Impulse and Momentum: Understanding Forces on a Baseball

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 1/26 Homework 5-1 Due: 11:57pm on Tuesday, February 13, 2024 To understand how points are awarded, read the Grading Policy for this assignment. Impulse on a Baseball Learning Goal: To understand the relationship between force, impulse, and momentum. The effect of a net force acting on an object is related both to the force and to the total time the force acts on the object. The physical quantity impulse is a measure of both these effects. For a constant net force, the impulse is given by . The impulse is a vector pointing in the same direction as the force vector. The units of are or . Recall that when a net force acts on an object, the object will accelerate, causing a change in its velocity. Hence the object's momentum ( ) will also change. The impulse-momentum theorem describes the effect that an impulse has on an object's motion: . So the change in momentum of an object equals the net impulse, that is, the net force multiplied by the time over which the force acts. A given change in momentum can result from a large force over a short time or a smaller force over a longer time. In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 to the ball for a time of . Part A Assuming that this force is constant, what is the magnitude of the impulse on the ball? Enter your answer numerically in newton seconds using two significant figures. ANSWER: Correct We often visualize the impulse by drawing a graph showing the force versus time. For a constant net force such as that used in the previous part, the graph showing the magnitude of the force versus time will look like the one shown in . = 8.4

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 2/26 Part B The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have physical meaning. ANSWER: Correct The assumption of a constant net force is idealized to make the problem easier to solve. A real force, especially in a case like the one presented in Parts A and B, where a large force is applied for a short time, is not likely to be constant. A more realistic graph showing the magnitude of the force that the swinging bat applies to the baseball will show the force building up to a maximum value as the bat comes into full contact with the ball. Then as the ball loses contact with the bat, the graph will show the magnitude of the force decaying to zero. It will look like the graph in . For this graph, the length of the rectangle corresponds to the impulse. For this graph, the height of the rectangle corresponds to the impulse. For this graph, the area of the rectangle corresponds to the impulse. For this graph, the slope of the rectangle corresponds to the impulse.

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 3/26 Part C If both the graph representing the constant net force and the graph representing the variable net force represent the same impulse acting on the baseball, which geometric properties must the two graphs have in common? ANSWER: Correct When the net force varies over time, as in the case of the real net force acting on the baseball, you can simplify the problem by finding the average net force acting on the baseball during time . This average net force is treated as a constant force that acts on the ball for time . The impulse on the ball can then be found as . Graphically, this method states that the impulse of the baseball can be represented by either the area under the net force versus time curve or the area under the average net force versus time curve. These areas are represented in the figure as the areas shaded in red and blue respectively. The impulse of an object is also related to its change in momentum. Once the impulse is known, it can be used to find the change in momentum, or if either the initial or final momentum is known, the other momentum can be found. Keep in mind that . Because both impulse and momentum are vectors, it is essential to account for the direction of each vector, even in a one-dimensional problem. Part D Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the + x direction. The impulse on the ball caused by the bat will be in the ANSWER: maximum force area slope

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 4/26 Correct Part E Now assume that the pitcher in Part D throws a 0.145- baseball parallel to the ground with a speed of 32 in the + x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's x - component of velocity just after leaving the bat if the bat applies an impulse of to the baseball? Enter your answer numerically in meters per second using two significant figures. ANSWER: Correct The negative sign in the answer indicates that after the bat hits the ball, the ball travels in the opposite direction to that defined to be positive. 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: 1. The blocks can move in only one dimension, namely, along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. positive x direction. negative x direction. = -26

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 5/26 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. Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Correct Part B Find the time derivative of the x component of the system's total momentum. Express your answer in terms of , , , and . Hint 1. Finding the derivative of momentum for one block Consider the momentum of block 1: . Take the derivative of this expression with respect to time, noting that velocity is a function of time, and mass is a constant: . Hint 2. The relationship between velocity and acceleration Recall the definition of acceleration as . ANSWER: Correct 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: = =

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 6/26 Correct Part D Acceleration is due to which of the following physical quantities? ANSWER: Correct 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? Hint 1. Force and acceleration Since the system is closed, the only object that can affect block 1 is the other block in the system, block 2. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 7/26 Correct Part F What could be the reason for the acceleration of block 2? ANSWER: Correct 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: Correct Note that both and are internal forces. 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? 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 a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 and and and and

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 8/26 Hint 1. Which of Newton's laws is useful here? Newton's third law! ANSWER: Correct Newton's third law states that forces and are equal in magnitude and opposite in direction. Therefore, their x components are related by 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 . Hint 1. What is ? , the total (internal) force on the system (as a whole). Use the information from the last part to simplify the right-hand side of the above equation. ANSWER: Correct The derivative of the total momentum is zero; hence the total momentum is a constant function of time. We have just shown that for the special case of a closed two-block system, the internal forces do not change the total momentum of the system. It can be shown that in any system, the internal forces do not change the total momentum: It is conserved. In other words, total momentum is always conserved in a closed system of objects. ± The Impulse-Momentum Theorem Both forces have equal magnitudes. 0 nonzero constant

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 9/26 Learning Goal: To learn about the impulse-momentum theorem and its applications in some common cases. Using the concept of momentum, Newton's second law can be rewritten as , (1) where is the net force acting on the object, and is the rate at which the object's momentum is changing. If the object is observed during an interval of time between times and , then integration of both sides of equation (1) gives . (2) The right side of equation (2) is simply the change in the object's momentum . The left side is called the impulse of the net force and is denoted by . Then equation (2) can be rewritten as . This equation is known as the impulse-momentum theorem . It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force acting along the direction of motion, the impulse-momentum theorem can be written as . (3) Here , , and are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately. The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass moving along the x axis. The net force is acting on the particle along the x axis. is a constant force. Part A The particle starts from rest at . What is the magnitude of the momentum of the particle at time ? Assume that . Express your answer in terms of any or all of , , and . ANSWER: Correct Part B The particle starts from rest at . What is the magnitude of the velocity of the particle at time ? Assume that . Express your answer in terms of any or all of , , and . ANSWER: = =

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 10/26 Correct Part C The particle has momentum of magnitude at a certain instant. What is , the magnitude of its momentum seconds later? Express your answer in terms of any or all of , , , and . ANSWER: Correct Part D The particle has momentum of magnitude at a certain instant. What is , the magnitude of its velocity seconds later? Express your answer in terms of any or all of , , , and . ANSWER: Correct Let us now consider several two-dimensional situations. A particle of mass is moving in the positive x direction at speed . After a certain constant force is applied to the particle, it moves in the positive y direction at speed . Part E Find the magnitude of the impulse delivered to the particle. Express your answer in terms of and . Use three significant figures in the numerical coefficient. Hint 1. How to approach the problem This is a two-dimensional situation. It is helpful to find the components and separately and then use the Pythagorean theorem to find . Hint 2. Find the change in momentum Find , the magnitude of the change in the x component of the momentum of the particle. Express your answer in terms of and . ANSWER: = =

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 11/26 ANSWER: Correct Part F Which of the vectors below best represents the direction of the impulse vector ? ANSWER: Correct Part G What is the angle between the positive y axis and the vector as shown in the figure? = = 1 2 3 4 5 6 7 8

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 12/26 ANSWER: Correct Part H If the magnitude of the net force acting on the particle is , how long does it take the particle to acquire its final velocity, in the positive y direction? Express your answer in terms of , , and . If you use a numerical coefficient, use three significant figures. ANSWER: Correct So far, we have considered only the situation in which the magnitude of the net force acting on the particle was either irrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on a particle. Part I A particle of mass kilograms is at rest at seconds. A varying force is acting on the particle between seconds and seconds. Find the speed of the particle at seconds. Express your answer in meters per second to three significant figures. Hint 1. Use the impulse-momentum theorem In this case, and . Therefore, 26.6 degrees 30 degrees 60 degrees 63.4 degrees =

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3/1/24, 11:40 PM Homework 5-1 13/26 . Hint 2. What is the correct antiderivative? Which of the following is an antiderivative ? ANSWER: ANSWER: Correct ± PSS 11.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 11.1 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 11.1 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 = 43.0 https:// session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 14/26 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: Correct Part B Which of the following best describes why you can analyze the interaction described in this problem using conservation of momentum? ANSWER: Reset Help In system Not in system

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 15/26 Correct The quarterback-football system is not an isolated system because gravity, an external force, acts on the system. However, since the interaction between the objects (the quarterback and the ball) is very short, this external force can be ignored, and conservation of momentum can be applied in the horizontal direction. Visualize Now, draw a before-and-after pictorial representation. Define symbols representing mass and velocity of all the objects in the system before and after the interaction. Then, list known values and make sure to clearly identify what you are trying to find. Your final sketch might look like the picture shown to the left. Solve Part C How fast, , will the quarterback be moving backward just after releasing the ball? Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Since the football is thrown horizontally, and you are interested in the speed with which the quarterback is moving backward, only the horizontal component of momentum is relevant. Thus, you need only to find the horizontal momentum of the quarterback and football before and after the throw and substitute these into the conservation-of- momentum equation: . Hint 2. Find the initial momentum Assuming that the quarterback jumps straight up, what is the initial horizontal momentum, , of the ball and quarterback system? Express your answer in kilogram-meters per second. The throwing action is quick enough that external forces may be ignored. There are no external forces acting on the system. External forces don't act on the system during the jump. Conservation of momentum is always the best way to analyze motion.

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 16/26 ANSWER: Hint 3. Find the ball's final momentum What is the magnitude of the final horizontal momentum, , of the ball? Express your answer in kilogram-meters per second. ANSWER: Hint 4. Find the quarterback's final momentum What is the magnitude of the horizontal component of the quarterback's final momentum, ? Express your answer in kilogram-meters per second. ANSWER: ANSWER: Correct Assess Part D Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far will he move horizontally, assuming his speed is constant? Express your answer in meters to two significant figures. Hint 1. How to approach the problem Once the football leaves the quarterback's hand, he undergoes projectile motion. Recall that in projectile motion, the horizontal and vertical components of an object's motion are treated as independent of one another. In the vertical direction, the object moves under the inﬂuence of the acceleration due to gravity. In the horizontal direction, we assume that there is no force acting on the object, and so the object moves with a constant speed in the horizontal direction. ANSWER: = 0 = 6.45 = 6.45 = 8.1×10 −2

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 17/26 Correct This is a reasonable result. Had you found that the horizontal distance were much larger than a few centimeters, common sense would have told you that your calculations were wrong. When you throw a ball after jumping straight up in the air, you never land very far from the point you took your jump. By applying the conservation of momentum to a situation, you can obtain information about the speed of objects immediately before or after an interaction. However, you will often need to use kinematics equations to study the motion of these objects some time after the interaction. 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: 1. The original mass of Chuck and his cart does not include the 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. Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . Hint 1. How to approach the problem All this question is asking is: "How fast are Chuck and the ball moving away from each other?" If two objects are moving at the same speed (with respect to the ground) in the same direction, their relative speed is zero. If they are moving at the same speed, , in opposite directions, their relative speed is . In this problem, you are given variables for the speed of Chuck and the ball with respect to the ground, and you know that Chuck and the ball are moving directly away from each other. ANSWER: = 2.4×10 −2 =

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 18/26 Correct Make sure you understand this result; the concept of "relative speed" is important. In general, if two objects are moving in opposite directions (either toward each other or away from each other), the relative speed between them is equal to the sum of their speeds with respect to the ground. If two objects are moving in the same direction, then the relative speed between them is the absolute value of the difference of the their two speeds with respect to the ground. 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 . Hint 1. How to approach the problem Apply conservation of momentum. Equate the initial (before the ball is thrown) and final (after the ball is thrown) momenta of the system consisting of Chuck, his cart, and the ball. Use the result from Part A to eliminate from this equation and solve for . Hint 2. Initial momentum of Chuck, his cart, and the ball Before the ball is thrown, Chuck, his cart, and the ball are all at rest. Therefore, their total initial momentum is zero. Hint 3. Find the final momentum of Chuck, his cart, and the thrown ball What is the total momentum of Chuck, his cart, and the ball after the ball is thrown? Express your answer in terms of , , , and . Remember that and are speeds, not velocities, and thus are positive scalars. ANSWER: ANSWER: Correct Part C What is Chuck's speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . Hint 1. How to approach the problem Use the answer to Part B to eliminate from the equation derived in Part A. Then solve for . ANSWER: = =

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 19/26 Correct Part D Find Jackie's speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . Hint 1. How to approach the problem Apply conservation of momentum. Equate the initial (before Jackie catches the ball) and final (after the ball is caught) momenta of the system consisting of Jackie, her cart, and the ball, and solve for . Hint 2. Initial momentum Just before Jackie catches the ball, the momentum of the system consisting of Jackie, her cart, and the ball is equal to the momentum of the ball as it ﬂies through the air: . Hint 3. Find the final momentum What is the final momentum of the system after Jackie catches the ball? Express your answer in terms of , , and . ANSWER: ANSWER: Correct Part E Find Jackie's speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . Hint 1. How to approach the problem In Part B, you found an expression for in terms of . You can substitute this expression for into the equation you found in Part D, which will give you an expression for in terms of the desired quantities. ANSWER: = = =

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 20/26 Correct 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: Correct 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? Hint 1. How to approach the problem Think about how you would calculate the final speed of the two cars with the information provided and using the idea of conservation of momentum. Better yet, try the calculation out. What do you get? ANSWER: = the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually

3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 21/26 Correct 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? Hint 1. How to approach the problem Think about how you would calculate the final momentum of the two cars using the information provided and the idea of conservation of momentum. Better yet, try the calculation out. What do you get? Keep in mind that momentum is a vector, but you are asked about the magnitude of the momentum, which is a scalar. ANSWER: Correct 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? The answer depends on the directions in which the cars were moving before the collision. The answer depends on the directions in which the cars were moving before the collision.

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 22/26 Hint 1. Momentum is a vector Momentum is a vector quantity, and conservation of momentum holds for two-dimensional and three-dimensional collisions as well as for one-dimensional collisions. ANSWER: Correct You can decompose the vector equation that states the conservation of momentum into individual equations for each of the orthogonal components of the vectors. 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? Hint 1. How to approach the problem mathematically Momentum is a vector quantity. It is impossible to make exact predictions about the direction of motion after a collision if nothing is known about the direction of motion before the collision. However, one can put some bounds on the values of the final momentum. Start with the expression for from Part D: . Therefore, , where is the angle between and . (To derive the above, you would have to break each vector into components.) So the value of is controlled by . Hint 2. How to approach the problem empirically Consider the directions for the initial momenta that will give the largest and smallest final momentum. ANSWER:

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 23/26 Correct When the two cars collide, the magnitude of the final momentum will always be at most (a value attained if the cars were moving in the same direction before the collision) and at least (a value attained if the cars were moving in opposite directions before the collision). 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 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 . Hint 1. Conservation of momentum Recall that conservation of linear momentum may be expressed as a vector equation, . Each vector component of linear momentum is conserved separately.

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3/1/24, 11:40 PM Homework 5-1 24/26 Hint 2. x and y components of momentum The momentum of the two-car system immediately after the collision may be written as , where the x and y directions are the eastward and northward directions, respectively. Find and Express the two components, separated by a comma, in terms of , , and . ANSWER: Hint 3. A vector and its components Recall that the square of the magnitude of a vector is given by the Pythagorean formula: . Hint 4. Velocity and momentum Find , the magnitude of the final velocity. Express in terms of the magnitude of the final momentum and the masses and . ANSWER: ANSWER: Correct Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Correct Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 , = , = = =

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 25/26 Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: Correct Problem 11.40 A 500 cart is released from rest 1.00 from the bottom of a frictionless, 30.0 ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure shows the force during the collision. Part A After the cart bounces, how far does it roll back up the ramp? Express your answer with the appropriate units. ANSWER: Correct Problem 11.30 The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. = 0.498

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3/1/24, 11:40 PM Homework 5-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 26/26 A ball of mass and another ball of mass are placed inside a smooth metal tube with a massless spring compressed between them. When the spring is released, the heavier ball ﬂies out of one end of the tube with speed . Part A With what speed does the lighter ball emerge from the other end? Express your answer in terms of . ANSWER: Correct Score Summary: Your score on this assignment is 100%. You received 9 out of a possible total of 9 points. =

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