South Pasadena A.P. Physics Chapter 7Study List

South Pasadena A.P. Physics Chapter 7Study List

South Pasadena A.P. Physics Chapter 7Study List

Linear Momentum

Main Topics:

7-1Momentum and Its Relation to Force 7-6 Inelastic Collisions

7-2Conservation of Momentum7-7Collisions in Two or Three Dimensions

7-3Collisions and Impulse7-8Center of Mass (CM)

7-4Conservation of Energy and Momentum 7-9*CM for the Human Body

in Collisions7-10*Center of Mass and Translational Motion

7-5Elastic Collisions in One Dimension

Key Terms:

Linear Momentum (p) – is defined as the product of an object’s mass (m) and its velocity (v).

Law of Conservation of Momentum– states that in any collision between two or more objects, the vector sum of the momenta before impact equals the vector sum of the momenta after impact. The total momentum remains constant, i.e. the momentum is conserved.

Impulse – is defined as the product of the force (F) acting on an object and the time (t) during which the force acts.

Perfectly elastic collisions– occur when the kinetic energy as well as the linear momentum is conserved.

Completely inelastic collisions–occur when objects stick together after impact. Most of the kinetic energy is converted into heat. This type of problem can be solved by applying the law of conservation of momentum.

Center of gravity(cg)– of an object is the point where the entire weight of the object can be considered to be concentrated. At the center of gravity the entire weight of the object can be balanced by a single vertical force equal to the object’s weight.

Center of mass(CM)–is the point where all of the mass of an object can be considered to be concentrated. For almost all objects, the center of mass is at the same point in the object as the center of gravity.

Translational motion– is where, at any instant, the motion of all points of a moving object have the same velocity and direction of motion. This type of motion, as well as the concept of center of mass, has been implied in solving problems in previous chapters.

Summary of Momentum/Impulse Formulas:

momentum / p = m v
Newton’s Second Law / F = p/t
Conservation of Momentum / m1v1 + m2v2 = m1 v1’+ m2 v2’
Impulse / F ●t
Impulse and Momentum / F ●t =  p = mvf − mvi
Coordinate positions of the center of mass / Xcm = (Σmi xi) / Σmi)
Y cm = (Σmi yi) / Σmi)
Translational velocity of the center of mass / M v cm = mivi + m2 v2 + m3 v3

□Know how to calculate momentum as the product of m● v and its units.

□Know thatmomentum is a separately conserved quantity, different from energy.□ Understand thatan unbalanced force on an object produces a change in its momentum.

□How to solve problems involving elastic and inelastic collisions in one dimension using the principles of conservation of momentum and energy.

□Knowexamples from our everyday life where momentum plays a role?

□Knowhow Impulsedefined, its units and how it relates to momentum.

□How can an object’s momentum can change? E.g. an object is moving then an impulse causes it to slow down or stop, or an object is at rest and an impulse causes it to start moving.

□Be able to compare the impulse for an object moving and colliding with something in-elastically or elastically. In other words, is there more impulse when bouncing occurs or when it just stops?

□Know how to calculate the impulse from a collision.

□How does anelastic collision differ from an inelastic collision?

□Know how to predict (calculate) the velocity after to objects collideelastically or in-elastically using the law of conservation of momentum.

□How do you calculate the momentum or velocity of objects in an “explosion”?

□What does it mean to say that momentum is a VECTOR and momentum vectors

are conserved? Example: The fragments in a fireworks display.

Energy and Momentum Conservation in Collisions

In collisions between two (or more) objects the force each object exerts on the other is usually very large compared to any other force acting and the time interval during which the interaction occurs is usually very short. Usually, both the magnitude of the force and the time interval remain unknown; however, the impulse can be determined if it is possible to determine the change in momentum.

In perfectly elastic collisions,the kinetic energy as well as the linear momentum is conserved. The sum of the kinetic energies of the objects before the collision is equal to the sum of the kinetic energies of the objects after the collision. No heat energy results from the collision. However, if the collision is not perfectly elastic, heat and other forms of energy may result. Certain perfectly elastic collision problems can be solved by applying both the law of conservation of energy and the law of conservation of momentum.

In completely inelastic collisions,the objects stick together after impact. Most of the kinetic energy is converted into heat. Therefore, the kinetic energy is NOT conserved. This type of problem can be solved by applying the law of conservation of momentum.

The center of gravity (cg) of an object is the point where the entire weight of the object can be considered to be concentrated. At that point the entire weight of the object can be balanced by a single vertical force equal to the object’s weight. For example, you can test for the center of gravity of a textbook by attempting to balance the book with one finger. The center of gravity of a textbook should be very close to the center of the book. For a baseball bat, the balance point and therefore the center of gravity is displaced from the center toward the fat end of the bat.

The center of mass, which is abbreviated CM, is the point where all of the mass of an object can be considered to be concentrated. For almost all objects, the center of mass is at the same point in the object as the center of gravity. An extreme example to show the difference is a thin, uniform rod that extends from the surface of the Earth vertically upward. The center of mass of the rod would be at its center. However, since the value of g decreases with altitude and the rod is extremely long, the lower half of the rod would weigh more than the upper half. In this case the center of gravity would fall below the center of mass.

This formula shows how to determine the distance from one end to the center of mass of a mass-

less rod along which a number of point masses have been attached.

XCM = (m1 x1 + m2 x2 + … + mn xn) / (m1 + m2 + … + mn )

3 M’s of Momentum: Mysteries, Magic and Myths

(1)Why is it safer to drive a large vehicle than a small one?

(2)Why is it not a good idea to hold a rifle out in front of you when you fire it?

(3)What if the rifle were firing blanks? Would it be as dangerous to hold the rifle

out in front of you?

(4) How can a ping pong ball bounce higher than a super ball?

(5) Considering the popular “Executive Office Toy”, when one ball is released, why

don’t two balls come out?

(6) Which do you choose in this quick thinking situation? Driving into an

approaching car of identical mass and speed as yours or hitting a large, solid tree?

Videos:

Mechanical Universe: Conservation of Momentum

Demonstrations:

Newton’s Cradles

Impulse: Knocking Over A Wooden Board

Happy/Sad Balls and Clacking Balls

Air Track with Gliders

Throwing a raw egg into a bed-sheet

Transferring momentum with super-balls and other balls

Catching a heavy object in a short time or longer time.

Homework Assignments

Assignment 7AQuestions:1 −8

Problems:1 − 11 (odd)

Assignment 7 BQuestions:9 −16, 19

Problems:14 – 16, 18 – 19

Assignment 7 CProblems:21 – 23, 37, & 46