1. This Week  Momentum Is momentum in basketball physics?  Rockets and guns How do spaceships work?  Collisions of objects They get impulses!  Practical Propulsion 6/13/2016Physics 214 Fall 20101

2. 6/13/2016Physics 214 Fall 20102 Momentum What happens when a force acts on an object We know v = v 0 + at So that mv – mv o = mat mv – mv o = Ft The quantity mv is known as the momentum p and it is a vector quantity in the same direction as the velocity. The result of applying a force for a time t results in a change of momentum. So a body moving with velocity v has kinetic energy = 1/2mv 2 and momentum = mv F

3. 6/13/2016Physics 214 Fall 20103 Conservation of Momentum F1F1 F2F2 F 1 = -F 2 and Δp 1 + Δp 2 = 0 Total momentum is conserved For a short interval of time  t we have mv – mv o = F  t and  p = F  t This is the impulse equation If we have a collision between two objects then in general the velocity of each object changes and kinetic energy is lost in the form of heat, sound or in a permanent deformation of the bodies.

4. 6/13/2016Physics 214 Fall 20104 Isolated systems In each case m 1 v 1 = - m 2 v 2 anim0010.mov http://www.physics.purdue.edu/academic_programs/courses/phys214/lectures/anim0010.mov

5. 6/13/2016Physics 214 Fall 20105 Momentum conservation The conservation of momentum also is connected to the fundamental physical laws The laws of physics do not change under translation or rotation in space

6. 6/13/2016Physics 214 Fall 20106 Impulse We have seen that a force F acting for a time Δt changes momentum FΔt = Δp. FΔt is called an impulse There are many situations where it is more useful to use the impulse when two objects collide than equating the change in momentum of each. In the case shown below the momentum of the earth changes but that is too difficult to calculate. It is better to use the fact that the earth exerts a force for a short time.

7. 6/13/2016Physics 214 Fall 20107Collisions In a closed, isolated system containing a collision, the linear momentum of each colliding object can change but the total momentum of the system is a constant. This statement is true even if energy is lost by the colliding bodies *

8. 6/13/2016Physics 214 Fall 20108 Types of collision  Elastic - no energy is lost  Inelastic - Energy is lost (transformed)  Perfectly inelastic – objects stick together  In two dimensions momentum is conserved along the x and y axes separately y x

9. 6/13/2016Physics 214 Fall 20109 Perfectly Inelastic Let initial velocity be v i and the mass of a car be m. Then mv i = 3mv final and v final = v i /3 Kinetic Energy before = 1/2mv i 2 Kinetic energy after = ½ x 3mv 2 final So KE before /KE final = 3

10. 6/13/2016Physics 214 Fall 201010 Head on Elastic collisions Pool ball collisions are close to being elastic vv Pool balls v Bowling ball hits tennis ball Tennis ball hits bowling ball http://www.physics.purdue.edu/class/applets/phe/collision.htm

11. 6/13/2016Physics 214 Fall 201011 Collisions of Particles Both energy and momentum and charge are conserved in elementary particle interactions and are a powerful tool in analyzing the fundamental physics. There are also other important conservation laws that for example prevent the proton from decaying.

12. 6/13/2016Physics 214 Fall 201012 Summary of Chapter 7 The action of a force changes the momentum of an object mv – mv o = Ft p = mv and is a vector When two bodies interact they feel equal and opposite forces F 1 = -F 2 and Δp 1 - Δp 2 = 0 Total momentum is conserved

13. 6/13/2016Physics 214 Fall 201013 Summary: Impulse FΔt = Δp is the impulse equation and is used to determine the momentum change when a force acts

14. Practical Propulsion A space vessel has to have a propulsion system and this requires an engine and fuel or an external force. It has to be able to maneuver and be able to escape from the gravitational attraction of all objects that affect it’s path. Or, for example, to leave the moon after landing.  Conventional rockets are ~90% fuel by weight most of which is used escaping from the earth.  Very small satellites might be put into orbit using a powerful laser beam  Nuclear engines are used in deep space probes. In this radioactive decay of the fuel emits particles which eject “backwards” and give the probe momentum. In this case it takes a long time to achieve high velocities  Solar sails have been tested which use enormous sails pushed by the solar wind of particles. This technique has limited application  Remember the nearest star is 4 light years away that is ~ 2 x 10 13 miles or ~200,000 times the distance to the sun (ONE WAY!!) 6/13/2016Physics 214 Fall 201014

15. 6/13/2016Physics 214 Fall 201015 1N-02 Collision of Two Large Balls In practice some energy is always lost. You can hear the noise when they hit and there will be some heat generated at impact Conservation of Energy (Elastic) ½ mv 1 2 + ½ mv 2 2 = ½ mv 1 2 + ½ mv 2 2  v 1 2 = v 1A 2 + v 2A 2 v 1A = 0 & v 2A = v 1 Can we predict the velocities of each ball after a collision ? What happens when two large balls of equal mass collide one is at rest at the other has velocity v 1 ? Completely Inelastic collision (stick together) mv 1 + mv 2 = (m + m)v A (v 2 =0)  v 1 = 2v A v A = ½v 1 Conservation of momentum mv 1 + mv 2 = mv 1A + mv 2A  v 1 = v 1A + v 2A

16. 6/13/2016Physics 214 Fall 201016 1N-04 Conservation of Linear Momentum IF THE MASSES ARE IN INVERSE RATIO TO THE INITIAL DISTANCES, THE CARTS WILL ARRIVE AT THE STOPS SIMULTANEOUSLY. TT Is momentum conserved in this system ? d A /d B = (v A t A ) / (v B t B ) If, t A = t B d A /d B = m B / m A v A / v B = m B / m A Two carts move under tension of weight on frictionless track The initial momentum of the carts is zero and they each feel equal and opposite forces. So at any time the net momentum will be zero. 0 = m A v A – m B v B  m A v A = m B v B

17. 6/13/2016Physics 214 Fall 201017 1H-01 Action – Reaction NO MATTER HOW THE PULLING IS DONE, THE CARTS END UP IN THE SAME PLACE. NO EXTERNAL FORCES ACT ON SYSTEM. THE CENTER OF MASS STAYS WHERE IT IS. SO THE CARTS ALWAYS MEET AT THE CENTER OF MASS. CM Two Carts, Supporting different masses, are pulled and pushed together in different ways. Note where the carts come together.

18. 6/13/2016Physics 214 Fall 201018 1N-05 Elastic Collision (Magnets) MOMENTUM TRANSFER AND CONSERVATION REQUIRE ONLY THAT THERE BE A MUTUAL INTERACTION. AT THE MICROSCOPIC LEVEL, ALL CONTACT INVOLVES ELECTROMAGNETIC INTERACTIONS. We know that the magnets repel each other, so they will not touch. So is this a collision? SS v Rest SS v What does it mean for objects to ‘touch’ ? Two Magnets collide with like poles facing each other

19. 6/13/2016Physics 214 Fall 201019 1N-06 Equality of Momentum THE SPRING FORCE THAT DRIVES THEM APART IS INTERNAL TO THE SYSTEM, SO THE NET MOMENTUM REMAINS ZERO. SINCE THE METAL CYLINDER IS HEAVIER IT FLIES AWAY WITH A SMALLER VELOCITY TO CONSERVE MOMENTUM What happens when the spring is released ? First case: Both identical Second case: One much heavier Two cylinders are exploded apart by a spring Use Momentum conservation m 1 v 1 = m 2 v 2 The height reached is an indication of the initial speed since 1/2mv 2 = mgh v = sqrt (2gh)

20. 6/13/2016Physics 214 Fall 201020 1N-10 Elastic & Inelastic Collisions WE CAN MEASURE THE SPEED BY TIMING THE CARTS ACROSS A FIXED DISTANCE. For THE INELASTIC CASE HALF THE VELOCITY IMPLIES IT SHOULD TAKE TWICE THE TIME. Elastic and Inelastic collisions of Two identical Carts on a Frictionless Track How do the collisions compare? Conservation of momentum mv A + mv B = mv A ’ + mv B ’ Conservation of Energy (Elastic) ½ mv A 2 + ½ mv B 2 = ½ mv A ’ 2 + ½ mv B ’ 2 If v B = 0 then v A ’ = 0 and v B ’ = v A Completely inelastic (two carts stick) if v B = 0 then v AB = ½ v A

21. 6/13/2016Physics 214 Fall 201021 1N-12 Fun Balls NO MATTER HOW MANY BALLS ARE PULLED BACK, THE SAME NUMBER RECOIL AT THE SAME SPEED. What happens when more of the balls are pulled back than are left at rest? An enlarged version of the Classic Toy - The Array of Steel Balls The collision is nearly elastic so we can use both momentum conservation and kinetic energy conservation What happens if we use the big ball? First case pull back one ball and release

22. 6/13/2016Physics 214 Fall 201022 Questions Chapter 7 Q5 Are impulse and momentum the same thing? Explain. Q6 If a ball bounces off a wall so that its velocity coming back has the same magnitude that it had prior to bouncing: A. Is there a change in the momentum of the ball? Explain. B. Is there an impulse acting on the ball during its collision with the wall? Explain. A. Yes momentum is a vector B. Yes a force acts for a short time No impulse changes momentum

23. 6/13/2016Physics 214 Fall 201023 Q9 What is the advantage of an air bag in reducing injuries during collisions? Explain using impulse and momentum ideas. Q11 If you catch a baseball or softball with your bare hand, will the force exerted on your hand by the ball be reduced if you pull your arm back during the catch? Explain. It increases the time over which the force acts. It also spreads the force over a larger area Yes. The impulse is the same but the impact time is longer. From a work point of view the kinetic energy = Fd so increasing d reduces F

24. 6/13/2016Physics 214 Fall 201024 Q17 A compact car and a large truck have a head-on collision. During the collision, which vehicle, if either, experiences: A. The greater force of impact? Explain. B. The greater impulse? Explain. C. The greater change in momentum? Explain. D. The greater acceleration? Explain. Q22 Is it possible for a rocket to function in empty space (in a vacuum) where there is nothing to push against except itself? Yes. It ejects material at high velocity and momentum conservation means the rocket recoils A. The forces are equal and opposite B. The impulse for each is the same C. The momentum changes are equal and opposite D. F = ma so a is larger for the compact car

25. 6/13/2016Physics 214 Fall 201025 Q23 Suppose that you are standing on a surface that is so slick that you can get no traction at all in order to begin moving across this surface. Fortunately, you are carrying a bag of oranges. Explain how you can get yourself moving. Q24 A railroad car collides and couples with a second railroad car that is standing still. If external forces acting on the system are ignored, is the velocity of the system after the collision equal to, greater than, or less than that of the first car before the collision? Throw the oranges opposite to the direction you wish to move The velocity after is exactly half

26. 6/13/2016Physics 214 Fall 201026 What is the momentum of a 1200 kg car traveling at 27 m/s? P= mv = (1200 kg)(27 m/s) P = 32400 kg m/s M v Ch 7 E 2

27. 6/13/2016Physics 214 Fall 201027 A ball experiences a change in momentum of 9.0 kg∙m/s. a)What is the impulse? b) If the time of interaction = 0.15 s, what is the magnitude of the average force on the ball? a) Impulse =  p = 9.0 kg m/s b) Impulse = F  t, F = 9/0.15 = 60N Ch 7 E 6

28. 6/13/2016Physics 214 Fall 201028 A ball has an initial momentum = 2.5 kg m/s, it bounces off a wall and comes back in opposite direction with momentum = -2.5 kg m/s a) What is the change in momentum of the ball? b) What is the impulse? b) Impulse = Δp = - 5kgm/s a) Δp = p f – p i = -2.5 – (+2.5) = - 5kgm/s Ch 7 E 8 P i = 2.5kgm/s P f = -2.5kgm/s F +

29. 6/13/2016Physics 214 Fall 201029 M 1 and M 2 collide head on a) Find initial momentum of M 1 and M 2 b) What is the total momentum of the system before collision? c) Ignore external forces, if they stick together after collision, which way do the masses travel? a) p 1 = -100 x 3.5 = 350kgm/s p 2 = 80 x 6 = 480kgm/s b) Total momentum = 480 – 350 = 130kgm/s east c) The masses will travel east with p = 130kgm/sec 6.0m/s 3.5m/s M 1 = 100kg M 2 = 80kg east west Ch 7 E 10

30. 6/13/2016Physics 214 Fall 201030 M 1 = 4000kg v 1 = 10m/s due north M 2 = 1200kg v 2 = 20m/s due south Masses collide and stick. a)Find initial momentum of each mass. b)Find size and direction of momentum after collision. a)Call due north the +x direction p 1 = m 1 v 1 = 4000 kg (10m/s) = 40000 kg m/s p 2 = m 2 v 2 = 1200 kg (-20m/s) = - 24000 kg m/s b)p = p 1 + p 2 = +16000 kg m/s This is momentum of system before collision, but momentum is conserved. So after masses stick: p = 16000 kg m/s due North. N S M2M2 M1M1 v2v2 v1v1 +x Ch 7 E 16

31. 6/13/2016Physics 214 Fall 201031 y x a)p 1 = 40000kgm/s + y p 2 = 30000kgm/s + x A truck of mass 4000kg and speed 10m/s collides at right angles with a car of mass 1500kg and a speed of 20m/s. a)Sketch momentum vectors before collision b)Use vector addition to get total momentum of system before collision. b) p 1 = 40000 p 2 = 30000 p p 2 = p 1 2 + p 2 2 P = 50000kgm/s Ch 7 E 18

32. 6/13/2016Physics 214 Fall 201032 A bullet is fired into block sitting on ice. The bullet travels at 500 m/s with mass 0.005 kg. The wooden block is at rest with a mass of 1.2 kg. Afterwards the bullet is embedded in the block. a) Find the velocity of the block and bullet after the impact (assume momentum is conserved). b) Find the magnitude of the impulse on the block of wood. c) Does the change in momentum of the bullet equal that of wood? m M No friction (ice) b) Impulse = Δp = p final – p initial = (1.2 kg)(2.07 m/s) – 0 = 2.50 kg m/s c)Δp for bullet = (0.005 kg)(500 m/s) – (0.005 kg)(2.07 m/s) = 2.50 kg m/s Momentum is conserved, so momentum lost by bullet is gained by wood. Ch 7 CP 2 a)p final = p initial = (0.005 kg)(500 m/s) p final = (M bullet + M wood )v = 2.5 kg m/s v = (2.5 kg m/s)/(1.205 kg) = 2.07 m/s

33. 6/13/2016Physics 214 Fall 201033 Car travels 18 m/s and hits concrete wall. M driver = 90 kg. a)Find change in momentum of driver. b)What impulse produces this change in momentum? c)Explain difference between wearing and not wearing a seat belt. a) When driver comes to a stop his p = 0.  p = 0 – (18 m/s)(90 kg) = -1620 kg m/s b) Impulse =  p = -1620 kg m/s c)Impulse = F  t With seat belt:  t is large and F is spread over torso of driver Without seat belt:  t is small. This makes F much larger (F = Impulse/  t). Ch 7 CP 4 v pp