Faculty of Science Technology and Engineering Department of Physics Senior Laboratory Current balance Objectives When a steady electric current flows perpendicularly across a uniform magnetic field it experiences a force. This experiment aims to investigate this effect, and to determine the direction of the force relative to the current and magnetic field. You will design and perform a series of experiments to show how the magnitude of the force depends upon the current and the length of the conductor that is in the field. Task You are provided with a current balance apparatus (Figure 1), power supply and a magnet. This current balance consists of five loops of conducting wire supported on a pivoted aluminium frame. Current may be made to flow in one or up to five of the loops at a time in either direction. If the end of the loop is situated in a perpendicular magnetic field, when the current is switched on, the magnetic force on the current will unbalance the apparatus. By moving the sliding weights to rebalance it, the magnitude of this magnetic force may be measured. A scale is etched on one arm of the balance, so that the distance moved by the slider can be measured. The circuitry of the balance cannot cope currents greater than 5 Amps, so please do not exceed this level of current. Figure 1: Schematic diagram of current balance apparatus and circuitry. Start by familiarising yourself with the apparatus. Use the two sliding weights to balance the apparatus, then apply a magnetic field to either end of the loop. Pass a current through just one of the conducting loops and observe the direction of the resulting magnetic force, relative to the direction of the current and the applied field. Change the magnitude and direction of the current, observe qualitatively the effect this has on the magnetic force. Having familiarised yourself with the apparatus, you should design and perform a series of quantitative experiments aiming to: (1) determine how the size of the magnetic force is dependant on the size of the current flowing in the conductor. (2) determine how the size of the force is dependant on the length of the conductor which is in the field. (3) measure the value (in Tesla) of the field of the magnet provided. For each of these, the balance should be set up with the magnet positioned at the end of the arm that has the distance scale, and orientated so that the magnetic force will be directed upwards when a current is passed through the conductor. The sliding weight on this arm should be positioned at the zero-mark. The weight on the opposite arm should be adjusted to balance the apparatus in the absence of a current. When a current is applied, you should re-balance the apparatus by moving the weight on the scaled arm outwards, while keeping the opposite weight fixed in position. The distance moved by the weight is directly proportional to the force applied by the magnetic field to the end of the balance. In your report, make sure you discuss why this is the case. Use the position of the sliding weight to quantify the magnetic force as a function of the current applied to the conductor, and of the number of conducting loops through which the current flows. For tasks (1) and (2) you can use the position of the sliding weight as a measure of the force. Look up the relationship that relates the force to the applied field, current and length of conductor in the field. Is this consistent with your data? To complete task (3) you need to determine the magnitude (in Newtons) of the magnetic force from the measurement of the position of the sliding weight. To do this, what other information do you need to know? When you have determined a value for the field, you can measure the field directly using the laboratory’s Gaussmeter for comparison.

Faculty of Science Technology and Engineering Department of Physics Senior Laboratory Current balance Objectives When a steady electric current flows perpendicularly across a uniform magnetic field it experiences a force. This experiment aims to investigate this effect, and to determine the direction of the force relative to the current and magnetic field. You will design and perform a series of experiments to show how the magnitude of the force depends upon the current and the length of the conductor that is in the field. Task You are provided with a current balance apparatus (Figure 1), power supply and a magnet. This current balance consists of five loops of conducting wire supported on a pivoted aluminium frame. Current may be made to flow in one or up to five of the loops at a time in either direction. If the end of the loop is situated in a perpendicular magnetic field, when the current is switched on, the magnetic force on the current will unbalance the apparatus. By moving the sliding weights to rebalance it, the magnitude of this magnetic force may be measured. A scale is etched on one arm of the balance, so that the distance moved by the slider can be measured. The circuitry of the balance cannot cope currents greater than 5 Amps, so please do not exceed this level of current. Figure 1: Schematic diagram of current balance apparatus and circuitry. Start by familiarising yourself with the apparatus. Use the two sliding weights to balance the apparatus, then apply a magnetic field to either end of the loop. Pass a current through just one of the conducting loops and observe the direction of the resulting magnetic force, relative to the direction of the current and the applied field. Change the magnitude and direction of the current, observe qualitatively the effect this has on the magnetic force. Having familiarised yourself with the apparatus, you should design and perform a series of quantitative experiments aiming to: (1) determine how the size of the magnetic force is dependant on the size of the current flowing in the conductor. (2) determine how the size of the force is dependant on the length of the conductor which is in the field. (3) measure the value (in Tesla) of the field of the magnet provided. For each of these, the balance should be set up with the magnet positioned at the end of the arm that has the distance scale, and orientated so that the magnetic force will be directed upwards when a current is passed through the conductor. The sliding weight on this arm should be positioned at the zero-mark. The weight on the opposite arm should be adjusted to balance the apparatus in the absence of a current. When a current is applied, you should re-balance the apparatus by moving the weight on the scaled arm outwards, while keeping the opposite weight fixed in position. The distance moved by the weight is directly proportional to the force applied by the magnetic field to the end of the balance. In your report, make sure you discuss why this is the case. Use the position of the sliding weight to quantify the magnetic force as a function of the current applied to the conductor, and of the number of conducting loops through which the current flows. For tasks (1) and (2) you can use the position of the sliding weight as a measure of the force. Look up the relationship that relates the force to the applied field, current and length of conductor in the field. Is this consistent with your data? To complete task (3) you need to determine the magnitude (in Newtons) of the magnetic force from the measurement of the position of the sliding weight. To do this, what other information do you need to know? When you have determined a value for the field, you can measure the field directly using the laboratory’s Gaussmeter for comparison.

Abstract   The present experiment aims to investigate the effect … Read More...
http://www.constitution.org/mac/prince17.htm How does Machiavelli feel about cruelty versus clemency? A. Machiavelli equates clemency with being loved and cruelty with being despised, and suggests that being despised is acceptable. B. Machiavelli suggests that cruelty doesn’t always result in being despised and winning the love of your subjects is the most important thing. C. Machiavelli says that cruelty, when applied in a prudent manner, will be held in more esteem than too much mercy. D. Cruelty and clemency are identical; being merciful to one person means that you must be cruel to another. E. Clemency is equated with happiness, and a happy set of subjects is the ultimate goal of a successful leader. What is the difference between hatred and fear? A. Fear makes people respect you. Hatred makes them work against you. B. Fear and hatred follow one another. If you create fear you will eventually create hatred. All leaders should avoid this. C. Hatred from external powers breeds nationalism within your country, causing people to fear external powers. D. Fear and hatred are opposites. E. Hatred follows love; fear follows clemency. http://www.constitution.org/mac/prince23.htm According to Machiavelli, what is a flatterer? A. Someone who wants to lavish gifts upon you in exchange for power. B. An external power that wants to ally with you. C. Someone who will tell you what you think rather than giving their own opinion. D. Someone who tests your opinions against their own to make a good argument. E. An external power that wants to make strong trade alliances to weaken you over time. According to Machiavelli, what is the best way to seek truth from advisers? A. Advisors should present any complaints to you as a group. B. Advisors should be called upon when the leader has a question, otherwise they are to remain silent. C. An advisor should involve the public, allowing them to call the leader to court to listen to their opinions. D. The leader should listen carefully to one private advisor with whom he always disagrees. E. Machiavelli thinks that advisors are not helpful because they will always try to flatter their leader.

http://www.constitution.org/mac/prince17.htm How does Machiavelli feel about cruelty versus clemency? A. Machiavelli equates clemency with being loved and cruelty with being despised, and suggests that being despised is acceptable. B. Machiavelli suggests that cruelty doesn’t always result in being despised and winning the love of your subjects is the most important thing. C. Machiavelli says that cruelty, when applied in a prudent manner, will be held in more esteem than too much mercy. D. Cruelty and clemency are identical; being merciful to one person means that you must be cruel to another. E. Clemency is equated with happiness, and a happy set of subjects is the ultimate goal of a successful leader. What is the difference between hatred and fear? A. Fear makes people respect you. Hatred makes them work against you. B. Fear and hatred follow one another. If you create fear you will eventually create hatred. All leaders should avoid this. C. Hatred from external powers breeds nationalism within your country, causing people to fear external powers. D. Fear and hatred are opposites. E. Hatred follows love; fear follows clemency. http://www.constitution.org/mac/prince23.htm According to Machiavelli, what is a flatterer? A. Someone who wants to lavish gifts upon you in exchange for power. B. An external power that wants to ally with you. C. Someone who will tell you what you think rather than giving their own opinion. D. Someone who tests your opinions against their own to make a good argument. E. An external power that wants to make strong trade alliances to weaken you over time. According to Machiavelli, what is the best way to seek truth from advisers? A. Advisors should present any complaints to you as a group. B. Advisors should be called upon when the leader has a question, otherwise they are to remain silent. C. An advisor should involve the public, allowing them to call the leader to court to listen to their opinions. D. The leader should listen carefully to one private advisor with whom he always disagrees. E. Machiavelli thinks that advisors are not helpful because they will always try to flatter their leader.

http://www.constitution.org/mac/prince17.htm   How does Machiavelli feel about cruelty versus clemency? … Read More...
Use each link provided below to answer the questions below. http://www.constitution.org/mac/prince19.htm That One Should Avoid Being Despised and Hated What are the two threats a leader should fear? A. Terrorism and war B. Threats from within (the subjects) and from without (external powers) C. A weak economy and the overall health of subjects D. Govermental stability and economic prosperity E. Conspirators and the military When should a leader fear conspiracy? A. Leaders should always be worried about conspiracies. B. Leaders who do not lavish gifts upon their subjects should fear conspiracies. C. Leaders who are not held in high esteem by their subjects should fear conspiracies. D. Leaders never need to fear conspiracy if they have a strong military. E. Leaders who have large militaries should always be watchful of military conspiracy.

Use each link provided below to answer the questions below. http://www.constitution.org/mac/prince19.htm That One Should Avoid Being Despised and Hated What are the two threats a leader should fear? A. Terrorism and war B. Threats from within (the subjects) and from without (external powers) C. A weak economy and the overall health of subjects D. Govermental stability and economic prosperity E. Conspirators and the military When should a leader fear conspiracy? A. Leaders should always be worried about conspiracies. B. Leaders who do not lavish gifts upon their subjects should fear conspiracies. C. Leaders who are not held in high esteem by their subjects should fear conspiracies. D. Leaders never need to fear conspiracy if they have a strong military. E. Leaders who have large militaries should always be watchful of military conspiracy.

Use each link provided below to answer the questions below. … Read More...
Individual case study Due date: 1:00pm AEST, Thursday, Week 11 All students are to submit electronically – max file size is 2Mb. ASSESSMENT Weighting: 35% Length: No set length 2 I…Assignment 2 SPECIFICATIONS CIS8011_Digital Innovation Assignment 2 (30%) (1500 words maximum) This assignment continues from the first assignment and your task is to write a report on the following a…1 CSE2DES/CSE5DES – Assignment 1 Due Date: 10 am Monday 22nd September 2014 Assessment: This assignment 1 is worth 15% of the final mark for CSE2DES/CSE5DES. This is an individual assignment. Copying,…All questions are from the textbook: Fatseas, Victor & Williams, John, Cost Management (2013) 3rd edition, McGraw HillMLC 703: PRINCIPLES OF INCOME TAX LAW INSTRUCTIONS Please note that the following will not form part of the word count: ? References, including statute and cases; ? Diagrams; ? Tables; ? Calculations….WRITTEN ESSAY Outline This assessment has been written to develop your understanding of Human Resource Management, assessing learning outcomes a, b, c, h and i: “The external environmental (e.g. econo…Subject: INTERNATIONAL MARKETING B01ITMK208 Assessment item 2: International Marketing Analysis Weighting: 30% Due: Week 10. A daily penalty of 5% will be applied to late assignments. Task: You are a …B01ITMK208 INTERNATIONAL MANAGEMENT ASSIGNMENT INSTRUCTIONS KEY INFORMATION Maximum Length: 2500 words Due: Week 8. Note that late submission will attract a penalty. Weighting: 30% Instructions: Read …Subject: Advertising Management BO1ADMG207 Assessment item 2: IMC Report Weighting: 30% Due: Week 8. A daily penalty of 5% will be applied to late assignments. Task: You are the Australian-based Marke…Attached are two Memos, please have a lookgetEconomics topic Assignment 2 Value: 40% Due date: 01-Sep-2014 Return date: 22-Sep-2014 Length: about 1500-2000 words each Submission method options Alternative submission method Task Analytical essays…Accounting for Business Decisions –HI5001 Trimester 2 2014 The assignment allows students to exhibit their knowledge and understanding of the subject matter of Accounting. The students will use the sk…HOLMES INSTITUTE FACULTY OF HIGHER EDUCATION HI6007 SPSS Assignment 02 Due Friday 4pm week 11 WORTH 30% The data set you need to do the assignment can be found on Blackboard in the folder “Assignments…Assignmnet of Economic Assignment (Written report): 25% 1. Organize yourselves into groups. Each group is to have Four or Five members. 2. Groups need to choose a topic from the list of topics provide…2. Rio Tinto Annual Report Financial Analysis [10 marks] Consider the sources below and answer the following questions. Source 1: Rio Tinto Annual Report 2012 (see report uploaded on the portal) Sourc…Quantitative Methods for Business Business Statistics Assignment – Semester, 2 2014 Total Marks: 60, Worth: 20% of final assessment This assignment requires a considerable amount of computer work and …BUACC 2613 Management Accounting 1 Semester 2, 2014 Assignment Contribution to overall assessment: 25% Due date: 26/09/2014 • This assignment has two parts: o Part 1

Individual case study Due date: 1:00pm AEST, Thursday, Week 11 All students are to submit electronically – max file size is 2Mb. ASSESSMENT Weighting: 35% Length: No set length 2 I…Assignment 2 SPECIFICATIONS CIS8011_Digital Innovation Assignment 2 (30%) (1500 words maximum) This assignment continues from the first assignment and your task is to write a report on the following a…1 CSE2DES/CSE5DES – Assignment 1 Due Date: 10 am Monday 22nd September 2014 Assessment: This assignment 1 is worth 15% of the final mark for CSE2DES/CSE5DES. This is an individual assignment. Copying,…All questions are from the textbook: Fatseas, Victor & Williams, John, Cost Management (2013) 3rd edition, McGraw HillMLC 703: PRINCIPLES OF INCOME TAX LAW INSTRUCTIONS Please note that the following will not form part of the word count: ? References, including statute and cases; ? Diagrams; ? Tables; ? Calculations….WRITTEN ESSAY Outline This assessment has been written to develop your understanding of Human Resource Management, assessing learning outcomes a, b, c, h and i: “The external environmental (e.g. econo…Subject: INTERNATIONAL MARKETING B01ITMK208 Assessment item 2: International Marketing Analysis Weighting: 30% Due: Week 10. A daily penalty of 5% will be applied to late assignments. Task: You are a …B01ITMK208 INTERNATIONAL MANAGEMENT ASSIGNMENT INSTRUCTIONS KEY INFORMATION Maximum Length: 2500 words Due: Week 8. Note that late submission will attract a penalty. Weighting: 30% Instructions: Read …Subject: Advertising Management BO1ADMG207 Assessment item 2: IMC Report Weighting: 30% Due: Week 8. A daily penalty of 5% will be applied to late assignments. Task: You are the Australian-based Marke…Attached are two Memos, please have a lookgetEconomics topic Assignment 2 Value: 40% Due date: 01-Sep-2014 Return date: 22-Sep-2014 Length: about 1500-2000 words each Submission method options Alternative submission method Task Analytical essays…Accounting for Business Decisions –HI5001 Trimester 2 2014 The assignment allows students to exhibit their knowledge and understanding of the subject matter of Accounting. The students will use the sk…HOLMES INSTITUTE FACULTY OF HIGHER EDUCATION HI6007 SPSS Assignment 02 Due Friday 4pm week 11 WORTH 30% The data set you need to do the assignment can be found on Blackboard in the folder “Assignments…Assignmnet of Economic Assignment (Written report): 25% 1. Organize yourselves into groups. Each group is to have Four or Five members. 2. Groups need to choose a topic from the list of topics provide…2. Rio Tinto Annual Report Financial Analysis [10 marks] Consider the sources below and answer the following questions. Source 1: Rio Tinto Annual Report 2012 (see report uploaded on the portal) Sourc…Quantitative Methods for Business Business Statistics Assignment – Semester, 2 2014 Total Marks: 60, Worth: 20% of final assessment This assignment requires a considerable amount of computer work and …BUACC 2613 Management Accounting 1 Semester 2, 2014 Assignment Contribution to overall assessment: 25% Due date: 26/09/2014 • This assignment has two parts: o Part 1

info@checkyourstudy.com
Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $ p $ p1p2 −−−− ” p1 +p2 $ p $ p1+p2 2 p1 + p2 $ p $ |p1 − p2 | p1 + p2 $ p $ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

please email info@checkyourstudy.com
Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

please email info@checkyourstudy.com
1 | P a g e Lecture #2: Abortion (Warren) While studying this topic, we will ask whether it is morally permissible to intentionally terminate a pregnancy and, if so, whether certain restrictions should be placed upon such practices. Even though we will most often be speaking of terminating a fetus, biologists make further classifications: the zygote is the single cell resulting from the fusion of the egg and the sperm; the morula is the cluster of cells that travels through the fallopian tubes; the blastocyte exists once an outer shell of cells has formed around an inner group of cells; the embryo exists once the cells begin to take on specific functions (around the 15th day); the fetus comes into existence in the 8th week when the embryo gains a basic structural resemblance to the adult. Given these distinctions, there are certain kinds of non-fetal abortion—such as usage of RU-486 (the morning-after “abortion pill”)—though most of the writers we will study refer to fetal abortions. So now let us consider the “Classical Argument against Abortion”, which has been very influential: P1) It is wrong to kill innocent persons. P2) A fetus is an innocent person. C) It is wrong to kill a fetus. (Note that this argument has received various formulations, including those from Warren and Thomson which differ from the above. For this course, we will refer to the above formulation as the “Classical Argument”.) Before evaluating this argument, we should talk about terminology: A person is a member of the moral community; i.e., someone who has rights and/or duties. ‘Persons’ is the plural of ‘person’. ‘Person’ can be contrasted with ‘human being’; a human being is anyone who is genetically human (i.e., a member of Homo sapiens). ‘People’ (or ‘human beings’) is the plural of ‘human being’. Why does this matter? First, not all persons are human beings. For example, consider an alien from another planet who mentally resembled us. If he were to visit Earth, it would be morally reprehensible to kick him or to set him on fire because of the pain and suffering that these acts would cause. And, similarly, the alien would be morally condemnable if he were to propagate such acts on us; he has a moral duty not to act in those ways (again, assuming a certain mental resemblance to us). So, even though this alien is not a human being, he is nevertheless a person with the associative rights and/or duties. 2 | P a g e And, more controversially, maybe not all human beings are persons. For example, anencephalic infants—i.e., ones born without cerebral cortexes and therefore with severely limited cognitive abilities—certainly do not have duties since they are not capable of rational thought and autonomous action. Some philosophers have even argued that they do not have rights. Now let us return to the Classical Argument. It is valid insofar as, if the premises are true, then the conclusion has to be true. But maybe it commits equivocation, which is to say that it uses the same word in multiple senses; equivocation is an informal fallacy (i.e., attaches to arguments that are formally valid but otherwise fallacious). Consider the following: P1) I put my money in the bank. P2) The bank borders the river. C) I put my money somewhere that borders the river. This argument equivocates since ‘bank’ is being used in two different senses: in P1 it is used to represent a financial institution and, in P2, it is used to represent a geological feature. Returning to the classical argument, it could be argued that ‘person’ is being used in two different senses: in P1 it is used in its appropriate moral sense and, in P2, it is inappropriately used instead of ‘human being’. The critic might suggest that a more accurate way to represent the argument would be as follows: P1) It is wrong to kill innocent persons. P2) A fetus is a human being. C) It is wrong to kill a fetus. This argument is obviously invalid. So one way to criticize the Classical Argument is to say that it conflates two different concepts—viz., ‘person’ and ‘human being’—and therefore commits equivocation. However, the more straightforward way to attack the Classical Argument is just to deny its second premise and thus contend that the argument is unsound. This is the approach that Mary Anne Warren takes in “On the Moral and Legal Status of Abortion”. Why does Warren think that the second premise is false? Remember that we defined a person as “a member of the moral community.” And we said that an alien, for example, could be afforded moral status even though it is not a human being. Why do we think that this alien should not be tortured or set on fire? Warren thinks that, intuitively, we think that membership in the moral community is based upon possession of the following traits: 3 | P a g e 1. Consciousness of objects and events external and/or internal to the being and especially the capacity to feel pain; 2. Reasoning or rationality (i.e., the developed capacity to solve new and relatively complex problems); 3. Self-motivated activity (i.e., activity which is relatively independent of either genetic or direct external control); 4. Capacity to communicate (not necessarily verbal or linguistic); and 5. Possession of self-concepts and self-awareness. Warren then admits that, though all of the items on this list look promising, we need not require that a person have all of the items on this list. (4) is perhaps the most expendable: imagine someone who is fully paralyzed as well as deaf, these incapacities, which preclude communication, are not sufficient to justify torture. Similarly, we might be able to imagine certain psychological afflictions that negate (5) without compromising personhood. Warren suspects that (1) and (2) are might be sufficient to confer personhood, and thinks that (1)-(3) “quite probably” are sufficient. Note that, if she is right, we would not be able to torture chimps, let us say, but we could set plants on fire (and most likely ants as well). However, given Warren’s aims, she does not need to specify which of these traits are necessary or sufficient for personhood; all that she wants to observe is that the fetus has none of them! Therefore, regardless of which traits we want to require, Warren thinks that the fetus is not a person. Therefore she thinks that the Classical Argument is unsound and should be rejected. Even if we accept Warren’s refutation of the second premise, we might be inclined to say that, while the fetus is not (now) a person, it is a potential person: the fetus will hopefully mature into a being that possesses all five of the traits on Warren’s list. We might then propose the following adjustment to the Classical Argument: P1) It is wrong to kill all innocent persons. P2) A fetus is a potential person. C) It is wrong to kill a fetus. However, this argument is invalid. Warren grants that potentiality might serve as a prima facie reason (i.e., a reason that has some moral weight but which might be outweighed by other considerations) not to abort a fetus, but potentiality alone is insufficient to grant the fetus a moral right against being terminated. By analogy, consider the following argument: 4 | P a g e P1) The President has the right to declare war. P2) Mary is a potential President. C) Mary has the right to declare war. This argument is invalid since the premises are both true and the conclusion is false. By parity, the following argument is also invalid: P1) A person has a right to life. P2) A fetus is a potential person. C) A fetus has a right to life. Thus Warren thinks that considerations of potentiality are insufficient to undermine her argument that fetuses—which are potential persons but, she thinks, not persons—do not have a right to life.

1 | P a g e Lecture #2: Abortion (Warren) While studying this topic, we will ask whether it is morally permissible to intentionally terminate a pregnancy and, if so, whether certain restrictions should be placed upon such practices. Even though we will most often be speaking of terminating a fetus, biologists make further classifications: the zygote is the single cell resulting from the fusion of the egg and the sperm; the morula is the cluster of cells that travels through the fallopian tubes; the blastocyte exists once an outer shell of cells has formed around an inner group of cells; the embryo exists once the cells begin to take on specific functions (around the 15th day); the fetus comes into existence in the 8th week when the embryo gains a basic structural resemblance to the adult. Given these distinctions, there are certain kinds of non-fetal abortion—such as usage of RU-486 (the morning-after “abortion pill”)—though most of the writers we will study refer to fetal abortions. So now let us consider the “Classical Argument against Abortion”, which has been very influential: P1) It is wrong to kill innocent persons. P2) A fetus is an innocent person. C) It is wrong to kill a fetus. (Note that this argument has received various formulations, including those from Warren and Thomson which differ from the above. For this course, we will refer to the above formulation as the “Classical Argument”.) Before evaluating this argument, we should talk about terminology: A person is a member of the moral community; i.e., someone who has rights and/or duties. ‘Persons’ is the plural of ‘person’. ‘Person’ can be contrasted with ‘human being’; a human being is anyone who is genetically human (i.e., a member of Homo sapiens). ‘People’ (or ‘human beings’) is the plural of ‘human being’. Why does this matter? First, not all persons are human beings. For example, consider an alien from another planet who mentally resembled us. If he were to visit Earth, it would be morally reprehensible to kick him or to set him on fire because of the pain and suffering that these acts would cause. And, similarly, the alien would be morally condemnable if he were to propagate such acts on us; he has a moral duty not to act in those ways (again, assuming a certain mental resemblance to us). So, even though this alien is not a human being, he is nevertheless a person with the associative rights and/or duties. 2 | P a g e And, more controversially, maybe not all human beings are persons. For example, anencephalic infants—i.e., ones born without cerebral cortexes and therefore with severely limited cognitive abilities—certainly do not have duties since they are not capable of rational thought and autonomous action. Some philosophers have even argued that they do not have rights. Now let us return to the Classical Argument. It is valid insofar as, if the premises are true, then the conclusion has to be true. But maybe it commits equivocation, which is to say that it uses the same word in multiple senses; equivocation is an informal fallacy (i.e., attaches to arguments that are formally valid but otherwise fallacious). Consider the following: P1) I put my money in the bank. P2) The bank borders the river. C) I put my money somewhere that borders the river. This argument equivocates since ‘bank’ is being used in two different senses: in P1 it is used to represent a financial institution and, in P2, it is used to represent a geological feature. Returning to the classical argument, it could be argued that ‘person’ is being used in two different senses: in P1 it is used in its appropriate moral sense and, in P2, it is inappropriately used instead of ‘human being’. The critic might suggest that a more accurate way to represent the argument would be as follows: P1) It is wrong to kill innocent persons. P2) A fetus is a human being. C) It is wrong to kill a fetus. This argument is obviously invalid. So one way to criticize the Classical Argument is to say that it conflates two different concepts—viz., ‘person’ and ‘human being’—and therefore commits equivocation. However, the more straightforward way to attack the Classical Argument is just to deny its second premise and thus contend that the argument is unsound. This is the approach that Mary Anne Warren takes in “On the Moral and Legal Status of Abortion”. Why does Warren think that the second premise is false? Remember that we defined a person as “a member of the moral community.” And we said that an alien, for example, could be afforded moral status even though it is not a human being. Why do we think that this alien should not be tortured or set on fire? Warren thinks that, intuitively, we think that membership in the moral community is based upon possession of the following traits: 3 | P a g e 1. Consciousness of objects and events external and/or internal to the being and especially the capacity to feel pain; 2. Reasoning or rationality (i.e., the developed capacity to solve new and relatively complex problems); 3. Self-motivated activity (i.e., activity which is relatively independent of either genetic or direct external control); 4. Capacity to communicate (not necessarily verbal or linguistic); and 5. Possession of self-concepts and self-awareness. Warren then admits that, though all of the items on this list look promising, we need not require that a person have all of the items on this list. (4) is perhaps the most expendable: imagine someone who is fully paralyzed as well as deaf, these incapacities, which preclude communication, are not sufficient to justify torture. Similarly, we might be able to imagine certain psychological afflictions that negate (5) without compromising personhood. Warren suspects that (1) and (2) are might be sufficient to confer personhood, and thinks that (1)-(3) “quite probably” are sufficient. Note that, if she is right, we would not be able to torture chimps, let us say, but we could set plants on fire (and most likely ants as well). However, given Warren’s aims, she does not need to specify which of these traits are necessary or sufficient for personhood; all that she wants to observe is that the fetus has none of them! Therefore, regardless of which traits we want to require, Warren thinks that the fetus is not a person. Therefore she thinks that the Classical Argument is unsound and should be rejected. Even if we accept Warren’s refutation of the second premise, we might be inclined to say that, while the fetus is not (now) a person, it is a potential person: the fetus will hopefully mature into a being that possesses all five of the traits on Warren’s list. We might then propose the following adjustment to the Classical Argument: P1) It is wrong to kill all innocent persons. P2) A fetus is a potential person. C) It is wrong to kill a fetus. However, this argument is invalid. Warren grants that potentiality might serve as a prima facie reason (i.e., a reason that has some moral weight but which might be outweighed by other considerations) not to abort a fetus, but potentiality alone is insufficient to grant the fetus a moral right against being terminated. By analogy, consider the following argument: 4 | P a g e P1) The President has the right to declare war. P2) Mary is a potential President. C) Mary has the right to declare war. This argument is invalid since the premises are both true and the conclusion is false. By parity, the following argument is also invalid: P1) A person has a right to life. P2) A fetus is a potential person. C) A fetus has a right to life. Thus Warren thinks that considerations of potentiality are insufficient to undermine her argument that fetuses—which are potential persons but, she thinks, not persons—do not have a right to life.

Assignment 4 Due: 11:59pm on Wednesday, February 26, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy ± Two Forces Acting at a Point Two forces, and , act at a point. has a magnitude of 9.80 and is directed at an angle of 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.20 and is directed at an angle of 54.1 below the negative x axis in the third quadrant. Part A What is the x component of the resultant force? Express your answer in newtons. Hint 1. How to approach the problem The resultant force is defined as the vector sum of all forces. Thus, its x component is the sum of the x components of the forces, and its y component is the sum of the y components of the forces. Hint 2. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . F 1 F  2 F  1 N  F  2 N  F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 Hint 2. Find the direction of is directed at an angle of 56.0 above the x axis in the second quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 56.0 . ANSWER: ANSWER: Hint 3. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , F 1 F 1  F  1 F  1  =   180 −  180 +  90 +  -5.48 N F 2 F  2 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Typesetting math: 100% and if . Hint 2. Find the direction of is directed at an angle of 54.1 below the x axis in the third quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 54.1 . ANSWER: ANSWER: ANSWER: Correct Part B What is the y component of the resultant force? Express your answer in newtons. Ax < 0 Ay <  <  < 3 2 F 2 F 2  F 2 F  2  =   180 −   − 180 −90 −  -3.05 N -8.53 N Typesetting math: 100% Hint 1. How to approach the problem Follow the same procedure that you used in Part A to find the x component of the resultant force, though now calculate the y components of the two forces. Hint 2. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: Hint 3. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 8.12 N F 2 F  2 Typesetting math: 100% Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: ANSWER: Correct Part C What is the magnitude of the resultant force? Express your answer in newtons. Hint 1. Magnitude of a vector Consider a vector , whose components are and . The magnitude of is . A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 -4.21 N 3.91 N A Ax Ay A A = A + 2 x A2 y −−−−−−−  Typesetting math: 100% ANSWER: Correct Enhanced EOC: Problem 5.9 The figure shows acceleration-versus-force graphs for two objects pulled by rubber bands. You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Finding the Slope of a Line from a Graph Part A What is the mass ratio ? Express your answer using two significant figures. 9.38 N m1 m2 Typesetting math: 100% Hint 1. How to approach the problem How are the acceleration and the force on an object related to its mass? How is the slope of each line in the figure related to each object's mass? For each line, what two points are easy to measure accurately to determine the slope of line? How is the slope determined from the x and y coordinates of the two points you chose for each line? ANSWER: Correct A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude . Part A How much horizontal force must a sprinter of mass 54 exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Hint 1. Newton's 2nd law of motion According to Newton's 2nd law of motion, if a net external force acts on a body, the body accelerates, and the net force is equal to the mass of the body times the acceleration of the body: . ANSWER: = 0.36 m1 m2 15 m/s2 F kg Fnet m a Fnet = ma F = 810 N Typesetting math: 100% Correct Part B Which body exerts the force that propels the sprinter, the blocks or the sprinter? Hint 1. How to approach the question To start moving forward, sprinters push backward on the starting blocks with their feet. Newton's 3rd law tells you that the blocks exert a force on the sprinter of the same magnitude, but opposite in direction. ANSWER: Correct To start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward. Problem 5.12 The figure shows an acceleration-versus-force graph for a 600 object. the blocks the sprinter g Typesetting math: 100% Part A What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Part B What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Free-Body Diagrams Learning Goal: To gain practice drawing free-body diagrams Whenever you face a problem involving forces, always start with a free-body diagram. a1 a1 = 1.67 m s2 a2 a2 = 3.33 m s2 Typesetting math: 100% To draw a free-body diagram use the following steps: Isolate the object of interest. It is customary to represent the object of interest as a point 1. in your diagram. Identify all the forces acting on the object and their directions. Do not include forces acting on other objects in the problem. Also, do not include quantities, such as velocities and accelerations, that are not forces. 2. Draw the vectors for each force acting on your object of interest. When possible, the length of the force vectors you draw should represent the relative magnitudes of the forces acting on the object. 3. In most problems, after you have drawn the free-body diagrams, you will explicitly label your coordinate axes and directions. Always make the object of interest the origin of your coordinate system. Then you will need to divide the forces into x and y components, sum the x and y forces, and apply Newton's first or second law. In this problem you will only draw the free-body diagram. Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor (see the figure). The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? To solve this problem you should start by drawing a free-body diagram. Part A Determine the object of interest for the situation described in the problem introduction. Hint 1. How to approach the problem You should first think about the question you are trying to answer: What is the acceleration of the piano? The object of interest in this situation will be the object whose acceleration you are asked to find. ANSWER: Typesetting math: 100% Correct Part B Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano. Check all that apply. ANSWER: Correct Now that you have identified the forces acting on the piano, you should draw the free-body diagram. Draw the length of your vectors to represent the relative magnitudes of the forces, but you don't need to worry about the exact scale. You won't have the exact value of all of the forces until you finish solving the problem. To maximize your learning, you should draw the diagram yourself before looking at the choices in the next part. You are on your honor to do so. Part C For this situation you should draw a free-body diagram for the floor. Chadwick. the piano. acceleration of the piano gravitational force acting on the piano (piano's weight) speed of the piano gravitational force acting on Chadwick (Chadwick's weight) force of the floor on the piano (normal force) force of the piano on the floor force of Chadwick on the piano force of the piano pushing on Chadwick Typesetting math: 100% Select the choice that best matches the free-body diagram you have drawn for the piano. Hint 1. Determine the directions and relative magnitudes of the forces Which of the following statements best describes the correct directions and relative magnitudes of the forces involved? ANSWER: ANSWER: The normal force and weight are both upward and the pushing force is horizontal. The normal force and weight are both downward and the pushing force is horizontal. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a greater magnitude than the weight. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force and weight have the same magnitude. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a smaller magnitude than the weight. Typesetting math: 100% Typesetting math: 100% Correct If you were actually going to solve this problem rather than just draw the free-body diagram, you would need to define the coordinate system. Choose the position of the piano as the origin. In this case it is simplest to let the y axis point vertically upward and the x axis point horizontally to the right, in the direction of the acceleration. Chadwick now needs to push the piano up a ramp and into a moving van. at left. The ramp is frictionless. Is Chadwick strong enough to push the piano up the ramp alone or must he get help? To solve this problem you should start by drawing a free-body diagram. Part D Determine the object of interest for this situation. ANSWER: Correct Now draw the free-body diagram of the piano in this new situation. Follow the same sequence of steps that you followed for the first situation. Again draw your diagram before you look at the choices For this situation, you should draw a free-body diagram for the ramp. Chadwick. the piano. Typesetting math: 100% below. Part E Which diagram accurately represents the free-body diagram for the piano? ANSWER: Typesetting math: 100% Typesetting math: 100% Correct In working problems like this one that involve an incline, it is most often easiest to select a coordinate system that is not vertical and horizontal. Instead, choose the x axis so that it is parallel to the incline and choose the y axis so that it is perpendicular to the incline. Problem 5.18 The figure shows two of the three forces acting on an object in equilibrium. Part A Redraw the diagram, showing all three forces. Label the third force . Draw the force vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: F  3 Typesetting math: 100% Correct Problem 5.25 An ice hockey puck glides across frictionless ice. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the ice hockey puck. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Normal force ; Gravity Normal force ; Gravity ; Kinetic friction Tension ; Weight Thrust ; Gravity n F  G n F  G fk  T  w Fthrust  F  G Typesetting math: 100% Correct Problem 5.26 Your physics textbook is sliding to the right across the table. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the object. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Weight ; Kinetic friction Thrust ; Kinetic friction Normal force ; Weight ; Kinetic friction Normal force ; Weight ; Static friction w fk  Fthrust  fk  n w fk  n w fs  Typesetting math: 100% Correct Enhanced EOC: Problem 5.35 A constant force is applied to an object, causing the object to accelerate at 13 . You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Proportions I Proportions II Part A m/s2 Typesetting math: 100% What will the acceleration be if the force is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is halved but the mass remains the same? ANSWER: Correct Part B What will the acceleration be if the object's mass is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the mass is halved but the force remains the same? ANSWER: Correct Part C a = 6.50 m s2 a = 26.0 m s2 Typesetting math: 100% What will the acceleration be if the force and the object's mass are both halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if both the force and mass are reduced by a factor of two? ANSWER: Correct Part D What will the acceleration be if the force is halved and the object's mass is doubled? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is decreased by a factor of two and the mass is increased by a factor of two? Check your answer by choosing numerical values of the force and mass, and then halve the force and double the mass. ANSWER: Correct a = 13.0 m s2 a = 3.25 m s2 Typesetting math: 100% Problem 5.44 A rocket is being launched straight up. Air resistance is not negligible. Part A Which of the following is the correct motion diagram for the situation described above? Enter the letter that corresponds with the best answer. ANSWER: Correct Part B Draw a free-body diagram. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Score Summary: Your score on this assignment is 99.7%. You received 63.82 out of a possible total of 64 points. Typesetting math: 100%

Assignment 4 Due: 11:59pm on Wednesday, February 26, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy ± Two Forces Acting at a Point Two forces, and , act at a point. has a magnitude of 9.80 and is directed at an angle of 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.20 and is directed at an angle of 54.1 below the negative x axis in the third quadrant. Part A What is the x component of the resultant force? Express your answer in newtons. Hint 1. How to approach the problem The resultant force is defined as the vector sum of all forces. Thus, its x component is the sum of the x components of the forces, and its y component is the sum of the y components of the forces. Hint 2. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . F 1 F  2 F  1 N  F  2 N  F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 Hint 2. Find the direction of is directed at an angle of 56.0 above the x axis in the second quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 56.0 . ANSWER: ANSWER: Hint 3. Find the x component of Find the x component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , F 1 F 1  F  1 F  1  =   180 −  180 +  90 +  -5.48 N F 2 F  2 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Typesetting math: 100% and if . Hint 2. Find the direction of is directed at an angle of 54.1 below the x axis in the third quadrant. When you calculate the components of , however, the direction of the force is commonly expressed in terms of the angle that the vector representing the force forms with the positive x axis. What is the angle that forms with the positive x axis? Select an answer from the following list, where 54.1 . ANSWER: ANSWER: ANSWER: Correct Part B What is the y component of the resultant force? Express your answer in newtons. Ax < 0 Ay <  <  < 3 2 F 2 F 2  F 2 F  2  =   180 −   − 180 −90 −  -3.05 N -8.53 N Typesetting math: 100% Hint 1. How to approach the problem Follow the same procedure that you used in Part A to find the x component of the resultant force, though now calculate the y components of the two forces. Hint 2. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: Hint 3. Find the y component of Find the y component of . Express your answer in newtons. Hint 1. Components of a vector F 1 F  1 A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 8.12 N F 2 F  2 Typesetting math: 100% Consider a vector that forms an angle with the positive x axis. The x and y components of are, respectively, and , where is the magnitude of the vector. Note that and if , and if . ANSWER: ANSWER: Correct Part C What is the magnitude of the resultant force? Express your answer in newtons. Hint 1. Magnitude of a vector Consider a vector , whose components are and . The magnitude of is . A  A Ax = Acos  Ay = Asin  A Ax < 0 Ay > 0  <  <  2 Ax < 0 Ay < 0  <  < 3 2 -4.21 N 3.91 N A Ax Ay A A = A + 2 x A2 y −−−−−−−  Typesetting math: 100% ANSWER: Correct Enhanced EOC: Problem 5.9 The figure shows acceleration-versus-force graphs for two objects pulled by rubber bands. You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Finding the Slope of a Line from a Graph Part A What is the mass ratio ? Express your answer using two significant figures. 9.38 N m1 m2 Typesetting math: 100% Hint 1. How to approach the problem How are the acceleration and the force on an object related to its mass? How is the slope of each line in the figure related to each object's mass? For each line, what two points are easy to measure accurately to determine the slope of line? How is the slope determined from the x and y coordinates of the two points you chose for each line? ANSWER: Correct A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude . Part A How much horizontal force must a sprinter of mass 54 exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Hint 1. Newton's 2nd law of motion According to Newton's 2nd law of motion, if a net external force acts on a body, the body accelerates, and the net force is equal to the mass of the body times the acceleration of the body: . ANSWER: = 0.36 m1 m2 15 m/s2 F kg Fnet m a Fnet = ma F = 810 N Typesetting math: 100% Correct Part B Which body exerts the force that propels the sprinter, the blocks or the sprinter? Hint 1. How to approach the question To start moving forward, sprinters push backward on the starting blocks with their feet. Newton's 3rd law tells you that the blocks exert a force on the sprinter of the same magnitude, but opposite in direction. ANSWER: Correct To start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward. Problem 5.12 The figure shows an acceleration-versus-force graph for a 600 object. the blocks the sprinter g Typesetting math: 100% Part A What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Part B What must equal in order for the graph to be correct? Express your answer with the appropriate units. ANSWER: Correct Free-Body Diagrams Learning Goal: To gain practice drawing free-body diagrams Whenever you face a problem involving forces, always start with a free-body diagram. a1 a1 = 1.67 m s2 a2 a2 = 3.33 m s2 Typesetting math: 100% To draw a free-body diagram use the following steps: Isolate the object of interest. It is customary to represent the object of interest as a point 1. in your diagram. Identify all the forces acting on the object and their directions. Do not include forces acting on other objects in the problem. Also, do not include quantities, such as velocities and accelerations, that are not forces. 2. Draw the vectors for each force acting on your object of interest. When possible, the length of the force vectors you draw should represent the relative magnitudes of the forces acting on the object. 3. In most problems, after you have drawn the free-body diagrams, you will explicitly label your coordinate axes and directions. Always make the object of interest the origin of your coordinate system. Then you will need to divide the forces into x and y components, sum the x and y forces, and apply Newton's first or second law. In this problem you will only draw the free-body diagram. Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor (see the figure). The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? To solve this problem you should start by drawing a free-body diagram. Part A Determine the object of interest for the situation described in the problem introduction. Hint 1. How to approach the problem You should first think about the question you are trying to answer: What is the acceleration of the piano? The object of interest in this situation will be the object whose acceleration you are asked to find. ANSWER: Typesetting math: 100% Correct Part B Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano. Check all that apply. ANSWER: Correct Now that you have identified the forces acting on the piano, you should draw the free-body diagram. Draw the length of your vectors to represent the relative magnitudes of the forces, but you don't need to worry about the exact scale. You won't have the exact value of all of the forces until you finish solving the problem. To maximize your learning, you should draw the diagram yourself before looking at the choices in the next part. You are on your honor to do so. Part C For this situation you should draw a free-body diagram for the floor. Chadwick. the piano. acceleration of the piano gravitational force acting on the piano (piano's weight) speed of the piano gravitational force acting on Chadwick (Chadwick's weight) force of the floor on the piano (normal force) force of the piano on the floor force of Chadwick on the piano force of the piano pushing on Chadwick Typesetting math: 100% Select the choice that best matches the free-body diagram you have drawn for the piano. Hint 1. Determine the directions and relative magnitudes of the forces Which of the following statements best describes the correct directions and relative magnitudes of the forces involved? ANSWER: ANSWER: The normal force and weight are both upward and the pushing force is horizontal. The normal force and weight are both downward and the pushing force is horizontal. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a greater magnitude than the weight. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force and weight have the same magnitude. The normal force is upward, the weight is downward, and the pushing force is horizontal. The normal force has a smaller magnitude than the weight. Typesetting math: 100% Typesetting math: 100% Correct If you were actually going to solve this problem rather than just draw the free-body diagram, you would need to define the coordinate system. Choose the position of the piano as the origin. In this case it is simplest to let the y axis point vertically upward and the x axis point horizontally to the right, in the direction of the acceleration. Chadwick now needs to push the piano up a ramp and into a moving van. at left. The ramp is frictionless. Is Chadwick strong enough to push the piano up the ramp alone or must he get help? To solve this problem you should start by drawing a free-body diagram. Part D Determine the object of interest for this situation. ANSWER: Correct Now draw the free-body diagram of the piano in this new situation. Follow the same sequence of steps that you followed for the first situation. Again draw your diagram before you look at the choices For this situation, you should draw a free-body diagram for the ramp. Chadwick. the piano. Typesetting math: 100% below. Part E Which diagram accurately represents the free-body diagram for the piano? ANSWER: Typesetting math: 100% Typesetting math: 100% Correct In working problems like this one that involve an incline, it is most often easiest to select a coordinate system that is not vertical and horizontal. Instead, choose the x axis so that it is parallel to the incline and choose the y axis so that it is perpendicular to the incline. Problem 5.18 The figure shows two of the three forces acting on an object in equilibrium. Part A Redraw the diagram, showing all three forces. Label the third force . Draw the force vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: F  3 Typesetting math: 100% Correct Problem 5.25 An ice hockey puck glides across frictionless ice. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the ice hockey puck. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Normal force ; Gravity Normal force ; Gravity ; Kinetic friction Tension ; Weight Thrust ; Gravity n F  G n F  G fk  T  w Fthrust  F  G Typesetting math: 100% Correct Problem 5.26 Your physics textbook is sliding to the right across the table. Part A Identify all forces acting on the object. ANSWER: Typesetting math: 100% Correct Part B Draw a free-body diagram of the object. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Weight ; Kinetic friction Thrust ; Kinetic friction Normal force ; Weight ; Kinetic friction Normal force ; Weight ; Static friction w fk  Fthrust  fk  n w fk  n w fs  Typesetting math: 100% Correct Enhanced EOC: Problem 5.35 A constant force is applied to an object, causing the object to accelerate at 13 . You may want to review ( pages 127 - 130) . For help with math skills, you may want to review: Proportions I Proportions II Part A m/s2 Typesetting math: 100% What will the acceleration be if the force is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is halved but the mass remains the same? ANSWER: Correct Part B What will the acceleration be if the object's mass is halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the mass is halved but the force remains the same? ANSWER: Correct Part C a = 6.50 m s2 a = 26.0 m s2 Typesetting math: 100% What will the acceleration be if the force and the object's mass are both halved? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if both the force and mass are reduced by a factor of two? ANSWER: Correct Part D What will the acceleration be if the force is halved and the object's mass is doubled? Express your answer with the appropriate units. Hint 1. How to approach the problem How is the acceleration of an object related to its mass and the force applied? Expressing the acceleration in terms of the force and mass, what happens to the acceleration if the force is decreased by a factor of two and the mass is increased by a factor of two? Check your answer by choosing numerical values of the force and mass, and then halve the force and double the mass. ANSWER: Correct a = 13.0 m s2 a = 3.25 m s2 Typesetting math: 100% Problem 5.44 A rocket is being launched straight up. Air resistance is not negligible. Part A Which of the following is the correct motion diagram for the situation described above? Enter the letter that corresponds with the best answer. ANSWER: Correct Part B Draw a free-body diagram. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Score Summary: Your score on this assignment is 99.7%. You received 63.82 out of a possible total of 64 points. Typesetting math: 100%

please email info@checkyourstudy.com