Chapter 9: Impulse and Momentum

Transcription

1 Conservation Laws Much of the rest of Phys 221 will be concerned with conservation of energy and momentum This new perspective of conservation laws requires us to compare situations BEFORE and AFTER an interaction. We start with momentum. Energy comes in chapter 10 1

2 9.1 Momentum and Impulse p Solving Momentum and Impulse Problems p Conservation of Momentum p Explosions p Inelastic Collisions p

3 9.1 Momentum and Impulse p

4 The momentum of a particle of mass m and velocity v is defined as p = m v NOTE: Momentum is a VECTOR Its units are kg.m/s Like any vector, p can be decomposed into x- and y-components p = m v x x p = m v y y p has the same sign as v x x p has the same sign as v y y 4

5 Cliction 9.1 The cart s change of momentum is kg m/s kg m/s kg m/s kg m/s kg m/s. 5

6 The change in momentum is related to the (net) force: This means that the change of momentum is given by an integral over the net force: or, explicitly written with the components, = F = p - p f i Change of momentum in the x-direction is due to F x Change of momentum in the x-direction 6 is due to F y

7 DEFINITION OF AN IMPULSE Object A INTERACTING with Object B Understanding an IMPULSE requires thinking about a force exerted over an INTERVAL OF TIME! 7

8 We call the integral over the force impulse J, Or stated in another way, the impulse is the area under the force-time graph 8

9 Impulse is a VECTOR with components AVERAGE force during impulse This area can also be expressed as the duration of the interaction (when a force is applied) times the average force 9

10 IMPULSE-MOMENTUM THEOREM Δp x = J x An impulse delivered to a PARTICLE changes the particle s momentum! Particle s momentum after the INTERACTION p = p + J f i x Particle s momentum JUST before the INTERACTION Impulse that arises from the INTERACTION p = p + area under the force curve f i 10

11 Chapter 9: Momentum Impulse-momentum theorem and velocity change Impulse-momentum theorem gave us p = p + area under the force curve f i But since p = mv, it is equivalent to v = v + area under the force curve Comparing to what we have learned in earlier chapters: For a particle or an object f i m Force Impulse causes a change in velocity (which we know is an acceleration) causes a change in momentum 11

12 9.2 Solving Momentum and Impulse Problems p

13 A typical application of the impulse-momentum theorem: Δp x = J x is the collision of an elastic ball with a wall, or a collision between two particles The impulse momentum theorem is useful to estimate the average force acting on a particle given that we know the momentum change 13

14 The momentum change associated with an impulse is sometimes represented with a momentum bar chart: EXAMPLE 1 EXAMPLE 2 In both examples J x >0, but for a) the initial momentum is >0, for b) it is <0 14

15 How to use the momentum-impulse theorem: Example 9.2 If this force-time graph is given, how high does this 100g ball bounce? NOTE: you are not given the ball s momentum JUST before interaction; YOU need to find That FIRST! 15

16 Initial height: 2m, initial velocity: 0 m/s Find velocity v 1y immediately before impact: The leading minus sign is because the ball is falling down Momentum before impact: p 1y = m v 1y = kg m/s Impulse J 16 y = area = 4 ms 300 N = 1.2 N s = 1.2 kg m/s

17 Momentum after bouncing off the ground: p 2y = p 1y + J y = kg m/s >0, the direction of the ball has been inverted. Speed v 2y = p 2y / m = 5.74 m/s Final height: 2 g y = v 2y 2 y = v 2y 2 / (2g) = 1.68 m The difference to the initial height is due to energy loss (e.g., breaking some molecular bonds of the Earth's surface). We will come back to this later For your own consideration: The dropped ball has a LONG impulse due to gravity as it falls, then a SHORT impulse due to both the normal force and gravity as it bounces! 17

18 Cliction 9.2 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? 1. The clay ball exerts a larger impulse because it sticks. 2. The rubber ball exerts a larger impulse because it bounces. 3. They exert equal impulses because they have equal momenta. 4. Neither exerts an impulse on the wall because the wall doesn t move. 18

19 9.3 Conservation of Momentum p

20 FROM SINGLE PARTICLES/OBJECTS TO SYSTEMS The most important consequence of the momentumimpulse theorem is the conservation of momentum for isolated systems (no environment or external forces) The Impulse-momentum theorem is used in the context of SINGLE-PARTICLE dynamics. Consider now two particles which are interacting by an action/reaction pair of forces, with no other forces acting on them. Their total momentum is P = p 1 + p 2. Then This means that the total momentum of the two particles doesn't change, regardless of the details of the interaction forces 20

21 This means that interaction forces between particles do not change the total momentum of the system In general for an isolated system (no external forces) the total momentum is conserved, dp/dt =0, regardless of the details of the interaction between the particles Law of conservation of momentum: The total momentum of an isolated system is a constant P f = P i Interactions within the system do not change the system's total momentum 21

22 Let us assume the motion is one-dimensional along the x-axis: (P x ) + (P x ) = constant 1 2 Particle 1 Particle 2 This leads to Law of Momentum Conservation for 2 interacting particles (P x ) + (P x ) 1 2 FINAL = (P x ) + (P x ) 1 2 INITIAL 22

23 Applications of momentum conservation Example 9.3 in textbook: Known: Initial total momentum P i =0 final velocity (v fx ) 2 = 6 m/s What is (v fx ) 1? Hence, 23

24 Abbreviated Example 9.4 in textbook: Bob jumps on the resting cart with a velocity of (v 1x ) B = 4.0 m/s. In the longer version you first need to calculate the Boy s momentum (thus speed) just before he jumps on the cart. How fast is the combined system afterwards? Momentum conservation P = P ix fx Equating equations (1) and (2) (1) (2) 24

25 Cliction 9.3 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? 1. v A > v C 2. v A = v C 3. v A < v C 25

26 This result can be generalized to an arbitrary number of INTERACTING particles. Consider N particles: Total momentum: However, if system is isolated, then 0 26

27 Problem-Solving Strategy The trick in using the law of momentum conservation is to choose your combined system wisely If possible, choose a system with F net =0 (no external net force). This makes the analysis much easier 27

28 End of Week 9 28

29 9.4 Inelastic Collisions p

30 Inelastic collisions are another example for momentum conservation A collision is inelastic if some or all of the initial momentum of two objects is used to deform the objects. Examples: chemical reactions, earthquakes, bruises after hitting something Perfectly inelastic collisions correspond to the case when the two colliding objects stick together 30

31 Simple example (9.7): Two gliders stick together after colliding Initial total momentum: P ix = m 1 (v ix ) 1 + m 2 (v ix ) 2 Final total momentum: P fx = (m 1 + m 2 ) v fx Momentum conservation 31

32 Cliction 9.4 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? 1. v f is greater than v v f = v v f is less than v v f = v v f is greater than v 2, but less than v 1. 32

33 9.5 Explosions p Particles of the system move apart from each other after a BRIEF, INTENSE interaction. The EXPLOSIVE FORCES are internal forces. 33

34 Explosions are ideally suited to apply momentum conservation After a short time of interaction the particles move apart. Examples include guns, rocket engines, and (Example 9.6) radioactivity: Known: initial mass m, final velocities (v fx ) 1, (v fx ) 2 Isotope of Uranium with Atomic mass of 238 u (92 protons+146 neutrons) Find masses m 1 and m 2 34

35 Momentum conservation The above is an equation for two unkowns m 1 and m 2. We need a 2 nd equation which is mass conservation: m=m 1 +m 2 Helium! = 234 u m 2 = m - m1 = 238 u u = 4 u 35

36 Chapter 9 Reading Quiz 36

37 Impulse is 1. a force that is applied at a random time. 2. a force that is applied very suddenly. 3. the area under the force curve in a forceversus-time graph. 4. the time interval that a force lasts. 37

38 The total momentum of a system is conserved 1. always. 2. if the system is isolated. 3. if the forces are conservative. 4. never; it s just an approximation. 38

39 In an inelastic collision, 1. impulse is conserved. 2. momentum is conserved. 3. force is conserved. 4. energy is conserved. 5. elasticity is conserved. 39

40 Selected Problems 40

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45 End of Chapter 9 IMPORTANT: Print a copy of the SUMMARY page (p. 260) and add it here to your lecture notes. It will save you crucial time when trying to recall: Concepts, Symbols, and Strategies 45

46 End of Week 10 46