Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 9. Impulse and Momentum Chapter Goal: To introduce the ideas of impulse and momentum and to learn a new problem-solving strategy based on conservation laws.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Ch. 9 Student Learning Objectives To understand interactions from the new perspective of impulse and momentum. To learn what is meant by an isolated system. To apply conservation of momentum in simple situations. To understand the basic ideas of inelastic collisions, explosions, and recoil.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulsive Forces and Newton’s 2nd Law An impulsive force can be defined as a large force exerted during a small interval of time. Collisions and explosions are examples of impulsive forces. Until now, we’ve been using Newton’s Law (and kinematic equations) for constant forces and accelerations. Impulsive forces are not constant.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulsive Forces and Newton’s 2nd Law F net-x = ma x This law applies to non-constant forces as well F x (t) = ma x (t) so F x (t) = m dv x /dt multiplying through by dt: F x (t) dt = m dv x. Now take the integral of both sides: =

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulsive Forces and Newton’s 2nd Law This can be written as: = mv f – mv i which is how Newton actually wrote the law in the first place You can solve for the velocity of an object, even if the acceleration is not constant.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulse The impulse upon a particle is defined as: Impulse has units of N-s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulse Problem (EOC #5) What value of F max gives an impulse of 6 N-s?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulse Problem - Answer What value of F max gives an impulse of 6 N-s? Answer: 1.5 x 10 3 N

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Impulse Momentum Theorem (an alternate version of Newton’s 2 nd Law) J x = ∆p x where change in momentum = ∆p x = mv f - mv i

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The cart’s change of momentum is A. 30 kg m/s. B. 10 kg m/s. C.–10 kg m/s. D.–20 kg m/s. E.–30 kg m/s.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The cart’s change of momentum is A. 30 kg m/s. B. 10 kg m/s. C.–10 kg m/s. D.–20 kg m/s. E.–30 kg m/s.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. They exert equal impulses because they have equal momenta. B. The clay ball exerts a larger impulse because it sticks. C. Neither exerts an impulse on the wall because the wall doesn’t move. D. The rubber ball exerts a larger impulse because it bounces. A 10 g rubber ball and a 10 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. They exert equal impulses because they have equal momenta. B. The clay ball exerts a larger impulse because it sticks. C. Neither exerts an impulse on the wall because the wall doesn’t move. D. The rubber ball exerts a larger impulse because it bounces. A 10 g rubber ball and a 10 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A graphical interpretation of impulse

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

A bouncing ball (EOC #29) a 200 g ball is dropped from a height of 2.0 m, bounces and rebounds to a height of 1.5 m. The figure shows the impulse received from the floor. What is F max ? Force of floor on ball

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Simplifying assumptions – impulse approximation The figure shows the force on the ball due to the floor. Is it reasonable to neglect the force on the ball due to the earth. This would be a impulse of: -F G ∆t = (-1.96 N)(.005s) J G = N Force of floor on ball

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Sketch a before, after (and if necessary during) representation to show objects before/after interaction. Coordinate system, symbols for knowns and unknowns. Time usually isn’t important. Position is necessary only if needed to find velocity (as in this case). 0

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What is the appropriate Impulse Momentum theorem (using correct subscripts)?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What is the appropriate Impulse Momentum Theorem (using correct subscripts)? = m (v 2y -v 1y ) ½ F max ∆t = m (v 2y -v 1y ) Many unknowns lurk in this equation. Do we have the technology to obtain them?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Answers: v iy = m/s v 2y = 5.42 m/s F max = 9.3 x 10 2 N It was reasonable to ignore the weight impulse.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Momentum The change in momentum of ball 1 is due to the impulse of ball 2 on ball 1. The change in momentum of ball 2 is due to the impulse of ball 1 on ball 2. If we added the impulse of ball 1 on ball 2 to the impulse of ball 2 on ball 1, what would we get? Why?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Momentum If we added the change in momentum of ball 1 to the change in momentum of ball 2, what would we get? Why?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Momentum Stated mathematically, the law of conservation of momentum for an isolated system is The total momentum after an interaction is equal to the total momentum before the interaction.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. v f = v 2. B. v f is less than v 2. C. v f is greater than v 2, but less than v 1. D. v f = v 1. E. v f is greater than v 1. The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick together and continue on with velocity v f. Which of these statements is true?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. v f = v 2. B. v f is less than v 2. C. v f is greater than v 2, but less than v 1. D. v f = v 1. E. v f is greater than v 1. The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick together and continue on with velocity v f. Which of these statements is true?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Collisions Problem A 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Collisions Problem - Answer A 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity (include direction)? 1.5 m/s west

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An explosion in a rigid pipe shoots out three pieces. A 6 g piece comes out the right end. A 4 g piece comes out the left end with twice the speed of the 6 g piece. From which end does the third piece emerge? A. Right end B. Left end

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. An explosion in a rigid pipe shoots out three pieces. A 6 g piece comes out the right end. A 4 g piece comes out the left end with twice the speed of the 6 g piece. From which end does the third piece emerge? A. Right end B. Left end

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Explosion Problem - Radioactivity The nucleus of an atom travels east at 1.0 x 10 6 m/s. The nucleus undergoes radioactive decay and breaks into two masses the smaller being 1/10 th the mass of the original nucleus. The smaller fragment travels east at 2.0 x 10 7 m/s. Determine the velocity of the larger fragment.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Explosion Problem - Radioactivity The nucleus of an atom travels east a 1.0 x 10 6 m/s. The nucleus undergoes radioactive decay and breaks into two masses the smaller being 1/10 th the mass of the original nucleus. The smaller fragment travels east at 2.0 x 10 7 m/s. Determine the velocity of the larger fragment. Answer: 1.1 x 10 6 m/s going west.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Momentum Race Identical constant net forces continuously push blocks A and B from the starting point to the finish. Both blocks start at rest. Block A has 4 times the mass of Block B. Which block has the larger change in momentum? a.A, since it has a greater m b.B, since it will have a greater v c.Equal, the mass and velocity change will make the quantities equal. d.Too little information given to determine

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Momentum Race - Answer Which block has the larger change in momentum? A: ∆p = F ∆t, and both forces are equal. However, F = ma, so the larger mass will have less acceleration. It will take more time to reach the finish line.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Objects A and C are made of different materials, with different “springiness,” but they have the same mass and are initially at rest. When ball B collides with object A, the ball ends up at rest. When ball B is thrown with the same speed and collides with object C, the ball rebounds to the left. Compare the velocities of A and C after the collisions. Is v A greater than, equal to, or less than v C ? A. v A > v C B. v A < v C C. v A = v C

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Compare the velocities of A and C after the collisions. Is v A greater than, equal to, or less than v C ? A. v A > v C B. v A < v C C. v A = v C ∆p Ball is greater after collision with C and ∆p Box = ∆p ball