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Momentum p= mv Impulse Momentum Change Theorem • The sports announcer says "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter." Sports vs. Physics • Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team which is "on the move" has the momentum. If an object is in motion ("on the move") then it has momentum. Defining Momentum • Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. The Equations • Momentum = mass * velocity In physics, the symbol for the quantity momentum is the small case "p"; thus, the above equation can be rewritten as • p=m*v • where m = mass and v=velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity Defining the Units of Measure • The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg*m/s. While the kg*m/s is the standard metric unit of momentum, there are a variety of other units which are acceptable (though not conventional) units of momentum; examples include kg*mi/hr, kg*km/hr, and g*cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum. Vector or Scalar? • Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity which is fully described by both magnitude and direction. To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg*m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. Defining Direction • The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction which an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg*m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction. Which variable (mass or velocity) defines momentum? • From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest; for the momentum of any object which is at rest is 0. Objects at rest do not have momentum they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects. Conceptualizing Momentum • The momentum equation can help us to think about how a change in one of the two variables might effect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg*m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg*m/s. A doubling of the mass results in a doubling of the momentum. Problem Solving Momentum • Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg*m/s (instead of 4.0 kg*m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p=m*v serves as a "guide to thinking" and not merely a "recipe for algebraic problem-solving." 1. Determine the momentum of a ... • 60-kg halfback moving eastward at 9 m/s. • 1000-kg car moving northward at 20 m/s. • 40-kg freshman moving southward at 2 m/s. 1 Answers A) p = m * v (60 kg)(9m/s) = 540 kg * m /s B) p = m * v (1000 kg)(20 m/s) 20,000 kg* m/s C) p = m * v 40 kg * 2 m/s 80 kg * m/s 2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ... • its velocity were doubled. • its velocity were tripled. • its mass were doubled (by adding more passengers and a greater load) • both its velocity were doubled and its mass were doubled. 2 answers A) 40, 000 units (doubling the velocity will double the momentum) B) 60, 000 units (tripling the velocity will triple the momentum C) 40, 000 units (doubling the mass will double the momentum) D) 80, 000 units (doubling the mass and doubling the velocity will quadruple the momentum) 3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below. • Compare the velocities of these three players. How many times greater is the velocity of the halfback and the velocity of the tight end than the velocity of the lineman? Connecting Momentum and Impulse • As mentioned in the previous part of this lesson, momentum is a commonly used term in sports. When a sports announcer says that a team has the momentum they mean that the team is really on the move and is going to be hard to stop. An object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum which an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time (or both) to bring an object with more momentum to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed. Momentum in sports • The concepts in the above paragraph should not seem like abstract information to you. You have observed this a number of times if you have watched the sport of football. In football, the defensive players apply a force for a given amount of time to stop the momentum of the offensive player who has the ball. You have also experienced this a multitude of times while driving. As you bring your car to a halt when approaching a stop sign or stoplight, the brakes serve to apply a force to the car for a given amount of time to stop the car's momentum. An object with momentum can be stopped if a force is applied against it for a given amount of time. Picture Illustration Revisiting Force Balance • A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object either speeding it up or slowing it down. If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed. Back to Newton’s Second Law • These concepts are merely an outgrowth of Newton's second law as discussed in an earlier unit. Newton's second law (Fnet=m*a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a=change in velocity/time), the following equalities result. Formulas of Newton’s 2nd Law Multiply both sides by t Examining the Equation • This equation is one of two primary equations to be used in this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force*time is known as the impulse. And since the quantity m*v is the momentum, the quantity m*"Delta "v must be the change in momentum. The equation really says that the Impulse = Change in momentum Collisions • One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first law which we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. The law can be expressed this way: Expressions of collisions • In a collision, an object experiences a force for a specific amount of time which results in a change in momentum (the object's mass either speeds up or slows down). The impulse experienced by the object equals the change in momentum of the object. In equation form, F * t = m * Delta v. Objects in collisions with impulse • In a collision, objects experience an impulse; the impulse causes (and is equal to) the change in momentum. Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed (and thus his momentum). If the motion was represented by a ticker tape diagram, it might appear as follows: Putting numbers in… • At approximately the tenth dot on the diagram, the collision occurs and lasts for a certain amount of time; in terms of dots, the collision lasts for approximately nine dots. In the halfback-defensive back collision, the halfback experiences a force which lasts for a certain amount of time to change his momentum. Since the collision causes the rightward-moving halfback to slow down, the force on the halfback must have been directed leftward. If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N*s. This impulse would cause a momentum change of 720 kg*m/s. In a collision, the impulse experienced by an object is always equal to the momentum change. Tennis Ball Collision • Now consider a collision of a tennis ball with a wall. Depending on the physical properties of the wall (its elastic nature), the speed at which the ball rebounds from the wall upon colliding with it will vary. The diagrams below depict the changes in velocity of the same ball. For each representation (vector diagram, v-t graph, and ticker tape pattern), indicate which case (A or B) has the greatest change in velocity, greatest acceleration, greatest momentum change, and greatest impulse. Support each answer. Vector Diagrams Tabulate your Answers Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse? Answers Greatest velocity change? B (changes from +30 m/s to -28 m/s which is a change of 58 m/s and A only changes -15 m/s) Greatest acceleration? B, because it has the greatest velocity change and acceleration is dependent on velocity change. Greatest momentum change? B, because momentum is dependent on velocity and the change in velocity is greatest in B. Greatest Impulse? B, impulse is momentum change and the momentum change is greatest in B. Graphs Setting up your answers Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse? Answers Greatest velocity change? Greatest in A because it changes from +5 m/s to -3 m/s which is a change of 8 whereas B only changes 4 m/s Greatest A, acceleration is greatest because acceleration? velocity change is greatest in A Greatest momentum change? A, momentum is dependent on velocity change and that is greatest in A Greatest Impulse? A, Impulse equals the change in momentum Using Ticker Tape Diagrams Organizing the Answers Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse? Answers Greatest velocity change? B, vi is the same for both, but object B rebounds with a greater speed so it goes from -10 to +5 m/s whereas A goes from -10 m/s to +2 m/s. Greatest acceleration? B, acceleration is dependent on velocity change Greatest momentum change? B, Momentum is dependent on velocity Greatest Impulse? B, impulse is defined as the change in momentum Recapping the Examples • Observe that each of the collisions above involved the rebound of a ball off a wall. Observe that the greater the rebound effect, the greater the acceleration, momentum change, and impulse. A rebound is a special type of collision involving a direction change; the result of the direction change is large velocity change. Elastic Collisions • On occasions in a rebound collision, an object will maintain the same or nearly the same speed as it had before the collision. Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force. Momentum and collisions • Use the impulse-momentum change principle to fill in the blanks in the following rows of the table. As you do, keep these three major truths in mind: • the impulse experienced by an object is the force*time • the momentum change of an object is the mass*velocity change • the impulse equals the momentum change Create this Chart Force (N) time (s) 1. 0.010 2. 0.100 3. 0.010 4. -20 000 5. -200 Mom. Impuls Chg. e (kg*m (N*s) /s) -40 Mass (kg) Vel. Chg. (m/s) 10 -4 10 -200 50 -200 1.0 -8 50 Answers Force (N) time (s) Mom. Impuls Chg. Mass e (kg*m (kg) (N*s) /s) 1. -4000 N 0.010 -40 -40 10 -4 2. -400 N 0.100 -40 -40 10 -4 -200 -200 50 -4 4. -20 000 .010 s -200 -200 25 -8 5. 1.0 -200 50 -4 3. -20, 000 N 0.010 -200 -200 Vel. Chg. (m/s) Reflections on the Table • There are a few observations which can be made in the above table which relate to the computational nature of the impulse-momentum change theorem. First, observe that the answers in the table above reveal that the third and fourth columns are always equal; that is, the impulse is always equal to the momentum change. Observe also that the if any two of the first three columns are known, then the remaining column can be computed; this is true because the impulse=force*time. Knowing two of these three quantities allows us to compute the third quantity. And finally, observe that knowing any two of the last three columns allows us to compute the remaining column; this is true since momentum change = mass*velocity change. Additional Thoughts • There are also a few observations which can be made which relate to the qualitative nature of the impulsemomentum theorem. An examination of rows 1 and 2 show that force and time are inversely proportional; for the same mass and velocity change, a tenfold increase in the time of impact corresponds to a tenfold decrease in the force of impact. An examination of rows 1 and 3 show that mass and force are directly proportional; for the same time and velocity change, a fivefold increase in the mass corresponds to a fivefold increase in the force required to stop that mass. Finally, an examination of rows 3 and 4 illustrate that mass and velocity change are inversely proportional; for the same force and time, a twofold decrease in the mass corresponds to a twofold increase in the velocity change. Examples Express your understanding of the impulse-momentum change theorem by answering the following questions. Example 1 • 1. A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds. Which cart (#1 or #2) has the greatest acceleration? Explain. Answer 1 • Cart 2 has the greatest acceleration. Recall that force is dependent on acceleration and mass. Both carts have the same mass but cart 2 has the greatest force. Example 2 • Which cart (#1 or #2) has the greatest impulse? Explain. Answer 2 • The impulse is the same for both carts. Impulse is force * time which calculates each cart to 1 N * s for each. Example 3 • Which cart (#1 or #2) has the greatest change in momentum? Explain. Answer 3 • The momentum change is the same for both carts. Momentum change equals the impulse. If both carts have the same impulse then they both have the same momentum change. Example 4 • 2. In a phun physics demo, two identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below. Question 4a • Which balloon (A or B) has the greatest acceleration? Explain. Answer 4a • Balloon B has the greatest acceleration The rate at which the velocity changes is greatest for balloon B, this is shown by the fact that the speed (distance/time) changes most rapidly. Example 4B • Which balloon (A or B) has the greatest final velocity? Explain. Answer 4B • Balloon B has the greatest final velocity. At the end of the diagram, the distance traveled in the latest interval is greatest for Balloon B. Question 4C • Which balloon (A or B) has the greatest momentum change? Explain. Answer 4C • Balloon B has the greatest momentum change. The final velocity is greatest for Balloon B, its velocity change is also the greatest. Momentum change depends on velocity change. The balloon with the greatest velocity change will have the greatest momentum change. Question 4D • Which balloon (A or B) experiences the greatest impulse? Explain. Answer 4D • The impulse is the same for each car. The impulse equals the momentum change. If the momentum change is the same for each car, then so must be the impulse. Question 5 • The diagram to the right depicts the before- and aftercollision speeds of a car which undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car "sticks" to the wall. Question 5A • In which case (A or B) is the change in velocity the greatest? Explain. Answer 5A • Case A has the greatest velocity change. The velocity change is -9 m/s in case A and only -5 m/s in case B. Question 5B • In which case (A or B) is the change in momentum the greatest? Explain. Answer 5B • Case A has the greatest momentum change. The momentum change is dependent on the velocity change; the object with the greatest velocity change has the greatest momentum change. Question 5C • In which case (A or B) is the impulse the greatest? Explain. Answer 5C • The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for car A then the impulse is greatest. Question 5D • In which case (A or B) is the force which acts upon the car the greatest (assume contact times are the same in both cases)? Explain. Answer 5D • The impulse is greatest for car A. The force is related to impulse (I = F * t). The bigger impulse for car A is attributed to the greatest force upon car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds. Question 6 • Rhonda, who has a mass of 60.0 kg, is riding at 25.0 m/s in her sports car when she must suddenly slam on the brakes to avoid hitting a dog crossing the road. She strikes the air bag, which brings her body to a stop in 0.400 s. What average force does the seat belt exert on her? Answer 6 • F = m x change in velocity)/time • F = (60 kg x 25 m/s)/ 0.400 s • F = 3750 N Question 6B • If Rhonda had not been wearing her seat belt and not had an air bag, then the windshield would have stopped her head in 0.001 s. What average force would the windshield have exerted on her? Answer 6B • F = (m x D v ) / t • F = (60 kg x 25 m/s) / 0.001 s • F = 1,500,000 N Question 7 • A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck. Answer 7 • Impulse = Force x time • I = 80 N x 0.1 s • I=8Nxs Ex. 8 • If a 5-kg object experiences a 10-N force for a duration of 0.1 seconds, then what is the momentum change of the object? Question 8 • Momentum change = 1.0 kg m/s • The momentum change = mass x velocity change • But since velocity change is unknown, another strategy must be used to find the momentum change. The strategy involves first finding the impulse (I = F x t). Since impulse = momentum change, the answer is 1 N x s.