Linear Momentum And Collisions - Weebly
9Linear Momentum and CollisionsCHAPTER OUTLINE9.19.29.39.49.59.69.7Linear Momentum and ItsConservationImpulse and MomentumCollisions in One DimensionTwo-Dimensional CollisionsThe Center of MassMotion of a System ofParticlesRocket Propulsion(c)ANSWERS TO QUESTIONSQ9.1No. Impulse, F t , depends on the force and the time for whichit is applied.Q9.2The momentum doubles since it is proportional to the speed.The kinetic energy quadruples, since it is proportional to thespeed-squared.Q9.3The momenta of two particles will only be the same if themasses of the particles of the same.Q9.4(a)It does not carry force, for if it did, it could accelerateitself.(b)It cannot deliver more kinetic energy than it possesses.This would violate the law of energy conservation.It can deliver more momentum in a collision than it possesses in its flight, by bouncing fromthe object it strikes.Q9.5Provided there is some form of potential energy in the system, the parts of an isolated system canmove if the system is initially at rest. Consider two air-track gliders on a horizontal track. If youcompress a spring between them and then tie them together with a string, it is possible for thesystem to start out at rest. If you then burn the string, the potential energy stored in the spring willbe converted into kinetic energy of the gliders.Q9.6No. Only in a precise head-on collision with momenta with equal magnitudes and oppositedirections can both objects wind up at rest. Yes. Assume that ball 2, originally at rest, is strucksquarely by an equal-mass ball 1. Then ball 2 will take off with the velocity of ball 1, leaving ball 1 atrest.Q9.7Interestingly, mutual gravitation brings the ball and the Earth together. As the ball movesdownward, the Earth moves upward, although with an acceleration 10 25 times smaller than that ofthe ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system isconserved.Q9.8(a)Linear momentum is conserved since there are no external forces acting on the system.(b)Kinetic energy is not conserved because the chemical potential energy initially in theexplosive is converted into kinetic energy of the pieces of the bomb.251
252Linear Momentum and CollisionsQ9.9Momentum conservation is not violated if we make our system include the Earth along with theclay. When the clay receives an impulse backwards, the Earth receives the same size impulseforwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than theacceleration of the clay, but the planet absorbs all of the momentum that the clay loses.Q9.10Momentum conservation is not violated if we choose as our system the planet along with you.When you receive an impulse forward, the Earth receives the same size impulse backwards. Theresulting acceleration of the Earth due to this impulse is significantly smaller than your accelerationforward, but the planet’s backward momentum is equal in magnitude to your forward momentum.Q9.11As a ball rolls down an incline, the Earth receives an impulse of the same size and in the oppositedirection as that of the ball. If you consider the Earth-ball system, momentum conservation is notviolated.Q9.12Suppose car and truck move along the same line. If one vehicle overtakes the other, the fastermoving one loses more energy than the slower one gains. In a head-on collision, if the speed of them 3m ctimes the speed of the car, the car will lose more energy.truck is less than T3mT m cQ9.13The rifle has a much lower speed than the bullet and much less kinetic energy. The butt distributesthe recoil force over an area much larger than that of the bullet.Q9.14His impact speed is determined by the acceleration of gravity and the distance of fall, inv 2f vi2 2 g 0 yi . The force exerted by the pad depends also on the unknown stiffness of the pad.Q9.15The product of the mass flow rate and velocity of the water determines the force the firefightersmust exert.Q9.16The sheet stretches and pulls the two students toward each other. These effects are larger for afaster-moving egg. The time over which the egg stops is extended so that the force stopping it isnever too large.Q9.17(c) In this case, the impulse on the Frisbee is largest. According to Newton’s third law, the impulseon the skater and thus the final speed of the skater will also be largest.Q9.18Usually but not necessarily. In a one-dimensional collision between two identical particles with thesame initial speed, the kinetic energy of the particles will not change.Q9.19g downward.Q9.20As one finger slides towards the center, the normal force exerted by the sliding finger on the rulerincreases. At some point, this normal force will increase enough so that static friction between thesliding finger and the ruler will stop their relative motion. At this moment the other finger startssliding along the ruler towards the center. This process repeats until the fingers meet at the center ofthe ruler.Q9.21The planet is in motion around the sun, and thus has momentum and kinetic energy of its own. Thespacecraft is directed to cross the planet’s orbit behind it, so that the planet’s gravity has acomponent pulling forward on the spacecraft. Since this is an elastic collision, and the velocity of theplanet remains nearly unchanged, the probe must both increase speed and change direction for bothmomentum and kinetic energy to be conserved.bg
Chapter 9253Q9.22No—an external force of gravity acts on the moon. Yes, because its speed is constant.Q9.23The impulse given to the egg is the same regardless of how it stops. If you increase the impact timeby dropping the egg onto foam, you will decrease the impact force.Q9.24Yes. A boomerang, a kitchen stool.Q9.25The center of mass of the balls is in free fall, moving up and then down with the acceleration due togravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the timewhen the juggler is engaged in catching and tossing, the center of mass must accelerate up with asomewhat smaller average acceleration. The center of mass moves around in a little circle, makingthree revolutions for every one revolution that one ball makes. Letting T represent the time for onecycle and Fg the weight of one ball, we have FJ 0.60T 3 Fg T and FJ 5 Fg . The average force exertedby the juggler is five times the weight of one ball.Q9.26In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn,because no outside force acts on this system. According to the text’s ‘basic expression for rocketpropulsion,’ the change in speed of the rocket body will be larger than the speed of the exhaustrelative to the rocket, if the final mass is less than 37% of the original mass.Q9.27The gun recoiled.Q9.28Inflate a balloon and release it. The air escaping from the balloon gives the balloon an impulse.Q9.29There was a time when the English favored position (a), the Germans position (b), and the Frenchposition (c). A Frenchman, Jean D’Alembert, is most responsible for showing that each theory isconsistent with the others. All are equally correct. Each is useful for giving a mathematically simplesolution for some problems.SOLUTIONS TO PROBLEMSSection 9.1P9.1Linear Momentum and Its Conservatione(a)(b)jv 3.00 i 4.00 j m sm 3.00 kg ,ejp mv 9.00 i 12.0 j kg m sThus,p x 9.00 kg m sandp y 12.0 kg m sa9.00f a12.0f 15.0 kg m sF p I tan a 1.33f 307 GH p JK2p p x2 p y2 θ tan 1yx 12
254P9.2Linear Momentum and Collisions(a)At maximum height v 0 , so p 0 .(b)Its original kinetic energy is its constant total energy,Ki af b11mvi2 0.100 kg 15.0 m s22g2 11.2 J .At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational andthe other half is kinetic:v bgbbg10.100 kg v 222 5.62 J 10.6 m s0.100 kgK 5.62 J gThen p mv 0.100 kg 10.6 m s jp 1.06 kg m s j .P9.3I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground withspeed given bydiv 2f vi2 2 a x f x i :jae0 vi2 2 9.80 m s 2 0.250 mfvi 2.20 m sTotal momentum of the system of the Earth and me is conserved as I push the earth down andmyself up:j begb0 5.98 10 24 kg v e 85.0 kg 2.20 m sgv e 10 23 m sP9.4(a)For the system of two blocks p 0 ,orpi p fTherefore,0 Mv m 3 M 2.00 m sSolving givesv m 6.00 m s (motion toward thea fbgleft).(b)a f1 2 112 3 M v 32M 8. 40 Jkx Mv M222FIG. P9.4
Chapter 9P9.5(a)The momentum is p mv , so v (b)K Section 9.2*P9.61mv 2 implies v 2FG IJH Kpp11and the kinetic energy is K mv 2 mmm222K2K, so p mv m mm2 Impulse and Momentuma fFrom the impulse-momentum theorem, F t p mv f mvi , the average force required to holdF di b12 kg gb0 60 mi hg F 1 m s I 6.44 10GH 2.237 mi h JK0.050 s 0a t fm v f vi3N.Therefore, the magnitude of the needed retarding force is 6.44 10 3 N , or 1 400 lb. A personcannot exert a force of this magnitude and a safety device should be used.*P9.8p2.2m2mK .onto the child isP9.7255(a)zI Fdt area under curvejbgI 11.50 10 3 s 18 000 N 13.5 N s2(b)F 13.5 N s 9.00 kN1.50 10 3 s(c)From the graph, we see that Fmax 18.0 kNeFIG. P9.711mv12 mgy1 . The rebound speed is given by mgy 2 mv 22 . The22impulse of the floor is the change in momentum,The impact speed is given byb megmv 2 up mv1 down m v 2 v1 upj2 gh2 2 gh1 upe 0.15 kg 2 9.8 m s 2je 1.39 kg m s upwardj0.960 m 1.25 m up
256P9.9Linear Momentum and Collisions p F tfej a ma v sin 60.0 v sin 60.0 f 2mv sin 60.0 2b3.00 kg gb10.0 m sga0.866f p y m v fy viy m v cos 60.0 mv cos 60.0 0 p x 52.0 kg m sFave P9.10P9.11 p x 52.0 kg m s 260 N0.200 s tFIG. P9.9Assume the initial direction of the ball in the –x direction.ga f bbga fe j(a)Impulse, I p p f pi 0.060 0 40.0 i 0.060 0 50.0 i 5.40 i N s(b)Work K f K i b10.060 02g a40.0f a50.0f22 27.0 JTake x-axis toward the pitcher(a)b0.200 kg gb15.0 m sga cos 45.0 f I b0.200 kggb40.0 m sg cos 30.0 pix I x p fx :xI x 9.05 N sb0.200 kg gb15.0 m sga sin 45.0 f I b0.200 kg gb40.0 m sg sin 30.0 I e9.05 i 6.12 jj N spiy I y p fy :(b)byfgaaeFm P9.12fa110 Fm 4.00 ms Fm 20.0 ms Fm 4.00 ms22Fm 24.0 10 3 s 9.05 i 6.12 j N sI fje377 i 255 jj NIf the diver starts from rest and drops vertically into the water, the velocity just before impact isfound fromK f U gf K i U gi12 0 0 mgh v impact 2 ghmv impact2With the diver at rest after an impact time of t , the average force during impact is given byF em 0 v impact tj m2 gh tor F m 2 gh t(directed upward).Assuming a mass of 55 kg and an impact time of 1.0 s , the magnitude of this average force isb55 kg g 2e9.8 m s ja10 mf 770 N , orF 21.0 s 10 3 N .
Chapter 9P9.13The force exerted on the water by the hose isF bgbg0.600 kg 25.0 m s 0 p water mv f mvi 15.0 N .1.00 s t tAccording to Newton's third law, the water exerts a force of equal magnitude back on the hose.Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting waterstream) to hold the hose stationary.*P9.14(a)Energy is conserved for the spring-mass system:K i U si K f U sf :1 2 1kx mv 2 022kv xm0 klarger.m(b)From the equation, a smaller value of m makes v x(c)I p f p i mv f 0 mx(d)From the equation, a larger value of m makes I x km larger.(e)For the glider, W K f K i k x kmm11mv 2 0 kx 222The mass makes no difference to the work.Section 9.3P9.15Collisions in One Dimensionb200 g gb55.0 m sg b46.0 g gv b200 g gb40.0 m sgv 65.2 m s*P9.16bm vg bm v m v g22.5 g b35 m sg 300 g b 2.5 m sg 22.5 gv1 1v1 f m2 v2i1 12 2 f1f 037.5 g m s 1.67 m s22.5 gFIG. P9.16257
258P9.17Linear Momentum and CollisionsMomentum is conserved10.0 10 3 kg v 5.01 kg 0.600 m sj begbgv 301 m sP9.18(a)mv1i 3mv 2 i 4mv f where m 2.50 10 4 kgvf P9.19a f4.00 3 2.00 2.50 m s4a fLMNa f OPQ ejaf1114m v 2f mv12i 3m v 22i 2.50 10 4 12.5 8.00 6.00 3.75 10 4 J222(b)K f Ki (a)The internal forces exerted by the actor donot change the total momentum of thesystem of the four cars and the movie actora4mfv a3mfb2.00 m sg mb4.00 m sgi6.00 m s 4.00 m s 2.50 m svi 4(b)(c)gbgfb ga fb11223m 2.00 m s m 4.00 m s 4 m 2.50 m s222.50 10 4 kg2 12.0 16.0 25.0 m s 37.5 kJ2Wactor K f K i Wactora fbjaFIG. P9.19eg2The event considered here is the time reversal of the perfectly inelastic collision in theprevious problem. The same momentum conservation equation describes both processes.P9.20v1 , speed of m1 at B before collision.1m1 v12 m1 gh2a fa fv1 2 9.80 5.00 9.90 m sv1 f , speed of m1 at B just after collision.m m21v1 f 1v1 9.90 m s 3.30 m sm1 m 23At the highest point (after collision)a fm1 ghmax a1m1 3.302fFIG. P9.20b 3.30 m sg 2e9.80 m s j22hmax 20.556 m
Chapter 9P9.21259(a), (b) Let v g and v p be the velocity of the girl and the plankrelative to the ice surface. Then we may say that v g v p isthe velocity of the girl relative to the plank, so thatv g v p 1.50(1)But also we must have m g v g m p v p 0 , since totalmomentum of the girl-plank system is zero relative to theice surface. Therefore45.0 v g 150 v p 0 , or v g 3.33 v pPutting this into the equation (1) above givesFIG. P9.21 3.33 v p v p 1.50 or v p 0.346 m safThen v g 3.33 0.346 1.15 m s*P9.22For the car-truck-driver-driver system, momentum is conserved:p 1i p 2 i p 1 f p 2 f :b4 000 kg gb8 m sg i b800 kg gb8 m sge ij b4 800 kg gv ifvf 25 600 kg m s 5.33 m s4 800 kgFor the driver of the truck, the impulse-momentum theorem isF t p f pi :af bgbg bgbgF 0.120 s 80 kg 5.33 m s i 80 kg 8 m s ie jF 1.78 10 3 N i on the truck driverFor the driver of the car,af bgbg bgbge jF 0.120 s 80 kg 5.33 m s i 80 kg 8 m s iF 8.89 10 3 N i on the car driver , 5 times larger.P9.23(a)According to the Example in the chapter text, the fraction of total kinetic energy transferredto the moderator isf2 4m1 m 2bm1 m2g2where m 2 is the moderator nucleus and in this case, m 2 12m1f2 b4m1 12m1b13m g12g 48 1690. 284 or 28.4%of the neutron energy is transferred to the carbon nucleus.(b)a fejJj a0.716 fe1.6 10K C 0.284 1.6 10 13 J 4.54 10 14 JKn 131.15 10 13 J
260P9.24Linear Momentum and CollisionsEnergy is conserved for the bob-Earth system between bottom andtop of swing. At the top the stiff rod is in compression and the bobnearly at rest.1Mv b2 0 0 Mg 2 A2v b2 g 4A so v b 2 gAK i Ui K f U f :FIG. P9.24Momentum of the bob-bullet system is conserved in the collision:mv mP9.25v M 2 gA2ejv 4MmgAAt impact, momentum of the clay-block system is conserved, so:bgmv1 m1 m 2 v 2After impact, the change in kinetic energy of the clay-block-surfacesystem is equal to the increase in internal energy:bbggbgge1m1 m 2 v 22 f f d µ m1 m 2 gd210.112 kg v 22 0.650 0.112 kg 9.80 m s 2 7.50 m2v 22 95.6 m 2 s 2v 2 9.77 m se12.0 10P9.26 3bj bgbkg v1 0.112 kg 9.77 m sjagfFIG. P9.25v1 91.2 m sWe assume equal firing speeds v and equal forces F required for the two bullets to push wood fibersapart. These equal forces act backward on the two bullets.For the first,K i Emech K fFor the second,pi p f17.00 10 3 kg v 2 F 8.00 10 2 m 027.00 10 3 kg v 1.014 kg v feejj be7.00 10 jv ejg 3vf1.014bejK i Emech K f :Substituting for v f ,117.00 10 3 v7.00 10 3 kg v 2 Fd 1.014 kg221.014ebje 311 7.00 10 32Fd 7.00 10 v 221.014eSubstituting for v,eg117.00 10 3 kg v 2 Fd 1.014 kg v 2f22Again,jjFGHFd F 8.00 10 2 m 1 7.00 10 31.014IJKjgFGHIJK22v2d 7.94 cm
Chapter 9*P9.27(a)261c ph c ph , givesa4.0 10 3.0f kg v b4.0 kggb5.0 m sg b10 kg gb3.0 m sg b3.0 kggb 4.0 m sg .Using conservation of momentum,afterbeforeTherefore, v 2.24 m s , or 2. 24 m s toward the right .(b)No . For example, if the 10-kg and 3.0-kg mass were to stick together first, they wouldmove with a speed given by solvingb13 kggv b10 kg gb3.0 m sg b3.0 kg gb 4.0 m sg , or v11 1.38 m s .Then when this 13 kg combined mass collides with the 4.0 kg mass, we haveb17 kggv b13 kggb1.38 m sg b4.0 kg gb5.0 m sg , and v 2.24 m sjust as in part (a). Coupling order makes no difference.Section 9.4P9.28(a)Two-Dimensional CollisionsFirst, we conserve momentum for the system of two football players in the x direction (thedirection of travel of the fullback).b90.0 kg gb5.00 m sg 0 b185 kg gV cosθwhere θ is the angle between the direction of the final velocity V and the x axis. We findV cos θ 2.43 m s(1)Now consider conservation of momentum of the system in the y direction (the direction oftravel of the opponent).b95.0 kg gb3.00 m sg 0 b185 kggaV sinθ fwhich gives,V sin θ 1.54 m sDivide equation (2) by (1)tan θ V 2.88 m sThen, either (1) or (2) givesbbgbgb190.0 kg 5.00 m s21K f 185 kg 2.88 m s2Ki 1.54 0.6332.43θ 32.3 From which(b)(2)ggbgb195.0 kg 3.00 m s22 2 7.67 10 2 Jg2 1.55 10 3 JThus, the kinetic energy lost is 783 J into internal energy.
262P9.29Linear Momentum and Collisionsp xf p xibmvO cos 37.0 mv Y cos 53.0 m 5.00 m s0.799 vO 0.602 v Y 5.00 m sg(1)p yf p yimvO sin 37.0 mv Y sin 53.0 00.602 vO 0.799 v Y(2)Solving (1) and (2) simultaneously,vO 3.99 m s and v Y 3.01 m s .P9.30p xf p xi :afmvO cos θ mv Y cos 90.0 θ mvivO cos θ v Y sin θ vip yf p yi :FIG. P9.29a(1)fmvO sin θ mv Y sin 90.0 θ 0vO sin θ v Y cos θ(2)From equation (2),vO v YFG cosθ IJH sinθ K(3)FIG. P9.30Substituting into equation (1),vYsoF cos θ I vGH sinθ JK2eYsin θ vijv Y cos 2 θ sin 2 θ vi sin θ , and v Y vi sin θ .Then, from equation (3), vO vi cos θ .We did not need to write down an equation expressing conservation of mechanical energy. In theproblem situation, the requirement of perpendicular final velocities is equivalent to the condition ofelasticity.
Chapter 9P9.31The initial momentum of the system is 0. Thus,a1.20mfvandBib m 10.0 m sgv Bi 8.33 m sbb gagafbgFG eH11122m 10.0 m s 1. 20m 8.33 m s m 183 m 2 s 2222111 122K f m vG 1.20m v B m 183 m 2 s 2222 2Ki orfb gvG2 1.20 v B2 91.7 m 2 s 2ejIJKj(1)From conservation of momentum,afmvG 1.20m v BorvG 1.20 v B(2)Solving (1) and (2) simultaneously, we findvG 7.07 m s (speed of green puck after collision)andP9.32v B 5.89 m s (speed of blue puck after collision)We use conservation of momentum for the system of two vehiclesfor both northward and eastward components.For the eastward direction:bgM 13.0 m s 2 MV f cos 55.0 For the northward direction:Mv 2i 2 MV f sin 55.0 Divide the northward equation by the eastward equation to find:bgv 2 i 13.0 m s tan 55.0 18.6 m s 41.5 mi hThus, the driver of the north bound car was untruthful.FIG. P9.32263
264P9.33Linear Momentum and CollisionsBy conservation of momentum for the system of the two billiardballs (with all masses equal),bg5.00 m s 0 4.33 m s cos 30.0 v 2 fxv 2 fx 1.25 m sbg0 4.33 m s sin 30.0 v 2 fyv 2 fy 2.16 m sv 2 f 2.50 m s at 60.0 FIG. P9.33Note that we did not need to use the fact that the collision is perfectly elastic.P9.34(a)pi p fsop xi p xfandp yi p yfFrom (2),mvi mv cos θ mv cos φ(1)0 mv sin θ mv sin φ(2)sin θ sin φθ φFurthermore, energy conservation for the systemof two protons requires111mvi2 mv 2 mv 2222viv so2so(b)Hence, (1) gives vi b2 vi cos θgFIG. P9.34θ 45.0 2φ 45.0 a fv e3.00 i 1.20 jj m s3.00 5.00 i 6.00 j 5.00 vP9.35m1 v 1i m 2 v 2i m1 m 2 v f :P9.36x-component of momentum for the system of the two objects:p1ix p 2ix p1 fx p 2 fx : mvi 3mvi 0 3mv 2 xy-component of momentum of the system:0 0 mv1 y 3 mv 2 yby conservation of energy of the system: also1111mvi2 3mvi2 mv12y 3m v 22 x v 22 y22222 viv2x 3v1 y 3 v 2 ySo the energy equation becomes4vi2 9 v 22 y we haveorcontinued on next page8 vi2 12 v 22 y32 viv2y 3e4vi2 3 v 22 y3j
Chapter 9(a)The object of mass m has final speedv1 y 3 v 2 y and the object of mass 3 m moves atv 22 x v 22 y 2 viv 22 x v 22 y (b)θ tan 1Fv IGH v JK2yθ tan 12xP9.37FGHm 0 17.0 10 27 kgv i 0 (the parent nucleus)m1 5.00 10 27 kgv 1 6.00 10 6 j m sm 2 8.40 10 27 kgv 2 4.00 10 6 i m s(a)4vi2 2 vi2 992vi3IJK2 vi 3 35.3 3 2 vim1 v 1 m 2 v 2 m 3 v 3 0where m 3 m 0 m1 m 2 3.60 10 27 kgFIG. P9.37 27 27 276 6 5.00 106.00 10 j8.40 104.00 10 i3.60 10v3 0ev3 (b)jej ejej eje 9.33 10 i 8.33 10 jj m s6111m1 v12 m 2 v 22 m 3 v 322221 275.00 106.00 10 6E 26E LMeNjej e8.40 10 je4.00 10 j e3.60 10 je12.5 10 j OQP2 276 2 276 2E 4.39 10 13 JSection 9.5P9.38The Center of MassThe x-coordinate of the center of mass isx CM m i xi mi b0 0 0 02.00 kg 3.00 kg 2.50 kg 4.00 kggxCM 0and the y-coordinate of the center of mass isyCM m i yi miyCM 1.00 m b2.00 kgga3.00 mf b3.00 kg ga2.50 mf b2.50 kg ga0f b4.00 kg ga 0.500 mf2.00 kg 3.00 kg 2.50 kg 4.00 kg265
266P9.39Linear Momentum and CollisionsTake x-axis starting from the oxygen nucleus and pointing toward themiddle of the V.yCM 0Thenx CM andx CM mi x i mia faf0 1.008 u 0.100 nm cos 53.0 1.008 u 0.100 nm cos 53.0 15.999 u 1.008 u 1.008 uFIG. P9.39xCM 0.006 73 nm from the oxygen nucleus*P9.40Let the x axis start at the Earth’s center and point toward the Moon.ej24228m1 x1 m 2 x 2 5.98 10 kg 0 7.36 10 kg 3.84 10 m m1 m 26.05 10 24 kgx CM 4.67 10 6 m from the Earth’s centerThe center of mass is within the Earth, which has radius 6.37 10 6 m.P9.41Let A1 represent the area of the bottom row of squares, A 2the middle square, and A3 the top pair.A A1 A 2 A 3M M1 M 2 M 3M1 M A1AA1 300 cm 2 , A 2 100 cm 2 , A3 200 cm 2 , A 600 cm 2A300 cm 2MM1 M 1 M 2A2600 cmM2M3x CMx CMFG IJH KF A IJ 100 cm MGH A K 600 cmF A IJ 200 cm MGH A K 600 cm2222M M6FIG. P9.41M3x M x 2 M 2 x 3 M 3 15.0 cm 1 1 M 11.7 cmyCM 312af2M c M h 5.00 cmc Mh 10.0 cmc MhayCM 13.3 cm16Mf c Mha25.0 cmf 13.3 cmM 5.00 cm 16 M 15.0 cm M121313
Chapter 9*P9.42(a)267Represent the height of a particle of mass dm within the object as y. Its contribution to thegravitational energy of the object-Earth system is dm gy . The total gravitational energy is1Ug gy dm g y dm . For the center of mass we have yCM y dm , so U g gMyCM .Mall massz(b)a fzaefaf13.6 m 15.7 m 64.8 m 1.83 10 3 m 3 . Its mass is21.83 10 3 m 3 6.96 10 6 kg . Its center of mass is above its base by one-The volume of the ramp isρV 3 800 kg m3fazjej1third of its height, yCM 15.7 m 5.23 m . Then3U g MgyCM 6.96 10 6 kg 9.8 m s 2 5.23 m 3.57 10 8 J .eP9.43M (a)zz0.300 m0 .300 m00λdx j50.0 g m 20.0 x g m 2 dxM 50.0 x g m 10.0 x 2 g m 2x CM (b)x CM*P9.44zxdmall massM LMMN1Mz0.300 mλ xdx 00.300 m01Mz 15.9 g0.300 m50.0 x g m 20.0 x 2 g m 2 dx0320 x g m 2125.0 x 2 g m 15.9 g3OPPQ0.300 m 0.153 m0Take the origin at the center of curvature. We have L r 2Lπ12πr ,4. An incremental bit of the rod at angle θ from the x axis hasdm MMr , dm dθ where we have used therdθ LLdefinition of radian measure. Nowmass given byyCM zz135 11Mrr2y dm r sin θdθ M all massM θ 45 LLFG 2L IJ 1 a cosθ fHπK L2135 45 4Lπ22Lπ 4 2Lπ2 F1 2 2 I L π GHπ JK2zθx135 sin θ dθFIG. P9.4445 FG 1 1 IJ 4 2LH 2 2K πThe top of the bar is above the origin by r byy0.063 5L .22Lπ, so the center of mass is below the middle of the bar
268Linear Momentum and CollisionsSection 9.6P9.45Motion of a System of Particles(a)v CM v CM P9.46bm1 v 1 m 2 v 2M2.00 kg 2.00 i m s 3.00 j m s 3.00 kg 1.00 i m s 6.00 j m sM gej bgee1.40 i 2.40 jj m sbgeje7.00 i 12.0 jj kg m sp Mv CM 5.00 kg 1. 40 i 2.40 j m s (a)See figure to the right.(b)Using the definition of the position vector at the center of mass,rCM 2.00 kg 3.00 kge 2.00 i 1.00 jj mFIG. P9.46The velocity of the center of mass isv CM v CM (d)m1 r1 m 2 r2m1 m 2b2.00 kg ga1.00 m, 2.00 mf b3.00 kg ga 4.00 m, 3.00 mfrCM (c)j5.00 kg(b)rCM P9.47 mi v ibgbg bgbg2.00 kg 3.00 m s , 0.50 m s 3.00 kg 3.00 m s , 2.00 m sP m1 v 1 m 2 v 2 Mm1 m 22.00 kg 3.00 kgbe3.00 i 1.00 jj m sThe total linear momentum of the system can be calculated as P Mv CMor asP m1 v 1 m 2 v 2Either givesP e15.0 i 5.00 jj kg m sLet x distance from shore to center of boatA length of boatx ′ distance boat moves as Juliet moves toward RomeoThe center of mass stays fixed.Before:h M cx hdM M M iM ax x ′f M cx x ′h M c x x ′h xdM M M iF 55.0 77.0 IJ x ′a 80.0 55.0 77.0f A a55.0 77.0fAG H 2 2K255.0 A 55.0a 2.70fx′ 0.700 mx CM cMb x M J x BAfter:BA2RJRJA2A2RCMB212212JRA2FIG. P9.47g
Chapter 9P9.48(a)269Conservation of momentum for the two-ball system gives us:bgbg0.200 kg 1.50 m s 0.300 kg 0.400 m s 0.200 kg v1 f 0.300 kg v 2 fRelative velocity equation:v 2 f v1 f 1.90 m sd0.300 0.120 0.200 v1 f 0.300 1.90 v1 fThenv1 f 0.780 m sv 2 f 1.12 m sv 1 f 0.780 i m s(b)v CM Before,iv 2 f 1.12 i m sb0.200 kg gb1.50 m sgi b0.300 kg gb 0.400 m sg ibg0.500 kgv CM 0.360 m s iAfterwards, the center of mass must move at the same velocity, as momentum of the systemis conserved.Section 9.7P9.49Rocket PropulsiondMdtejejThrust 2.60 10 3 m s 1.50 10 4 kg s 3.90 10 7 N(a)Thrust v e(b) Fy Thrust Mg Ma :ja f eej3.90 10 7 3.00 10 6 9.80 3.00 10 6 aa 3.20 m s 2*P9.50(a)dM 12.7 g 6.68 10 3 kg sdt1.90 sThe fuel burns at a rateThrust v edM:dte5.26 N v e 6.68 10 3 kg sjv e 787 m s(b)F M I:GH M JKv f vi v e lng FGH 53.553g . 5 25g . 5 25g . 512g .7 g IJKbiv f 0 797 m s lnfv f 138 m sP9.51v v e lnMiMf(a)Mi e v v e M fMi e 5 3.00 10 3 kg 4.45 10 5 kgThe mass of fuel and oxidizer is M Mi M f 445 3.00 10 3 kg 442 metric tons(b)aeajff M e 2 3.00 metric tons 3.00 metric tons 19.2 metric tonsBecause of the exponential, a relatively small increase in fuel and/or engine efficiency causesa large change in the amount of fuel and oxidizer required.
F M I v lnF M IGH M JKGH M JKF M kt IJ v lnFG 1 k tIJ M kt , so v v lnGH M KH M KFrom Equation 9.41, v 0 v e lniNow, M fiifefeieiiMWith the definition, Tp i , this becomeskFGHIJKF t IJ 144 s , v b1 500 m sg lnG 1 H 144 s Kafv t v e ln 1 With v e 1 500 m s, and Tpa f vbm sgv (m/s)ts1220178012026901323730FHdtpIK OP F 1Q v GGH 1 etTpIF 1 I F v IF 1JJ GH T JK GH T JK GG 1 KHepptTpIJJ , orKveTp t1 500 m s144 s ta f aem s 5continued on next page2a (m/s )140120100806040200t (s)FIG. P9.52(d)140ts120With v e 1 500 m s, and Tp 144 s , a 100(d)LMNd v e ln 1 Tt80afat FIG. P9.52(b)60afdvat dtt ar Momentum and Collisions0270
Chapter 9(e)LM F t I OPLM1 t OPF dt I vln1dtvTlnGJz z MN H T K PQ z MN T PQGH T JKLF t I F t I F t I Oxat f v T MG 1 J lnG 1 J G 1 J PMNH T K H T K H T K PQF tIxat f v eT t j lnG 1 J v tH TKtafx t 0 vdt 0ttee pp00ppte peppp0epWith v e 1 500 m s 1.50 km s , and Tp 144 s ,f FGHIJKt 1.50t144a f t (s)FIG. P9.52(f)*P9.53The thrust acting on the spacecraft is F b3 500 kg ge2.50 10 6 je9.80 F ma :thrust FG dM IJ v :H dt Ke8.58 10 2 N M 4.41 kgF M I b70 m sgGH 3 600 s JKjm s 2 8.58 10 2 N1409.23120401401201002.19802016060000x (km)40ax 1.50 144 t ln 1 20(f)p271
272Linear Momentum and CollisionsAdditional ProblemsP9.54(a)When the spring is fully compressed, each cart moves with same velocity v. Applyconservation of momentum for the system of two glidersbpi p f :(b)gv m1 v 1 m 2 v 2 m1 m 2 vm1 v 1 m 2 v 2m1 m 2bg1111m1 v12 m 2 v 22 m1 m 2 v 2 kx m22222Only conservative forces act, therefore E 0 .Substitute for v from (a) and solve for x m .x m2bm xm (c)1gbgb m 2 m1 v12 m1 m 2 m 2 v 22 m1 v1bk m1 m 2em1 m 2 v12 v 22 2 v1 v 2bk m1 m 2gj bv1 v2gg bm v g22 22 2m 1 m 2 v1 v 2g kbmm m m g1212m1 v 1 m 2 v 2 m1 v 1 f m 2 v 2 fdidiConservation of momentum:m1 v 1 v 1 f m 2 v 2 f v 2Conservation of energy:1111m1 v12 m 2 v 22 m1 v12 f m 2 v 22 f2222which simplifies to:Factoring givesem dvj eji dv v i m dv(1)m1 v12 v12 f m 2 v 22 f v 2211 v1 f11f22fid v2 v2 f v2iand with the use of the momentum equation (equation (1)),this reduces todvorv1 f v 2 f v 2 v11i d v1 f v 2 f v 2i(2)Substituting equation (2) into equation (1) and simplifying yields:v2 f FG 2m IJ v FG mHm m K Hm1112IJK m1v2 1 m22Upon substitution of this expression for v 2 f into equation 2, one findsv1 f FG mHmIJKFGHIJK m22m 2v1 v2m1 m 21 m21Observe that these results are the same as Equations 9.20 and 9.21, which should have beenexpected since this is a perfectly elastic collision in one dimension.
P9.55(a)b60.0 kg g4.00 m s a120 60.0f kgvChapter 9273fv f 1.33 m s i(b)b Fy 0 :gn 60.0 kg 9.80 m s 2 0aff k µ k n 0.400 588 N 235 Nf 235 N iFIG. P9.55k(c)For the person, pi I p fmvi Ft mv fb60.0 kg g4.00 m s a235 Nft b60.0 kg g1.33 m st 0.680 s(d)afperson:mv f mv i 60.0 kg 1.33 4.00 m s 160 N s icart:120 kg 1.33 m s 0 160 N s ibgaf(e)x f xi 114.00 1.33 m s 0.680 s 1.81 mvi v f t 22(f)x f xi 11vi v f t 0 1.33 m s 0.680 s 0.454 m22(g)111mv 2f mvi2 60.0 kg 1.33 m s222(h)111mv 2f mvi2 120.0 kg 1.33 m s222(i)ddibigbbg2 g2b160.0
9.1 Linear Momentum and Its Conservation 9.2 Impulse and Momentum 9.3 Collisions in One Dimension 9.4 Two-Dimensional Collisions 9.5 The Center of Mass 9.6 Motion of a System of Particles 9.7 Rocket Propulsion Linear Momentum and Collisions ANSWERS TO QUESTIONS Q9.1 No. Impulse, Ft , depends on the force and the time for which it is applied.
Impulse, Momentum 2011, Richard White www.crashwhite.com ! This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring a knowledge . Conservation of linear momentum cannot be used to find the final .
Momentum is conserved in all collisions Elastic collisions: no deformation occurs Kinetic energy is also conserved Inelastic collisions: deformation occurs Kinetic energy is "lost" Perfectly inelastic collisions Objects stick together; kinetic energy is "lost" Explosions Reverse of perfectly inelastic collisions;
momentum is kg·m/s. Linear Momentum Linear momentum is defined as the product of a system's mass multiplied by its velocity: p mv. (8.2) Example 8.1Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player's momentum with the
4. Impulse - Momentum 3. This area is called Impulse F.dt r. 5/14/2008 Momentum - 1 25 Force, Time and Momentum The area under the force-time graph equals the change in linear momentum. If the LHS of this . Collisions - Linear Momentum is always conserved Is the collision force gravity or springs? - If YES, then Kinetic .
AP Physics Practice Test: Impulse, Momentum 2011, Richard White www.crashwhite.com This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring a knowledge of basic calculus. Part I. Multiple Choice 1.
1. Impulse and Momentum: You should understand impulse and linear momentum so you can: a. Relate mass, velocity, and linear momentum for a moving body, and calculate the total linear momentum of a system of bodies. Just use the good old momentum equation. b. Relate impulse to the change in linear momentum and the average force acting on a body.
CHAPTER 3 MOMENTUM AND IMPULSE prepared by Yew Sze Ling@Fiona, KML 2 3.1 Momentum and Impulse Momentum The linear momentum (or "momentum" for short) of an object is defined as the product of its mass and its velocity. p mv & & SI unit of momentum: kgms-1 or Ns Momentum is vector quantity that has the same direction as the velocity.
2020 Manual for Railway Engineering (MRE) – Individual/Downloadable Chapters in PDF format. Visit www.arema.org Publication Title Member Price Non-Member Price S & H Fee Schedule Quantity Total Cost 2020 Manual for Railway Engineering (MRE) – Annual Publication released every April Complete Print Set 960 1,470 1
