1. Momentum Introduction Section 0 Lecture 1 Slide 1 Lecture 15 Slide 1 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Physics of Technology PHYS 1800 Lecture 15 Momentum

2. Introduction Section 0 Lecture 1 Slide 2 Lecture 15 Slide 2 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 PHYSICS OF TECHNOLOGY Spring 2009 Assignment Sheet *Homework Handout

3. Momentum Introduction Section 0 Lecture 1 Slide 3 Lecture 15 Slide 3 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Physics of Technology PHYS 1800 Lecture 15 Momentum Introduction

4. Momentum Introduction Section 0 Lecture 1 Slide 4 Lecture 15 Slide 4 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Describing Motion and Interactions Position—where you are in space (L or meter) Velocity—how fast position is changing with time (LT -1 or m/s) Acceleration—how fast velocity is changing with time (LT -2 or m/s 2 ) Force— what is required to change to motion of a body (MLT -2 or kg-m/s 2 or N) Energy—the potential for an object to do work. (ML 2 T -2 or kg m 2 /s 2 or N-m or J) Work is equal to the force applied times the distance moved. W = F d Kinetic Energy is the energy associated with an object’s motion. KE=½ mv 2 Potenital Energy is the energy associated with an objects position. Gravitational potential energy PE gravity =mgh Spring potential energy PE apring = -kx In this chapter we will develop the concept of…MOMENTUM…and and its associated law of Conservation of Momentum and apply this to collisions.

5. Momentum Introduction Section 0 Lecture 1 Slide 5 Lecture 15 Slide 5 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Work is equal to the force applied times the distance moved. –Work = Force x Distance: W = F d –Work output = Work input units: 1 joule (J) = 1 Nm = 1 kg m 2 / s 2 [ML 2 T -2 ] Defining Work

6. Momentum Introduction Section 0 Lecture 1 Slide 6 Lecture 15 Slide 6 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Momentum and Collisions How can we describe the change in velocities of colliding football players, or balls colliding with bats? How does a strong force applied for a very short time affect the motion? Can we apply Newton’s Laws to collisions? What exactly is momentum? How is it different from force or energy? What does “Conservation of Momentum” mean? Δp=F Δt with Δt=t final – t initial Δt=t final – t initial Initial time Final time ΔW=F Δd with Δd=d final – d initial

7. Momentum Introduction Section 0 Lecture 1 Slide 7 Lecture 15 Slide 7 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What happens when a ball bounces? When it reaches the floor, its velocity quickly changes direction. There must be a strong force exerted on the ball by the floor during the short time they are in contact. This force provides the upward acceleration necessary to change the direction of the ball’s velocity.

8. Momentum Introduction Section 0 Lecture 1 Slide 8 Lecture 15 Slide 8 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What happens when a ball bounces? Forces like this are difficult to analyze: Strong forces that act for a very short time. Forces that may change rapidly during the collision. It will help to write Newton’s second law in terms of the total change in velocity over time, instead of acceleration:

9. Momentum Introduction Section 0 Lecture 1 Slide 9 Lecture 15 Slide 9 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Momentum and Impulse Multiply both sides of Newton’s second law by the time interval over which the force acts: The left side of the equation is impulse, the (average) force acting on an object multiplied by the time interval over which the force acts. How a force changes the motion of an object depends on both the size of the force and how long the force acts. The right side of the equation is the change in the momentum of the object. The momentum of the object is the mass of the object times its velocity.

10. Momentum Introduction Section 0 Lecture 1 Slide 10 Lecture 15 Slide 10 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Momentum and Impulse A bowling ball and a tennis ball can have the same momentum, if the tennis ball with its smaller mass has a much larger velocity.

11. Momentum Introduction Section 0 Lecture 1 Slide 11 Lecture 15 Slide 11 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Impulse-Momentum Principle The impulse acting on an object produces a change in momentum of the object that is equal in both magnitude and direction to the impulse. In analogy, work = change in energy = ΔE

12. Momentum Introduction Section 0 Lecture 1 Slide 12 Lecture 15 Slide 12 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Impulse-Momentum Principle When a ball bounces back with the same speed, the momentum changes from mv to -mv, so the change in momentum is 2mv.

13. Momentum Introduction Section 0 Lecture 1 Slide 13 Lecture 15 Slide 13 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Look here carefully! A Closer Look at Collisions

14. Momentum Introduction Section 0 Lecture 1 Slide 14 Lecture 15 Slide 14 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Bonds between atoms in a compressed solid can be treated as compressed springs. Ultimately the forces come from electrostatic interactions between electrons and protons (and a little quantum mechanics). Compression on an Atomic Scale F spring =-k Δx + + + + + + + + +

15. Momentum Introduction Section 0 Lecture 1 Slide 15 Lecture 15 Slide 15 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What Happens During the Collision? Does Newton’s third law still hold? –For every action, there is an equal but opposite reaction. –The defensive back exerts a force on the fullback, and the fullback exerts an equal but opposite force on the defensive back.

16. Momentum Introduction Section 0 Lecture 1 Slide 16 Lecture 15 Slide 16 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Conservation of Momentum Does Newton’s third law still hold? –For every action, there is an equal but opposite reaction. –The defensive back exerts a force on the fullback, and the fullback exerts an equal but opposite force on the defensive back.

17. Momentum Introduction Section 0 Lecture 1 Slide 17 Lecture 15 Slide 17 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Conservation of Momentum –The impulses on both are equal and opposite. –The changes in magnitude for each are equal and opposite. –The total change of the momentum for the two players is zero.

18. Momentum Introduction Section 0 Lecture 1 Slide 18 Lecture 15 Slide 18 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Conservation of Momentum If the net external force acting on a system of objects is zero, the total momentum of the system is conserved.

19. Momentum Introduction Section 0 Lecture 1 Slide 19 Lecture 15 Slide 19 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 A 100-kg fullback moving straight downfield collides with a 75-kg defensive back. The defensive back hangs on to the fullback, and the two players move together after the collision. What is the initial momentum of each player?

20. Momentum Introduction Section 0 Lecture 1 Slide 20 Lecture 15 Slide 20 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What is the initial momentum of each player? Fullback: p = mv = (100 kg)(5 m/s) = 500 kg·m/s 2 Defensive back: p = mv = (75 kg)(-4 m/s) = -300 kg·m/s 2

21. Momentum Introduction Section 0 Lecture 1 Slide 21 Lecture 15 Slide 21 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What is the total momentum of the system? Total momentum: p total = p fullback + p defensive back = 500 kg·m/s - 300 kg·m/s = 200 kg·m/s

22. Momentum Introduction Section 0 Lecture 1 Slide 22 Lecture 15 Slide 22 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 What is the velocity of the two players immediately after the collision? Total mass: m = 100 kg + 75 kg = 175 kg Velocity of both: v = p total / m = (200 kg·m/s) / 175 kg = 1.14 m/s

23. Momentum Introduction Section 0 Lecture 1 Slide 23 Lecture 15 Slide 23 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Recoil Why does a shotgun slam against your shoulder when fired, sometimes painfully? How can a rocket accelerate in empty space when there is nothing there to push against except itself?

24. Momentum Introduction Section 0 Lecture 1 Slide 24 Lecture 15 Slide 24 INTRODUCTION TO Modern Physics PHYX 2710 Fall 2004 Physics of Technology—PHYS 1800 Spring 2009 Physics of Technology Next Lab/Demo: Energy & Oscillations Momentum and Collisions Thursday 1:30-2:45 ESLC 53 Ch 6 and 7 Next Class: Wednesday 10:30-11:20 BUS 318 room Review Ch 6 Read Ch 7