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Conceptual Physics. Chapter Eight Notes: Momentum. 8.1 Momentum. Momentum
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Conceptual Physics Chapter Eight Notes: Momentum
8.1 Momentum • Momentum • 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." • Momentum is a commonly used term in sports. A team that has the momentum is onthe moveand is going to take some effort to stop. A team that has a lot of momentum is really on the moveand 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 movehas the momentum. If an object is in motion (on the move) then it has 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. • Momentum = mass • velocity • In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as • p = m • v • where m is the mass and vis the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.
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. • 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. 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. • 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. 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.
The momentum equation can help us to think about how a change in one of the two variables might affect 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. • 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 "plug-and-chug recipe for algebraic problem-solving.“
8.2 Inpulse Changes Momentum • 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 moveand is going to be hard to stop. The term momentum is a physics concept. Any 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 such an object 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.
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 change the car's momentum. An object with momentum can be stopped if a force is applied against it for a given amount of time.
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. • 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.
If both sides of the equation to • The right are multiplied by the • quantity “t”, a new equation results. • This equation represents one of two primary principles to be used in the analysis of collisions during 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 impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the • Impulse = Change in momentum
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: • In a collision, an object experiences a force for a specific amount of time which results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, • F • t = m • v. • 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: • 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 a time equivalent to 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.
Now consider a collision of a tennis ball with a wall. Depending on the physical properties of the ball and wall, 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, velocity-time 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. Click the mouse to move to the next page and check your answer. • Vector Diagram • Greatest velocity change? • Greatest acceleration? • Greatest momentum change? • Greatest Impulse?
Quick Quiz Answers • a. The velocity change is greatest in case B. The velocity changes from +30 m/s to -28 m/s. This is a change of 58 m/s (-) and is greater than in case A (-15 m/s). • b. The acceleration is greatest in case B. Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above) • c. The momentum change is greatest in case B. Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above). • d. The impulse is greatest in case B. Impulse equals momentum change and the momentum change is greatest in case B (as stated above).
Velocity-Time Graph • Ticker Tape Diagram • Answers on next page
Ticker Tape Diagram • a. The velocity change is greatest in case B. In each case the initial velocity is the same. In case B, the object rebounds in the opposite direction with a greater speed than in case A. This is equivalent to a change from +10 m/s to -5 m/s; whereas, case A has a change from +10 m/s to -2 m/s. • b. The acceleration is greatest in case B. Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above) • c. The momentum change is greatest in case B. Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above). • d. The impulse is greatest in case B. Impulse equals momentum change and the momentum change is greatest in case B (as stated above) Velocity-Time Graph • a. The velocity change is greatest in case A. The v changes from +5 m/s to -3 m/s. This is a change of 8 m/s (-) and is greater than in case B (-4 m/s). • b. The acceleration is greatest in case A. Acceleration depends on velocity change and the velocity change is greatest in case A (as stated above). • c. The momentum change is greatest in case A. Momentum change depends on velocity change and the velocity change is greatest in case A (as stated above). • d. The impulse is greatest in case A. Impulse equals momentum change and the momentum change is greatest in case A (as stated above).
Observe that each of the collisions above involve 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 adirection change in addition to a speed change. The result of the direction change is a large velocity change. 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.
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 • Click the mouse to view answers. • 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. -4000 -40 -40 -400 -40 -4 -20,000 -200 -4 0.010 -200 25 -200 -200 -4
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. • There are also a few observations which can be made which relate to the qualitative nature of the impulse-momentum 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.
8.4 Conservation of Momentum • One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows. • For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. • The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof.
Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows. • The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as
Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as • But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as
The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.
Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand. The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West (review the section on adding vectors if necessary). Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.
goal-line stand: • Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram. Later, we will use the momentum conservation principle to solve problems in which the after-collision velocity of objects is predicted. • Law of conservation of momentum: • In the absence of an external force, the momentum of a system remains unchanged!
8.5 Collisions • Elastic Collision • Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object. • Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision. • The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.
In the collision between the truck and the car, total system momentum is conserved. Before the collision, the momentum of the truck is 60000 kg*m/s and the momentum of the car is 0 kg*m/s; the total system momentum is 60000 kg*m/s. After the collision, the momentum of the truck is 30000 kg*m/s and the momentum of the car is 30000 kg*m/s; the total system momentum is 60000 kg*m/s. The total system momentum is conserved. The momentum lost by the truck (30000 kg*m/s) is gained by the car.
Inelastic Collision • Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object. • Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.
The animation below portrays the inelastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables. • In the collision between the truck and the car, total system momentum is conserved. Before the collision, the momentum of the truck is 60000 kg*m/s and the momentum of the car is 0 kg*m/s; the total system momentum is 60000 kg*m/s. After the collision, the momentum of the truck is 45000 kg*m/s and the momentum of the car is 15000 kg*m/s; the total system momentum is 60000 kg*m/s. The total system momentum is conserved. The momentum lost by the truck (15000 kg*m/s) is gained by the car.
8.6 Momentum Vectors • The animation below portrays the inelastic collision between two 1000-kg cars. The before- and after-collision velocities and momentum are shown in the data tables. • In the collision between the two cars, total system momentum is conserved. Yet this might not be apparent without an understanding of the vector nature of momentum. Momentum, like all vector quantities, has both a magnitude (size) and a direction.
When considering the total momentum of the system before the collision, the individual momentum of the two cars must be added as vectors. That is, 20 000 kg*m/s, East must be added to 10 000 kg*m/s, North. The sum of these two vectors is not 30 000 kg*m/s; this would only be the case if the two momentum vectors had the same direction. Instead, the sum of 20 000 kg*m/s, East and 10 000 kg*m/s, North is 22 361 kg*m/s at an angle of 26.6 North of East. Since the two momentum vectors are at right angles, their sum can be found using the Pythagorean theorem; the direction can be found using SOH CAH TOA (specifically, the tangent function). The value 22 361 kg*m/s is the total momentum of the system before the collision; and since momentum is conserved, it is also the total momentum of the system after the collision. Since the cars have equal mass, the total system momentum is shared equally by each individual car. In order to determine the momentum of either individual car, this total system momentum must be divided by two (approx. 11 200 kg*m/s). Once the momentum of the individual cars are known, the after-collision velocity is determined by simply dividing momentum by mass (v=p/m).
VECTOR ANALYSIS: • Vector sum: From a parallelogram with sides of 20,000 (kg*m/s) horizontal and 10,000 (kg*m/s) vertical the diagonal is 22,400 (kg*m/s) @ angle θ = arctan(10/20) N of E = 27° N of E • Mass (kg) 1000 • Vel. (m/s) 20.0 East BEFORE • Mom. (kg*m/s) 20,000 East • Mass (kg) 1000 • Vel. (m/s) 10.0 North • Mom. (kg*m/s) 10,000 North Mass (kg) 1000 Vel. (m/s) 11.2 27° N of E AFTER Mom. (kg*m/s) 11,200 27° N of E Mass (kg) 1000 Vel. (m/s) 11.2 27° N of E Mom. (kg*m/s) 11,200 27° N of E