. Use a consistent scale throughout each motion. Set the lowest point of the motion as the zero-point of gravitational potential energy\r\n\r\n \r\n\r\n a. Tired of walking up the stairs, an 80 kg engineering student designs an ingenious device for reaching his third floor dorm room. An 84 kg block is attached to a rope that passes over a pulley.<\/em> The student holds the other end of the rope. When the 84 kg block is released, the student is pulled up to his dorm room, 8.0 m off the ground.<\/em>\r\n\r\n![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210317\/ww-300x203.png\") \r\n<\/div>\r\n \r\n\r\n \r\n\r\n b. Tired of walking down the stairs, a 75 kg engineering student designs an ingenious device for reaching the ground from his third floor dorm room. A 60 kg block at rest on the ground is attached to a rope that passes over a pulley.<\/em> The student grabs the other end of the rope and steps out of his window. <\/em>\r\n\r\n ![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2146\/2017\/06\/21210319\/xx-300x222.png\") <\/em>\r\n\r\n \r\n\r\nFor the following problems,\r\n\r\n \r\n\r\nA 100 kg man concerned about his weight decides to weigh himself in an elevator. He stands on a bathroom scale in an elevator that is moving upward at 3.0 m\/s. As the elevator reaches his floor, it slows to a stop.<\/em>\r\n\r\n \r\n\r\na. If the elevator slows to a stop over a distance of 2.0 m, what is the reading on the bathroom scale?<\/em>\r\n\r\nb. If the elevator slows to a stop in 1.5 s, what is the reading on the bathroom scale?<\/em>\r\n\r\n \r\n\r\n\r\n\r\nMotion Diagram<\/strong>\r\n\r\n\u00a0<\/strong><\/td>\r\na. Motion Information<\/strong>\r\n\r\nEvent 1:\r\n\r\n \r\n\r\nKE1 =\r\n\r\nGE1 =\r\n\r\nW12 =<\/td>\r\n| \r\n\r\nEvent 2:\r\n\r\n \r\n\r\nKE2 =\r\n\r\nGE2 =\r\n\r\n <\/td>\r\n | Mathematical Analysis[i]<\/a><\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n\r\n\r\nFree-Body Diagram<\/strong>\r\n\r\n\u00a0<\/strong>\r\n\r\n\u00a0<\/strong><\/td>\r\nb. Motion Information<\/strong>\r\n\r\nEvent 1:\r\n\r\n \r\n\r\nP1 =\r\n\r\nJ12 =<\/td>\r\n| \r\n\r\nEvent 2:\r\n\r\n \r\n\r\nP2 =\r\n\r\n <\/td>\r\n | Mathematical Analysis<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\u00a0<\/strong>\r\n\r\n \r\n\r\n\u00a0<\/em>\r\n\r\nA 70 kg student is 120 m above the ground, moving upward at 3.5 m\/s, while hanging from a rope hanging from a 280 kg helium balloon. The lift on the balloon due to the buoyant force is 3000 N. <\/em>\r\n\r\n \r\n\r\na. With what speed does the student hit the ground?<\/em>\r\n\r\nb. How long does it take the student to reach the ground?<\/em>\r\n\r\n \r\n\r\n\r\n\r\nMotion Diagram<\/strong>\r\n\r\n\u00a0<\/strong><\/td>\r\na. Motion Information<\/strong>\r\n\r\nEvent 1:\r\n\r\n \r\n\r\nKE1 =\r\n\r\nGE1 =\r\n\r\nW12 =<\/td>\r\n| \r\n\r\nEvent 2:\r\n\r\n \r\n\r\nKE2 =\r\n\r\nGE2 =\r\n\r\n <\/td>\r\n | Mathematical Analysis[ii]<\/a><\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n\r\n\r\nFree-Body Diagram<\/strong>\r\n\r\n\u00a0<\/strong>\r\n\r\nstudent & balloon<\/em>\r\n\r\n\u00a0<\/em><\/td>\r\nb. Motion Information<\/strong>\r\n\r\nEvent 1:\r\n\r\n \r\n\r\nP1 =\r\n\r\nJ12 =<\/td>\r\n| \r\n\r\nEvent 2:\r\n\r\n \r\n\r\nP2 =\r\n\r\n <\/td>\r\n | Mathematical Analysis<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\u00a0<\/strong>\r\n\r\n \r\n\r\nA 4000 kg rocket\u2019s engine produces a thrust of 70,000 N for 15 s. The rocket is fired vertically upward.<\/em>\r\n\r\n\u00a0<\/strong>\r\n\r\na. What is the speed of the rocket when its engine turns off?<\/em>\r\n\r\nb. How long does it take the rocket to reach its maximum height?<\/em>\r\n\r\n \r\n\r\n\r\n\r\nMotion Diagram<\/strong>\r\n\r\n\u00a0<\/strong><\/td>\r\na. Motion Information<\/strong>\r\n\r\nEvent 1:\r\n\r\n \r\n\r\nP1 =\r\n\r\nJ12 =<\/td>\r\n| \r\n\r\nEvent 2:\r\n\r\n \r\n\r\nP2 =\r\n\r\n <\/td>\r\n | Mathematical Analysis[iii]<\/a><\/strong>\r\n\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n \r\n\r\n \r\n\r\n\r\n\r\n| Free-Body Diagrams<\/strong>\r\n\r\n\u00a0<\/strong>\r\n\r\nwhile engine fires<\/em>\r\n\r\n \r\n\r\n | | | | | | | | | | | | | | | | | | | | | | | | | | | | |