Develop a 4 page-500 word précis on Chapter 7 “How to Monitor & Control a TPM Project” of the Wysocki 7th Ed. text.”

Develop a 4 page-500 word précis on Chapter 7 “How to Monitor & Control a TPM Project” of the Wysocki 7th Ed. text.”

Summary of ‘How to Monitor and Control a TPM Project’ … Read More...
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

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In case the body stay in lower temperature for short period (less than 20 minutes), explain how the body response to it.

In case the body stay in lower temperature for short period (less than 20 minutes), explain how the body response to it.

Sweat stops being formed. The minute muscles under the exterior … Read More...