6 th week Lectures Feb. 12. Feb
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1 Momentum Rockets and guns 6 th week Lectures Feb. 12. Feb How do spaceships work? Collisions of objects They get impulses! Practical Propulsion 2/11/2018 Physics 214 Spring
2 Announcements Exam 1 Feb Tuesday 8:00-10:00pm in Phys. Room 112 Exam Calculator: When taking a Physics 214 Exam, there is only one calculator model that is acceptable: The CASIO fx-260 SLRSC FRACTION. NO OTHER BRAND or TYPE WILL BE ALLOWED! Please bring your I.D. card and #2 pencil TA's office hours changed Tuesday 2:00 5:00pm Physics Room 12 2/11/2018 Physics 214 Spring
3 Momentum What happens when a force F acts on an object of mass m? We known v = v + at thus v v at after multiplying both side by m f i f i mv f mv i mat. Since F ma we have: Let us introduce the concept mvf mvi Ft of momentum vector as p mv p p p f pi F t F In the limit t t The quantity m v is known as the momentum p, p mv and it is a vector quantity points in the same direction as the velocity. The result of applying a force for a time t on mass m results in a change of momentum p of mass m, as exhibited in the equations above. So a body moving with velocity v has kinetic energy = 1/2mv 2 and momentum 0 F we obtain m dp dt F p mv 2/11/2018 Physics 214 Spring
4 Principle of Conservation of Linear Momentum The total linear momentum of an isolated system remains constant (is conserved). An isolated system is one for which the vector sum of the external forces acting on the system is zero. Examples: 1. Exploding system p 0 p p 0 i f fragments Before explosion Before pushing each others After explosion 2. Skaters pushing each others at rest. p p left right p 0 p 0 i f After pushing each other 2/11/2018 Physics 214 Spring
5 3. Gun before firing after firing p f pi 0 prifle pbullet 0 When you fire a rifle do not rest your shoulder against any solid object, because the recoiling rifle will crush the bones in your shoulder. 4. Spacecraft firing a rocket spacecraft i p s rocket p f s 0 at rest rocket moving p r p i s Before firing After firing 2/11/2018 Physics 214 Spring
6 5. Elastic Linear Collision with the target m 2 at rest, momentum and energy is conserved: i i f f p1 p2 0 p1 p2 m m m1 m2 1 2 Before Collision After Collision Imposing Energy and momentum conservation we find: m m v ( )v f 1 2 i 1 1 m1 m2 2/11/2018 Physics 214 Spring
7 Incoming mass larger Incoming mass smaller Head on Elastic collisions Subscript : R = red, B = Blue, Y = yellow Before Collision Masses are equal: m V R R m V 0 V 0 B B R After Collision m m m m V Y V R VY 0 V 0 R V B V R Tennis ball hits tennis ball Bowling ball hits tennis ball Tennis ball hits bowling ball Projectiles Targets at rest 2/11/2018 Physics 214 Spring
8 1N-06 Equality of Momentum Two cylinders are exploded apart by a spring What happens when the spring is released? First case: Both identical Second case: One much heavier 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) 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 2/11/2018 Physics 214 Spring
9 1N-12 Fun Balls An enlarged version of the Classic Toy - The Array of Steel Balls First case pull back one ball and release What happens if we use the big ball? What happens when more of the balls are pulled back than are left at rest? The collision is nearly elastic so we can use both momentum conservation and kinetic energy conservation NO MATTER HOW MANY BALLS ARE PULLED BACK, THE SAME NUMBER RECOIL AT THE SAME SPEED. 2/11/2018 Physics 214 Spring
10 1N-04 Conservation of Linear Momentum Two carts move under tension of weight on frictionless track T Is momentum conserved in this system? T 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 v A / v B = m B / m A d A /d B = (v A t A ) / (v B t B ) If, t A = t B d A /d B = m B / IF THE MASSES ARE IN INVERSE RATIO TO THE INITIAL DISTANCES, THE CARTS WILL ARRIVE AT THE STOPS SIMULTANEOUSLY. m A 2/11/2018 Physics 214 Spring
11 Isolated systems In each case m 1 v 1 = - m 2 v 2 anim0010.mov 2/11/2018 Physics 214 Spring
12 Momentum conservation The conservation of momentum is one of the fundamental physical laws The laws of physics do not change under translation or rotation in space 2/11/2018 Physics 214 Spring
13 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. 2/11/2018 Physics 214 Spring
14 Collisions 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 * 2/11/2018 Physics 214 Spring
15 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 2/11/2018 Physics 214 Spring
16 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 2/11/2018 Physics 214 Spring
17 1N-10 Elastic & Inelastic Collisions 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 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. 2/11/2018 Physics 214 Spring
18 1N-02 Collision of Two Large Balls What happens when two large balls of equal mass collide one is at rest at the other has velocity v 1? Conservation of momentum mv 1 + mv 2 = mv 1A + mv 2A v 1 = v 1A + v 2A Can we predict the velocities of each ball after a collision? Conservation of Energy (Elastic) ½ mv ½ mv 2 2 = ½ mv ½ mv 2 2 v 1 2 = v 1A 2 + v 2A 2 v 1A = 0 & v 2A = 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 In practice some energy is always lost. You can hear the noise when they hit and there will be some heat generated at impact 2/11/2018 Physics 214 Spring
19 1N-05 Elastic Collision (Magnets) Two Magnets collide with like poles facing each other We know that the magnets repel each other, so they will not touch. So is this a collision? Rest v S S S S v Rest What does it mean for objects to touch? MOMENTUM TRANSFER AND CONSERVATION REQUIRE ONLY THAT THERE BE A MUTUAL INTERACTION. AT THE MICROSCOPIC LEVEL, ALL CONTACT INVOLVES ELECTROMAGNETIC INTERACTIONS. 2/11/2018 Physics 214 Spring
20 Collisions of Proton Particles p incident proton _ p p p p p target proton recoil proton p 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, while pions ( ) will ultimately decay: meson p recoil proton 2/11/2018 Physics 214 Spring meson
21 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 2/11/2018 Physics 214 Spring
22 Summary: Impulse FΔt = Δp is the impulse equation and is used to determine the momentum change when a force acts 2/11/2018 Physics 214 Spring
23 Practical Propulsion A space vessel requires an engine and fuel or an external force. It has to be able to maneuver and be able to escape from the gravity 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 the 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 miles or ~200,000 times the distance to the sun (ONE WAY!!) Voyager just left the solar system after 36 years!! 2/11/2018 Physics 214 Spring
24 Questions Chapter 7 Q5 Are impulse and momentum the same thing? Explain. No impulse changes momentum 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 2/11/2018 Physics 214 Spring
25 Q9 What is the advantage of an air bag in reducing injuries during collisions? Explain using impulse and momentum ideas. It increases the time over which the force acts. It also spreads the force over a larger area 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. 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 2/11/2018 Physics 214 Spring
26 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. 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 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 2/11/2018 Physics 214 Spring
27 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. Throw the oranges opposite to the direction you wish to move 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? The velocity after is exactly half 2/11/2018 Physics 214 Spring
28 Ch 7 E 2 What is the momentum of a 1200 kg car traveling at 27 m/s? P= mv = (1200 kg)(27 m/s) P = kg m/s M v 2/11/2018 Physics 214 Spring
29 Ch 7 E 6 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 2/11/2018 Physics 214 Spring
30 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? Ch 7 E 8 a) Δp = p f p i = -2.5 (+2.5) = - 5kgm/s P i = 2.5kgm/s b) Impulse = Δp = - 5kgm/s P f = -2.5kgm/s F + 2/11/2018 Physics 214 Spring
31 M 1 and M 2 collide head on Ch 7 E 10 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? west M 2 = 80kg 6.0m/s 3.5m/s M 1 = 100kg east a) p 1 = -100 x 3.5 = - 350kgm/s p 2 = 80 x 6 = 480kgm/s b) Total momentum = = 130kgm/s east c) The masses will travel east with p = 130kgm/sec 2/11/2018 Physics 214 Spring
32 Ch 7 E 16 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. N M 2 +x v 2 a) Call due north the +x direction p 1 = m 1 v 1 = 4000 kg (10m/s) = kg m/s v 1 p 2 = m 2 v 2 = 1200 kg (-20m/s) = kg m/s S M 1 b) p = p 1 + p 2 = kg m/s This is momentum of system before collision, but momentum is conserved. So after masses stick: p = kg m/s due North. 2/11/2018 Physics 214 Spring
33 Ch 7 E 18 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. a) p 1 = 40000kgm/s + y p 2 = 30000kgm/s + x b) y p p p 1 2 p 1 p 1 = car 2 p 1 p p p 1 2 p 2 p 2 x p 2 = car 1 Pythagoras theorem:p 2 = p p 2 2 P = 50000kgm/s 2/11/2018 Physics 214 Spring
34 Ch 7 CP 2 A bullet is fired into block sitting on ice. The bullet travels at 500 m/s with mass 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? a) p final = p initial = (0.005 kg)(500 m/s) p final = (M bullet + M wood )v = 2.5 kg m/s m v v = (2.5 kg m/s)/(1.205 kg) = 2.07 m/s 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) 2/11/2018 Physics 214 Spring M No friction (ice) = 2.50 kg m/s Momentum is conserved, so momentum lost by bullet is gained by wood.
35 Ch 7 CP 4 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) = kg m/s b) Impulse = p = kg m/s v p 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). 2/11/2018 Physics 214 Spring
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