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Chapter 2b 2b6 2b7 2b9 More on the Momentum Principle Derivative form of the Momentum Principle 2b approximate result F ma Momentum not changing 2b21 Static equilibrium 2b22 Uniform motion 2b23 Momentarily at rest vsstatic equilibrium 2b24 Finding the rate of change of momentum 2b25 Example Hanging block static equilibrium 2b26 Preview of the Angular Momentum Principle Curving motion 2b31 Example The Moon around the Earth curving motion 2b32 Example Tarzan swings from avine curving motion 2b33 The Momentum Principle relates different things 2b34 Example Sitting in an airplane curving motion 2b35 Example A turning car curving motion A special case Circular motion at constant speed 2b41 The initial conditions required for a circular orbit 2b42 Nongravitational situations 2b43 Period 2b44 Example Circular pendulum curving motion 2b45 Comment Forward reasonin 2b46 Example An amusement park ride curving motion Systems consisting of several objects 2b51 Collisions Negligible external forces 2b52 Momentum flow within a system Conservation of momentum Predicting the future of complex gravitating systems 2b71 The threebody problem 2b72 Sensitivity to initial conditions Determinism 2b81 Practical limitations 2b82 Chaos 2b83 Breakdown of Newton s laws 2b84 Probability and uncertainty Derivation of the multiparticle Momentum Principle 2b10 Summary 2b11 Review questions 2b12 Problems 2b13 Answers to exercises 114 115 116 116 116 117 118 119 120 120 121 122 123 123 125 126 127 128 129 129 131 131 132 133 134 136 136 137 137 138 138 139 139 140 142 143 144 114 Chapter 2 The Momentum Principle Chapter 2b More on the Momentum Principle In this chapter we continue to apply the Momentum Principle to a variety of systems The major topics in this chapter are 39 The derivative form of the Momentum Principle with applications 39 Applying the Momentum Principle to multiparticle systems 39 Conservation of momentum 2b1 Derivative form of the Momentum Principle The forms of the Momentum Principle given so far Ap PnetAt and pf pi netAt are particularly useful when we know the momentum at a particular time and want to predict what the momentum will be at a later time We use these forms in repetitive computer calculations to predict fu ture motion Another important form is obtained by dividing by the time interval At 17 net At Just as we did in finding the instantaneous velocity ofan object see Section 176 we can nd the instantaneous rate of change of momentum by let ting the time interval Atapproach zero an infinitesimal time interval The ratio of the infinitesimal momentum change written as dp to the infini tesimal time interval written as dt is the instantaneous rate of change of the momentum As the time interval gets very small A t gt 0 the ratio of the momentum change to the time interval approaches the ratio of the infini tesimal quantities lim E E A H 0 At dt and we obtain the following form THE MOMENTUM PRINCIPLE DERIVATIVE FORM A EB net dt In words the instantaneous time rate ofchange ofthe momen tum ofan object is equal to the net force acting on the object Or in calculus terms since the limit of a ratio of infinitesimal quantities is called a derivative we can say that the derivative of the momentum with respect to time is equal to the net force acting on the object This form of the Momentum Principle is useful when we know something about the rate ofchange of the momentum at a particular instant Knowing the rate of change of momentum we can use this form of the Momentum Principle to deduce the net force acting on the object which is numerically equal to the rate of change of momentum Knowing the net force we may be able to figure out particular contributions to the net force 2171Dnivativef07m of the Momentum Principle Ex 2b 1 At a certain instant the z component of the momentum of an object is changing at a rate of 4 kgms per second At that instant what is the 2 component of the net force on the object Ex 2b 2 lfan object is sitting motionless what is the rate ofchange ofits momentum What is the net force acting on the object Ex 2b If an object is moving with constant momentum lt10 712 78gt kgms what is the rate of change of momentum dp dt What is the net force acting on the object 2b11 An approximate result F ma In simple cases the derivative form of the Momentum Principle reduces to a form that may be familiar to you from a previous course If the mass is con stant the usual situation and the speed is small enough compared to the speed oflight that the momentum is well approximated by p w my we have this A 3 dt 77 mg Pnet nonrelativistic form constant mass 3 The quantity g is called acceleration and is often given the symbol a This form ofthe Momentum Principle Newton s second law says that mass times acceleration time rate of change of velocity dv dt is equal to the net force or in simplified scalar form maF or Fma The approximate form is not valid in situations where an object s mass isn t constant One example is a rocket with exhaust gases ejecting out the back as a result the rocket has decreasing mass In such cases the momen tumbased formula dp dt Pna gives the correct results whereas the con stantmass formula mdv dt Pnet cannot be used The approximate form is also not valid for objects moving at speeds close to the speed of light because in these circumstances the approximation p u my is not valid When the principle is written in terms of momentum it is valid even for objects whose mass changes or which are moving at speeds close to the speed of light as long as we use the relativistically correct definition of mo mentum So dp dt quota is the more general and more powerful form The extra symbols on dp dt Pna compared to the simpli ed version Fma are important 39 The Momentum Principle is a vector principle so the arrows over the symbols are extremely important they remind us that there are really three separate component equations for x y and z 39 It is also important to remember that vie have to add up all vector forc es to give the net force so we write Fnet notjust F Ex 2b4 The velocity of a 80 gram ball changes from 5 073 ms to 502 07304gt ms in 001 s due to the gravitational attraction of the Earth and to air resistance What is the acceleration of the ball What is the rate of change of momentum of the ball What is the net force acting on the ball 115 116 Figure 2btl A block hangs motionless from a spring static equilibriumt Chapter 2 The Momentum Principle 2b2 Momentum not changing We saw in Chapter 1 that ifa system is in uniform motion its momentum is constant and does not change with time A special case of uniform motion is a situation in which object is at rest and remains at rest this situation is called static equilibrium 2b21 Static equilibrium 9 if If the system never moves what IS dt The momentum of the system isn t changing so the rate of change of the momentum is zero A 0 lt000gt static equilibrium 7 What does this imply about the net force acting on the system From the derivative form of the Momentum Principle dp dt Fna we deduce that the net force acting on a system that never moves must be zero Fnet 0 lt000gt static equilibrium For example consider a spring whose stiffness is kI 80 Nm You sus pend the spring vertically and hang a block of mass 05 kg from the bottom end At first the block oscillates up and down but eventually air resistance brings it to a stop and the block hangs motionless never moving Figure 2b1 How much has the spring stretched Use the Momentum Principle to prove your result rigorously Choose the block as the system If the block no longer moves at all the rate of change ofits momentum dp dt is zero From the derivative form of the Momentum Principle we conclude that the net force acting on the block must be zero 7 What objects in the surroundings exert forces on the block The spring pulls up and the Earth pulls down 7 Apply the derivative form of the Momentum Principle to find the stretch The vector sum of these two forces must equal zero is Fnet FspringFEanh lt0 0 0gt lt0 kxs 0gt lt07mg 0gt lt0 kssi mg 0gt 0 kxsi mg which gives 5 mg k S 05 kg98 Nkg 0061 m 80 Nm 2b22 Uniform motion Suppose the spring and hanging mass are on an elevator that goes up at a constant speed of 4 ms How does putting the apparatus into uniform mo tion affect our analysis A 7 What is for the block which moves upward at 4 ms 2172 Momentum not changing Now the momentum ofthe block is not zero but constant and dp dt is still zero which means that the net force Fnet must also be zero The spring will be stretched 0061 mjust as when there was no motion This is an example of the principle of relativity that physical laws in this case the Momentum Principle work in the same in all reference frames that are in uniform mo tion that is constant velocity with respect to the cosmic microwave back ground which is the stage on which phenomena play themselves out 2b23 Momentarin at rest vsstatic equilibrium Pull the block down aways and release it and it oscillates up and down at the end of the spring During this oscillating motion whenever the block reaches the bottom of its travel its speed is momentarily zero At first glance this situation may seem similar to the case ofblock hanging at restwhichwe considered in Section 2b2l but there are important differences 7 At this instant when the speed is momentarily zero is the rate of change of the block s momentum zero No The momentum just before hitting bottom points downward Figure 2b andjust after hitting bottom the momentum poins upward Figure 2b4 The momentum is changing all during this time and the momentum is zero only momentarily Figure 2b3 Apply the derivative form of the Momentum Principle What can we deduce about the magnitude and direction of the net force on the block at the instant the block is at rest at is lowest position Since dp dt is nonzero the net force must be nonzero as well Since the initial momentum was downward and the final momentum is upward the direction of dp dt must be upward This implies that the net force is also upward at the instant the block is at rest 7 Does an upward net force make sense physically If the net force at the bottom of the motion were zero the momentum couldn t change from its current value zero and the blockwould stay for ever at the bottom point of the oscillation But that s notwhat happens The block slows to a stop but then starts back upwards In order for the y com ponent of momentum to change from zero to a positive quantity there must be an upward net force to cause this change 7 At the bottom of the oscillating motion which force is larger the force exerted upward by the spring or the force exerted downward by the Earth How do you know Since the net force must be upward in order to produce the observed change in the momentum it must be that the spring force is larger than the gravitational force exerted by the Earth This conclusion is also consistent with the fact that the block when headed downward slows to a stop which means that the net force must point upward to oppose the downward mo tion 7 How does the stretch of the spring when the block is momentarily at rest at its lowest point compare to the stretch of the spring in the previous situation where the block hung motionless in equilibrium The stretch of the spring is longer in this case block momentarily at rest than it is in the previous case static equilibrium Let s look at this numerically When a block of 05 kg hangs motionless from a spring whose stiffness is 80 Nm we found that in static equilibrium the spring is stretched 0061 m 61 cm Now suppose the blockis oscillat 117 pl Figure 2b just before the block comes to a momentary stop is momentum is down ward P2 lt000gt Figure 2b The blocks momentum is zero only at the instant it reaches is lowest posi tion p3 Figure 2bi4 As the block starts back upward is momentum is upwardi 118 Figure 2br6 The momentum of a block oscillating on a spring at an instant before it comes to a momentary rest at the bottom of is path 1 and an instant after it starts back upward 2 Ar le is 131 Figure 2br7 The direction of nipalt is the direction of A39p 1327131 Chapter 2 The Momentum Principle ing up and down in such away that at the bottom ofits motion the spring is stretched 01 m 10 cm 7 What is the net momentarily zero force when the block s momentum is The y component of the net force is Fm 80 Nm0l m705 kg98 Nkg 31 N kxsi mg 0 which gives Fnew As expected the net force points upward positive ycomponent consistent with the rate of change of momentum being upward Avoiding confusion One of the hardest things to keep straight when first learning about motion and the causes of motion is that being momentarily at rest is not at all like being permanently at rest static equilibrium When you encounter a situ ation where the momentum is zero always ask yourselfwhether it is staying zero permanently If not the momentum is changing and the net force is nonzero There are many other situations in which the net force on an object is nonzero at the instantwhen the object s momentum is momentarily zero This happens whenever an object reverses direction Throw a ball up in the air When the ball reaches its maximum height 1 is momentarily zero but the gravitational force still points downward and makes 1 decrease If the net force dropped to zero when 1 dropped to zero the ball would stay mo tionless at the top ofits trajectory When a tennis ball hits a wall and rebounds the momentum of the ball momentarily goes to zero but aforce due to the wall acts on the ball chang ing the momentum of the ball to go in the opposite direction If the force exerted by the wall went to zero when the speed of the ball went to zero the ball wouldjust remain motionless squashed against the wall 2b24 Finding the rate of change of momentum In order to apply the derivative form of the Momentum Principle it is im portant to be able to decide if dp dt the instantaneous rate of change of momentum is zero or nonzero and if it is nonzero what its direction is The procedure for finding dp dt is based on the procedure for finding Ap which you practiced in Chapter 1 and requires that you draw a simple dia gram as shown in Figure 2b5 1 Draw two arrows one representing E the momentum of the object ashort time be fore the instant of interest and a second arrow representing pf the momentum of the object a short time after the instant ofinterest 2 Graphically find Ap fe by placing the arrows tail to tail and drawing the resultant arrow starting at the tail of t and going to the tip of pf 3 This arrow indicates the direction of Ap which is the same as the direction of dp dt provided the time interval involved is suffi ciently small The diagrams in Figure 2b6 and Figure 2b7 show the result of this proce dure when applied to the oscillating block on the end ofa spring at the in stant it is at rest at the bottom The direction of dp dt at this instant is upward 2172 Momentum not changing Ex 2b5 A block oscillating up and down on a spring comes to the top ofits path where it is momentarily at rest Follow the procedure outlined above to answer the following questions At the instant it is at rest is dp dt zero or nonzero lf nonzero what is its direction Ex 2b6 You throw a ball straight up into the air Follow the procedure outlined above to answer the following questions At the instant the ball reaches its highest location is dp dt zero or nonzero lf nonzero what is its direction Ex 2b 7 A ping pong ball hits a paddle and bounces straight back While the ball is in contact with the paddle is dpdt zero or nonzero lf nonzero what is its direction 2b25 Example Hanging block static equilibrium A block of mass m hangs from a string Figure 2b8 Then you pull the block to the side by pulling horizontally with a second string What is the magnitude F2 of the force you must apply to the second string in order for the first string to be at an angle of 9 to the vertical What is the tension F1 in the first string The tension in a string is the force it exerts on another object Because a string is exible the force a string exerts is always in the direction of the string a string cannot exert a force perpendicular to the string 1 Choose a system System the block 2 List external objects that interact with the system with diagram string 1 string 2 the Earth 3 Apply the Momentum Principle A Fnet ltF2 7 F1 cose F1 sineimg 0 d dt lt0 0 0gt 4 No need to apply the position update tormula no motion 5 Solve for the unknowns algebraic manipulation 0 inFlcose so Flcose F2 0 Flsineimg so F1 sine mg which gives F1 sm gcos9 m F2 cmose Sine 119 Figure 2b8 A block hangs from a string Then you pull the block to the side with another string 120 M28 Mig d1 d2 Figure 2bl9 Two children sit on a seesaw in static equilibriuml Chapter 2 The Momentum Principle 6 Check We can make some checks on the reasonableness of the results If 9 90 the first string is vertical and we expect the tension in this string to be mg Since sin90 l we do indeed find F1 mg Also F2 0 as it should in this case On the other hand if 9 0 sin0 0 the first string is hor izontal and F2 mgO which is infinitely large and impossi ble This makes sense if both strings are horizontal there is no y component to support the block We could also choose the system ofthe ball alone and show that the tension in the vertical portion of the first string must be equal to mg since the net force must be zero since the ball is constantly at rest 2b26 Preview of the Angular Momentum Principle In many static equilibrium situations the Momentum Principle is not suf cient for carrying out a full analysis and we also need the Angular Momen tum Principle which we will discuss in a later chapter For example consider two children sitting motionless on a seesaw Figure 2b9 Choose as the system the two children and the lightweight board For this choice of system the objects in the surroundings that exert signi cant external forces on the system are the Earth pulling down and the support pivot pushing up The momentum of the system isn t changing so the Momentum Principle correctly tells us that the net force must be ze ro and the support pivot pushes up with a force equal to the weight of the children we are neglecting the lowmass board However the Momentum Principle alone doesn t tell us where the chil dren have to sit in order to achieve balance and static equilibrium The force on a child exerts a twist the technical term is torque about the piv otwhich tends to make the board turn The torque associated with the forc es on the two children have to add up to zero one tends to twist the board clockwise the other counterclockwise Torque is defined as force times le ver arm and a nonzero net torque causes changes in a quantity called angu lar momentumjust as a nonzero netforce causes changes in the ordinary momentum we have been studying 2b3 Curving motion Recall from Chapter 1 that the rate ofchange of momentum along a curving path has a component toward the center of the kissing circle PERPENDICULAR COMPONENT OF dpdt FOR MOTION ALONG A CURVING PATH 35 The direction is toward the center of the kissing circle of radius R EA Rlpl Perpendicular component of if is given by If an object is moving along a curved path and you know the speed and the radius ofcurvature of the path you know dp dti and therefore you also know the component ofthe net force acting toward the center ofthe kissing circle since dp dtL Fneu 2173 Curvng motion 121 2b31 Example The Moon around the Earth curving motion The first part of this example was discussed in Chapter 1 The Moon which has a mass of 7gtltlO22 kg orbits the Earth once every 28 days a lunar month following a path which is nearly circular Figure 2b10 The dis tance from the Earth to the Moon is 4gtlt108 m What is the magnitude ofthe rate of change of the Moon s momentum Calculate the gravitational force exerted on the Moon by the Earth whose mass is 6gtlt 10 kg Compare these two results 21t4gtlt108 m I I lxlOgms 28 days24 hrday60 m1nhr60 sm1n lvl 7nle 7X1022kg1gtlt103ms 73x1025 kgms 3 M mS73gtlt1025 kgms 18xi020 M 4X108m 52 The Moon s speed is not changing so dp dt is perpendicular to p as Figure Qb lo The Moon s orbit around shown in Figure 2bI10I the Earth is nearly circularl Not to scale The magnitude of the gravitational force of the Earth on the Moon is the 5225 of the Earth and Moon are exag gera e l L M 4 dt Rlpl 2 24 22 IFnal GMm 6397X10711 N m 6x10 kg7gtlt10 kg R2 k 2 8 2 g 4X10 m ling 18xi020 N The net force is equal in magnitude to the observed magnitude of the rate of change of the Moon s momentum The directions are also the same toward the center of the kissing circle This comparison was first made by Newton who realized that the fall of an apple and the orbit of the Moon were due to the same fundamental cause the gravitational attraction of the Earth for the apple and for the Moon Until Newton had this insight no one had made this connection Ex 2b 8 A child of mass 40 kg sits on awooden horse on a carousel The wooden horse is 5 meters from the center of the carousel which completes one revolution every 90 seconds What is the rate of change of the momentum of the child both magnitude and direction What is the net force acting on the child Ex 2b 9 A person rides on aFerris wheel at constant speed Figure 2bll When the person is at the location shown what is the direction of the person s rate of change of momentum dp dt What must be the direction of the net force acting on the person The net force is due to the Earth s gravitational pull plus the force of the seat on the person angry Ex 2b0 The orbit of the Earth around the Sun is approximately Figure 2b A Person rides on a Ferris circular and takes one year to complete The Earth s mass is 11661 3 constantI Speeg lExerIase Eb g 6X10 4 kg and the distance from the Earth to the Sun is SI t e perm 15 at t e ocatlon S Olm 1 what 15 the directlon of the net force acting l5gtlt10 m What 1s the magn1tude of the rate of change of the I I I I on the person Earth s momentum What 1s the d1rectlon of the rate of change of the Earth s momentum Aglgat is the magnitude of the gravitational force the Sun mass 2X10 kg exerts on the Earth What is the direction of this force 122 Figure 2b12 Tarzan swings from a vine The kissing circle has radius L the length of the vine We ll neglect air resistance Chapter 2 The Momentum Principle 2b32 Example Tarzan swings from a vine curving motion Tarzan swings from a vine and the radius of his circular motion is L the length ofthe vine Figure 2b12 We will calculate how strong the vine must be so that at the bottom of the swing with the vine momentarily vertical the vine doesn t break Tarzan s mass is Mand his speed at the bottom of the swing is 1 Because Tarzan s motion is circular we know that the compo nent of the rate ofchange ofhis momentum that is perpendicular to his mo tion has magnitude and is directed toward the center of the kissing circle the top of the vine 1 Choose a system System Tarzan 2 List external objects that interact with the system with diagram the vine tension FV the Earth and the air represent the system by a circle 3 Apply the Momentum Principle A m Fnet d dt lt0lfol0gt lt0FwMg0gt 4 No need to apply the position update formula 5 Solve for the unknowns FWMg so FV 3MvMg 6 Check Mg has units of N and vLMv has units of ls kg ms or kgm52 which are the units of the rate ofchange of momen tum which are equivalent to N So the units check Is the result reasonable The result shows that the vine has to ex ert an upward force that is greater than Tarzan s weight Mg That makes sense If Tarzan were hanging motionless static equilibrium the vine would only have to exert an upward force of Mg net force would be zero But to make Tarzan s momen tum turn from horizontal to upward requires that the net force be upward As a numerical example Tarzan s mass might be 100 kg about 220 pounds and the vine might be 10 m long Suppose his speed at the bottom of the motion is 15 ms His weight Mg is 980 N but the vine has to provide a much larger force FV 100k 15ms 2980N 10m FV 2250 N 980 N 3230 N If the maximum tension this vine can support without breaking is less than 3230 N Tarzan is in trouble Note that the vine has to support a force about three times as big as Tarzan s weight 2173 Curvng motion 2b33 The Momentum Principle relates different things The equation dp dt PM does not mean that dp dt and PM are the same thing They are not In the analysis of Tarzan and the vine note carefully that we knew the rate ofchange of the momentum dp dt because we knewwhat kind ofcurving path Tarzan was following This result dp dt 0 vLMv 0 does not involve the strength of the vine nor g the magnitude of the gravitational field itjust depends on the way Tarzan s motion is curving On the other hand we know that the net force is 0 FWMg 0 the vec tor sum ofall the external forces This expression for the net force doesn t contain anything about how Tarzan is moving It is the derivative form of the Momentum Principle which links these two kinds of information together telling us that the vector rate of change of the momentum is numerically equal to the vector net force We set the two vectors equal to each other and solve for the quantity of interest The result ldp dtil is essentially a geometrical result whichwe proved in Chapter 1 using a geometrical argument It is important to understand that it doesn t involve anything at all concerning the partic ular force that is causing the change of direction For example the gravita tional force law constant C does not appear in this result nor is there anything about electric charges or electric forces in the result Yet the Momentum Principle says that dp dt is equal to Pnet These two quantities are very different in kind They even have different units kg mss and newtons although the Momentum Principle indicates that these two physical units must be equivalent in some sense The Momentum Principle dp dt Pnet is powerful precisely because it re latex two very different things to each other Ifyou know something about the net force you know how rapidly the momentum must be changing Or if you know how rapidly the momentum is changing you know what the net force must be The Momentum Principle relates an e ect rate of change of mo mentum to its cause net force due to interactions with other objects This is very different from a mere definition such as the definition ofmo mentum p2 mv ll 7 v c 2 We introduce the symbol p and let it stand for the quantity mv 1 7 IvIc2 as a convenience but there is no physics in this because there is no difference betweenwhat we mean by p and what we mean by mv Il 7IvIc2 2b34 Example Sitting in an airplane curving motion An airplane makes a curving motion at constant speed 1 as shown in Figure 2b13 where the radius of the kissing circle at the top of the path is R Choose a passenger as the system of interest and determine the force that the seat exerts on the passenger s bottom 1 Choose a system System passenger in airplane 2 List external objects that interact with the system with diagram the seat and the Earth F5 represent the system by a circle 3 Apply the Momentum Principle 1THEN d dt lt07lrol0gt Mg lt0 Fs Mgs 0gt 123 Figure 2bt13 An airplane goes over the top of is path with speed vi The kissing circle has radius Rat the top of the path We ll neglect air resistance 124 Figure 2bl4 An airplane goes through the bottom of is path Chapter 2 The Momentum Principle The rate of change of the momentum dp dt as usual points toward the center of the kissing circle so the y component of dp dt is negative 4 No need to apply the position update formula 5 Solve for the unknowns Fsng so FS Mgi 9M1 6 Check The units check see previous example on these units Let s explore the physical significance of this result There is an interesting and important minus sign in our result The upward force of the seat FS on the passenger s bottom is lass than the passenger s weight Mg 7 That means that the net force points downward Does that make sense Yes because the net force should point toward the center of the kissing cir cle which is below the airplane But since FS Mgi MvzR if the airplane s speed 1 is big enough the force exerted by the seat could be zero Does that make any sense Yes that s the case where the airplane changes direction so quickly that the seat is essentiallyjerked out from under the passenger and there is no long er contact between the seat and the passenger No contact no force of the seat on the passenger In this case what does the situation look like to the passenger From the passenger s point of View it is natural to think that the passenger has been thrown upward away from the seat But actually the seat has been yanked out from under the passenger in which case the passenger is falling toward the Earth being no longer supported by the seat but changing velocity less rapidly than the airplane and so the ceiling comes closer to the passen ger In the extreme case if the airplane s maneuver is very fast it may seem as though the passenger is thrown up against the cabin ceiling In actual fact however it is the cabin ceiling that is yanked down and hits the passenger You can see the value ofwearing a seat belt to prevent losing contactwith the seat When the airplane yanks the seat downward the seat belt yanks the passenger downward This may be uncomfortable but it is a lot better than having the cabin ceiling hit you hard in the head 7 How would this analysis change if the airplane goes through the bottom ofa curve rather than the top Figure 2bl4 Now the center of the kissing circle is above the airplane so dp dt points upward That means that the net force must also point upwar 7 Which is larger the force of the seat on the passenger or the gravitational force of the Earth on the passenger The force of the seat is greater than Mg since the net force is upward The passenger sinks deeper into the seat Or to put it another way the airplane yanks the seat up harder against the passenger s bottom squeezing the seat against the passenger 2173 Curvng motion The feeling of weight and of weightlessness As you sit reading this text you feel a sensation that you associate with weight If someone suddenly yanks the chair out from under you you feel weightless which feels funny and odd It also feels scary because you know from experience that whenever you lose the support of objects under you chair airplane seat floor mountain path something bad is about to happen In reality what we perceive as weight is actually not the gravitational force at all but rather the forces of atoms in the chair or airplane seat or floor or mountain path on atoms in your skin You have nerve endings that sense the compression ofinteratomic bonds in your skin and you interpret this as evidence for gravity acting on you But if you were placed in a space ship that was accelerating going faster and faster its floor would squeeze against you in a way that would fool your nerve endings and brain into thinking you were subjected to gravity even if there are no stars or planets nearby If you lose contact with the seat in an airplane that is going rapidly over the top of a curving path your nerve endings no longer feel any contact forces and you feel weightless Yet this is a moment when the only force acting on you is in fact the Earth s gravitational force Weightlessness near the Earth paradoxically is associated with being subject only to your weight Nor is itjust nerve endings in your skin that give you the illusion of weight As you sit here reading your internal organs press upward on oth er organs above them making the net force zero You feel these internal contact forces If you are suddenly weightless a main reason for feeling funny is that the forces one organ exerts on another inside your body are suddenly gone To train astronauts NASA has a cargo plane that is deliberately own in a curving motion over the top so that people in the padded cargo bay lose contactwith the floor and seem to float freely in actual fact they are accel erating toward the Earth due to the Mgforces acting on them but the plane is also deliberately accelerating toward the Earth rather than flying level so the people don t touch the walls and appear to be floating All the nerve endings are crying out for the usual comforting signals and not getting them This airplane is sometimes called the vomit comet in honor of the effects it can have on trainees 2b35 Example A turning car curving motion Suppose you re riding as the passenger in a convertible with left hand drive American or continental European so you are sitting on the right Figure 2b15 Assume you have foolishly failed to fasten your seat belt The driver makes a sudden right turn Before studying physics you would probably have said that you are thrown to the left and in fact you do end up closer to the driver 7 There s something wrong with this thrown to the left idea What object in the surroundings exerted a force to the left to make you move to the left There is no such object To understand what s happening watch the turn ing car from a fixed vantage point above the convertible and observe care fully what happens to the passenger In the absence of forces to change the passenger s direction of motion the passenger keeps moving ahead in a straight line at the same speed as before Figure 2bl6 It is the car that is yanked to the right out from under the passenger The driver moves closer to the passenger it isn t that the passenger moves toward the driver 125 Figure 2bl15 You are a passenger in a car that makes a sudden right turn 126 Driver run into door Figure 2brl6 The passenger moves straight ahead and is now closer to the driver who was thrown to the rights The driver was thrown to right by the door pushing the driver to the rights Chapter 2 The Momentum Principle The driver may also feel thrown to the left against the door But it is the door that runs into the driver forcing the driver to the right not to the left Sometimes people invent fictitious forces such as the socalled centrifu gal force to account for the passenger and driver being thrown to the left But thisjust adds a second confusion to the first The passenger and driver are thrown to the right in a right turn not to the left and there are real forc es to the right that make them go to the right for example the force of the door on the driver s left shoulder Moreover forces are associated with interactions between objects in a chosen system and objects in the surroundings The centrifugal force is just made up and is not associated with any real object 2b4 A special case Circular motion at constant speed Motion in a circle at constant speed is a common kind ofmotion This is ap proximately the situation for the planets in our Solar System Although the orbits are actually ellipses except for Pluto these elliptical orbits are very close to circular In the case oft e arth the ellipse is so close to a circle that on a drawing the size of this page you couldn t tell the differ ence The orbit of our Moon around the Earth is also nearly circular at con stant speed Also each day you are carried around in a circle at constant speed around the Earth s axis by the rotation of the Earth A Ferris wheel or a merrygoround moves its passengers in a circle at con stant speed Some physical properties of the hydrogen atom can be under stood by a simple model of an electron going around in a circle around the proton at constant speed though this model is too simple to explain many properties of hydrogen atoms and quantum mechanics is needed for a full analysis For a planet of mass m moving around a star ofmass M in a circular orbit at constant speed we can write E dt l netl l l IFgmi 0 Planets move at very high speeds compared to baseballs but at speeds that are very small compared to the speed of light So approximately pm my and we can write E R s M mlvl m CF Canceling the common factors We and R we have the following result Ile m GA g special case see following paragraph This is such a specialcase formula that it is not a good idea to memorize it If you do you are likely to misuse it using it in situations where it does not apply It applies only to 39 circular motion at constant speed 39 with a gravitational force 39 between a massive nearly motionless star and 39 a comparatively lowmass planet You should be able to derive this result from the Momentum Principle and the gravitational force law as done above Only then can you say that you really understand the result Note how short the derivation is 2174 A special case Circular motion at constant speed The gravitational force is unusual because the force is proportional to the mass m ofthe object so when the gravitational force appears in the Momen tum Principle the mass m cancels This is not the case when the force is an electric force which depends on electric charge not on mass In Einstein s general theory of relativity which describes gravity in a very different way than Newton did the star s mass alters the space around the star and all small masses move in that space in the same way no matterwhat their small mass We see an echo of this Einstein view in the cancellation of min the result above 2b41 The initial conditions required for a circular orbit What can we do with our result Ile m GMR for a circular orbit at constant speed We know the mass Mof the Sun and we know the gravitational con stant C so for a circular orbitwhose radius of curvature is Rwe can calculate the speed that a planet must have in such a circular orbit Conversely for a particular speed we can calculate the radius that will give a circular orbit What if the planet starts out in a direction perpendicular to the line between Sun and planet but the planet has less than the calculated speed for a circular orbit In that case the planet will not go in a circle its orbit will be a short ellipse Figure 2bl7 If the planet has more than this calculated speed the orbit will be a long ellipse or a parabola or a hyperbola not a circle The circu lar orbit is very special and requiresjust the right combination of radius of curvature and speed For a given radius only one value of the speed gives a circular orbit Like wise if you know the speed of an object in a circular orbit you can figure out what the radius ofits orbit has to be Newton himself published a diagram that shows very clearly how the ini tial speed determines the type oforbit He imagined standing on avery high mountain on the Earth so high as to be above the atmosphere and throw ing a rock horizontally at various speeds In Figure 2b18 you see that if New ton throws the rock with a low speed the rock simply falls toward the Earth and hits the ground near the base of the mountain If he throws a bit faster the rock goes further as it falls hitting the ground a bit farther from the base of the mountain With a somewhat higher initial speed the rock goes a significant distance around the Earth falling all the time before hitting the ground But if Newton throws the rockjust right the rock falls and falls and falls and falls in a circle and comes around and hits him in the back of the head This is the magic special speed for circular motion and there is only one speed that will produce a circle It is a speed which you can calculate as Newton did from what you now know and it is the speed of those satellites that are in circular orbits just above our atmosphere The speed is thou sands of meters per second so this is a thought experiment as far as an unaided human thrower is concerned If Newton throws the rock even faster than the magic speed for circular motion the rock moves in a long elliptical orbit And for even higher speeds still small compared to the speed of light the orbit is a parabola or hyper bola and the rock never comes bac Example The Earth s elliptical orbit can be well approximated for many purposes by a circular orbit of radius l5gtlt1011 In What would we predict for the orbital speed of the Earth Start from the 127 J Short ellips Figure 2bt17 The initial speed 10 has to be just right to initiate a circular orbit If the speed is more or less than this value you get an ellipse or for very high speeds a parabola or hyperbolat Circle Figure 2bt18 Newton throws a rock hori zontally from a high mountain Only one particular speed makes the rock move in a circular orbit Chapter 2 The Momentum Principle Momentum Principle as applied to circular motion Solar system data are given in the inside back cover of the textbook Solution LEE IPnetl Momentum Principle t F Mm GMm circular motion gravitational force 1 ltlt 0 0 mass m of the Earth cancels lvl2 2 GA g solve for speed squared 30 GA g 67x10 11Nm2kg2 2X10 11k 3x104ms 15gtlt10 m Ex 2b 11 What is the speed ofa satellite in a circular orbit near the Earth You can use the radius of the Earth as the approximate radius of the orbit because the atmosphere is only about 50 km thick whereas the radius ofthe Earth is 64X106 m The mass of the Earth is 6gtltlO24 kg Follow the analysis in the example above but think about the meaning of Mand Rfor this situation 2b42 Nongravitational situations Our analysis is not limited to gravitational forces Any time an object moves at constant speed in a circular orbit when subjected to a net force of con stant magnitude directed radially inward we have from the Momentum Principle x 3 anetl circular motion at constant speed For example if you swing a rock at the end of a string in a horizontal circu lar orbit at constant speed the net force is the horizontal component of the tension in the string For this reason rather than memorizing the very special result Ivl 2R GM R2 for the case ofcircular motion due to a gravitational force it is better to memorize the more general result for the time rate of change of the perpendicular component of momentum ldp dti and then equate this perpendicular component of the rate ofchange of mo mentum to whatever the perpendicular component of the net force hap pens to be Ex 2b 12 In outer space a ball of mass 5 kg at the end ofa spring moves in a circle of radius 2 m at a constant speed of 3 ms What is the magnitude of the force exerted by the spring on the rock ln words in what direction does the force act 2b4 A Whose Cimbzr motion at constant speed 2b43 Period Often we know or would like to calculate the time it mkes a planet to go around is star once called the period T F r example the period ofthe Earth39s orbit around the Sun is one calendar year Ifa particle goes all the way around a circle ofradius Rin a time T at consmnt speed what is the speed T e distance around once is the circumference 27ER so the speed is this distance divided by the time it takes 39339 27 T This makes it possible to predict a relationship between the period Tofa planet and is dismnce Rfrom the Sunust put this value for the speed into the result we obmined above the gravitational force law was a triumph ofthe Newtonian approach Ex 21713 0n the previous page we predicted that the speed ofthe Earth around the Sun should be 3x10 ms Do you get the same result for the speed from the simple geometrical formula v 27tRT HintYou need the periodinsecondsHow many seconds in a year Compare with our prediction from the Momentum Principle 2b44 Example Circularpendulum cuning motion A another 39 39 39 f 39 39 39 ofthe Momentum Principle we will analyze a ball hanging at the end ofa 39 39 otion There are many possible motions ofthis all could move so violently that the string goes slack for part of the time with abrupt changes of momentum every time the string nntireahl th ere can be asizable length Essentially we would need the effective spring stiffness for this stiff n i g tension force that the string applies to the ball can vary a great deal This makes it very difficult to do a comput numerical integration because a tiny error in position makes change in the orce here is however one particular motion ofthe ball that is simple enough er a huge mOLIOn with a xed stretch of the string This is a circular pendulum Figure 2b19 As you discovered when studying planetary orbim circular motion doesn39t Figure 2b19 A circular pendulum 129 130 Figure 2b20 A diagram of the circular pendulum showing the main geometrical aspecm of the situation E Figure 2b21 We show known and unknown forces to create a useful force diagram 2 71 E Figure 2b22 There is no centrifugal force There is no interaction associated with such a fictitious force Moreover if the net force were zero the ball would have to move in a straight line Chapter 2 The Momentum Principle just happen To get the ball moving in a circle you have to start it moving withjust the right initial conditions We tackle a problem that we can solve by imposing the requirement that we only consider the case of circular mo tion A large circular pendulum was used by Newton to measure g the magni tude of the gravitational field with considerable accuracy You could ob serve and measure the motion of a circular pendulum yourself and your analysis of the motion would let you deduce a value for g which you could compare with the accepted value A physics diagram Figure 2b20 shows a snapshot of a side view of the circular pendulum at an instant when the string of length L to the center of the ball lies in the xy plane at an angle 9 to the vertical and the ball is at the leftmost point in its circular path of radius r Lsine Let s assume that the ball is moving counterclockwise as seen from above Here is the analysis 1 Choose a system System the ball 2 List external objects that interact with the system with diagram the string the Earth and the air but neglect air resistance see Figure Qb l 3 Apply the Momentum Principle fuet d dt l l 0 0gt ltFTsine FTcoseimg 0gt 4 No need to apply the position update formula 5 Solve for the unknowns 2 FTsine so FT my where r Lsine r rsmG 0 FTcoseimg so F mg cose From the two expressions for the tension in the string FT m lie so g m Lsine2 Cose Lsin92 Measuring the speed 1 length L and angle 9 determines g lfthe time to go around once is T adistance 21w Qanine the speed is v 21thin9 T so you can measure L 9 and Tto get g 6 Check The units can be shown to be correct Also note that if 9 0 the string hangs vertically and there is no motion so the tension in the string ought to be mg which is consistent with F m T cosO g No centrifugal force We point out that in this analysis there is no role for a radially outward cen trifugal force Figure 2b There is no interaction associated with such a fictitious force Moreover if the net force were zero Newton s first law says that the ball would have to move in a straight line 2174 A special case Circular motion at constant speed 2b45 Comment Forward reasoning A common technique for solving problems is to start by considering what you want to know or what you have been asked to determine and to work backward from that to the solution We have approached the analysis of physical systems in the opposite way We have begun by organizing our knowledge about the system according to a fundamental principle the Pr39 39p d 39 recording our knowledge in equations and diagrams After we have recorded all the information we have whatever is missing is what we need to figure out At this point it is usu ally quite clear how to figure out this missing information This way ofworking problems sometimes called forward reasoning is typical of the way expert physicists solve problems of this kind starting each new problem from the fundamental principles This approach works For the student however approaching problems in this way may require a bit offaith it is not necessarily obvious how the desired answerwill emerge from the analysis lfyou practice approaching problems in this way you will develop confidence in the approach and will gain important experience in explaining complex phenomena starting from fundamental principles 2b46 Example An amusement park ride curving motion There is an amusement park ride that some people love and others hate in which a bunch of people stand against the wall ofa cylindrical room of ra dius R and the room starts to rotate at higher and higher speed Figure 2b23 When a certain critical speed is reached the floor drops away leav ing the people stuck against the whirling wall Explain why the people stick to the wall without falling down Include a carefully labeled force diagram ofa person Figure 2b24 and discuss how the person s momentum changes and why 1 Choose a system System person 2 List external objects that interact with the system with diagram the wall the Earth and the air neglect air resistance represent the system by a circle 3 Apply the Momentum Principle dpdt fquot immoo 7F 7 0 ltRpgt lt Mfmgm 4 No need to apply the position update formula 5 Solve for the unknowns 7F so F 3 mv R N N R 0 f7mgsof mg 6 Check The units check Also the faster the ride the bigger the inward force of the wall which sounds right The wall exerts an unknown force which must have a y component fbe cause the person isn t falling and an as component FN normal to the wall because the person s momentum is changing direction Note that there 131 A Figure 2b23 An amusement park ride f J LFN X mg R Figure 2b24 Physics diagram of the per son At this instant the person is moving in the z direction 132 P tom Figure 2b25 The total momentum of the system of four objecm is the sum of the individual momenta of each object Chapter 2 The Momentum Principle is no outwardgoing centrifugal force There is a momentum change in ward if the net force were zero the person would move in a straight line The vertical componentfof the wall force is a frictional force If the wall has friction that is too low the person won t be supported In Chapter 5 we will see that f HFN where the coefficient of friction H has a value for many materials between about 01 and 10 Basically the more strongly two objects are squeezed together the harder it is to slide one along the other The speed of the ride has to be large enough that f HFN so we have quotre t mv This predicts that the lower the speed 1 the greater the coef cient of fric tion must be to hold the person up on the wall which sounds right Con versely if the ride spins very fast the wall doesn t have to have a very large coefficient of friction to hold the people against the wall 2b5 Systems consisting of several objects In our applications of the Momentum Principle up to this point we have usually chosen a single object as the system of interest We have examined the change of momentum of this singleobject system either instantaneous ly using the derivative form of the Momentum Principle or over a finite time interval using the Momentum Principle in the finite difference form pf panama In some situations however it can be very useful to choose a system that consists of two or more interacting objects This is a legitimate choice of sys tem the Momentum Principle applied to a system of two or more objects says that the change in the totalmomentum of the system Figure 2b25 can be determined by nding the net external impulse applied to the system MOMENTUM PRINCIPLE FOR MULTIPARTICLE SYSTEMS APmtai Pmmr Pmmu PnetextAt Where A A A A ptotal E p1 p2 p3 sum of momenta of all objects 1n the system A A A A Fnegextg F1 F2 F3 sum of all external forces on the system This is an interesting statement of the Momentum Principle it implies that ifwe know the external forces acting on a multiobject system we can draw conclusions about the change of momentum of the system over some time interval without worrying about any of the details of the interactions of the objects with each other This can greatly simplify the analysis of the motion of some very complex systems We have implicitly been using the multiparticle version of the Momentum Principle when we have treated macroscopic objects like humans space craft and planets as if they were single pointlike objects In Section 2b9 there is a full derivation of the multiparticle version of the Momentum Prin ciple in which it is shown that because ofreciprocity forces between objects inside the system do not change the total momentum of the system 2175 Systems consisting of several oly39ects 2b51 Collisions Negligible external forces An event is called a collision ifit involves an interaction that takes place in a short time and has a large effect on the momenta of the objects compared to the effects ofother interactions A collision does not necessarily involve actual physical contact between objects which may be interacting via long distance forces like the gravitational force or the electric force Often it can be useful to analyze collisions by choosing a system that includes all of the colliding objects As a simple example consider the case of two sticky balls traveling through the air toward each other at speeds much less than the speed of light Figure 2b6 Choose both balls as the system What is the total momentum of the system at the instant shown in Figure 2b26 Since 1 ltlt 0 the total momentum of the twoball system is s Ptomu 1 71 szzi Suppose that when they collide the balls stick together What can we conclude about their nal velocity lfwe can assume that the effects of gravity and air resistance have a negligi ble effect during the short time of the actual collision we know that the mo mentum of the stucktogether balls the final total momentum of the system must be the same as the initial total momentum of the system Ptomlf Ptomlj quot 1 m27f ml s71i m2 372139 quotMP1 2 vf m1 m2 If the mass of ball 1 is 2 kg and it initial velocity is 6 0 0 ms and the mass of ball 2 is 5 kg and its initial velocity is 75 4 0 ms then the final velocity of the stucktogether balls will be 3 2 kg600gt5 kg7540gt 186 286 0 v m s 7 m s f 7 kg lt gt We reached this conclusion without needing to know anything about the details ofthe complex forces that the balls exerted on each other during the collision Some problems require more than one principle Because the balls in the previous example had the same final speed we had enough information to solve for the final velocity of the stucktogether balls we had one equation with only one unknown vector quantity In more complex situations we will nd that we sometimes do not have enough information to solve for all of the unknown quantities by applying only the Momentum Principle For example consider a collision between two balls that balls bounce off each other as shown in Figure 2b28 The Momentum Principle tells us that ptomlj A A P1aP2i totalf PM P21 butwe are leftwith two unknown vector quantities p and pzf and only one equation We will not be able to analyze problems ofthis kind fully until we can invoke the Energy Principle Chapter 4 and the Angular Momen tum Principle Chapter 9 along with the Momentum Principle 133 ml 9139 39lgt 6239 W2 Figure 2bi26 Two sticky balls traveling through the airjust before colliding kl 7 ml Figure 2b27 The momentum of the stuck together balls just after the collision is equal to the sum of the initial momenta of the two balls M m 61quot 62 M m l v v1f gt 2f Figure 2bi28 The initial momenta of the colliding objecm are known but the nal momenta after the collision are unknown B B initiall at rest Y A I now at rest V I A initial state I final state before collision a er collision Figure 21129 Head on collision between two balls of equal mass Initially left side ball A has Velocity y and B is at rest After 39 39 A is at rest and B The tom momentum ofthe twoeball system has not changed Figure 2b 50 A binary soar of the smrs as the system T e m of the system changes due to the external force choose j ust on e h om ent Figure 2b 51 A binary soar choose both smrs as the s stem The momentum of e combined system doesn t change Chapm 2 m Momentum Printipb 2b52 Momentum flow within a system Su pose a bowling ball ball A rolls along asmooth surface at almost con stant velocity and hits another bowling ball ball B ofequal mass which is initially at rest Ifthe collision is a headeon collision we will see ball A come ball B roll along with a velocity nearly equal to the initial vee p29 choose asystem including both balls we nd that the initial momene tum and the final momentum ofthe system are the same 5 2 Q a U gt E W E r ru N p m 0 P despite the fact that the momentum of each object within the system has c anged Assuming that the e ecm ofair resismnce and friction are negligie ble during the short time when the balls interact with each other so the net external force applied to the system is approximately zero this is consistent w39 h the momentum principle 07 3r lt000gtAt lt0 10gt Ex 21714 You and afriend each hold a lump ofwet clay Each lump has a mass of20 msYou e ch p ofclay into the air and the velocity of the as lt73 0 72gt ms What was the total momentum of the lumpsjust before impact What is the momentum of the stucke together lump just after the impact What is in velocity Em 21715 In outer space far from other ob and stick together Before t e collision their momenta were 10 20 e5 kgms and 8 76 12gt kgms What was their total momentum before the collision What must be the momentum of the combined object after the collision Ex 21216 At a certain instant the momentum of a proton is lt34x10 71 0 0gtkg ms as it approaches another proton which is initially at rest The two protons repel each other electrically without coming close enough to touch When they are once again far apart one o the protons now has momentum lt24x10 71 16x10 710gtkgms At this instant what is the momentum ofthe other proton jects two rocks collide 2b6 Conservation of momentum The choice ofsystem affecm the detailed form of the Momentum Principle Consider the c the surroundings consist of other stars which may be so far away as to have negligible effecm on the binary star In that case the net external force acte 39 on the system is nearly zero What happens to the toml momentum ofthe isolated binary smr as time goes by 217 6 Commation of momentum The Momentum Principle pf pi lT netAt reduces in this case to pf pi which predicts that the total momentum of the system p p1 p2 remains constant in magnitude and direction This is an important special case the total momentum of an isolated system a system with negligible interactions with the surroundings doesn t change but stays constant In a later chapter we will see that the total momentum can be expressed as Mm cem mum where the center of mass is a mathematical point be tween the two stars closer to the more massive star The velocity of the cen ter of mass does not change but is constant as the binary star drifts through space or is zero if the binary star s total momentum is zero One way to think about this result is to say that momentum gained by one star is lost by the other because as we saw in the previous chapter the grav itational forces and impulses are equal in magnitude but opposite in di rection The effectis that the total momentum doesn t change Let F be the force exerted on star 1 by star 2 so fl is the force exerted on star 2 by star 1 After a short time interval At the new total momentum is this 31FAt 2 FAt 1 2 Whatever momentum is lost by one star is gained by the other star This is a simple example of an important principle the conservation of momen tum which says that the change ofmomentum in a system plus the change of momentum in the surroundings adds up to zero CONSERVATION OF MOMENTUM quot Ap 5 Apsysuem surroundings In the case of the two stars there are no objects in the surroundings and no external forces so the momentum of the system doesn t change Zero net external impulse lTnaAt zero net momentum change One of the stars can gain momentum due to a force acting on it but only if the other star loses the same amount Relativistic momentum conservation Particle accelerators produce beams of particles such as electrons protons and pions at speeds very close to the speed of light When these highspeed particles interact with other particles experiments show that the total mo mentum is conserved but only if the momentum ofeach particle is defined in the way Einstein proposed p2 mv All 7 Ivl c2 It is found that using the lowspeed approximation pm my there is no conservation of momen tum when the speed v approaches the speed of light 0 Ex 2b7 Consider the headon collision of two identical bowling balls discussed above and shown in Figure 2b29 a Choose a system consisting only of ball A What is the momen tum change of the system during the collision What is the momen tum change of the surroundings b Choose a system consisting only of ball B What is the momen tum change of the system during the collision What is the momen tum change of the surroundings c Choose a system consisting of both balls What is the momen tum change of the system during the collision What is the momen tum change of the surroundings 135 Local clmter ie of galax Solar System Figure 232 Our Solar System orbits around the center of our galaxy which interacts with other galaxies Figure 2mg In the three body problemquot each of the three objects interacts With two other objects Chapm 2 m Momentum Printipb 2b7 Predicting the future of complex gravitating systems We can use the Newtonian method and numerical integration to try to pre dict the future of a group of objects that interact with each other mainly gravimtionally such as the Solar System consisting of the Sun planets moons asteroids comes etc There are other nonegravimtional interace tions present Radiation pressure from sunlight makes a comet s tail sweep away from the sun Streams ofcharged particles the solar wind from the Sun hit the Earth and contribute to the Northern and Southern Lighm aue roras But the main interactions within the S 1 directions mke place very slowly compared to those within the solar system C can L A ing the inner workings of the Solar System itself odeled the motion of planets and smrs using the Momentum that is necessary is to include the forces associated with all pairs of objecm In principle 39 39 39 39 391 quot 39 C tire Solar System Of course our prediction would be only approximate be cause we mke finite time steps which introduce numerical errors and we are ene u i in V i cure the prediction ofour model is a reasonable approximation to the real mor tion ofour real Solar 5 Stem Our Solar System contains a huge number of objects Most of the nine lanets have moons and Jupiter and Saturn have many moons There are thousands of asteroids and comes not to mention an uncounted number 539 r 39 H Tquot 39 39 Iimitminnquot would take a very long time to carry out even one time step for all the pieces ofthe Solar System 2b71 The threebody problem and planet We simply use the superposition principle to include the inter h action with 7713 as well as the interaction wit m1 quotlblzrz39l Glbzlm We use this force to update the momentum of object 17 Similar computar tions apply to the other masses m1 and 7713 but let the computer do them We can carry out a numerical integration for each mass iven the current locations ofall the masses we can calculate the vector net force exerted on each m n we can calculate the new values of momentum and position at a slightly later time t At 217 8 Determinism While there exist analytical nonenuinerical solutions for tworbody grave itational motion except for 39 reerbody prob lem has not been solved analytically However in a numerical integration the computer doesn t mind that there are loLs of quantities and a lot ofrer petitive calculations to be done You yourself can solve the threerbody probe em numerically by adding additional calculations to your tworbody computations see the threerbody homework problem in Chapter 2 r 1 L or LJ o a ing each other with l a circle ellipse parabola hyperbola or straight line The motion ofa threer body system can be vastly more complex and diverse Figure 2b34 shows the numerical integration ofa complicated threerbody trajectory for one particr ular set ofinitial conditions Addingjust one more object opens up a vast range of complex behaviors Imagine how complex the motion ofa galaxy can be 2b72 Sensitivity to initial conditions For a tworbody system slight changes in the initial conditions make only slight changes in the orbit such as changing a circle into an ellipse or an ellipse into a slightly different ellipse but you never get anything other than a simple trajectory The situation is very different in systems with three or more interacting objecm Consider a lowrmass object orbiting two massive objecLs which we imagine somehow to be nailed down so that they cannot move Figure 2b35 This could also represent two positive charges that are fixed in por sition with a lowrmass negative charge orbiting around them In Figure 2b36 are two very different orbiLs one shown in black the 0th er in gray starting from only slightly different initial conditions In both of these cases the lowrmass object was released from rest but at slightly differ ent initial locations The trajectories are wildly different The mass of the object on the left is twice the mass of the object on the right in this compur tation This sensitivity to initial conditions generally becomes more extreme as you add more objecm and the Solar System contains loLs of objects big and small Also we can anticipate that ifsmall errors in specifying the initial con ditions can have large effecLs so too it is likely that failing to take into ac count the 39 quot 39 difference after a long integration time The Solar System is actually fairly predictable because there is one giant mass the Sun and the ot er muc smaller masses are very far apart This is unlike the example shown here which deliberately emphasizes the sensie tivity observed when there are large masses near each other might make 2 hi 2b8 Determinism Ifyou know the net force on an object as a function of is position and mor mentum or velocity you can predict the future motion of the object by simple steprbyrstep calculations based on the momentum principle Newr ton s demonstration of the power of this approach induced many sevenr teenth and eighteenth century philosophers and scientisLs to adopt the view already proposed by Descartes Boyle and others that the Universe is a gir 39 39 r detenuiued by the present positions and momenta of all the macroscopic and microscopic objecLs in it just turn the crank and predict the future This point ofview is called determinism and taken 0 is extreme it raises the question of whether humans actually have any free will or whether all of our actions are preder 137 Figure 2b34 One example of three bodies interacting gravimtionally 0 Figure 2b35 A low mass object orbiting tw massive objects Figure 2b36 Two different orbits one gray the other black smting from rest but from slightiy different positions 138 Chapter 2 The Momentum Principle termined Scientific and technological advances in the twentieth century however have led us to see that although the Newtonian approach can be used to predict the longterm future ofa simple system or the shortterm fu ture of a complicated system there are both practical and theoretical limi tations on predictions in some systems 2b81 Practical limitations One reason we may be unable to predict the longterm future of even a sim ple system is a practical limitation in our ability to measure its initial condi tions with sufficient accuracy Over time even small inaccuracies in initial conditions can lead to large cumulative errors in calculations This is a prac tical limitation rather than a theoretical one because in principle if we could measure initial conditions more accurately we could perform more accurate calculations Another practical limitation is our inability to account for all interactions in our model Every object in the Universe interacts with every other object In constructing simplified models we ignore interactions whose magnitude is extremely small However over time even very small interactions can lead to significant effects Even the tiny radiation pressure exerted by sunlight een shown to affect noticeably the motion of an asteroid over a long time With larger and faster computers we can include more and more in teractions in our models but our models can never completely reflect the complexity of the real wor A branch of current research that focuses on how the detailed interac tions ofa large number ofatoms or molecules lead to the bulk properties of matter is called molecular dynamics For example one way to test our un derstanding of the nature of the interactions ofwater molecules is to create a computational model in which the forces between molecules and the ini tial state of a large number of molecules several million are described and then to see if running the model produces a virtual fluid that actually behaves like water However even with an accurate force law for the elec tric interactions between atoms or molecules it is not feasible to compute and track the motions of the 1025 molecules in a glass ofwater Significant efforts are underway to build faster and faster computers and to develop more efficient computer algorithms to permit realistic numerical integra tion of more complex systems 2b82 Chaos A second kind of limitation occurs in systems that display an extreme sensi tivity to initial conditions We saw something like this in the example of the threebody problem in gravitating systems small changes in initial condi tions produced large changes in behavior In recent years scientists have dis covered systems in which a change in the initial conditions no matter how small infinitesimal can lead to complete loss of predictability the differ ence in the two possible future motions of the system diverges exponentially with time Such systems are called chaotic It is thought that over a long time period the weather may be literally unpredictable in this sense An interesting popular science book a out this new field of research is Chaos Making a New Science byJames Gleick Penguin 1988 The issue of small changes in initial conditions is a perennial concern of time travellers in science fiction stories Perhaps the most famous such story is Ray Brad bury s The Sound of Thunder in which a time traveller steps on a butter fly during the Jurassic era and returns to an eerily changed present time 2178 Determinism 2b83 Breakdown of Newton s laws There are other situations of high interest where we cannot usefully apply Newton s laws of motion because these classical laws do not adequately de scribe the behavior of physical systems To model systems composed of very small particles such as protons and electrons quarks and photons it is nec essary to use the laws of quantum mechanics To model in detail the gravi tational interactions between massive objects it is necessary to apply the principles of general relativity Given these principles one might assume that it would be possible to follow a procedure similar to the one we have followed in this chapter and predict in detail the future of these systems However there appear to be more fundamental limitations on what we can know about the future 2b84 Probability and uncertainty Our understanding of the atomic world of quantum mechanics suggests that there are fundamental limits to our ability to predict the future be cause at the atomic and subatomic levels the Universe itself is nondetermin istic We cannot know exactly what will happen at a given time but only the probability that certain events will occur within a given time frame A simple example may make this clear A free neutron one not bound into a nucleus is unstable and eventually decays into a positively charged proton a negatively charged electron and an electrically neutral antineu tr1no ngtpfv The average lifetime of the free neutron is about 15 minutes Some neu trons survive longer than this and some last a shorter amount of time All ofour experiments and all of our theory are consistent with the notion that it is not possible to predict when a particular neutron will decay There is only a known probability that the neutron will decay in the next microsecond or the next or the next If this is really how the Universe works then there is an irreducible lack of predictability and determinism of the Universe itself As far as our own predictions using the momentum principle are concerned consider the fol lowing simple scenario An electron is traveling with constant velocity through nearly empty space but there is a free neutron in the vicinity Fig ure 237 We might predict that the electron will move in a straight line and some fraction of the time we will turn out to be right the electrically uncharged neutron exerts no electric force on the electron there is a magnetic inter action but it is quite small and the gravitational interaction is tiny But if the neutron happens probabilistically to decayjust as our electron passes nearby suddenly our electron is subjected to large electric forces due to the decay proton and electron and will deviate from a straight path Moreover the directions in which the decay products move are also only probabilistic so we can t even predictwhether the proton or the electron will come clos est to our electron Figure 238 This is a simple but dramatic example of how quantum indeterminacy can lead to indeterminacy even in the context of the Momentum Principle The Heisenberg uncertainty principle The Heisenberg uncertainty principle states that there are actual theoreti cal limits to our knowledge of the state ofphysical systems One quantitative 139 H 0 Figure 237 The electron may continue to travel in a straight line but we cannot be certain that it will because we do not know when the neutron will decayt c I Figure 238 If the neutron does decay when the electron is near it the trajectory of the electron will depend on the direc tions of the momenta of the decay prod ucmt 140 2 Figure 2bl39 External and internal forces acting on a system of three particlesr Chapter 2 The Momentum Principle formulation states that the position and the momentum ofa particle cannot both be simultaneously measured exactly AxApxzh This relation says that the product of the uncertainty in position Ax and the uncertainty in momentum Apx is equal to a constant k called Planck s con stant Planck s constant is tiny I 66X10734 kgm s so this limitation is not noticeable for macroscopic systems but for objects small enough to re quire a quantum mechanical description the uncertainty principle puts fundamental limits on how accurately we can know the initial conditions and therefore how well we can predict the future 2b9 Derivation of the multiparticle Momentum Principle In a threeparticle system Figure 21239 we show all of the forces acting on each particle where the lower case f s are forces the particles exert on each other socalled internal forces and the upper case F s are forces exert ed by other objects that are not shown and are not part of our chosen system these are so called external forces such as the gravitational attraction of the Earth or a force that you exert by pulling on one of the particles We will use the following shorthand notation flyg will denote the force exerted on particle l by particle 3 F275 will denote the force on particle 2 exerted by external objects and so on Here is the Momentum Principle for each of the three particles dquot A A A E F1 ext f12 f13 dt dA A A A g F2ext f21 f23 dt dA A A A E F3 ext f31 32 dt Nothing new so far But now we add up these three equations That is we create a new equation by adding up all the terms on the left sides of the three equations and adding up all the terms on the right sides and setting them equal to each other dil L132 e A A A A A A A A A dt dt dt Flext f12 f13 172 ext f21 f23 F3ext f31 f32 Many of these terms cancel By the principle of reciprocity Newton s third law of motion which is obeyed by gravitational and electric interactions we have the following A A f12 421 A A f13 7f31 A A f23 32 Thanks to reciprocity all that remains after the cancellations is this dA dpg A A A W W FlextF2ext F3ext A A A The total momentum of the system IS Pm p1 p2 p3 so we have die A A A dt Fnetyext or in terms of impulse APLOL FnayextAt 2179 Dnivatitm of the multiparticle Momentum Principle The importance of this equation is that reciprocity has eliminated all of the internal forces the forces that the particles in the system exert on each oth er internal forces cannot affect the motion of the system as awhole All that matters in determining the rate of change of total momentum is the net external force The equation has exactly the same form as the Momen tum Principle for a single particle Moreover if the object is a sphere whose density is only a function of ra dius the object exerts a gravitational force on other objects as though the sphere were a point particle We can predict the motion ofa star or a planet or an asteroid as though it were a single point particle oflarge mass 141 1 42 Chapter 2 The Momentum Principle 2b10 Summary THE MOMENTUM PRINCIPLE DERIVATIVE FORM dp net dt In words the instantaneous time rate ofchange of the momentum ofan object is equal to the net force acting on the object Or in cal culus terms since the limit of a ratio of infinitesimal quantities is called a derivative we can say that the derivative of the momentum with respect to time is equal to the net force acting on the object Finding dp dt 1 Draw two arrows Figure 2b40 one representing pi the momentum of the object ashort time be fore the instant of interest and a second arrow representing pf the momentum of the object a short time after the instant ofinterest A 2 Graphically find Ap pfi t by placing the arrows tail to tail and drawing the resultant arrow starting at the tail of E and going E to the tipofpf 3 This arrow indicates the direction of Ap which is the same as the direction of dp dt provided the time interval involved is suffi ciently small 7 p 1 MOMENTUM PRINCIPLE FOR MULTIPARTICLE SYSTEMS V Athal meu istomlj FneueXtAt H p fnz 21 1 Where if pmal E p1 p2 p3 sum of momenta of all objects in the system ix i A Fnegextz F1 F2 F3 sum of all external forces on the system 7 34 13 A 3 F4 CONSERVATION OF MOMENTUM P2 131 I r APsysuem APsurmunaungs 0 1 tom Figure 2bi4l The total momentum of the system of four objecm is the sum of the individual momenta of each object 21711 Mien questions 2b11 Review questions Circular motion RQ 2b You are driving an American car sitting on the left side of the front seat You make a sharp right turn You feel yourself thrown to the left and your left side hits the left door Is there a force that pushes you to the left What object exerts that force What really happens Draw a diagram to il lustrate and clarify your analysis Circular motion RQ 2b2 A 30 kg child rides on a playground merrygoround 2 m from the center The merrygoround makes one complete revolution every 4 sec onds How large is the net force on the child In what direction does the net force act Determinism RQ 2b3 List as many limitations as you can on our ability to predict the fu ture 143 144 m2 m1 3 Figure 2bl42 Push on a block Problem Chapter 2 The Momentum Principle 2b12 Problems Problem 2b A springlike device A certain springlike device with uneven windings has the property that to stretch it an amount 5 from its relaxed length requires a force that is given by F bs3 You suspend this device vertically and its unstretched length is 20 cm a You hang a mass of 15 grams from the device and you observe that the length is now 24 cm What is b including units Start your analysis from the Momentum Principle b You hold the 15 gram mass and throw it downward releasing it when the length of the springlike device is 27 cm and the speed of the mass is 4 ms One millisecond later 10 3 5 what is the stretch of the device and what is the speed of the mass Problem 2b2 Pushing on a block Two blocks ofmass m1 and M3 connected by a rod ofmass m2 are sitting on a lowfriction surface and you push to the left on the right block mass ml with a constant force Figure 2b42 a What is the acceleration dvxdt of the blocks b What is the compression force in the rod mass m2 near its right end Near its left end c How would these results change ifyou pull to the left on the left block mass M3 with the same force instead of pushing the right block Problem 2b3 Relating radius to period for circular orbits The planets in our Solar System have orbits around the Sun that are nearly circular and 1 ltlt 0 Calculate the period T a year the time required to go around the Sun once for a planetwhose orbit radius is r This is the re lationship discovered by Kepler and explained by Newton It can be shown by advanced techniques that this result also applies to elliptical orbits if you replace rby the semi major axis which is half the longer major axis of the ellipse Use this analytical solution for circular motion to predict the Earth s orbital speed using the data for Sun and Earth on the inside back cover of the textbook Problem 2b4 Experirnent determine gwith a circular pendululn Use a circular pendulum to determine g You can increase the accuracy of the time it takes to go around once by timing Nrevolutions and then divid ing by N This minimizes errors contributed by inaccuracies in starting and stopping the clock It is wise to start counting from zero 0 l 2 3 4 5 rath er than starting from 1 l 2 3 4 5 represents only 4 revolutions not 5 It also improves accuracy if you start and stop timing at awelldefined event such as when t e mass crosses in front ofan easily visible mar This was the method used by Newton to get an accurate value of g New ton was not only a brilliant theorist but also an excellent experimentalist For a circular pendulum he built a large triangular wooden frame mounted on a vertical shaft and he pushed this around and around while making sure that the string of the circular pendulum stayed parallel to the slanting side of the triangle Problem 2b5 Mass on a spring with circular motion A small block of mass m is attached to a spring with stiffness k5 and relaxed length L The other end of the spring is fastened to a fixed point on a low friction table The block slides on the table in a circular path ofradius Rgt L How long does it take for the block to go around once 217 I 2 Problems Problem 2b6 Circular motion in a magnetic field When a particle with electric charge qmoves with speed 1 in a plane perpen dicular to a magnetic field B there is a magnetic force at right angles to the motion with magnitude qu and the particle moves in a circle of radius 7 Figure 2b43 This formula for the magnetic force is correct even if the speed is comparable to the speed oflight Show that mv A17c2 even if v is comparable to 0 Remember that f m3 is not valid for high speeds This result is used to measure relativistic momentum if the charge q is known we can determine the momentum ofa particle by observing the ra qBr dius ofa circular trajectory in a known magnetic field Problem 2b7 Dark matter in the Universe In the 1970 s the astronomer Vera Rubin made observations of distant gal axies that she interpreted as indicating that perhaps 90 of the mass in a galaxy is invisible to us dark matter She measured the speed with which stars orbit the center of a galaxy as a function of the distance of the stars from the center The orbital speed was determined by measuring the Dop pler shift of the light from the stars an effect which makes light shift to ward the red end of the spectrum red shift if the star has a velocity component away from us and makes light shift toward the blue end of the spectrum if the star has a velocity component toward us She found that for stars farther out from the center of the galaxy the orbital speed of the star hardly changes with distance from the center of the galaxy as is indicated in the diagram below The visible components of the galaxy stars and illu minated clouds of dust are most dense at the center of the galaxy and thin out rapidly as you move away from the center so most of the visible mass is near the center roachm us bluershifted ligh About the same speed at a different radius a Predict the speed 1 ofa star going around the center of a galaxy in a circular orbit as a function ofthe star s distance rfrom the center of the gal axy assuming that almost all of the galaxy s mass M is concentrated at the center b Construct a logical argument as to why Rubin concluded that much of the mass ofa galaxy is not visible to us Reason from principles discussed in this chapter and your analysis of part a Explain your reasoning You need to address the following issues 39 Rubin s observations are not consistent with your prediction in a 39 Most of the visiblematter is in the center of the g ax 39 Your prediction in a assumed that most of the mass is at the center This issue has not yet been resolved and is still a current topic ofastrophys ics research Here is a reference to the original work DarkMatter in Spiral Galaxies by Vera C Rubin Scienti cmericanJune 1983 96108You can find several graphs of the rotation curves for spiral galaxies on page 101 of this article 145 Figure 2br43 A charge moving in a circle due to a magnetic force Problem 2br6r 146 Figure 2bi44 Three stars in a circular orbit Problem 2bl9i Figure 2bl45 Observed positions of a star from 1995 to 1999 near the center of our Milky Way Galaxy and an orbit fit to the data Problem 2bilOi Chapter 2 The Momentum Principle Problem 2b8 Orbital periods a Many communication satellites are placed in a circular orbit around the Earth at a radius where the period the time to go around the Earth once is 24 hours If the satellite is above some point on the equator it stays above that point as the Earth rotates so that as viewed from the rotating Earth the satellite appears to be motionless That is why you see dish anten nas pointing at a fixed point in space Calculate the radius of the orbit of such a synchronous satellite Explain your calculation in detail b Electromagnetic radiation including light and radio waves travels at a speed of 3gtltlO8 ms lfa phone call is routed through a synchronous satel lite to someone not very far from you on the ground what is the minimum delay between saying something and getting a response Explain Include in your explanation a diagram of the situation c Some humanmade satellites are placed in nearEarth orbit just high enough to be above almost all of the atmosphere Calculate how long it takes for such a satellite to go around the Earth once and explain any ap proximations you make d Calculate the orbital speed for a nearEarth orbit which must be pro vided by the launch rocket Recently large numbers ofnear Earth commu nications satellites have been launched Their advantages include making the signal delay unnoticeable but with the disadvantage of having to track the satellites actively and having to use many satellites to ensure that at least one is always visible over a particular region e When the first two astronauts landed on the Moon a third astronaut remained in an orbiter in circular orbit near the Moon s surface During half of every complete orbit the orbiter was behind the Moon and out of radio contactwith the Earth On each orbit how long was the time when ra dio contactwas lost Problem 2b9 Analytical 3body orbit There is no general analytical solution for the motion of a 3body gravita tional system However there do exist analytical solutions for very special initial conditions Figure 2b44 shows three stars eac 0 mass m which move in the plane of the page along a circle of radius 7 Calculate how long this system takes to make one complete revolution In many cases 3body orbits are not stable any slight perturbation leads to a breakup of the or bit Problem 2b10 The center of our galaxy Remarkable data indicate the presence ofamassive black hole at the center ofour Milky Way galaxy One of the two W M Keck 10 meter diameter tele scopes in Hawaii was used by Andrea Ghez and her colleagues to observe in frared light coming directly through the dust surrounding the central region of our galaxy visible light is multiply scattered by the dust blocking a direct view Stars were observed for several consecutive years to deter mine their orbits near a motionless center that is completely invisible black in the infrared but whose precise location is known due to its strong output of radio waves which are observed by radio telescopes The data were used to show that the object at the center must have a mass that is huge compared to the mass of our own Sun whose mass is a mere 2x103 kg a The speeds of the stars as a function of distance from the centerwere consistent with Newtonian predictions for orbits around a massive central object How should the speed of each star depend on its distance from the center and the mass of the central object Express your result symbolically b The star nearest the center of our galaxy was observed to be about lgtltlO14 m from the center with a period of about 15 years What is the speed of this star in ms Also express the speed as a fraction of the speed 217 I 2 Problems of light The actual orbit is an ellipse as reproduced in Figure 2b45 but for purposes of this analysis you may approximate it as being circular c This is an extraordinarily high speed for a star Is it reasonable to ap proximate the star s momentum as m1 d Calculate the mass of the massive black hole about which this star is orbiting e How many of our Suns does this represent It is thought that all galaxies may have such a black hole at their centers as a result of long periods of mass accumulation When many bodies orbit each other sometimes an object happens in an interaction to acquire enough speed to escape from the group and the remaining objects are left closer together Some simulations show that over time as much as half the mass may be ejected with agglomeration of the remaining mass This could be part of the mechanism for forming massive black holes For more information see the home page of Andrea Ghez ht tpwwwastrouclaedufacultyghezhtm The papers available there re fer to arcseconds which is an angular measure of how far apart objects appear on the sky and to parsecs which is a distance equal to 33 light years a lightyear is the distance light goes in one year Problem 2b 1 Looping roller coaster What is the minimum speed 1 that a roller coaster car must have in order to make it around an inside loop andjust barely lose contactwith the track at the top of the loop Figure 2b46 The center of the car moves along a cir cular arc of radius R Include a carefully labeled force diagram State briefly what approximations you make Design a plausible roller coaster loop in cluding numerical values for v and Problem 2bl2 Your weight on a bathroom scale If you stand on a bathroom scale at the North Pole the scale shows your weight as an amount Mg actually it shows the force FN that the scale ex erts on your feet At the North Pole you are 6357 km from the center ofthe Earth lf instead you stand on the scale at the equator the scale reads a dif ferent value due to two effects 1 The Earth bulges out at the equator due to its rotation and you are 6378 km from the center of the Earth 2 You are moving in a circular path due to the rotation of the Earth one rotation every 24 hours Taking into account both of these effects what does the scale read at the equator Problem 2b3 Space station An engineer whose mass is 70 kg holds onto the outer rim of a rotating space station whose radius is 14 m and which takes 30 s to make one com plete rotation What is the magnitude of the force the engineer has to exert in order to hold on What is the magnitude of the net force acting on the engineer Problem 2b 14 NEAR spacecraft orbits Eros After the NEAR spacecraft passed Mathilde on several occasions rocket propellant was expelled to adjust the spacecraft s momentum in order to follow a path that would approach the asteroid Eros the final destination for the mission After getting close to Eros further small adjustments made the momentum just right to give a circular orbit of radius 45 km 45X103 m around the asteroid So much propellant had been used that the final mass of the spacecraft while in circular orbit around Eros was only 500 kg The spacecraft took 104 days to make one complete circular orbit around Eros Calculate what the mass of Eros must be 147 Figure 2bi46 Make a loop in a roller coaster Problem Qbill 148 Figure 2bl47 Go over the top in a roller coaster Problem 2bll5al Figure 2bl48 Go through a dip in a roller coaster Problem 2bll5bl Chapter 2 The Momentum Principle Problem 2b 5 Weightlessness By weight we usually mean the gravitational force exerted on an object by the Earth However when you sit in a chair your own perception of your own weight is based on the contact force the chair exerts upward on your rear end rather than on the gravitational force The smaller this contact force is the less weight you perceive and if the contact force is zero you feel peculiar and weightless an odd word to describe a situation when the only force acting on you is the gravitational force exerted by the Earth Al so in this condition your internal organs no longer press on each other which presumably contributes to the odd sensation in your stomach a How fast must a roller coaster car go over the top ofa circular arc for you to feel weightless The center of the car moves along a circular arc of radius R Figure 2b47 Include a carefully labeled force diagram b How fast must a roller coaster car go through a circular dip for you to feel three times as heavy as usual due to the upward force of the seat on your bottom being three times as large as usual The center of the car moves along a circular arc of radius R Figure 2b48 Include a carefully labeled force diagram Problem 2b16 At the bottom of a Ferris wheel A Ferris wheel is a vertical circular amusement ride Riders sit on seats that swivel to remain horizontal as the wheel turns The wheel has a radius Rand rotates at a constant rate going around once in a time T At the bottom of the ride what are the magnitude and direction of the force exerted by the seat on a rider of mass m Include a diagram of the forces on the rider Problem 2bl7 Swing a ball on a spring A ball of unknown mass m is attached to a spring In outer space far from other objects you hold the other end of the spring and swing the ball around in a circle of radius 15 m at constant speed a You time the motion and observe that going around 10 times takes 688 seconds What is the speed of the ball b Is the momentum of the ball changing or not How can you tell c If the momentum is changing what interaction is causing it to change If the momentum is not changing why isn t it d The relaxed length of the spring is 12 m and its stiffness is 1000 Nm While you are swinging the ball since the radius of the circle is 15 m the length of the spring is also 15 In What is the magnitude of the force that the spring exerts on the ball e What is the mass m of the ball Problem 2b18 Car going over a hill A sports car and its occupants of mass M is moving over the rounded top ofa hill of radius R At the instantwhen the car is at the very top of the hill the car has a speed 1 You can safely neglect air resistance a Taking the sports car as the system of interest what objects exert nonnegligible forces on this system b At the instant when the car is at the very top ofthe hill draw a diagram showing the system as a dot with force vectors whose tails are at the location of the dot Label the force vectors that is give them algebraic names Try to make the lengths of the force vectors be proportional to the magnitudes of the forces c Starting from the momentum principle calculate the force exerted by the road on the car d Under what conditions will the force exerted by the road on the car be zero Explain 217 I 2 Problems Problem 2b19 Ferris wheel A person ofmass 70 kg rides on a Ferris wheel whose radius is 4 m The per son s speed is constant at 03 ms What is the magnitude of the net force acting on the person at the instant shown Draw the net force vector on the diagram at this instant with the tail of the vector on the person shown as a dot on the diagram Show your work clearly Problem 2b20 Outer space collision ln outer space a small rock with mass 5 kg traveling with velocity lt0 1800 0gt ms strikes a stationary large rock headon and bounces straight back with velocity lt0 71500 0gt ms After the collision what is the vector momentum of the large rock Problem 2b2l Two rocks collide Two rocks collide in outer space Before the collision one rock had mass 9 kg and velocity lt410072600 2800gt ms The other rock had mass 6 kg and velocity lt7450 1800 3500gt ms A 2 kg chunk of the first rock breaks off and sticks to the second rock After the collision the rockwhose mass is 7 kg has velocity lt1300 2001800gt ms After the collision what is the ve locity of the other rock whose mass is 8 kg Problem 2b22 Two rocks collide Two rocks collide with each other in outer space far from all other objects Rock1 with mass 5 kghas velocity lt30 45 720gt ms before the collision and lt710 50 75gt ms after the collision Rock 2 with mass 8 kg has velocity lt79 5 4gt ms before the collision Calculate the final velocity of rock 2 Problem 2b23 Two rocks stick together In outer space two rocks collide and stick together Here are the masses and initial velocities of the two rocks Rock 1 mass 15 kg initial velocity lt10 730 0gt ms Rock 2 mass 32 kg initial velocity lt1512 0gt ms What is the velocity of the stucktogether rocks after the collision Problem 224 The car and the mosquito A car ofmass Mmoving in the x direction at high speed 1 strikes a hovering mosquito of mass m and the mosquito is smashed against the windshield a What is the approximate momentum change of the mosquito Give magnitude and direction Explain any approximations you make 10 At a particular instant during the impact when the force exerted on the mosquito by the car is E what is the magnitude of the force exerted on the car by the mosquito c What is the approximate momentum change of the car Give magni tude and direction Explain any approximations you m e d Qualitatively why is the collision so much more damaging to the mos quito than to the car Problem 225 The bouncing ball A steel ball of mass mfalls from a height h onto a scale calibrated in newtons The ball rebounds repeatedly to nearly the same height h The scale is slug gish in its response to the intermittent hits and displays an averageforce Favg such that Fang FAt where FAt is the briefimpulse that the ball imparts to the scale on every hit and Tis the time between hits Calculate this average force in terms of m h and physical constants Com pare your result with what the scale reads if the ball merely rests on the scale Explain your analysis carefully but brie y 149 150 39 e M R Figure 2bl49 Spacejunk strikes a spinning satellite Problem 2b29i v meV yl x Figure 2bl50 Launch a package from a space station Problem 2bl3li M Chapter 2 The Momentum Principle Problem 2b26 Meteor hits a spinning satellite A satellite which is spinning clockwise has four lowmass solar panels stick ing out as shown A tiny meteor traveling at high speed rips through one of the solar panels and continues in the same direction but at reduced speed Afterwards calculate the 126 and 1 components of the centerofmass veloc ity of the satellite Initial data are provided on the diagram Z V Problem 227 Moving the Earth Suppose all the people of the Earth go to the North Pole and on a signal alljump straight up Estimate the recoil speed of the Earth Thegmass of the Earth is 6X10 kg and there are about 6 billion people 6X10 Problem 2b28 Bullet embeds in block A bullet ofmass m traveling horizontally at avery high speed 1 embeds itself in a block of mass M that is sitting at rest on a nearly frictionless surface What is the speed of the block after the bullet embeds itself in the block Problem 2b29 Space A tiny piece of space junk of mass m strikes a glancing blow to a spinning W satellite Figure 2b49 After the collision the spacejunk is traveling in a new direction and moving more slowly The spacejunk had negligible rota tion both before and after the collision The velocities of the spacejunk be fore and after the collision are shown in the diagram The satellite has mass M and radius R Before the collision the satellite was moving and rotating as shown in the diagram Just after the collision what are the components of the centerofmass velocity of the satellite vC and 12y Problem 2b30 Two balls collide A ball of mass 005 kg moves with a velocity 17 0 0gt ms It strikes a ball of mass 01 kg which is initially at rest After the collision the heavier ball moves with a velocity of lt3 3 0gt ms a What is the velocity of the lighter ball after impact b What is the impulse delivered to the 005 kg ball by the heavier ball c If the time of contact between the balls is 003 sec what is the force exerted by the heavier ball on the lighter ball Problem 2b31 Space station A space station has the form of a hoop of radius R with mass M Figure 2b50 Initially its center of mass is not moving but it is spinning Then a small package of mass m is thrown by a springloaded gun toward a nearby spacecraft as shown the package has a speed 1 after launch Calculate the centerofmass velocity of the space station vC and vy after the launch 21713 Answns to exercise 151 2b13 Answers to exercises 2b page 115 2b2 page 115 2b3 page 115 2b4 page 115 2b5 page 119 2b6 page 119 2b7 page 119 2b8 page 121 2b9 page 121 2b10 page 121 2b11 page 128 2b12 page 128 2b13 page 129 2b14 page 134 2b15 page 134 2b16 page 134 2b17 page 135 4N a 00 0139 6 lt2074gt msz lt16 073 kg msgts lt16 0 73 N nonzero downward nonzero downward nonzero in direction of final momentum 0975 kg mss toward center 0975 N toward center Both d tit and Pnet point toward the center 357x1022 kg ms s toward the Sun 357x1022 N toward the Sun 8000 ms 8 kms 225 N toward center of circle 3x104 ms lt004 004 701gt kg ms lt004 004 701gt kg ms lt11 725gt ms lt1814 7gt kg ms lt18 14 7gt kg ms 1x10 21e16x10 21 0 kg ms a imx I mx 7 b mx 7 7m1 7c 00 152 Chapter 2 The Momentum Principle Chapter 2 The Momentum Principle 21 The Momentum Principle 52 11 Force 53 212 Impulse 55 213 Predictions using the Momentum Principle 56 22 The superposition principle 57 221 Net force 58 23 System and surroundings 58 24 Applying the Momentum Principle to a system 60 241 xample Position and momentum ofa ball 60 242 Example A fan cart 1D constant net force 61 243 Example A thrown ball 2D constant net force 63 244 Graphical prediction of motion 68 245 Example Block on spring 1D nonconstant net force 69 246 Example Fast proton 1D constant net force relativistic 72 25 Problems of greater complexity 73 251 Example Strike a hockey puck 73 252 Example Colliding students 76 253 Physical models 79 26 Fundamental forces 80 27 The gravitational force law 80 Understanding the gravitational force law 81 272 Calculating the gravitational force on a planet 83 273 Using the gravitational force to predict motion 86 274 Telling a computer what to do 89 275 Approximate gravitational force near the Earth s surface 91 28 The electric force law Coulomb s law 92 281 lnteratomic forces 92 29 Reciprocity 93 210 The Newtonian synthesis 94 211 Derivation of special average velocity result 95 212 Points and spheres 96 213 Measuring the universal gravitational constant C 97 214 The Momentum Principle is valid only in inertial frames 97 215 Updating position at high speed 98 216 Definitions measurements and units 98 217 Summary 101 218 Review questions 103 219 Problems 104 220 Answers to exercises 112 Copyright 2005 John Wiley 8c Sons Adopters of Matter 59quot Interactions by Ruth Chabay and Bruce Sherwood may provide this revised chapter to their students Chapter 2 The Momentum Principle Chapter 2 The Momentum Principle In this chapterwe introduce the Momentum Principle the first of three fun damental principles of mechanics which together make it possible to pre dict and explain avery broad range ofrealworld phenomena the other two are the Energy Principle and the Angular Momentum Principle In Chap ter 1 you learned how to describe positions and motions in 3D and we dis cussed the notion of interaction where change is an indicator that an interaction has occurred We introduced the concept of momentum as a quantity whose change is related to the amount of interaction occurring The Momentum Principle makes a quantitative connection between amount of interaction and change of momentum The major topics in this chapter are 39 The Momentum Principle relating momentum change to interaction 39 Force as a quantitative measure of interaction 39 The concept ofa system to which to apply the Momentum Principle 21 The Momentum Principle Newton s first law of motion the stronger the interaction the bigger the change in the momentum states a qualitative relationship between mo mentum and interaction The Momentum Principle restates this relation in a powerful quantitative form that can be used to predict the behavior ofob jects The validity of the Momentum Principle has been verified through a verywide variety ofobservations and experiments It is a summary of the way interactions affect motion in the real world THE MOMENTUM PRINCIPLE A EMA for a short enough time interval At In words change of momentum the effect is equal to the net force acting on an object times the duration of the interaction the cause As usual the capital Greek letter delta A means change of some thing or final minus initial The net force Fnet is the vector sum of all the forces acting on an object We will study forces in detail in this chapter Examples of forces include 39 the repulsive electric force a proton exerts on another proton 39 the attractive gravitational force the Earth exerts on you 39 the force that a compressed spring exerts on your hand 39 the force on a spacecraft of expanding gases in a rocket engine 39 the force of the air on the propeller of an airplane or swamp boat Time interval short enough We require a short enough time interv for the Momentum Principle to be valid in the sense that the net force shouldn t change very much during the time interval If the net force hardly changes during the motion we can use a very large time interval If the net force changes rapidly we need to 21 The Momentum Principle use a series of small time intervals for accuracy This is the same issue we met with the position update relation if 13 vanAt where we need to use a short enough time interval that the velocity isn t changing very much or else we need to know the average velocity during the time interval Since Ap p 7p PnetAt final minusinitial we can rearrange the Momentum Principle like this UPDATE FORM OF THE MOMENTUM PRINCIPLE f i PnetAt for a short enough time interval At or written out lt11 11y 14gt lms PM In ltFxs Fys FMN This update version of the Momentum Principle emphasizes the fact that if you know the initial momentum and you know the net force acting during a short enough time interval you can predict the final momentum It s an interesting fact ofnature that the as component ofa force doesn t affect the y or z components of momentum as you can see from these equations The Momentum Principle written in terms of vectors can be interpreted as three ordinary scalar equations for components of the motion along the x y and z axes pr pxiTFnengt pyf pyil39FneuyAt 112 1111 Fneg zAt Note how much information is expressed compactly in the vector form of this equation pf pi PnetAt In some simple situations for example if we know that the y and z components of an object s momentum are not changing we may choose to work onlywith the xcomponent of the momen tum update equation The Momentum Principle has been experimentally verified in averywide range of phenomena We will see later that it can be restated in a very gen eral form the change in momentum of an object plus the change in mo mentum of its surroundings is zero Conservation of Momentum In this form the principle can be applied to all objects from the very small atoms and nuclei to the very large galaxies and black holes though understand ing these systems in detail requires quantum mechanics or general relativi ty Historically the Momentum Principle is often called Newton s second law ofmotion We will refer to it as the Momentum Principle to emphasize the key role played by momentum in physical processes You are already familiar with change of momentum Ap and with time in terval At The new element is the concept of force 211 Force Scientists and engineers employ the concept of force to quantify interac tions between two objects The net force the vector sum of all the forces act ing on an object acting for some time At causes changes of momentum Figure 21 Like momentum or velocity force is described by a vector since a force has a magnitude and is exerted in a particular direction Mea suring the magnitude of the velocity of an object in other words measur ing its speed is a familiar task but how do we measure the magnitude of a force 53 Ap Figure 21 The bigger the net force the greater the change of momentumi Figure 22 Stretching ofa spring is a meae sure of force Figure 2 3 Com measure of fore pression ofa spring is also a e Chlme 2 m Momentum Principb Asimple way to measure force is to use the stretch or compression ofa a spring In Figure 22 we hang a block from a spring and note that the spring is stretched a distance 5 Then we hang two such blocks from the spring and we see that the spring is stretched twice as much By experimenmtion we nd that any spring made of the same material and produced to the same speci cations behaves in the same way Similarly we can observe how much the spring compresses when the same blocks are supported by it We find that one block compresses the spring by the same distance Id and two blocks compress it by 2I5I Figure 23 Compression is considered negar tive stretch because the length of the spring decreases We can use a pring ke a 39 39 39 39 39 39 erms of what force is required to produce a given stretch The SI unit of force is the newton abbreviated as One newton is a rather small force A newton is approximately the downward gravitational force of the Earth on a small apple or about a quarter ofa pound Ifyou hold a small apple at rest in your hand you apply an upward force of about one newton compensating for the downward pull of the Earth The spring force law A force law describes mathematically how a force depends on the situa tion In this chapter we39ll learn about various force laws including the grave imtional force law and the electric force law For a spring the magnitude of the force exerted by a spring on an object attached to the spring is given by the following force law THE SPRING FORCE LAW MAGNITUDE s iFspmgl kSISi Isl is the absolute value ofthe stretch formally s AL LeL0 is the length ofthe relaxedspring L is the length of the spring when stretched or compressed k is the spring stiffness The constant k is a positive number a d is a property of the particular spring the sti er the spring the larger the spring stiffness and th larg orce In thefollming man to walk tnmagn tni stop and tnin t activities Iti m itaa ing aneaa To ban the nateiiai you mid to engage actively witn the Ws lms pasta new natjiot itaa pmsiwly Suppose a certain spring has been calibrated so that we know that in spring stiffness k is 500 Nm You pull on the spring and observe that it is 001 m 1 cm longer than it was when relaxed What is the magnitude of the force exerted by the spring on your hand The force law gives I sprmgl 500 NmI001 m 5 N Note that the total length ofthe sprin doesn39t matter it39sjust the amount ofstretch or compression that matters Suppose that instead ofpulling on the spring you push on it so the spring becomes shorter than in relaxed length Ifthe relaxed length of the spring is 10 cm and you compress the spring to a length of9 cm what is the magnitude ofthe force exerted by the spring on your hand 21 The Momentum Principle The stretch of the spring in SI units is s LiL0 009 m7010 m 7001m The force law gives liprangl 500 Nm7001 ml 5 N The magnitude of the force is the same as in the previous case Of course the direction of the force exerted by the spring on your hand is now differ ent but we would need to write a full vector equation to incorporate this in formation Reciprocity In the preceding example you pushed on a spring compressing it and we calculated the force exerted by the compressed spring on your hand Of course your hand has to exert a force on the spring in order to keep it com pressed It turns out that the force exerted by your hand on the spring is equal in magnitude though opposite in direction to the force exerted by the spring on your hand This reciprocity offorces is a fundamental prop erty of the electric interaction between the electrons and protons in your hand and the electrons and protons making up the spring We will say more about the reciprocity of electric and gravitational forces later in this chap ter Ex 2 You push on a spring whose stiffness is 11 Nm compressing it until it is 25 cm shorter than its relaxed length What is the magnitude of the force the spring now exerts on your hand Ex 22 A spring is 017 m long when it is relaxed When aforce of magnitude 250 N is applied the spring becomes 024 m long What is the stiffness of this spring Ex 23 The spring in the previous exercise is now compressed so that its length is 015 In What magnitude of force is required to do this 212 Impulse The amount of interaction affecting an object includes both a measure of the strength of the interaction expressed as the net force rm and ofthe du ration At of the interaction Either a bigger force or a longer application of the force will cause more change of momentum The product of a force and a time interval is called impulse Figure 24 DEFINITION OF IMPULSE Impulse E TAt for small enough At Impulse has units of Ns newtonseconds With this definition of impulse we can state the Momentum Principle in words like t is The change of momentum of an object is equal to the net ianulse applied to it 55 Ap Figure 24 Net impulse net force times duration of the interaction Net impulse changes momentuml Figure 25 Apply a consmnt force to a block on a lowrfriction air trac Chatme 2 m Momentum Prian Aconstantnet force lt3 75 4gt N acts on an object for 10 sWhat is the net impulse applied to the object impulse Prom lt3e54gtN10s lt30e5040gt Ns Ex 24 A constant net force of 705702 08 N acm on an object for 2 minutes What is the impulse applied to the object in unis 213 Predictions using the Momentum Principle Experimenm can easily be done to verify the Momentum Principle Many in troductory physics laboratories have air tracks like the one illustrated in Fig ure 25 The long triangular base has many small holes in it and air under pressure is blown out through these holes The air forms a cushion under the glider allowing it to coast smoothly with very little friction Suppose we place a block on a glider sitting on a long air track Figure 25 and we ate tach a spring to it whose stiffness is 500 Nm like the one discussed earlier x We will choose the xaxis to point in the direction ofthe motion Suppose the block smrts from rest and you pull for 1 second with the spring stretched 4 cm so that you know that you are pulling with a force of magnitude 500 Nm004 m 20 N you will have to move forward in order to keep the spring stretched The block stars from rest so 1339 lt0 0 0 kgms What would the Momentum Principle predict the new momentum of the block to be after 1 second Since 39 39 t e iiuci39 39 39 L c the glider is just the force with which you pull The Momentum Principle up date form applied to the glider is p lt0 0 0gt kgms lt20 0 0gt N1 s Since the net force is in the x direction we know that the y and z compoe nenm of the glider39s momentum will not change and we can work withjust the xcomponent ofthe momentum to powwow 020 N1 s 20 kgms Ifyou do the experiment this is what you will observe You keep pulling for another second but now with the spring stretched halfas much 2 cm so you know that you39re pulling with half the original force 500 Nm002 m 10 N What would the Momentum Principle predict the new ad component ofthe momentum ofthe lock to e Momentum Principle would predict the following where we take the final momentum from the first pull and consider that to be the initial moe mentum for the second pull 231 pm medt 20 kgms 10 N1 s 30 kgms This is what is observed experimenmlly Note that the effecm ofthe interac tions in the two lesecond intervals add we add the two momentum changes 2 2 The supeqooxition primipb Instead ofvarying the net force we could try varying the duration of the in teraction Figure 26 Smrt over with the block initially at rest and pull for 2 seconds with the springstretched 4 cm so the force is 20 N The block smrts from rest pm 0What would the Momentum Principle predict the new xcomponent ofthe momentum ofthe block my omelet 020 mm s 40 kgms Here the final x component of momentum was 40 kgms after applying a force of 20 N for 2 s whereas in the previous experimenm we got 30 kgms 39 for is In our cal When the force changed from 20 Nfor a second to 10 Nfor asecond we had to treat the two time intervals separately 39 39 own the validity of the Momentum two springs to move the block we nd that it is indeed the vector sum of the two spring forces the net force that accounm for the change in momentum 22 The superposition principle 9 Dra M k b the table 39 39 39 nsmnt direction You39re applying a force yet the momentum isn39t changing Explain brie and co The force you apply to the book isn39t the only force acting on the book The H i r in L 439 39 raiieri 39 39 rr This frictional force is due to collisions between atoms in the bottom layer of the book and atoms in the top layer of the able see Figure 27 for an atomic picture ofslidingfriction Ifthe momentum isn39t changing it must e t a e m force on t e book is zero This is why the Momentum Princic ple involves ne rather than just the force you apply What do you have to do to make the book go faster You have to apply a force that is bigger than the frictional force so that the vector sum is nonzero 9 The Momentum Principle predicts that ifthe netforce is zero the momentum doesn39t change but says the same Yet if you push the book across the table at high speed then let go the book doesn39t keep moving but comes to a stop Explain brie y After you let go the net force on the book isjust the frictional force ofthe table which acm opposite to the momentum and ma es the momentum de crease What ifyou were in outer space far from mbles and other objeco and you pushed the book so that it was going fast then let go What would the book do In the absence of any forces zero net force the book would continue for ever with the vector momentum you initially gave it It would move in a 7 A 39 n 7 straightline at constantspeed pf plFnet t 7 Why it took so long to discover the Momentum Principle In our everyday world friction between objeco is common As a result for many centuries people quite naturally believed that a force was necessary to sustain motion just as you saw when dragging the book across the able at Fig lo 6cm 12 motoz A13 Frag ure 26 Duration of interaction the nger the net force acts the greater the change ofmomentum 1 g Figure 27 A computer simulation ofslidr ing friction a diamond 39 39 along a diamond sur of Judith A Harrison m ey fac e U tip is dragged Image courtesy S Naval Acadr Figure 28 The presence of the oable does not change the interaction of the glass with the Earth SU39RROUNDllNGS SU39RROWDWGS SYSTEM l SU39RROU39NDLNGS SURROUNDLNGS Figure 29 System and surroundings Inter tjons in the form of impulses ow across the system boundary and change the sys tem s momentum Chapm 2 m Momentum Prinripb constant speed This made it hard to undersmnd what made the planets keep moving Galileo and Newton nally realized that it is the mtforce that matters and that objecm free of interactions just naturally keep moving with no forces needed to keep them moving Newton39s first law of motion This represented a major revolution in how humans viewed the world 221 Net force How do we calculate the net force acting on an obj ct It turns out to be sure t 39 39 39 39 d h th up rp iti In principle THE SUPERPOSITION PRINCIPLE e net force on an object is the vector sum ofthe individual forces exerted on it by all other objects Each individual interaction is unaffected by the presence ofother interacting objecm The superposition principle is completely general It has been found experr exerte n on the arth at a particular distance will always be the a e 39quot ml Uulclr 4 39 quot4 the Sun and the Earth The presence of other objecm and interactions does not block or change the interactions between each pair of objects n intere used up an th d by the presence ofa third object To take a silly example hold a glass over a able and then let go Figure 28 The glass falls so evidently the table doesn39t block the gravitational in teraction with the Earth In fact the table adds a tiny additional gravimtione al force on the glass to that of the Earth without changing the Earth39 attraction for the g ass Ea 25 A balloon experiences a gravitational force of lt0 7005 0gt N and a force due to the wind of 0003 0 002gt N What is the net force acting on the balloon l asl Ex 39k t n i an What are all the forces acting on the boat Is the net force zero or nonzero Ea 2 7 You watch someone carry a heavy block leftetoright across the room walking at constant speed According to the Momentum Prinrinle L H A 39 L block is upward toward the right or zero 23 System and surroundings In order to use the Momentum Principle to predict motion we must choose a system whose momentum change we will calculate By the word system n n l n system the baseball and the air which exerm a force of air resistance 23 System and surroundings against the moving ball Figure 210 It is true that there are changes in the momentum of the Earth and the momentum of the air molecules due to in teractions with the baseball but once we decide to choose the baseball as the system of interestwe don t have to pay attention to what happens to the momentum of objects in the surroundings The role of the surroundings is described entirely by the forces they exert on our chosen system Only external forces matter It is an important rule that we do not include in the net force acting on the baseball any forces that the baseball exerts on itself The atoms of the base ball do exert forces on each other but as we ll discuss in more detail later interatomic forces and gravitational force come in equal and opposite pairs Figure 211 Therefore these internal force pairs forces between pairs ofatoms internal to our chosen system add up to zero and can safely be ig nored in calculating the net force that appears in the Momentum Principle pf E FnetAt Only external forces matter forces associatedwith inter actions between our chosen system and objects in the surroundings Of course objects in the surroundings experience forces exerted by ob jects inside the system but when we apply the Momentum Principlejust to the system we only care about the change of momentum of the system Al so there are equal and opposite changes of momentum in the surround ings but they typically don t interest us Neglecting small effects The surroundings of the system ofthe baseball includes the Sun the Moon Mars etc Do we have to consider all of the forces these objects exert on the baseball In practice no because these forces are extremely small com pared to the forces exerted on the system by the Earth and the air during the brief flight of the baseball But if we were trying to plot an accurate course for a spacecraft going to Mars we would need to include even small forces exerted by other planets because impulse is forces times time dura tion FA t and in the long time required to go to Mars even small forces can produce significant impulses and significant changes in the momentum of the spacecraft Systems consisting of several objects A system can consist of more than one object For example we might choose to consider a system consisting of the entire Solar System The sur roundings of this large system would include the rest of the Universe in cluding neighboring stars The total momentum of the Solar System can change due to the gravitational forces exerted by stars especially nearby stars Ex 28 In Section 213 we applied the Momentum Principle to predict the motion ofa glider on an air track What did we choose as the system in this analysis What objects were included in the surroundings 59 System is baseball l l 39 l 3 P l Air is part of surroundings Earth is part of surroundings Figure 210 Choose a baseball as the sys temt The surroundings are the Earth and the air which interact across the system boundary to change the momentum of the systemt System T Internal forces cancel no effect on system momentum l l Fon 1 by 2 l l Fon 2 by 1 l l External forces change the system momentum Object in surroundings Figure 211 Internal force pairs cancel so only external forces can the momentum ofa systemt c ange A labeled diagram step 3 gives a physics view of the situation and it de nes sym bols to use in writing an algebraic state ment of the Momentum Principle Assume that the force didn t change much during 3 ms Or to put it another way we assumed that a time interval of 3 ms was sufficiently short that we could update the momentum fairly well assuming a constant force during that short time interval Chapter 2 The Momentum Principle 24 Applying the Momentum Principle to a system To apply the Momentum Principle to analyze the motion ofa realworld sys tem several steps are required 1 Choose a system consisting ofa portion of the Universe The rest of the Universe is called the surroundings 2 List objects in the surroundings that exert significant forces on the chosen system and make a labeled diagram showing the external forces exerted by the objects in the surroundings 3 Apply the Momentum Principle to the chosen system 3 ath For each term in the Momentum Principle substitute any values you know 4 Apply the position update formula if necessary if t ngAt 5 Solve for any remaining unknown quantities of interest 6 Check for reasonableness units etc 241 Example Position and momentum of a ball Inside a spaceship in outer space there is a small steel ball of mass 025 kg At a particular instant the ball is located at position lt9 5 0gt m and has mo mentum lt78 3 0gt kgms At this instant the ball is being pulled by a string which exerts a net force lt20 50 0gt N on the ball What is the ball s approximate momentum and position 3 milliseconds later 3X1073 s What 1 J 39 39 or 39 1 quotf 39 J 39 didyou make inyour anal ysis 1 Choose a system System the steel ball 2 List external objects that interact with the system with diagram the string a circle represents the system string 3 Apply the Momentum Principle pf 3 PnaAt ball pf lt78 3 0gt kgms lt20 50 0gt N3xi0 3 s pf lt78 3 0gt kgms lt006 015 0gt Ns pf 794 315 0gt kgms 2 4 Applying we Momentum Primipb to a sysm 4 Apply the position update formula W F ngM where ya pm since 1 ltlt c s 7 77943150gtkgms to r 7 lt9 5 0gt m 025 kg 3x10 5 r lt9 5 0gt m 700953 00378 0gt m r 8905 5038 0gt m 5 There are no remaining unknowns 6 Check Unim check momentum kgms and position In 242 Example A fan cart 1 D constant net force An easy way to arrange to apply a nearly consmnt force is to mount an elece tric fan on a cart Figure 212 Ifthe fan blows backwards the interaction with the air r 39 a makin the cart39s momentum continually increase Swamp boao used in the veryshale low Florida Everglades are built in a similar way with large fans on top of the boao propelling them through the swamp When a fan cart or boat gets going very fast air resismnce becomes im portant and at high speeds is as big as the propelling force so that the net esn39t increase any force becomes zero at which point the momentum do more and the cart or boat travels at constant speed For simplicity we39ll con sider the motion at low speed with negligible air resistance so we can make the approximation that the net force is due solely to the fan and is nearly constant Note that the y component of the net force is zero because the down ward gravimtional force on the cart is exactly balanced by the upward force exerted by the track on the cart In a later chapter we will examine the in teraction between solid objects like the cart and the track in more detail Predict new position and new momentum Suppose you have afan cartwhose mass is 400 grams 04 kg andwith the fan turned on e net force acting on the cart due to the air and friction with the track is 02 0 0 N and consmnt You ive the cart a shove and you release the cart at position 05 0 0 m with initial velocity 12 0 0 ms What is the position ofthe cart 3 seconds later and what is its momene tum at that time 1 Choose a system System the cart including the fan 2 List external objects that interact with the system with diagram the Earth the track the air Zack represent the system by a circle We used the momentum at the end of the interval to update the position Note that at Very high speeds t s pm isn t valid for updating position See page 98 Figure 212 A fan can on a track 62 Chapter 2 The Momentum Principle Since v c we can use the approximation 339 Apply the Momentum Principle that 1 m mv A A A Since the y component of the cart s mo pf pi FnetAt pi Ftrack F133th FairA t mentum does not change we know that A A the y component of the net force must be pf lt04kg12ms 0 0 02 IFtrackI IFEarthls 0gtN3 5 zerO and lireekl l Eanhle fe 108 0 0 kg Ins We can use a large time interval At be faufe the form isn t chaflging Very much Approximating the average velocity m ember magmde or dlrecnon In order to use the position update formula we need the average velocity of the cart How do we find avg We know initial and final values for v e lt12 0 0ms Vm vx f m 17 lt10800gt kgms lt27 0 0mm 04kg Can we use these values to find an You may have guessed that the average vfx velocity is the arithmetic pronounced arithMETic average W v 3 3 Jr l I w v W Van N 2 l u t I I I I I I I t tf The arlthmetlc averagels often a good approx1mat10n but 1t IS not always exactly equal to the average velocity avg ArAt The arithmetic average FI 2 13 I h I II 1 Ith does lie between the two extremes For example the arithmetic average of lgm 3 15 C angmg teary W1 6 and Sis 682 142 7 halfway between 6 and 8 see Figure 213 time so the arithmetic average is equal to The arlthmetlc average does not g1ve the true average veloc1ty unless the velocity is changing at a constant rate which is the case only if the net force is constant as it happens to be for a fan cart For example if you drive 50 mihr for four hours and then 20 mihr for an hour you go 220 miles and your average speed is 220 mi5 hr 44 mihr whereas the arithmetic average is 50202 35 mihr In situations where the force is not con stant we have to choose short enough time intervals that the velocity is near ly constant during the brief At see Figure 214 APPROXIMATE AVERAGE VELOCITY x l I i v v I i tf ewe 1 22 1fvltlt c Figure 214 muc is not changing linearly with exactly true only if v changes at a constant rate lnet constant time In this case the arithmetic average is much higher than the average value of y The proof which is more complicated than one might expect is given in optional Section 211 at the end of this chapter 4 Apply the position update formula 3 23 lt1227 00 00gt 2 2 2 s The net force on the cart is constant so Vavg 2 this calculation of average velocity gives the correct value 6 195 0 0gtrns hwargm lt05 0 0gtm lt195 0 03s if 635 0 0gtm 5 There are no remaining unknowns 6 Check Position rn correct units xfgt xi as it should be 2 4 Applying we Momentum Principb to a sysm Ex 29 A hockey puck is sliding along the ice with nearly consmnt momentum lt10 0 m s when it is suddenly struck by a o ey stick with a force lt0 0 2000gtN that lass for only 3 milliseconds 3x10 8 5 What is the new momentum ofthe puck Ea 210 You were driving a car with velocity lt25 0 15gt ms You braked and your velocity became in Ap du attention to signs What was the vector impulse applied to the car by the ground Ex 211 In the previous exercise ifthe maneuver took 3 seconds what was the average net vector force PM that the ground exerted on the car Ex 212 A truck driver slams on the brakes and the momentum changes from lt9x10quot 0 0gt kg ms to lt5x10quot 0 0gt kg ms in 4 seconds due to a consmntforce ofthe road on the wheels ofcar As avector write the force exerted by the road 3At a certain insmnt a particle is moving in the 2 direction with momentum kgms During the next 01 s a consmnt force lt76 30gtN acts on the particle What is th momentum ofthe particle at the end ofthis 01 s interva 243 Example A thrown ball 2D constant net force Asecond example is the prediction ofthe motion ofa ball thrown through 39 n on the ball with a gravitational force and the small compared to the gravimtional force so we may be able to neglect air resismnce A lowedensity object such as a styrofoam ball experiences air resistance all so air resis it would in avacuum because air resistance is large at this high speed Fig ure 215 We39ll analyze the ight ofa ball with the assumption that we can neglect air resistance which means we are dealing with a highedensity object at low speed or motion in a vacuum Later we39ll develop techniques that make it of approximately 98 N T tional pull on a 2 kg block is 2 kg98 Nkg or 196 N In general the Earth exerts a force on a mass in like this Near the Earth39s surface him a mg where g 98 Nkg We39ll have more to say about gravitational forces later in this chapter but this is suf cient for analyzing the motion ofa thrown ball no an resistance air mutant We make the approximation that air resis tance is negligible compared to the gravita tional force 1 cso mmv Divide both sides of the equation by m The xand z componenm ofvelocity are not changing this makes sense because the net force has only a y component The y com ponent of velocity is decreasing continu ously this makes sense because the gravitational force on the ball by the Earth affecm the y component of the ball s veloc ity Chapter 2 The Momentum Principle Predict new velocity and new position of a thrown ball You throw the ball so thatjust after it leaves your hand at location ltxi yi 0gt it has velocity lt1 1 0gt with no component in the z direction Now that it has left your hand and we re neglecting air resistance the net force at all times is lt0 7mg 0gt since the Earth s gravitational force acts downward to ward the center of the Earth and we normally choose our axes so y points up Predict the velocity and position of the ball after a time At 1 Choose a system System the ball 2 List external objects that interact with the system with diagram the Earth the air Fair b3 represent the system by a circle 3 Apply the Momentum Principle A FEath f when mltvxf vyf 0gt mltvw v 0gt lt0 7mg 0gtAt lt1 12 0gt lt1 1 0gt lt0 7g 0gtAt ltvxi 0N 151 rgAt 0 0Atgt WW 7 gA t 0gt 4 Apply the position update tormula S vv vv7gAt 00 x1 x1 11 11 v lt 2 2 2 gt V 3 l arg WM 757 gA t 0gt rf 13 6mm 1 lt99 3 0gt We yes 0gt Wm vyi igAts 0gtAt l lt96 y 0gt ltxi valAt yi TatAte gmt 0gt Alternatively xf xi vmAt yf y vyme gAt2 zf 0 5 There are no remaining unknowns 6 Check The units check for the final position Note that g is Nkg so that gAt is Nskg units ofimpulsekg which is kgmskg or ms 7 Under what circumstances can we use this result to predict the trajectory of an object We assumed that air resistance was negligible that the object was traveling at a speed much less than the speed of light and that the net force was con stant If any of these three conditions are not met then these results do not apply and using them will give a wrong answer 24 Applying the Momentum Principle to a system Using this result lfa ball of mass 90 g is initially on the ground at location 0 0 0 m and you kick itwith initial velocity 3 7 0 ms where will the ball be halfa sec ond later xf 0 3ms05 s 15 m yf 0 7ms05 s 7 98Nkg05 s2 105 m 7 Can we use these equations to find the location of the ball 10 seconds after you kick it No Our result would be that the ball was far underground yf 7ms10 s798Nkg10 s2 7910 In which is not physically reasonable The ball would have hit the ground and stopped before 10 seconds had passed 7 What should you remember from the preceding example If you understand the basic method you can reproduce the specific results quickly and accurately Just memorizing the results won t help you much because you won t really understand what they mean or when they can be used For concreteness we analyzed the flight of a thrown ball but this same analysis would work for any situation where an object moves in two dimen sions under the influence of a constant net force For example just by changing the constant y component of the net force to be something other than 7mg we could analyze the twodimensional motion of a swamp boat if the direction of the thrust of the propeller doesn t change and we can neglect friction with the water or to the twodimensional motion of an electron between two large charged plates However in these cases the re sulting equations would be slightly different because the mass of the object would appear in the result the m cancels only in the case ofa force that is proportional to m such as the gravitational force Graphs ot the motion In Figure 216 we shows graphs of position and velocity components vs time and the actual path y vs at of the ball The first graph 1 vs t is sim ply a horizontal line because 1 doesn t change since there is no xcompo nent of force The graph of x is a straight line second graph rising if vx is positive Note that the slope of the xvs tgraph is equal to 19 The graph of 1 is a falling straight line third graph because the y com ponent ofthe force is 7mg which constantly makes the y component of mo mentum decrease At some point the y component ofmomentum decreases to zero at the top of the motion after which the ball heads downward with negative vy The graph ofyvs time tis an inverted parabola fourth graph since the equation for y is a quadratic function in the time Note that the slope of the yvs tgraph the fourth graph at any time is equal to v at that time In particular when the slope is zero at the maxi mum y vy is momentarily equal to zero Before that point the slope is pos itive corresponding to 1 gt 0 and after that point the slope is negative corresponding to 1 lt 0 The actual path of the ball the graph ofyvs x is also an inverted parab ola the bottom graph Since it increases linearly with t whether we plot y vs tor yvs xwe ll see a similar curve The scale factor along the horizontal axis is different of course meters instead of seconds 65 1 does not change 1 con tinually decreases y Rising y gt O Stops I isin g heads downward v lt O y Stops rising y iheads downward yf Falling Rising 3 e t tf This isyvs time t It is not the path 3 Actual path y vs 6 J f lt x yf 0 gt it lt x y o gt x1 xf Figure 216 Graphs for the thrown ballt Chapter 2 The Momentum Principle Understanding the results Let s explore the results a bit to understand what they tell us about the mo tion The as component of the motion is very simple it is completely unaf fected by the downwardpointing gravitational force so as simply increases at a constant rate 1 constant Ofcourse once the ball hits the ground xno longer increases The y component of the motion is more interesting The equation vyf vyiigAt says that if you throw the ball so that vyi is positive heading up at some time 1 will decrease to zero and then become negative 7 At what time At will the ball reach its highest point top Our result for y as a function of time yf y vyiAti gAt2 does not help us answer this question because we don t yet know how high the ball will go we don t know what to use for yf Can we use the equation for 1 as afunction oftime to find the time A twp at which the ball reaches its highest point Do we know the value of 1 when the ball reaches its highest point Just before the ball reaches its maximum height 1 is positive Just after it reaches its maximum height and begins to head downward 1 is negative At the instant the ball reaches its maximum height vy 0 Using this in formation we can solve for the elapsed time at that instant If vyf vyi gAtmp 0 then Atmp 7 How high is the ball when it turns around and heads downward Now that we have avalue for A twp we can use that in our general result for the final position 1 vi 1 vi 2 y yiT vyiAttop Q Attopz yi Ty21 rig 2 yfi y is the maximum height above your hand Check the algebra yourself to verify this result This result makes sense in that the bigger the initial y component of velocity the higher the ball will go Notice that it doesn t matter at all what as component of initial velocity you give the ball all that matters is 1211 However it takes more effort to give the ball some vm in addition to giving it some 1211 so for maximum height you want to throw the ball nearly straight up 7 Suppose you throw the ball so that it rises 2 m before falling back down If you double 1211 how far will it rise above your hand Since yfi yi viz2g doubling vy will make the ball go up 4 times as high to 8 m above your hand 9 Suppose you throw the ball so that it rises 2 m on Earth If you give the ball the same vyi on the Moon where gis about onesixth that on Earth how far will the ball rise above your hand 24 Applying the Momentum Principle to a system Since yfi yi via2g ifgdecreases by afactor of6 but vyi stays the same the rise of the ball will increase by a factor of 6 so the ball will rise to 12 m above your hand Because there is no air on the Moon our analysis works well there 7 On Earth what vyi must you give the ball so that it rises 10 m above your hand We 10 m via2amp8 Nkg and solving this we nd vyi 14 ms This is a high enough speed that air resistance might be sig nificant and our result inaccurate have 7 Suppose the ball is caught by a friend at the same height as you launched the ball What is the formula for how long the ball was in the air 1 In this case we have yf yi so yf yi vyii gAtAt gives us At panel down 221139 compare with Atmp g g The ball takes twice as long to go up and down as to go up so we conclude that the time to come down is the same as the time to go up In the pres ence of air resistance the speed at every height is slower on the way down so it takes longer to come down than to go up Again notice that the time to go up and down doesn t depend at all on the as component of velocity The x and y motions are independent of each other 7 How far from you is your friend when the ball is caught Note that you know how long the ball was in the air and you know the as component ofvelocity xf xi vmAtup and down so the distance is xfi x1 vxi2 This distance is called the range of the ball It is the distance the ball goes in the air if it returns to the same height It depends on 1 because that de termines how much time it spends in the air The result also depends on vx because that determines how far the ball will move in the x direction while the ball is in the air Initial speed and angle Sometimes you know the initial speed lvl and angle 9 for the launch of a ball Figure 217 shows how to calculate the sine and cosine of the angle Use these results to solve for the velocity components vm Ivlcose vyi Ivlsine Ex 214 A ball is kicked from a location lt9075gt m on the ground with initial velocity lt71013 75gt ms a What is the velocity of the ball 06 seconds after being thrown b What is the location of the ball 06 seconds after being thrown c What is the maximum height reached by the ball d At what time does the ball reach its maximum height e At what time does the ball hit the ground f What is the location of the ball when it hits the ground Ex 215 Apply the general results obtained in the full analysis on page 64 to answer the following questions You hold a small metal ball of mass m a height h above the oor You let go and the ball falls to the oor Choose the origin of the coordinate system to be on the floor where the ball hits with y up as usual Just after release 67 sine 33 hyp V vm lvllcose c059 E 1 hyp Ivl Figure 217 Converting speed and angle to velocity componenmi Graphical prediction of the motion of a ball for three successive time steps Chapter 2 The Momentum Principle much time At does it take for the ball to fall What is vyfjust before hitting the floor Express all results in terms of m g and It How would your results change if the ball had twice the mass Ex 216 A soccer ball is kicked at an angle of 60 to the horizontal with an initial speed of 10 ms Assume that we can neglect air resistance For how much time is the ball in the air How far does it go horizontal distance along the field How high does it go what are y and 1213 Just before hitting the floor what is yf How 244 Graphical prediction of motion It is instructive to apply the Momentum Principle qualitatively and graphi cally to the thrown ball to see visually how the Momentum Principle deter mines the motion The omentum Principle predicts that in a short time interval At the momentum of the ball will change by an amount Ap PnetAt which we can calculate because we know the force acting on the ball In the following diagram we show the initial momentum p1 the change in the momentum Ap PnetAt and the new momentum p2 p1 Ap Note that this vector addition corresponds to adding the ar rows p1 and Ap tip to tail We approximate the average momentum in the time interval At by the new momentum p2 and we advance the ball in the direction of p2 Ap 126m 132 4 a A13 12am P2 A a P3 x l P3 A final A FneiAt a Fuel A l Ap FnetAt Fnet During the short time interval At the ball will move an amount vAt in the direction ofits new velocity 62 pzm we assume that 1 ltlt 0 The ve locity is changing during this time interval but ifAtis quite small as it could be if we let a computer do the calculations the change in the velocity is quite small and it doesn t matter very much whether we use the velocity at the start or end of the time interval or some kind of average In this partic ular case ofconstant net force the average velocity is the arithmetic average but ifwe included air resistance the net force wouldn t be constant in mag nitude or direction yet this method would still be accurate as long as we use small At At the next position we repeat the procedure graphically adding the vec tors p2 and Ap PnetAt to obtain the new momentum p3 We then ad vance the ball in the direction of g As we do this repeatedly we trace out graphically the trajectory of the ball You can see that the trajectory looks like what is observed in the real world and you can even see the increasing magnitude of the ball as it falls corresponding to increasing speed The im portant point to see in the diagram is that the net impulse FnetAt changm the momentum 2 4 Applying we Momentum Primipb to a sysm 245 Example Block on spring 1 D nonconstant net force Next we39ll study how to predict motion when there is more than one force and the netforce isn39t constant The situation we39ll analyze is the motion of 2 t TL L relaedlenthof20 m 02 m and the spring stiffness is 8 Nm Figure 218You place a 60 gram block 006 kg on top ofthe spring compressing the spring and the block sim motionless on the spring Figure 219 How much is the spring compr ised Prove this using the Momentum Principle Remember Lhalstpnngl kslsl where s is the stretch ofthe spring a negative number ifthe spring is compressed Since the block sits motionless over a time interval At its momentum is not changing whic 39 39 C 39 the L L Figure 219 Ap netAt 0 0 0 lt000gtAt The net force is the vector sum of all the individual forces acting on the block the force due to the spring and the force due to a th ne SpringFEanh lt0 ka3170gt lt0rmgv 0gt lt0 kalsiomgt 0gt where k is the spring stiffness and as defined on page 54 s AL Le L0 is the compression of the spring the current length minus the relaxed len th which is a negative number or a compression so We39ve written i1 Consideringjust the y component we have 0 8 Nmi1 7006kg98 Nkg Fast 3 I W 00735 m or 735 cm 8 Nm We39ll choose the origin of our coordinate system to be at the bottom of the spring where it is attached to the oor As usual we take y to be up When the block is sitting motionless on the spring the bottom ofthe block or top 02 7 00735 in 01265 m mass down compressing the spring an additional because the spring as e downward gravitational force exerted by the Earth and upward with increasing momentum due 220 After release as the block passes a point 8 cm 008 m above the oor the block has an upward speed of ms Relaxed length 20 cm 0 2 m i i Figure 218 A relaxed vertical spring ale Compressed lel 7 35 cm Figure 219 A block sits motionless on the spring The net force must be zero Mei mg 7 l Figure 220 You compressed the spring then released the block heads upward with increasing speed because the net force is nonzero and upward We make the approximation that air resis tance is negligible compared to the other forces The net force is changing as the compres sion changes so the momentum update is approximate Chapter 2 The Momentum Principle We will predict the velocity and position of the block after 002 s 1 Choose a system System the block 2 List external objects that interact with the system with diagram the Earth the spring the air see force diagram on previous page 3 Apply the Momentum Principle 11y I pyi FnetyyAt w1th 1 ltlt 050 that Jym mvy and vym m Initially the spring is compressed 02 7 008m 012 m Fm kslsl 7 mg 8 Nkg012 m 7006 kg98 Nkg Fm 0372 N 1 006 kg03 ms 0372 N002 5 1 002544 kgms the new 1 Nl f 002544 kgms 1ny m 006 kg 0424 ms the new 1 4 Apply the position update tormula s s rf riv mg At At or yf yi vavgyy Make the approximation that the net force was nearly constant during the 002 s vzigvzl W 0362 ms vavgyy yf 008 m 0362 InS002 s 008724 m the newy 5 There are no remaining unknowns 6 Check The units check for the 7 both vy and y have increased as expected During the 002 s time interval the block is moving up and the length of the spring is changing so the stretch compression of the spring is changing That means that the spring force kllsl is not constant during the 002 5 Therefore when we use 1 1in FneUAt to update the momentum we re making an approximation that the force doesn t change very much during that time We can estimate how serious this issue is by calculating the net force at the new position yf 008724 m 7 What is the new net force You need to determine the new compression With the top of the spring now at yf 008724 rn the compression of the spring is Isl 02 7008724 m 011276 m and the net force has be come this Fm kslsl 7 mg 8 Nkg0ll276 m 7 006 kg98 Nkg Fm 0314 N 24 Applying the Momentum Principle to a system The net force at the start of the time interval was Fneg 0372 N So al though the net force did change during the 002 s it didn t change very much and our approximate analysis is therefore pretty good 7 How could our predictions be improved By taking shorter time steps Instead of taking one step of 002 s we could take two steps of 001 s or ten steps of 0002 s During each of these shorter time intervals the compression would change less so the force would be more nearly constant Unfortunately to achieve increased accuracy we have to do a lot more calculations Taking another step To continue predicting the motion of the block into the future we can take another step The final values ofmomentum and position after the original 002 s step become the initial values for the next 002 s step 3 Apply the Momentum Principle 13 pyf pyiFneUAt with vltlt cso that pym mvy and vym m Spring is compressed 02 7 008724m 011276 m Fnegy kxlsl mg 8 Nkg011276 m 7 006 kg98 Nkg Fm 0314 N pyf 002544 kgms 0314 N002 s pyf 003172 kgms the new py ml inf 003172 k mS 05287 ms the new 1 W 006 kg 4 Apply the position update formula if awargm or yf yi vavgyyAt Make the approximation that the net force was nearly constant during the 002 s 721 v 2 0424 025287 ms 2 04764 ms 0anle yf 008724 m 04764 ms002 s 009677 m Look through this calculation and make sure you understand how the fi nal results for position and momentum from the first 00 s time step have been used as the initial values for the second 002 s time step In principle we could continue taking more and more steps to predict farther and farther into the future Here is a summary of this iterative scheme 39 Start with initial positions and momenta of the interacting objects 39 Calculate the vector forces acting on each object 39 Update the momentum of each object pf p FnetAt s s 4 39 Update the pos1tions rf 5ng t 39 Repeat The initial momentum and position are taken from the nal momentum and posi tion in the previous step Chapter 2 The Momentum Principle Every time you repeat the final momentum and position become the ini tial momentum and position for the next step You have to use an approx imate value for mg either by using the velocity at the start or end of the time interval or by taking the arithmetic average of these two velocities as we did above While this scheme is very general doing it by hand is incredibly tedious It is possible to program a computer to do these calculations repetitively Computers are now fast enough that it is possible to get high accuracy sim ply by taking very short time steps so that during each step the net force and velocity aren t changing much We ll talk more about computer prediction of motion later in this chapter Ex 217 After a third time step of 002 seconds what will be the position and momentum of the block 246 Example Fast proton 1D constant net force relativistic A proton in a particle accelerator is moving with velocity lt096c 0 0gt so the speed is 096 X 3gtlt108 ms 288gtlt108 ms Aconstant electric force is applied to the proton to speed it up Fna lt5X10712 0 0gt N What is the proton s speed as a fraction of the speed of light after 20 nanoseconds 1 ns 1x10 gs 1 Choose a system System the proton 2 List external objects that interact with the system with diagram electric charges in the accelerator lt5X10 1Z00gt N a circle represents the system proton 3 Apply the Momentum Principle 3 Wham ltpr 00gt mva 0 0gt lt5x10 00gt N20x10 gs I 17gtlt10727 kg096 gtlt 3x108 ms 1xi0 19 Ns 140962 0 11V 175xi0 181ltgms1xi0 19 NS 185X10718kgms 2f L See Section 215 page 98 obtaining 1 from J c 1 292 when the speed is near the speed oflight m 718 Evaluate amp 185710 k itS 362 no units W 17x10 kg3gtlt10 ms 12 i 0964 0 A1 3622 The speed didn t increase very much because the proton s initial speed 0960 was already close to the cosmic speed limit 0 Because the speed hard ly changed the distance the proton moved during the 20 ns was approxi mately equal to 096 X 3x108 ms20xi0 9 s 58 m 25 Pmbbm of gram compbxity 73 25 Problems of greater complexity So far all of the examples we have considered have involved nding a change in momentum and position given a known force acting over a known time interval The following problems require you to find either the duration of an interaction time interval or the force exerted during an interaction These large problems involve several steps in reasoning 251 Example Strike a hockey puck In Figure 221 an 04 kg hockey puck is sliding along the ice with velocity lt20 0 0gt ms As the puck slides past location lt1 0 2gt m on the rink a r 39 uckwith quot 39 h 439 39 andthe hock ey stick breaks Some time later the puck39s position on the rink is lt13 0 21gt m When we pile weights on the side ofa hockey stick we find that the stick breaks under a force of about 1000 N this is roughly 250 39 f 39 39 i f f h arter of 1 a pound approximately the weight ofa small apple a For approximately how much time Atmm was the hockey stick in conmct with the puck Evidently the conmct time is quite short since you Figure 221 A hockey stick hits a puck as it hear a short sharp crack Be sure to show clearly the steps in your analysis slides by 39 39 39 A or simplifying r 39 M you make in your analysis 1 Choose a system System the hockey puck 2 List external objects that interact with the system with diagram Earm ice hOCkey sud air List all interacting objects even if you think some interactions will cancel out We39re looking down on the ice from above The y axis points out ofthe page toward you When possible make a 2D diagram which 39 much easier than trying to draw in 3D The symbol means out of the page the tip of an arrow pointing out at you the symbol 8 means into the page mil feathers of an arrow m pointing away from you 3 Apply the Momentum Principle We use the information that the puck slides without bouncing on the ice which lt1 07171 lt17 0 0gt lt flee fain Flee mg FsuckgtAtcontact means thath is zem will 39imes Write the three component equations separately 6 component Pa Pa T ee falxAtcontact The atoms in the top layer of the ice exert com Guam 0 F e m At which im lies F m upward forces on the atoms on the bottom 9 p quote g mmquot p quote g of the puck counteracting the downward z component 12 FsmkAtmm gravioational pull of the Earth Find new momentum ce e 0 andfai 0 negligible forces along xdirec39ion Since the puck slides long dismnces with little chan in Veloci friction and air re We pm 0Atmms pm so no change in pa 3 3 FY sismnce must be negligi e Pa EsuckAtcontact x gt initial lt 1 O 2 gtm b nallt13021gtm Figure 222 The A component of the momentum and velocity hardly changes but the 1 component of momentum and velocity changes quickly from zero to some nal value when the puck is hit Since the puck slides long distances with little change in velocity friction and air resistance must be negligible We assume that after the puck is struck the velocity is nearly constant with a new magnitude and direction Therefore the average ve locity is the same as the velocityjust after impact The contact time is very short a small fraction ofa secondl As a result you hear a short sharp crack when the stick him the puck Chapter 2 The Momentum Principle 7 Given these results from the Momentum Principle for 1 and 1 make a sketch of the path of the puck before and after it is hit The path of the puck must look something like that shown in Figure 222 In the result 1sz chkA twmm there are two unknowns 1sz and chk We need another equation in order to be able to solve for the unknown con tact time Atwnmt We can get additional information from the position up date formula t rivanAL 4 Apply the position update formula In B ix zargAt let Atslwle be the amount of time it takes the puck to slide from where it was struck to the known final position lt13 021gt m lt1 0 2gt m lt20 ms 0 vJAtsime xcomponent 13 m 1 m 20 msAtsllde so Atsrmle 12 m20 ms 06 s y component 0 0 0Atslmle so 0 0 not asurprise 06 s so 1 19 m06 s 317 ms zcornponent 21 m 2 m vz06 5 since Atslwle 5 Solve for the unknowns 1sz chkAtmnmm where 14 mvzf since 1 ltlt 0 04 kg317 ms 1000 NAt conmct At Contact 04 kg317 ms1000 N 0013 s 6 Check 39 Units check contact time is in seconds 39 Is the result reasonable The contact time is very short as expected If for example it had come out to be 300 s 5 minutes we should check our calculations How good were our approximations and 39 quot 39 We made the following 1 J 39 Ice exerts little force in the x or z directions low sliding friction 39 Negligible air resistance 39 Force ofstick is roughly constant during Atmnml and equal to 1000 N 39 The puck doesn t move very far during the contact time The neglect of sliding friction and air resistance is probably pretty good since a hockey puck slides for long distances on ice We know the hockey stick exerts a maximum force ong ck 1000 N be cause we observe that the stick breaks We approximate the force as nearly constant during contact Actually this force grows quickly from zero at first contact to 1000 N then abruptly drops to zero when the stick breaks The final approximation is somewhat questionable Although 0013 s is a short time the puck moves 20 ms 0013 s 026 m a bit less than one foot in the x direction during this time Also during this time 1 increases from 0 to 317 ms with an average value of about 158 ms so the z dis placement is about 158 ms 0013 s 02 m during contact On the oth er hand these displacements aren t very large compared to the displacement from lt1 0 2gt m to lt13 0 21gt m so our result isn t terribly 25 Problems of greater complexity 75 inaccurate due to this approximation Nevertheless a more accurate sketch of the path of the puck should show a bend as in Figure 223 X We can even calculate the radius of curvature of this bend the radius of the kissing circle discussed in the previous chapter We know that gtN A Fnet 3 2 2 On the left m M 0394 k 20 mS since pm my R R R 34 Z At first contact the velocity is in the x direction and has magnitude of 20 Figure 223 A more accurate OVerhead Vlew ms The kissing circle is tangent to the incoming velocity Of the Path Of the hOCkey Puck Showmg the bend during impactl On the right m 1000 N 2 Equate the left and right quantities 0394 k 130 mS 1000 N Solve R 0394 k 20 III92 016 m 1000 N This is a plausible result for the radius of the bend since we saw that as changes by about 026 m and z changes by about 02 m during contact It is important to see that even though our analysis of the stick contact time 0013 s isn t exact it is adequate to get a reasonably good determina tion of this short time something that we wouldn t know without using the Momentum Principle and the position update formula The short duration of the impact explains why we hear a sharp short crack Choice of system 7 We chose the hockey puck as the system to analyze Why not choose the system consisting of both the hockey puck and the hockey stick The problem with choosing both objects as the system is that the 1000 N force is now an internal force and doesn t show up in the Momentum Prin ciple so we aren t able to use this information By the reciprocity ofelectric forces including the interatomic forces between stick and puck the puck exerts a 1000 N force on the stick In the combined system the puck gains momentum from the stick and the stick loses momentumAto the puck The total momentum of the system doesn t change psymm 0 Review Let s review what we did to analyze this situation in the form of a general scheme for attacking problems We can summarize our workwith a diagram in the shape ofa diamond which emphasizes 39 starting from the Momentum Principle applied to a system 39 then expanding for the particular situation 39 then contracting down to solving for the quantity of interest 39 followed by checking for reasonableness Figure 224 Colliding studenmi The speed magnitude of velocity of each student is assumed to be the same and is represented by vi We also make the simplifying assump tion that the studenm have about the same mass m Expand the Momentum Principle by substituting any values you know Chapter 2 The Momentum Principle Link the lines in this diamond to the hockey stick analysis 1 Choose a system 2 List interacting objects with diagram 3 Apply the Momentum Principle pf irnam 4 You may also need to expand if iri v gAt 5 Solve for the unknowns 6 Check An important point implied by this diamond is that it is usually not useful to try to jump immediately to the step Solve for the unknowns by hunting for some formula that gives the unknown quantity directly Very often such a specialcase formula may not exist For example in the hockey puck prob lem the contact time emerged from applying the Momentum Principle and the position update formula There wasn t some readymade formula for the contact time for hockey sticks and pucks that you could use If you start by applying the Momentum Principle to a chosen system you can attack novel problems that you ve never before encountered The M0 mentum Principle is always valid whereas specialcase formulas aren t In previous studies you may have been taught a useful but restricted ap proach to solving problems which is to startwith a formula for the unknown quantity That is if you ve trying to find 1 startwith a formula for 1 We will help you learn a more powerful technique for solving problems which is to start from a fundamental physics principle in this case the Momentum Principle expand it by substituting known values then contract to solve for the unknown quantities In other words you derive the formula you need rather than hunt for it This is the only technique that can give you the pow er to solve novel problems ones that no one has previously encountered In an 39 39 cumple world gin r and i Mi t 39 being asked to meet new challenges A major goal of this course is to pre pare you to meet the novel challenges of the 2lst century 252 Example Colliding students Nextwe ll work through in detail a messy realworld situation Two students who are late for tests are running to classes in opposite directions as fast as they can Figure 224 They turn a corner run into each other headon and crumple into a heap on the ground Using physics principles estimate the force that one student exerts on the other during the collision You will need to estimate some quantities give reasons for your choices and provide checks showing that your estimates are physically reasonable This problem is rather illdefined and doesn t seem much like a text book prob em No numbers have been given yet you re asked to estimate the force of the collision This kind of problem is typical of the kinds of problems engineers and scientists encounter in their professional work For example suppose you are trying to design a crash helmet and you need to estimate the force it must withstand without breaking You don t know ex actly what the ultimate wearer will be doing at the time ofa crash so you have to make some reasonable estimates of typical human activities on which to base your analysis We ll make the simplifying assumption that the students have similar masses and similar speeds Figure 224 25 Problems of greater complexity Remember the diamond scheme as a guide to how to proceed We ll car ry out the analysis symbolically and plug in estimated values ofstudent mass and speed at the end That way we get a general solution that can be evalu ated for different values of these quantities 1 Choose a system System the student on the left 2 List external objects that interact with the system with diagram Earth ground other student air F b Aquot mg Fground more complex diagram shows point of application of each force fgro und 3 Apply the Momentum Principle lt0 0 0gt UM 0 0gt lt F fgmund fajrs Fground mg 0gtAt Write the three component equations separately as component 0 1in7 Ffground jlrAt F ground y component 0 i quotLth 50 Fgmund mg z component 0 0 which is true but uninformative Find new momentum Assume fground and fair are negligible compared to F 0 xi FAt so mv FAt sincevltlt c We could estimate the typical mass ofa student and the likely speed ofa student running at full speed but mv FAt is one equation in two un knowns the unknown force Fof the other student and the time duration At of the impact 7 Before trying to apply the position update formula if 13 ngAt try to estimate At of the impact simply by making a guess It is common to guess what seems a rather short time such as a second or halfa second However it is easy to show that such estimates are way off The record for the 100 m dash is about 10 seconds lfwe assume that these stu dents are very fast runners they each could be running 10 ms An alterna tive way of making the estimate is to note that you can easily walk about 4 miles in an hour which gives N 2i xlhrjt v4hr1396mi 1000km 36005 Nl39smS This moderate walking speed suggests that the students should be able to run between 5 ms and 10 ms Let s estimate that they each have a speed vm 5 ms During the collision the speed drops quickly from v to 0 so the average speed during the collision is approximately 0 02 or 25 ms lfwe guess that At is about halfa second during that half second how far does the student move List all interacting objecm even if you think some interactions will cancel out The student s heels are pushed to the left along the ground due to the collision and the resisting ground pushes to the right We ll assume that this force is much small er than the force exerted by the other stu dent because the student s shoes can slip Many of the momentum componenm are zero Do you see why my 0 In a vector equation the A component on the left must equal the xcomponent on the right similarly for y and z Estimate that the student is brought to an abrupt stop injust 5 cm about 2 inches Force has unim kg mss which has the unim of momentum kgms divid ed by s which is correct for a force change of momentum divided by time The collision time is 25 times smaller than our original guess of 05 s So the key to making a realistic estimate of a time interval is to estimate average speed and distance and then derive a time esti mate from that Chapter 2 The Momentum Principle At an average speed of25 ms in a half second the student would go 125 meters well over a yard This would imply that the two students pass right through each other coming out the other side Evidently an estimate of a half second is wildly too big and is inconsistent with the real world The problem is that it is very difficult to guess short time intervals accu rately On the other hand we can estimate short distances rather well and this gives us an indirect way to estimate short time intervals Each student gets squeezed during the collision Suppose each student s body gets pushed in a distance of about Axm 5 cm 005 m try pushing on your stomach and see how much deflection you can make 4 Apply the position update formula Apply if it vanAt or equivalently an AirAt ltv2 0 0gt ltAx 0 0gtAt xcomponent 12 AvaAt so At QAxv 5 Solve for the unknowns mv FAt RQAxv F quotLiz 60k 5 III92 15000N In 2M 2005 Atwoogs 1 ms 6 Check 39 Units check force is in newtons collision time is in seconds 39 Is the result reasonable The contact time is very short as expected Is the force reasonable or not See discussion below 15000 N is avery large force For example the gravitational force on a 60 kg student the weight is only about 60 kg98 Nkg m 600 N The force of the impact is more than 25 times the weight of the student It s like having a stack of25 students sitting on you If the students hit heads instead of stomachs the squeeze might be less than 1 cm and the force would be over 5 times as large This is why heads can break in such a collision So yes our result of 15000 N is plausible collisions involve very large forc es acting for very short times giving impulses of ordinary magnitude How good were our approximations We made the following 1 J 39 39 and 39 1 quotf 39 J 39 We estimated the running speed from known 100 m dash records 39 We estimated the masses of the students we could plug in other masses now that we have derived general results in terms of m an 1 39 We assumed that the horizontal component of the force of the ground on the bottom of the student s shoe was small compared to the force exerted by the other student Now that we find that the impact force is huge this assumption seems quite good 39 We made the approximation that the impact force was nearly constant during the impact so what we ve really determined is an average force 39 We assumed that the students had similar masses and similar running speeds to simplify the analysis If this is not the case the analysis is sig nificantly more complicated but we would still find that the impact force is huge 25 Problems of greater complexity You might object that with all these estimates and simplifying assumptions the final result of15000 N for the impact force is fatally flawed It is certainly the case that we don t have a very accurate result But nevertheless we gained valuable information that the impact force is very large Before do ing this analysis based on the Momentum Principle we had no idea of whether the force was small compared to the student s weight comparable or much bigger Now we have a quantitative result that the force is on the order of 25 times the weight of a student and we can appreciate why colli sions are so dangerous 7 What ifwe choose the system to include both students In this case there is almost no external impulse during the brief time ofcon tact The big internal forces the students exert on each other are equal in magnitude but opposite in direction due to the reciprocity of interatomic forces that come into play when the students make contact Therefore these forces don t change the momentum of the combined system Before the collision the total momentum of the combined system is zero mvi mv 0 and after the collision the total momentum is also zero they re not moving So there s no change in momentum which is consis tent with there being negligible external forces Simple analysis but with this choice of system we learn nothing about the force that one student ex erts on the other That s why we choosejust one student as the system 253 Physical models Our model of the colliding student situation is good enough for many pur poses However we left out some aspects ofthe actual motion For example we mostly ignored the flexible structure of the students how much their shoes slip on the ground during the collision etc We have also quite sensi bly neglected the gravitational force of Mars on the students because it is so tiny compared to the force of one student on the other Making and using models is an activity central to physics and the criteria for a good physical modelwill depend on howwe intend to use the mo e In fact one of the most important problems a scientist or engineer faces is deciding what interactions must be included in a model of a real physical chemical or biological system and what interactions can reasonably be ig nored Simplified or idealized models Ifwe neglect some hopefully small effects we say that we are constructing a simplified model of the situation A useful model should omit extraneous detail but retain the main features of the realworld situation We hope that the main results ofour analysis ofthe simplified model will apply adequately to the complex realworld situation Of course if we do a poorjob of modeling and make inappropriate ap proximations or neglect effects that are actually sizable we may get a rather inaccurate result though perhaps adequate for some purposes There is therefore a certain art to making a good model One goal of this course is to help you develop skill in formulating simplified but meaningful physical models of complex situations Another common way of describing a model is to say that it is an ideali zation by which we mean a simple clean strippeddown representation free of messy complexities Ideally a ball will roll forever on a level floor but a real ball rolling on a real floor eventually comes to a stop An ideal gas is a fictitious gas in which the molecules don t interactwith each other as opposed to a real gas whose molecules do interact but only when they come close to each other Figure 225 The electric force protons pel each repel each other electrons re other protons and electrons attract each ther lg 1 it Figure 226 The strong force the protons in the nucleus of an atom exert repulsive electric forces on each other but he strong interaction hols t th i J r39 quot 39 the nucleus Chapm 2 m Momentum Printipb TL l39r 39 ieanu39 somewhat from what actually happens in the real world Alternatively you can consider your result to be exact for a diffewnt realeworld situation one where the neglected in uences are not actually present For example the calculations we made concerning the colliding studenm could apply almost exactly to two rubber balls moving in outer space that compress by the amount we estimated An 39 engage 39 39 this course is making appropriate approximations to simplify the messy realeworld situation enough to permit approximate analysis using Newe ton39s laws Actually using Newton39s laws is imelf an example of modeling and making approximations because we are neglecting the effecm of quanr tum mechanics and of general relativity Einstein39s treatment of gravitae tion Newton39s laws are only an approximation to the way the world works though frequently an extremely good one 26 Fundamental forces There aieioui 39 an 39 lth JUul 39 39 39 elece tromagnetic nuclear also referred to as the strong interaction and the weak interaction o e gravitational interaction is responsible for an attraction every ob ject exem on every other object For example the Earth exem a rave itational force on the Moon and the Moon exerm a gravitational force th o 39 inc u es r 39L39 C sparks static cling arid the behavior of electronic circuits an g 39 39 o 39 of motors driven by electric ble for th a stretched or compressed spring is due to electric forces between the atoms that make up the sprin 0 The nuclear or smmginteracu39on holds protons and neutrons together in the nucleus of an atom despite the large mutual electric repulsion T not electrically o An example ofth tron Ifa neutron is removedfrom a nucleus with an average lifetime of about 15 minutes the neutron decays into a proton an electron and a ghostly particle called the antineutrino This change is brought about by the weak interaction We will 39 39 39 39 but we will e nuclear Ul iuLck action plays an impormnt role We will have little to say about the weak in teraction Nor will we deal with magnetism the other part of the 39 39 39 The second volume ofthis textbook deals exe tensively with both electric and magnetic interactions as nan ii um i LIUII 27 The gravitational force law To predict the motion of stars planes spacecraft comes satellites and other 39 J39 39 m we nee 39 39 39 force law together with the Momentum Principle As a matter ofgeneral cule 2 7 The gravitational force law ture it can be interesting to understand how the motion of objects in space can be predicted by physics principles If you are studying aerospace engi neering or astrophysics this may also be a professional interest What if your interests are in nanotechnology or chemistry or civil engi neering Studying how to predict the motion of stars and planets is one of the most direct ways to understand in general how the Momentum Princi ple determines the behavior ofobjects in the real world The motion ofstars and planets is in important ways simpler than other mechanical phenome na because there is no friction to worry about so this is a good place to start your study The basic ideas used to predict the motion of stars and planets can be applied later to awide range of everyday and atomic phenomena In the 1600 s Isaac Newton deduced that there must be an attractive force associated with a gravitational interaction between any pair of objects The gravitational force acts along a line connecting the two objects Figure 227 is proportional to the mass ofone object and to the mass of the other object and is inversely proportional to the square of the distance between the centers of the two objects not the gap between their surfaces Here is the formula for the gravitational force exerted on object 2 by object 1 THE GRAVITATIONAL FORCE LAW A m2 1 Fgmv on 2 byl eGs zrzr1 Irzrll in points from the center of object 1 to the center of object 2 2 Gis auniversal constant C 67gtlt10711 Nk I This force law looks pretty complicated but it could have been a lot more complicated than it is For example the gravitational force law does not de pend on the momentum or velocity of the objects It depends only on the masses and on the position of one object relative to the other object Large spheres An optional section at the end of this chapter page 96 shows that uniform density spheres interact gravitationally as t ough all of their mass were con centrated at the center of the sphere so the gravitational force law applies to large uniformdensity spheres as long as you use the centertocenter dis tance For examp e the force that the Earth exerts on you can be calculated as though all the mass of the Earth were at its center 64X106 m away from where you are sitting Measurement of G Another optional section at the end of this chapter page 97 tells how Cav endish in the late 1700 s measured by observing the tiny gravitational forces two lead balls exerted on each other 271 Understanding the gravitational force law We will spend some time understanding how to calculate forces between ob jects using Newton s gravitational force law Then we will use this force in the Momentum Principle to predict the future motion of objects The gravitational force law involves a lot of different symbols and may look pretty intimidating at first Let s take the law apart and look at the in dividual pieces to try to make sense of the formula The relative position vector in in the gravitational force law points from the center of objectl to the center of object 2 and so does the unit vector 271 rhat in Figure 228 Recall that in words in is the location of ob 81 fZI J 2 m2 m1 A lrzrll Fwy on 2 by 1 m1 Fwy on 1 by 2 Figure 227 The gravitational force exerted on object 2 by object lt The force exerted on object 1 by object 2 has the same magni tude but opposite direction Figure 228 The location of object 2 rela tive to object 1 final minus initialt w Figure 229 The gravitational force depends on the product of the two massesl gt m1 w Figure 230 The gravitational force is an inverse square laws mgm GO lr21l Figure 231 The direction of the gravita tional force on object 2 is in the opposite direction to the unit vector pointing toward object 2 Chapter 2 The Momentum Principle ject 2 relative to object 1 final minus initial The magnitude of in is the distance between the centers of the two objects We say that the gravitational constant C is universal because it is the same for any pair of interacting masses no matter how big or small they are orwhere they are located Because Gis universal it can be measured for any pair of objects and then used with other pairs of objects As we describe on page 97 Cavendish was the first person to make such a measurement As highlighted in Figure 229 the gravitational force is proportional to the product of the two masses mzml If you double either of these masses keeping the other one the same the force will be twice as big lfyou double both of the masses the force will be four times as big Since mzml mlmz the magnitude of the force exerted on object 1 by object 2 is exactly the same as the magnitude of the force exerted on object 2 by objectl but the direction is opposite The gravitational force is an inverse square law As highlighted in Figure 230 the square of the centertocenter distance appears in the denomina tor This means that the gravitational force depends very strongly on the dis tance between the objects For example ifyou double the distance between them the only thing that changes is the denominator which gets four times bigger 2 squared is 4 so the force is only 14 as big as before 7 If you move the masses 10 times farther apart than they were originally how does the gravitational force change The force goes down by a factor of 100 Evidently when two objects are very far apart the gravitational forces they exert on each otherwill be vanishing ly small big denominator small force The minus sign with the unit vector highlighted in Figure 231 gives the di rection of the force exerted on object 2 by object 1 The vector in points toward object 2 as does the unit vector 1 and the force acting on object 2 points in the opposite direction A useful way to think about the gravitational force law is to factor it into magnitude and direction like this 271 A A A 39 A vector IS a magn1tude t1mes a d1rect10n Fgmv ngmvngmv m2 m1 le 39 Direction unit vector Fgmv ifzrl It is usually simplest to calculate the magnitude and direction separately then combine them to get the vector force That way you can focus on one thing at a time rather than getting confused or intimidated by the full complexity of the vector force law 39 Magnitude IFgmvl Ex 218 Masses M and m attract each other with a gravitational force of magnitude F Mass m is replaced with a mass 3m and it is moved four times farther away Now what is the magnitude of the force Ex 219 A 3 kg ball and a 5 kg ball are 2 m apart center to center What is the magnitude of the gravitational force that the 3 kg ball exerts on the 5 kg ball What is the magnitude of the gravitational force that the 5 kg ball exerts on the 3 kg ball Ex 220 Measurements show that the Earth s gravitational force on a mass of 1 kg near the Earth s surface is 98 N The radius of the Earth is 6400 km 64X106 m From these data determine the mass of the Earth 2 7 7714 grizvihztiomzlfm km 33 272 Calculating the gravitational force on a planet We39ll go through a complete calculation to see in full gory detail how to evaluate the gravitational force law in the most general situation In Figure 232 is a smr o mass 4x1an k locate at position lt2X10111X101115X1011gt m and a planet ofmass 3x10 kg located at poe sition SXIOH35x1011e05X1011gt m These are typical values for stars and planets Notice that the mass ofthe smr is much greater than that ofthe planet We will calculate the gravitational force exerted on the planet by the smr In Figure 232 we show the y and z componenm of the positions to be multiplied by 1x1011 m Make sure you understand how the numbers on the diagram correspond to the positions given vectors Here is a summary ofthe steps we will take to calculate the force acting on the planet due to the smr Figure 232 A smr and a planet interact gravimtjonally We will calculate the gmv39r mtjonal forces The a y and 1 components are to be multiplied by 1gtlt10 1 m 0 Calculate r21 the position ofthe planet relative to the smr o Calculate r21 the scalar distance from star to planet o Calculate GmijIFHIZ the magnitude ofthe force o Calculate er1 e r271r271 the direction ofthe force 0 Multiply the magnitude times the direction to get the vector force The relative position vector An important quantity in the force law is the position ofthe center of the planet relative to the center of the star r271 so we start by calculating this relative position vector Think about what you know about calculating relative position vectors and try to calculate r21 before reading ahead As usual we just calculate nal minus initial 271 Fri 21 SXIOH35x1011e05x1011gtmelt2X10111x101115x1011gtm r271 lt1X101125x1011e2x1011gt m In Figure 233 the diagram shows that the signs ofthe componenm of 27 make sense positive xand y componenm negative 2 component which is an impormnt check that we haven39t made any sign errors The distance In order to calculate the magnitude of the force we39ll need the distance r271 between the centers of the smr and planet and a bit later we39ll also need r271 for calculating the unit vector Figure 233 The position vector of the Try to calculate the magnitude of r21 before reading ahead planemlmve m mesm As usual we calculate the dismnce which is a scalar by using the 3D version ofthe Pythagorean theorem rm AIago 25x10 7e2x10 Z m r21 335x1011 m s4 Chapm 2 m Momentum Printipb The magnitude of the force 1 Before readin ahead try to calculate the magnitude ofthe force The magnitude ofthe force on the planet by the star is this i quot 2 m1 IFmeuneeysml G lrzell 24 50 e Nm 3 10 k 4 10 k Fpampianetbysarl 67x1011 y X X 335x1011mZ 71 lemn planetbysarl 716x10 N This looks like a big force but it39s acting on a very big mass so it isn39t obvie ous whether this is really a big force in terms ofwhat it will do The direction of the force an you think how to calculate the vector direction of the gravitational force the unit vector FWm 91mm 5 Try it We39ve got the magnitude ofthe force We also need to calculate the direce tion ofthe force which involves the unit vector s e r21 Direction offorce meplmwym rm e I I 21 11 11 11 lt1X10 25X10 72X10 gtm Fg nzvonplanetbysmt i 11 335X10 m ngat on planetbysnr 0299 07467 0597gt Notice that the unit vector giving the direction has no units because the meters in the numerator and the denominator cancel Figure 234 shows the unit vectors pointing from star toward planet and from planet toward star The vector gravitational force Try You know L 39 A and on Figure 234 Unityectors know the direction Finally we can multiply the magnitude ofthe force times the unit vector die rection ofthe force to get the full vector force guz u on planetbystat l gnz u on planetbystarng7W on 913mm 5m gsaWnpianetbysm 716x1021 N 0299 70746 0597gt gsamnpianetbysta 7214X10217534x1021427X1021gt N mum Checking our result It is impormnt to check long calculations such as this because there are e many opportunities to make a mistake along the way There are several checks we can make o Di mAn important check is to make a diagram like Figure 235 and Figure 235 The grayioational force exerted see whether the direction ofthe calculatedforce makes sense Pay pare on the planet by the sear ticular attention to signs The diagram shows that the force acting on 2 7 m wwmmlfm km the planet points in the 7x 7y z direction andthis agrees with our calculation which has the same component signs Order ofmgnimde Make a rough order of magnitude check dropr ping all the detailed numbers In terms of order ofnmagnitude the dismnce between smr and pgoanetis very roughly 1x10 m the mass of the star is very roughly 1x10 kgthe mass ofthe planet is very roughly 1x10 kg and e gravitational constant is w roughly N mikgi Therefore we expect the magnitude ofthe force to be very roughly this 74 50 M10711 Nm21x10 kg111X102 kg moo N g x10 m k 1 In this very rough order ofmagnitude calculation ignoring numerical details we just need to add and subtract exponenm and one hardly even needs a calculator The fact that we getwithin an order ofmagnir tude of the result 716x1021 N is evidence that we haven39t made any huge mismkes 0 Unitx39An impormnt check is that the units came out correctly in new tons Unit mm Check to see whether the calculated unit vector does in deed have magnitude 1 pw on planabysml I702997 70746Z 05972 1001 The magnitude isn39t exactly 1 because we rounded offthe intermediate calculations to three signi cant gures Our result passes all these checks This doesn39t prove we haven39t made a mistake somewhere but at least we ve ruled out many possible errors The force on the star exerted by the planet We39vejust calculated the force exerted on the planet by the smr To calcur late the force exerted on the star by the planet we could redo all these lengthy calculations in the same way but that would be a lot ofwork Look at the form of the gravitational force law and think carefully Do you see a way to write down immediately the gravitational force on the star by the planet without doing any additional calculations Note that the magnitude is exactly the same because it involves exactly the same quantities The on y change is the direction of the force which is in the o posite direction Therefore we can immediately write the new result just by ipping the signs Essensmbypune 214x1021534X10217427x1021gt N Note L L 39 c L c I 39 39 Figure 236 You might be puzzled that the planet pullsjust as hard on the star as the star pulls on the planet We39ll discuss this in more detail later in this chapter How can you remember all these steps aicuiatin me ecto 39 39 l 39 r buteach step 39 r 39 Hn can p Think c L j 39 39 miirie and outline for What to do mzmlg Fgmvonibyl rGizrzr1 lrzrll Figure 236 The gravimtional force exerted on the star by the planet Chapter 2 The Momentum Principle 39 When you look at the formula you see that you need in so you calcu late it final minus initial 39 You also see that you need t271 so you calculate that 39 Now you see that you now have enough information to calculate the magnitude of the force 39 The formula tells you to calculate the negative ofthe unit vector to get the direction 39 The formula says to multiply magnitude times direction That s it With this scheme in mind look over the previous few pages to review how we carried out the evaluation of the gravitational force law Ex 221 The mass of the Earth is 6x1024kg and the mass of the Moon is 7gtltlO kg At a particular instant the Moon is at location lt28gtlt108 0 728X108gt m in a coordinate system whose origin is at the center of the Earth a What is rME the relative position vector from the Earth to the Moon b What is c What is the unit vector erVE d What is the gravitational force exerted by the Earth on the Moon Your answer should be avector 273 Using the gravitational force to predict motion You now know that this star exerts this force on this planet at this instant butwhat s the point in knowing this By itself this isn t particularly interest ing But ifwe use this calculated force in the Momentum Principle we can predict the future motion of the planet 39 We know the force at this instant and if we know the initial momen tum pi of the planetwe can use the update form of the Momentum Principle pf p FnetAt to predict the momentum of the planet a short time At later We need to choose At short enough that the force doesn t change much during this brief time interval due to changes in the positions of the interacting objects 39 We can also use the velocity to update the position of the planet using the position update formula if 13 vargAt Again we need to choose At short enough that the velocity doesn t change much during that brief time interval so that we can use the initial or final velocity as a good approximation to the average velocity After doing these momentum and position updates we have successfully predicted the new momentum and position a short time into the future Now we can repeat this process calculating a new gravitational force based on the new position updating the momentum and updating the position Step by step we can predict the motion of the planet into the future Although the massive star won t move very much we could also update its momentum and position repetitively As we ve seen it s easy to get the force on the starjust by ipping signs Here is a summary of this iterative scheme 39 Start with initial positions and momenta of the interacting objects 39 Calculate the vector gravitational forces acting on each object 39 Update the momentum of each object pf pi EnetAt 39 Update the positions if immght where v gm 39 Repeat 2 7 The gravitational force law In order to plan the trajectory for a spacecraft from Earth to a particular landing site on Mars NASA has to carry out avery large number of such up date calculations taking small steps small At to achieve high accuracy This includes calculating the net force as the vector sum of all the gravita tional forces exerted on the spacecraft along the way by the Sun Earth Moon Mars and other objects in the Solar System Doing all these force and update calculations by hand would be impossibly difficult so people at NASAwrite computer programs to do these repetitive calculations Laterwe will show you how to write such programs Updating the momentum and position of the planet To see in detail how this iterative scheme works we ll carry out one com plete step one update of the momentum and position In order to start the iteration we have to know the initial positions and momenta of the star and planet We already know their initial positions Let s suppose we also know their initial velocities m lt000gt ms 4 4 4 mm ltlgtlt10 2X10 l5gtlt10 gt ms The star is initially at rest and because it has a huge mass compared to the planet it won t move very much For simplicity we ll make the approxima tion that the star is fixed in position and never moves Choosing a sufficiently small time interval We need to decide how big a value of At we can get away with and yet pre serve adequate accuracy The smaller the time step At the more accurate the calculation because the force and velocity won t change much during the brief time interval However taking smaller steps means doing more cal culations to predict some time into the future Here is a way to choose a time interval The initial speed is about vm 3X10 ms Choose a At so that the distance the planetgoes in this time interval vAt is small compared to the distance dbetween star and planet which is about 3gtlt1011 m As the planet changes position the gravitational force will change in magnitude and direction as you can see in Figure 237 In the Momentum Principle pf pi lTnetAt we need for At to be small enough that Fnet is nearly constant during this time interval Let s choose as a tentative criterion that the distance the planet moves ought to be about 0001 times the distance from the star to the planet vA tm 0001d 0001d 00013gtlt1011m v 3x104 ms Am 1x104s This is about 3 hours If instead we choose a large time interval of lgtlt107 s say in the first update the planet would move a long distance about 3X104 mslgtlt10 s 3X10 m which is about as large as the distance to the star and during that step the actual gravitational force would change a great deal in magnitude and direction This would make our predictions very inaccurate Whenever you choose a time interval to use for updating momentum make sure it is sufficiently small that the net force doesn t change very much in magnitude or direction during your chosen time interval 87 mp i vAt I A V K Figure 237 If At is too big the gravitational force will change a lot in magnitude and direction during the time interval Chapter 2 The Momentum Principle On the other hand if the force happens to be constant in magnitude and direction you can use as large a At as you like without making an error in the momentum update Update the momentum of the planet 7 We know the force acting on the planet and we ve chosen a time interval to make one step in predicting the future of the planet Try to predict the momentum the planetwill have after lgtltlO s The planet is moving at a high speed about 3gtlt104 ms but this is small compared to the speed of light which is 3gtlt108 ms so the initial momen tum is pim mvi aw ma 3xi024 kgltlgtlt1042gtlt10415gtlt104gtms a lt3x1028 6x1028 45x1028gt kgms Use the Momentum Principle in its update form E Wham pf lt3x1028 6x1028 45x1028gt kgms lt7214gtlt10217534gtlt1021 427x1021gt N1xi04 s pf lt29979gtlt1028 59947x1028 45043x1028gt kgms Notice that the momentum didn t change very much because we deliber ately chose a small time interval to ensure accuracy in the momentum up date by making sure that the force remained nearly constant during the brief time interval Update the position of the planet 7 Think about how you would update the position of the planet We must use the position update formula if 13 ngAt but what is the average velocity v g The velocity and momentum changed during the lgtltlO s time step Should we use the initial velocity The final velocity The arithmetic average of these two velocities A key point is that the momentum changed very little so the velocity changed very little Therefore it hardly matters which velocity we use To be concrete about this and for simplicity in computer calculations we ll nor mally use the final velocity obtained from thejust calculated final momen tum to represent the average velocity This works well as long as the time step is small Since pfm mvf vfm m and the position update is this f lt3gtlt101135x1011705gtlt1011gt m lt29979x1028 59947x1028 45043x1028gt kgms 3xi024 kg 1X104s f lt3001gtlt1011 3502x1011704985gtlt1011gt m 2 7 m gruvihztiomzlfm hm What have we done Whew After all that compumtional drudgery what have we accomplished We39ve predicted the future ofthe p anet Knowi g the planet s initial position and momentum we predict that a 39 39 ition that we39ve calculated with a d So what Well we could do this a econd time step would involve exactly the same calculations we just did but the initial position and momentum would e the nal ones we just calculated We could keep going doing this over and over and over and predict as far ahead into the future as we are willing to do the tedious calculations ou say you39d rather not do all those calculations Right The thing to do is to instructa computer to do this for you need to do is learn how 39 39 39 uie computer 39 39 ove and over Nextwe will show you how to do this but it was important to guide you once through the gory demils ofthe calculation so that you could unr derstand exactly what it is you have to tell the computer to do 274 Telling a computer what to do Here we summarize how to organize a computer program in order to tell a computer how to apply the Momentum Principle repetitively to predict the u ure 0 De ne the values ofconsmnm such as Gto use in the program o Specify the masses initial positions and initial momenta ofthe interr acting objecm Specify an appropriate value for At small enough that the objecm don39t move very far during one up a e 0 Create a loop s ructure for repetitive calculations FmAt s pm o Update the positions 5 r vmgdt where vmg 0 Repeat The details of the actual program smtemenm depend on what programr ming language or system you use and your instructor will provide you with the demils ofthe particular tool you will use A computer calculation In the preceding sections you took one step in such a calculation by hand Figure 238 shows the orbit of the planet as predicted by a computer pror gram written using V39Python httpvpythonorg that continually updated L A 39 39 F L nlanet 39 39 39 al force law the Momentum Principle and the position update formula The smr and planet are shown much larger than they really are forvisibility in the largely empty space Using the gravitational force law Newton was able to explain the motion ofplanem and moons in the Solar System Later scientism have found that 39 39 quot39 r39 39 even L 39 at in our own andother galaxies with the same formula so it appears that this is truly a universal law of gravimtion applying to all pairs of masses everywhere in the universe Van the step size After you get a program running it is very important that you decrease the value ofthe step size At and make sure that the motion doesn39t change Of course the motion will take longer because you39re doing more calculations The motion of round the sur The scar and planet are shown much larger than they really are Figure 238 the planet Chapter 2 The Momentum Principle but the issue is that the shape of the trajectory shouldn t change If it does change significantly that means that At was originally too big leading to the net force and average velocity changing too much during the time At Keep decreasing At until you find that the shape of the trajectory hardly changes Now you have an adequately small step size At Define and use constants It is important to gather together at the start of the program the definitions of any constants such as G and use these defined symbols everywhere else in the program that is use Grather than 67E ll in later calculations You should avoid entering actual numbers such as 67E ll more than once in a program since if there is a mistake in a constant you have to correct it in many places Also it is easier to read a program that uses Geverywhere rath er than the number 67E ll Why not just use calculus You might wonder why we don t simply use calculus to predict the motion ofplanets and stars One answer is that we do use calculus Step by step we add up a large number of tiny increments of the momentum of a body to calculate a large change in its momentum over a long time and this corre sponds to an approximate evaluation ofan integral which is an in nite sum of infinitesimal amounts A more interesting answer is that the motion ofmost physical systems can not be predicted using calculus in any way other than by this stepby step ap proach In a few cases calculus does give a general result without going through this procedure For example an object subjected to a constant force has a constant rate ofchange of momentum and velocity and calculus can be used to obtain a prediction for the position as a function of time as we saw in the example ofa ball thrown through avacuum The elliptical or bits of two stars around each other can be predicted mathematically al though the math is quite challenging But the general motion of three stars around each other has never been successfully analyzed in this way The ba sic problem is that it is usually relatively easy to take the derivative of a known function but it is often impossible to determine in algebraic form the integral of a known function which is what would be involved in long term prediction lt isn ta question of taking more math courses in order to be able to solve the threebody problem there is no general mathematical solution How ever a stepby step procedure of the kind we carried out for the planet can easily be extended to three or more bodies Just calculate all the forces act ing between pairs of bodies update the momenta and update the positions This is why we study the stepby step prediction method in detail because it is a powerful technique of increasing importance in modern science and en gineering than to the availability of powerful computers to do the repet itive work for us Ex 222 The Earth goes around the Sun in 365 days in a nearly circular orbit In a computer calculation of the orbit which is actually an ellipse approximately how big can At be and still get good accuracy in the prediction of the motion Ex 2 23 A pendulum swings with a period time for one round trip of s In a computer calculation of the motion approximately how big can At be and still get good accuracy in the prediction of the motion 2 7 The gravitational force law 275 Approximate gravitational force near the Earth s surface Earlier we used the expression mg to represent the magnitude of the grav itational force on an object near the Earth s surface This is an approxima tion to the actual force but it is a good one The magnitude of the gravitational force that the Earth exerts on an object of mass m near the Earth s surface Figure 239 is MEm Fg G 2 RETy where y is the distance of the object above the surface of the Earth REis the radius of the Earth and ME is the mass of the Earth A spherical object of uniform density can be treated as if all its mass were concentrated at its center As a result we can treat the effect ofthe Earth on the object as though the Earth were a tiny very dense ball a distance RE y awa The gravitational force exerted by the Earth on an atom at the top of the object is slightly different from the force on an atom at the bottom of the object because each atom is a slightly different distance from the center of the Earth How much can this difference really matter in an analysis Sup pose the height of the object is a meter and the bottom of the ob sject is one meter above the surface The radius of the Earth is about 64gtltlO m Then MEm whereas Fbottom CW F G Em P 6400002 m2 For most purposes this difference is not significant In fact for all interac tions of objects near the surface of the Earth it makes sense to use the same approximate value RE for the distance from the object to the center of the Earth This simpli es calculation of gravitational forces by allowing us to combine all the constants into a single lumped constant g ME ME Fgm gm so that g GE E E The constant g called the magnitude ofthe gravitational field has the val ue g 98 newtonskilogram near the Earth s surface The gravitational field quotg at a location in space is defined to be the vector gravitational force that would be exerted on a 1 kg mass placed at that location A mass would experience a force twice as large In general a mass m will ex perience a force mg of magnitude mg Note that gis apositive number the magnitude of the gravitational field We can use this approximate formula for gravitational force in our analysis of any interactions occurring near the surface of the Earth Ex 2 24 Show that GMERIZE equals 98 Nkg The mass of the Earth is 6xi024 kg Ex 2 25 At what height above the surface of the Earth is there a 1 difference between the approximate magnitude of the gravitational field 98 Nkg and the actual magnitude of the gravitational field at that location That is at what height y above the Earth s surface is GMERE y2 099 GMER a 91 Figure 239 Determining the gravitational force of the Earth on an object a height y above the surface c Figure 240 Protons repel each other elec trons repel each other protons and elec trons attract each others F la 2b 1 7 1 12 11f e Lon y 7 if V 4n202712 21 12 Femon 1 byZ jfzrl Figure 241 The electric force exerted on object 2 by object ll The force exerted on object 2 by object 1 has the same magni tude but opposite directions Chapter 2 The Momentum Principle 28 The electric force law Coulomb s law There are electric interactions between charged particles such as the pro tons and electrons found in atoms It is observed that two protons repel each other as do two electrons while aproton and an electron attract each other Figure 240 where protons are said to have positive electric charge and electrons have negative electric charge T e force corresponding to this electric interaction is similar to the grav itational force law and it is known as Coulomb s law to honor the French scientist who established this law in the late 1700 s during the same period when Cavendish measured the gravitational constant C See Figure THE ELECTRIC FORCE LAW COULOMB39S LAW l q 91 A 2 1721200an 1 r Y s 271 41E80r242 9 N m2 IS a un1versal constant L 9X10 4180 4180 C2 where Both the gravitational and electric forces are proportional to the inverse square of the centertocenter distance inverse square laws The univer sal electric constant read as one over four pi epsilonzero is very much larger than the gravitational constant reflecting the fact that electric inter actions are intrinsically much stronger than gravitational interactions For example consider a heavy weight that hangs from a thin metal wire The small number of atoms interacting electrically in the wire have as big an ef fect on the hanging weight as a very much larger number of atoms in the huge Earth interacting gravitationally The charges ql and q2 must be meagsured in SI units called coulombs The protpgn has a charge of 16gtlt 10 C and the electron has a charge of 6X10 7 Study the three cases in Figure 240 Why is no minus sign needed in the force law unlike the case with the gravitational force law The difference is that two positively charged particles such as two protons repel each other whereas masses are always positive but the gravitational force is attractive Two negatively charged particles such as electrons also at tract each other and minus times minus gives plus Only if the two particles have opposite charges will they attract each other and then the factor q2q1 contributes the necessary minus signjust as in the gravitational force law 281 Interatomic forces When two objects touch each other they exert forces on each other At the microscopic level these contact forces are due to electric interactions be tween the protons and electrons in one object and the protons and elec trons in the other object In Chapter 3 we will examine interatomic forces in more detail Ex 226 Aproton and an electron are separated by lgtltlOi10 m the radius of a typical atom Calculate the magnitude of the electric force that the proton exerts on the electron and the magnitude of the electric force that the electron exerts on the proton 2 9 Mciln39ocity 29 Reciprocity An important aspect of the gravitational and electric interactions includ ing the electric forces of atoms in contactwith each other is that the force that objectl exerts on object 2 is equal and opposite to the force that object 2 exerts on object 1 Figure 242 That the magnitudes must be equal is clear from the algebraic form of the laws because mlmz mzml and qlqz qqu The directions of the forces are along the line connecting the centers and in opposite directions This property is called reciprocity or Newton s third law of motion RECIPROCITY A Fon 1 byz iFon 2 byl gravitational and electric forces The force that the Earth exerts on the massive Sun isjust as big as the force that the Sun exerts on t e Earth so in the same time interval the momen tum changes are equal in magnitude and opposite in direction equal and opposite for short A31 FonlbyzAt iA z However the velocity change Av Apm of the Sun is extremely small compared to the velocity change of the Earth because the mass of the Sun is enormous in comparisonwith the mass of the Earth The mass ofour Sun which is a rather ordinary star is 2gtltlO 30 kg This is enormous compared to the mass of the Earth 6x10 4 kg or even to the mass of the largest planet in our Solar System Jupiter 2gtltlO kg Nevertheless very accurate mea surements of small velocity changes of distant stars have been used to infer the presence of unseen planets orbiting those stars Magnetic forces do not have the property of reciprocity Two electrically charged particles that are both moving can interact magnetically as well as electrically and the magnetic forces that these two particles exert on each other need not be equal in magnitude nor opposite in direction Reciproc ity applies to gravitational and electric forces but not in general to magnet ic forces except in some special cases Why reciprocity The algebraic forms of the gravitational and electric force laws indicate that reciprocity should hold A diagram may be helpful in explaining why the forces behave this way Consider two small objects a 3 gram object made up of three 1 gram balls and a 2 gram object made up of two 1 gram balls Figure 243 The distance between centers of the two objects is large compared to the size of either object so the distances between pairs ofl gram balls are about the same for all pairs You can see that in the 3 gram object each ball has two gravitational forces exerted on it by the distant balls in the 2 gram object 7 How many forces act on the 3 gram object There is anet force of 3 X 2 6 times the force associated with one pair of balls 7 Similarly consider the forces acting on the 2 gram object How many forces act on the 2 gram object There is again a net force of 2 X 3 6 times the force associated with one pair of balls The effect is that the force exerted by the 3 gram object on the 2 gram object has the same magnitude as the force exerted by the 2 gram object on 93 2 out of page I Figure 242 The Sun and the Earth exert equal and opposite forces on each otheri 2 grams 3 grams Figure 243 Reciprocity there are 6 forces acting on the 3 gram object and 6 forces acting on the 2 gram objecti Chapter 2 The Momentum Principle the 3 gram object The same reciprocity holds for the same reasons for the electric forces between a lithium nucleus containing 3 protons and a helium nucleus containing 2 protons 210 The Newtonian synthesis Newton devised a particular explanatory scheme in which the analysis of motion is divided into two distinct parts 1 Quantify the interaction in terms ofa concept called force Specific examples are Newton s law of gravitation and Coulomb s law 2 Quantify the change of motion in terms of the change in a quantity le momentum The change in the momentum is equal to the force times At This scheme called the Newtonian synthesis has turned out to be extraor dinarily successful in explaining a huge variety of diverse physical phenom ena from the fall of an apple to the orbiting of the Moon Yetwe have no way ofasking whether the Universe really works this way It seems unlikely that the Universe actually uses the human concepts of force and momen tum in the unfolding motion ofan apple or the Moon We refer to the Newtonian synthesis both to identify and honor the par ticular highly successful analysis scheme introduced by Newton but also to remind ourselves that this is not the only possible way to view and analyze the universe Einstein s alternative view Newton stated his gravitational force law but could give no explanation for it He was contentwith showing that it correctly predicted the motion of the planets and this was a huge advance the real beginning ofmodern science Einstein made another huge advance by giving a deeper explanation for gravity as a part of his general theory of relativity He realized that the mas sive Sun bends space and time i in such away as to make the planets move the way they do The equations in Einstein s general theory of relativity make it possible to calculate the curvature ofspace and time due to massive objects and to predict how other objects will move in this altered space and time Moreover Einstein s theory of general relativity accurately predicts some tiny effects that Newton s gravitational law does not such as the slight bend ing of light as it passes near the Sun General relativity also explains some bizarre largescale phenomena such as black holes and the observed expan sion of the space between the galaxies Einstein s earlier special theory of relativity established that nothing not even information can travel faster than light Because Newton s gravitation al force formula depends only on the distance between objects not on the time something s wrongwith the formula since this implies that ifan object were suddenly yanked away its force on another objectwould vanish instan taneously thus giving in principle a way to send information from one place to another instantaneously Einstein s theory of general relativity doesn t have this problem Since the equations of general relativity are very difficult to work with and Newton s gravitational law works very well for most purposes in this course we will use Newton s approach to gravity But you should be aware that for the most precise calculations one must use the theory ofgeneral rel ativity For example the highly accurate atomic clocks in the satellites that make up the Global Positioning System GPS have to be continually cor rected using Einstein s theory of general relativity Otherwise GPS positions would be wrong by several kilometers afterjust one day of operation 211 Derivatitm of special average velocity result 211 Derivation of special average velocity result Here we offer two proofs one geometric and one algebraic using calcu lus for the following specialcase result concerning average velocity AVERAGE VELOCITY SPECIAL CASE 1 v I v pct pi only 1f 1 changes at a constant rate avgyx 2 x Fnew is constant 1 ltlt 0 similar results for vavgy and 12an Z If Fnew is constant pr JpnFneg xAt implies that 1x changes at a con stant rate At speeds small compared to the speed of light vxm 1 m so a graph of vx vs time is a straight line as in Figure 244 In Figure 244 we form narrow vertical slices each of height 1 and narrow width At Within each narrow slice 1 changes very little so the change in position during the brief time At is approximately Ax vat Therefore the change in x is approximately equal to the area of the slice of height 1 and width At Figure 245 If we add up the areas of all these slices we get approximately the area under the line in Figure 244 and this is also equal to the total displacement Ax1 sz Axg x 7 xi lfwe go to the limit ofan infinite number of slices each with infinitesimal width the sum of slices really is the area and this areawe have shown to be equal to the change in position This kind of sum of an infinite number of infinitesimal pieces is called an integral in calculus The area under the line is a trapezoid and from geometry we know that the area of a trapezoid is the average of the two bases times the altitude Turn Figure 244 on its side as in Figure 246 and you see that the top and bottom have lengths vm and vxf while the altitude of the trapezoid is the total time tfi ti Therefore we have the following result Trapezoid area xfi x1 tfi ti Dividing by tfi ti we have this xf xi vxiT vxf tfe t1 2 But by definition the xcomponent of average velocity is the change in addi vided by the total time so we have proved that vavgypc pm de only if 1 changes linearly with time Fnew is constant The proof depended critically on the straightline linear change in ve locity which occurs only if Fnew is constant and 1 ltlt 0 Otherwise we wouldn t have a trapezoidal area That s why the result isn t true in general it s only true in this important but special case Algebraic proof using calculus An algebraic proof using calculus can also be given We will use the as com ponent of the derivative version of the Momentum Principle more about this in the next chapter Fnet AA A FnaAt implies that XE 95 1 increases at constant rate Figure 244 In this graph of vx vsl t con stant force form narrow vertical slices each of height 10 and width Atl l l 39Ul 5 Areamvat l l T At Figure 245 One narrow slice has an area given approximately by vatl This is equal to Ax the displacement of the object Area WAltitude TOP v Altitude tfi t1 Bottom vxf Figure 246 The area of the whole trape zoid is equal to the total displacement The figure in Figure 244 has been turned on is sidel Equivalent ngWi G m1 2 Figure 247 A sphere whose density depends only on distance from the center acm like a point particle with all the mass at the centeri Chapter 2 The Momentum Principle d In the limit we have lim Fna and amp At oAt dt dt F netx IfF new is a constant the time derivative of 1 is a constant so we have 1x F tpxi since 1x pm when t 0 new You can check this by taking the derivative with respect to time t which gives the original equation dpx dt new t speeds small compared to the speed oflight vxm 1 m so we can write F vx newt 12m smce vx 1 when t 0 m But the as component of velocity is the rate at which n is changing F 1 6 na xt 1 dt m Now the question is can you think of a function of x that has this time de rivative Since the time derivative of t2 is 2t the following formula for xhas the appropriate derivative netx t2 m 2 vxit x1 since as act when t 0 You can check this by taking the derivative with respect to t which gives the equation for 19 since d1t2 dt t and d tdt 1 The average velocity which we seek is the change in position divided by the total time 9quotquot 1Fnetx 1 t 7tvxi vxf vxivxi vavgx where we have used the equation we previously derived for the velocity F newt 1 m vxf vx Simplifying the expression for vampc we have the proof 1 v pct 2 1 only 1f 1 changes at a constant rate FH 15 constant vavg x ec x 212 Points and spheres The gravitational force law applies to objects that are pointlike very small compared to the centertocenter distance between the objects In the sec ond volume of this textbook we will be able to show that any hollow spheri cal shell with uniform density acts gravitationally on external objects as though all the mass of the shell were concentrated at its center The density ofthe Earth is not uniform because the central iron core has higher density than the outer layers But by considering the Earth as layers ofhollow spher ical shells like an onion with each shell of nearly uniform density we get the result that the Earth can be modeled for most purposes as a point mass located at the center of the Earth for very accurate calculations one must take into account small irregularities in the Earth s density from place to place Similar statements can be made about other planets and stars In Fig ure 247 the gravitational force is correctly calculated using the centerto center distance r271 213 Mazxuring le universal gravihztiom mtht G This is not an obvious result After all in Figure 247 some ofthe atoms are closer and some farther apart than the centerctoccenter distance r274 auuiu upall 39 39 39 quot quot is as though the two objecm had collapsed down to poinm at their centers Tquot 39 39 172 orces L L 39 39 39 and is not true for forces that have a different dependence on dismnce 213 Measuring the universal gravitational constant G Inorder in 39 39 r A39t39 A l 39 ena 39 39 39 39 39 39 39 to kno the universal graviuational consmnt G In 179771798 Henry Cavendish performed the first experiment to determine a precise value for G Figure 248 In this kind of experiment a bar with metal s heres at each end is suspended from a thin 39 torsional spring From other measure quired to twist the fiber through a given angle Large balls are brought near the suspended spheres and one the fiber twism due to the gravimtional interactions between the large balls and the small spheres th in kilograms the dismnce in meters and the force in newtons the gravimtional consmnt Ghas been measured in such experiments to e measures how much e masses are measured Z G 67x10 11 N m kgZ TV 39 39 interaction are inherently very weak compared with electromagnetic interactions The only reason that gravitational interactions are significant in our daily lives is that objects interact with t e entire Earth which has a huge mass It m es sensitive measuremenm such as the Cavendish experiment to observe grave itational interactions between two ordinaryrsized objecm 214 The Momentum Principle is valid only in inertial frames Newton39s first law is valid only in an inertial frame ofreference one in unic vasive cosmic m not in a reference frame that is not in uniform motion Let39s check that this is true u view some objecm from a space ship that is moving uniformly with you are concerned For example a rock moving at the same velocity as your spacecraft would have vmk v 7 v5 0 in your reference frame it would appear to be stationary as it coasted along beside your spacecraft With a consmnt spaceship velocity we have Av 0 and the change of momentum of the moving object reduces to the following for speeds small compared to c Await5 Amvedmv5 Amy since Amt5 6 Therefore Amvrv5 Amy Rem Figure 248 The Cavendish experiment Chalzter 2 The Momentum Principle If the velocity v of the space ship doesn t change it represents an inertial frame of reference the form and validity of the Momentum Principle is unaffected by the motion of the space ship However if your space ship increases its speed or changes direction Avs 5 0 an object s motion relative to you changes without any force acting on it In that case the Momentum Principle is not valid for the object be cause you are not in an inertial frame Although the Earth is not an inertial frame because it rotates and goes around the Sun it is close enough to be ing an inertial frame for many everyday purposes 215 Updating position at high speed If 1 ltlt 0 p w my and v pm But at high speed it is more complicated to determine the velocity from the relativistic momentum Here is away to solve for v in terms 0 p 1 l l m J17c2 Irol2 lvl2 Divideb mands uare 7 y q in2 17lec2 5 A2 A2 A 3 Multiply by l 7Ivlc2 LzLL 7jlvlz IvI2 quot 2 Collect terms 1 Jill NW LP mzcz 4392 m2 M Iiim ED2 1 m But since p and 6 are in the same direction we can write this x pm 1 m THE RELATIVISTIC POSITION UPDATE FORMULA A f ti 26 for small At 1 m 2 Note that at low speeds m 7nle and the denominator is l j m l so the formula becomes the familiar if iri p mAt 216 Definitions measurements and units Using the Momentum Principle requires a consistent way to measure length time mass and force and a consistent set ofunits We state the def initions of the stande Systeme Internationale SI units and we briefly dis cuss some subtle issues underlying this choice of units Units meters seconds kilograms and newtons Originally the meterwas defined as the distance between two scratches on a platinum bar in a vault in Paris and a second was 186400th of a mean so 216 Definititms measurements and units lar day Now however the second is defined in terms of the frequency of light emitted by a cesium atom and the meter is defined as the distance light travels in 1299792458th ofa second or about 33X1079 seconds 33 nanoseconds The speed oflight is defined to be exactly 299792458 ms very close to 3X10 ms As a result of these modern redefinitions it is re ally speed oflight and time that are the internationally agreedupon basic units not length and time By international agreement one kilogram is the mass ofa platinum block kept in that same vault in Paris As a practical matter other masses are com pared to this stande kilogram by using a balancebeam or spring weighing scale more about this in a moment The newton the unit of force is de fined as that force which acting for 1 second imparts to l kilogram avelocity change ofl ms We could make a scale for force by calibrating the amount of stretch ofa spring in terms of newtons Some subtle issues What we have just said about SI units is sufficient for practical purposes to predict the motion ofobjects but here are some questions that might both er you ls it legitimate to measure the mass that appears in the Momentum Principle by seeing how that mass is affected by gravity on a balancebeam scale Is it legitimate to use the Momentum Principle to define the units of force when the concept of force is itself associated with the same law Is this all circular reasoning and the Momentum Principle merely a definition with little content Here is a chain of reasoning that addresses these issues Measuring inertial mass When we use balancebeam or spring weighing scales to measure mass what we re really measuring is the gravitational mass that is the mass that ap pears in the law of gravitation and is a measure of how much this object is affected by the gravity of the Earth In principle it could be that this grav itational mass would be different from the socalled inertial mass the mass that appears in the definition ofmomentum It is possible to compare the inertial masses of two objects and we find experimentally that inertial and gravitational mass seem to be entirely equivalent Here is a way to compare two inertial masses directly without involving gravity Starting from rest pull on the first object with a spring stretched by some amounts for an amount of time At and measure the increase of speed Av Then starting from rest pull on the second object with the same spring stretched by the same amounts for the same amount of time At and measure the increase of speed A122 We de ne the ratio of the inertial mass es as mlm A122 A121 Since one of these masses could be the standard kilogram kept in Paris we now have away of measuring inertial mass in ki lograms Having defined inertial mass this way we find experimentally that the Momentum Principle is obeyed by both of these objects in all situations notjust in the one special experiment we used to compare the two masses Moreover we find to extremely high precision that the inertial mass in ki lo rams measured by this comparison experiment is exactly the same as the gravitational mass in kilograms obtained by comparing with a standard kilo gram on a balancebeam scale or using a calibrated spring scale and that it doesn t matter what the objects are made of wood copper glass etc Thisjustifies the convenient use of ordinary weighing scales to determine inertial mass Is this circular reasoning The definitions of force and mass may sound like circular reasoning and the Momentum Principle may sound likejust a kind of de nition with no 99 100 Chapter 2 The Momentum Principle real content but there is real power in the Momentum Principle Forget for a moment the definition offorce in newtons and mass in kilograms The ex perimental fact remains that any object if subjected to a single force by a spring with constant stretch experiences a change of momentum and ve locity proportional to the duration of the interaction Note that it is not a change of position proportional to the time that would be a constant speed but a change of velocity That s real content Moreover we find that the change ofvelocity is proportional to the amount ofstretch of the spring That too is real content Then we nd that a different object undergoes a different rate ofchange of velocity with the same spring stretch but after we ve made one single comparison experiment to determine the mass relative to the stande kilo gram the Momentum Principle works in all situations That s real content Finally come the details of setting standards for measuring force in new tons and mass in kilograms and we use the Momentum Principle in helping set these standards But logically this comes after having established the law itself 21 7 Summary 101 217 Summary Fundamental Physical Principles THE MOMENTUM PRINCIPLE Ap netAt for a short enough time interval At Update form pf pi FnetAt THE SUPERPOSITION PRINCIPLE The net force on an object is the vector sum of the individual forces exerted on it by all other objects Each individual interaction is unaffected by the presence of other interacting objects Major new concepts To apply the Momentum Principle to analyze the motion ofa realworld sys tem several steps are required 1 Choose a system consisting ofa portion of the Universe The rest of the Universe is called the surroundings 2 List objects in the surroundings that exert signi cant forces on the chosen system an A labeled diagram step 3 gives a physics make a labeled diagram showing the external forces view of the situation and it de nes sym exerted by the objects in the surroundings bols to use in writing an algebraic state I ment of the Momentum Princip e 3 Apply the Momentum Pr1nc1ple to the chosen system pf 151 l l netAt For each term in the Momentum Principle substitute any values you know 4 Apply the position update formula if necessary if 3 amt 5 Solve for any remaining unknown quantities ofinterest 6 Check for reasonableness units etc System is a portion of the Universe acted on by the surroundings Fame is a quantitative measure of interactions units are newtons Impulse is the product of force times time lT At momentum change equals net impulse the impulse due to the net force Four fundamental types ofinteraction have been identified 39 gravitational interactions all objects attract each other gravitational 1y 39 electromagnetic interactions electric and magnetic interactions closely related to each other interatomic forces are electric in nature strong interactions inside the nucleus ofan atom 39 weak interactions neutron decay for example Physical models are tractable approximationsidealizations of the real world 1 02 Chapter 2 The Momentum Principle Computer prediction of motion 39 Define the values of constants such as G to use in the program 39 Specify the masses initial positions and initial momenta of the inter acting objects 39 Specify an appropriate value for At small enough that the objects don t move very far during one update 39 Create a loop structure for repetitive calculations 39 Calculate the vector forces acting on each object 39 Update the momentum of each object pf p FnetAt 39 Update the positions if 1 ngAt where 63 m m 39 Repeat Force laws V 2 THE GRAVITATIONAL FORCE LAW 21 m m E 2 1A mzmn Fgmonzbyi r 76739 llzrz1 Fgmv onuyl fax zrzrl 239 lrzrll m1 Fwy on 1 by z r24 points from the center of object 1 to the center of object 2 Figure 249 The gravitational force exerted on object 2 by object ll The force exerted on object 1 by object 2 has the same magni C Is a unlversa l conStant39 G 6397X10 tude but opposite direction 711 N m2 kg2 Near the Earth s surface Ifgruvl mg where g 98 Nkg A THE ELECTRIC FORCE LAW COULOMB39S LAW Feleeonibyl 1272971 4 alrzrll f H 1 L eeeon y 3 21 Almolrzrll2 12 l 9 N m2 1saun1versal constant 9X10 C2 1 4180 4180 where ileum 1 byZ jfzrl Figure 250 The electric force exerted on THE SPRING FORCE LAW MAGNITUDE object 2 by object ll The force exerted on object 2 by object 1 has the same magni lfspringl k 39539 tude but opposite direction 3 s is the stretch s AL L7 L0 relaxed length new length k5 is called the spring stiffness RECIPROCITY for 1 byz 7E0quot 2 byl gravitational and electric forces This is also called Newton s third law of motion Additional new concepts and results Uniformdensity spheres act as though all the mass were at the center AVERAGE VELOCITY SPECIAL CASE L N s i l39 21 vavg N 2 exactly true only 1f 1 ltlt cand 6 changes at a constant rate Fna constant 218 Mien questions 218 Review questions The Momentum Principle RQ 21 At a certain instant a particle is moving in the MC direction with mo mentum 10 kgms During the next 01 s a constant force acts on the par ticle FC 6 N and F3 3 N What is the magnitude of the momentum of the particle at the end of this 01 s interval RQ 22 At t 120 seconds an objectwith mass 2 kg was observed to have a velocity of lt10 35 8gt ms At t 123 seconds its velocity was lt20 30 41gt ms What was the average vector net force acting on the object RQ 23 A proton has mass 17gtlt10727 kg What is the magnitude of the im pulse required to increase its speed from 09900 to 09940 The gravitational force law RQ 24 Calculate the approximate gravitational force exerted by the Earth on a human standing on the Earth s surface Compare with the approxi mate gravitational force of a human on another human at a distance of 3 meters What 1 J 39 39 or 39 1 quotf 39 J 39 must you make See data on the inside back cover Reciprocity RQ 25 The windshield of a speeding car hits a hovering insect Compare the magnitude of the force that the car exerts on the bug to the force that the bug exerts on the car Which is bigger Compare the magnitude of the change of momentum of the bug to that of the car Which is bigger Com pare the magnitude of the change ofvelocity of the bug to that of the car Which is bigger Explain briefly Note the interatomic forces between bug and windshield are electric forces The superposition principle RQ 26 In order to pull a sled across a level field at constant velocity you have to exert aconstant force Doesn t this violate Newton s first and second laws of motion which imply that no force is required to maintain a constant velocity Explain this seeming contradiction 103 104 Chapter 2 The Momentum Principle 219 Problems Problem 21 In the space shuttle A space shuttle is in a circular orbit near the Earth An astronaut floats in the middle of the shuttle not touching the walls On a diagram draw and label a the momentum p1 of the astronaut at this instant b all of the forces if any acting on the astronaut at this instant c the momentum p2 of the astronaut a short time Atlater d the momentum change if any Ap in this time interval e Why does the astronaut seem to oat in the shuttle It is ironic that we say the astronaut is weightless despite the fact that the only force acting on the astronaut is the astronaut s weight that is the grav itational force of the Earth on the astronaut Problem 22 Two books attract each other Two copies of this textbook are standing right next to each other on a book shelf Make arough estimate of the magnitude of the gravitational force that the books exert on each other Explicitly list all quantities that you had to estimate and all simplifications and approximations you had to make to do this calculation Compare your result to the gravitational force on the book by the Earth Problem 23 Compare gravitational and electric forces Use data from the inside back cover to calculate the gravitational and elec tric forces two protons exert on each other when they are 1gtlt10710 m apart about one atomic radius Which interaction between two protons is stron ger the gravitational attraction or the electric repulsion If the two protons are at rest will they begin to move toward each other or away from each oth er Note that since both the gravitational and electric force laws depend on the inverse square distance this comparison holds true at all distances not just at a distance of 1gtlt10710 Problem 24 Crash test In a crash test a truckwith mass 2200 kg traveling at 25 ms about 55 miles per hour smashes headon into a concrete wall without rebounding The front end crumples so much that the truck is 08 m shorter than before What is the approximate magnitude ofthe force exerted on the truck by the wall Explain your analysis carefully and justify your estimates on physical grounds Problem 25 Pingpong ball A pingpong ball is acted upon by the Earth air resistance and a strong wind Here are the positions of the ball at several times Early time interval At t1235 s the position was lt 317 254 938 gt m At t1237 s the position was lt 325 250 940 gt m Late time interval At t1435 s the position was lt 1125 150 1140 gt m At t1437 s the position was lt 1127 186 1142 gt m a In the early time interval from t 1235 s to t 1237 s what was the average momentum of the ball The mass of the pingpong ball is 27 grams 27gtlt1073 kg Express your result as a vector b In the late time interval from t 1435 s to t 1437 s what was the average momentum of the ball Express your result as a vector c In the time interval from t 1235 s the start of the early time inter val to t 1435 s the start of the late time interval what was the average net force acting on the ball Express your result as a vector 2 I 9 Problems Problem 26 Kick a basketball A 06 kg basketball is rolling by you at 35 ms As it goes by you give it a kick perpendicular to its path Figure 251Your foot is in contactwith the ball for 0002 s The ball eventually rolls at a 20 angle from its original di rection The overhead view is approximately to scale The arrow represents the force your toe applies brie y to the basketball a 1n the diagram which letter corresponds to the correct overhead view of the ball s path b Determine the magnitude of the average force you applied to the ball Problem 27 Spacecraft and asteroid At t 5320 s after midnight a spacecraft of mass 1400 kg is located at posi tion 3x105 7x10574gtlt10 gt m and an asteroid ofmass 7gtlt10 kg is locat ed at position lt9X10573gtlt105712gtlt105gt m There are no other objects nearby a Calculate the vector force acting on the spacecraft b At t 5320 s the spacecraft s momentum was pi and at the later time t 5380 s its momentum was pf Calculate the vector change of momentum fe pi Problem 28 Proton and HCl molecule A proton interacts electricallywith a neutral HCl molecule located at the or igin At a certain time t the proton s position is l6gtlt107 0 0 m and the proton s velocity is 3200 800011 ms The force exerted on the proton by the HCl molecule is 7112gtlt10 0 0 N At a time t 2gtlt10 s what is the approximate velocity of the proton Problem 29 A free throw in basketball Determine two different possible ways for a player to make a free throw in basketball In both cases give the initial speed initial angle and initial height of the basketball The rim of the basket is 10 feet 30 m above the oor It is 14 feet 43 m along the oor from the freethrow line to apoint directly below the center of the basket Problem 210 A basketball pass You have probably seen a basketball player throw the ball to a teammate at the other end of the court 30 m away Estimate a reasonable initial angle for such a throw and then determine the corresponding initial speed For your chosen angle how long does it take for the basketball to go the length of the court What is the highest point along the trajectory relative to the thrower s hand Problem 211 The case of the falling flower pot You are a detective investigating why someone was hit on the head by a fall ing flowerpot One piece of evidence is a home video taken in a 4th oor apartment which happens to show the flowerpot falling past a tall window lnspection ofindividual frames of the video shows that in a span of 6 frames the flowerpot falls adistance that corresponds to 085 of the window height seen in the video note standard video runs at a rate of 30 frames per sec ond You visit the apartment and measure the window to be 22 m high What can you conclude Under what assumptions Give as much detail as you can Problem 212 Tennis ball hits wall A tennis ball has a mass of0057 kg A professional tennis player hits the ball hard enough to give it a speed of 50 ms about 120 miles per hour The ball hits a wall and bounces back with almost the same speed 50 ms As indicated in Figure 252 highspeed photography shows that the ball is 105 Figure 251 Kick a basketball Problem 2 l l I l Figure 252 A highspeed tennis ball deforms when it his a wall Problem 212 106 I Figure 253 A cathode 215 Chapter 2 The Momentum Principle crushed 2 cm 002 m at the instant when its speed is momentarily zero be fore rebounding Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact determine the approximate magnitude of this force Hint Think about the approximate amount of time it takes for the ball to come momentarily to rest For com parison note that the gravitational force on the ball is quite small only about 0057 kg98 Nkg m 06 N Aforce of5 N force is approximately the same as a force of one pound Problem 213 Mars probe small space probe of mass 240 kg is launched from a spacecraft near Mars lt travels toward the surface ofMars where it will land At a time 207 seconds after it is launched the probe is at the location 430X103 870gtlt102 0 m and at this same time its momentum is lt440x104 7760x103 0gt kg ms At this instant the net force on the probe due to the gravitationalzpull ofMars plus the air resistance acting on the probe is lt77x103 792gtlt10 0gt N a Assuming that the net force on the probe is approximately constant over this time interval what is the momentum of the probe 209 seconds af ter it is launche b What is the location of the probe 209 seconds after launch Problem 214 Spacecraft navigation Suppose you are navigating a spacecraft far from other objects The mass of the spacecraft is l5gtlt105 kg about 150 tons The rocket engines are shut off and you re coasting along with a constant velocity of 0 20 0 kms As you pass the location 12 15 0 km you briefly fire side thruster rockets so that your spacecraft experiences a net force of 6x104 0 0 N for 34 s The ejected gases have a mass that is small compared to the mass of the space craft You then continue coasting with the rocket engines turned off Where are you an hour later Also what approximations or simplifying assump tions did you have to make in your analysis Think about the choice of sys tem what are the surroundings that exert external forces on your system Problem 215 Electron motion in a CRT a cathode ray tube CRT used in oscilloscopes and televi sion sets a beam of electrons is steered to different places on a f I 39 phosphor screen which glows at locations hit by electrons Fig ure 253 The CRT is evacuated so there are few gas molecules present for the electrons to run into Electric forces are used to accelerate electrons of mass m to a speed 120 ltlt 0 after which they pass between positively and negatively charged metal plates which deflect the electron in the vertical direction up ward in Figure 253 or downward if the sign of the charges on the plates is reversed L quot While an electron is between the plates it experiences a uni form vertical force E butwhen the electron is outside the plates there is negligible force on it The gravitational force on the electron is negligibly small compared to the electric force in this situation The length of the metal plates is d and the phos phor screen is a distance L from the metal plates Where does the electron hit the screen That is what is yf ray tube Problem 2 I 9 Problems Problem 216 Star and planet A star of mass 7x1030 kg is located at 5x10122gtlt10120gt In A planet of mass 3X1024 kg is located at 3X1012 4X10 0 m and is moving with ave locityof03X10415gtlt1040gtms a At a time 1gtlt10 seconds later what is the new velocity of the planet b Where is the planet at this later time c Explain brie y why the procedures you followed in parts a and b were able to produce usable results but wouldn twork if the later time had been 1gtlt10 seconds instead of 1gtlt10 seconds after the initial time Explain brie y how could you use a computer to get around this difficulty Problem 217 Two stars At t 0 a star of mass 4x1030 kg has velocity 7x104 6x104 78x104gt ms and is located at lt200x1012e500x1012 400gtlt1012gt In relative to the cen ter ofa cluster ofstars There is only one nearby star that exerts a significant force on the first star The mass of the second star is 3X10 kg its velocity is 2x10471gtlt104 9x104 ms and this second star is located at 203x1012 7494x1012 395x1012gt rh relative to the center of the cluster of stars a At t 1gtlt105 s what is the approximate momentum of the first star b Discuss brie y some ways in which your result for a is approximate not exact c At t 1gtlt105 s what is the approximate position of the rst star d Discuss brie y some ways in which your result for b is approximate not exact Problem 218 The SLAC twomile accelerator SLAC the Stanford Linear Accelerator Center located at Stanford Univer sity in Palo Alto California accelerates electrons through a vacuum tube two miles long it can be seen from an overpass of the Junipero Serra free way that goes right over the accelerator Electrons which are initially at rest are subjected to a continuous force of 2gtlt1071 newton along the entire length of two miles one mile is 16 kilometers and reach speeds very near the speed of light a Determine how much time is required to increase the electrons speed from 0930to 0990 That is the quantity IvIc increases from 093 to 099 b Approximately how far does the electron go in this time What is ap proximate about your result Problem 219 Determining the mass of an asteroid anune 1997 the NEAR spacecraft Near Earth Asteroid Rendezvous see httpnearjhuapledu on its way to photograph the asteroid Eros passed within 1200 km of asteroid Mathilde at a speed of 10 kms relative to the asteroid Figure 254 From photos transmitted by the 805 kg space craft Mathilde s size was known to be about 70 km by 50 km by 50 km It is presumably made of rock Rocks on Earth have a density of about 3000 kgm3 3 gramscmg a Make a rough diagram to show qualitatively the effect on the space craft of this encounter with Mathilde Explain your reasoning b Make avery rough estimate of the change in momentum of the space craft that would result from encountering Mathilde Explain how you made your estimate c Using your result from part b make a rough estimate ofhow far off course the spacecraft would be one day after the encounter d From actual observations of the location of the spacecraft one day af ter encountering Mathilde scientists concluded that Mathilde is a loose ar 107 space craft I 10 kms 1200 km lt 7 Mathilde Figure 254 The NEAR spacecraft passes the asteroid Mathilde Problem 219 108 Chapter 2 The Momentum Principle rangement of rocks with lots of empty space inside What was it about the observations that must have led them to this conclusion Experimental background The position was tracked by very accurate measurements of the time that it takes for a radio signal to go from Earth to the spacecraft followed immediately by a radio response from the spacecraft being sent back to Earth Radio signals like light travel at a speed of 3gtlt108 ms so the time measurements had to be accurate to a few nanoseconds 1 ns 10 9 s The following problems are intended to introduce you to using a computer to model matter interactions and motion Some parts of these problems can be done with almost any tool spreadsheet math package etc Other parts are most easily done with a programming language for which we rec ommend the free 3D programming language VPython httpvpy thonorg Your instructor will introduce you to an available computational tool and assign problems or parts of problems that can be addressed using the chosen tool Problem 220 Planetary orbits In this problem you will study the motion of a planet around a star To start with a somewhat familiar situation you will begin by modeling the motion of our Earth around our Sun Write answers to questions either as com ments in your program or on paper as specified by your instructor Planning a The Earth goes around the Sun in a nearly circular orbit taking one year to go around Using data on the inside back cover what initial speed should you give the Earth in a computer model so that a circular orbit should result If your computer model does produce a circular orbit you have a strong indication that your program is working properly b Estimate an appropriate value for At to use in your computer model Remember that tis in seconds and consider your answer to part a in mak ing this estimate If your At is too small your calculation will require many tedious steps Explain briefly how you decided on an appropriate step size Circular orbit c Starting with speed calculated in a and with the initial velocity per pendicular to a line connecting the Sun and the Earth calculate and display the trajectory of the Earth Display the whole trail so you can see whether you have a closed orbit that is whether the Earth returns to its starting point each time around Include the Sun s position it ltx5 yl z5gt in your computations even ifyou set its coordinates to zero This will make it easier to modify the computation to let the Sun move Problem 2 Binary stars Computational acer d As a check on your computation is the orbit a circle as expected Run the calculation until the accumulated time tis the length ofayear does this take the Earth around the orbit once as expected What is the largest step size you can use that still gives a circular orbit and the correct length of a year What happens if you use a much larger step size Noncircnlar trajectories e Set the initial speed to 12 times the Earth s actual speed Make sure the step size is small enough that the orbit is nearly unaffected ifyou cut the step size What kind of orbit do you get What step size do you need to use to get results you can trust What happens ifyou use a much larger step size Produce at least one other qualitatively different noncircular trajectory for 2 I 9 Problems the Earth What difficulties would humans have in surviving on the Earth if it had a highly noncircular orbit Force and mommtum f Choose initial conditions that give a noncircular orbit Continuously display a vector showing the momentum of the Earth with its tail at the Earth s position and a different colored vector showing the force on the Earth by the Sun with its tail at the Earth s position You must scale the vectors appropriately to fit into the scene Away to gure out what scale fac tor to use is to print the numerical values of momentum and force and compare with the scale of the scene For example if the width of the scene is Wmeters and the force vector has a typical magnitude F in newtons you might scale the force vector by a factor 01 WE which would make the length of the vector be onetenth the width of the scene Is the force in the same direction as the momentum How does the momentum depend on the distance from the star Problem 221 Binary stars About a half of the visible stars are actually systems consisting of two stars orbiting each other called binary stars In your computer model of the Earth and Sun Problem 220 Planetary orbits replace the Earth with a star whose mass is half the mass of our Sun and take into account the grav itational effects that the second star has on the Sun a Give the second star the speed of the actual Earth and give the Sun zero initial momentum What happens Try a variety of other initial condi tions What kinds of orbits do you find b One of the things to try is to give the Sun the same magnitude of mo mentum as the second star but in the opposite direction If the initial mo mentum of the second star is in the y direction give the Sun the same magnitude of momentum but in the y direction so that the total momen tum of the binary star system is initially zero but the stars are not headed toward each other What is special about the motion you observe in this case Problem 222 Other force laws Modify your orbit computation to use a different force law such as a force that is proportional to lr or 173 or a constant force or a force propor tional to 72 this represents the force of a spring whose relaxed length is nearly zero How do orbits with these force laws differ from the circles and ellipses that result from a 11392 law If you want to keep the magnitude of the force roughly the same as before you will need to adjust the force con stant G Problem 223 The effect of the Moon and Venus on the Earth In Problem 220 you analyzed a simple model of the Earth orbiting the Sun in which there were no other planets or moons Venus and the Earth have similar size and mass At its closest approach to the Earth Venus is about 40 million kilometers away 4X10 0 m The Moon s mass is about 7gtlt1022 kg and the distance from Earth to Moon is about 4gtlt108 m 400000 km center to center a Calculate the ratio of the gravitational forces on the Earth exerted by Venus and the Sun Is it a good approximation to ignore the effect ofVenus when modeling the motion of the Earth around the Sun b Calculate the ratio of the gravitational forces on the Earth exerted by the Moon and the Sun Is it a good approximation to ignore the effect of the Moon when modeling the motion of the Earth around the Sun 109 110 Figure 255 The first Ranger 7 photo of the Moon NASA photograph Chapter 2 The Momentum Pn39mipk Problem 224 The threebody problem Cany out a numerical integration of the motion of a threebody gravimtion al system and plot the traj ectory leaving trails behind the objects Calculate all of the forces before using these forces to update the momenm and posi tions of the objects Otherwise the calculations of gravimtional forces would mix positions corresponding to different times Ti quot 39 39 39 1 39 39 ini al momenta Findat initial conditions that produces a longlasting orbit one set of initial condi tions that results in a collision 39th a massive object and one set of initial conditions that allows one of the objects to wander off without returning Report the masses and initial conditions that you used Problem 225 The Ranger 7 mission to the Moon The rst US spacecraft to photograph the Moon close up was the unmanned Ranger 7 photographic mission in 1964 The spacecraft shown in the illus tration at the right NASA photograph contained television cameras that trans mitted closeup pictures of the Moon back to Earth as the spacecraft ap proached the Moon The spacecraft did not have retrorockets to slow itself down and it eventually simply crashed onto the Moon s surface transmitting its last pho tos immediately before im act Figure 255 is the rst image of the Moon mken by a US spacecraft Rang er 7 onjuly 31 1964 about 17 minutes before impacting the lunar surface e large crater at center right is Alphonsus 108 km diameter above it and to the right is Ptolernaeus and below it is Arzachel The Ranger 7 im pact site is off the frame to the left of the upper left corner You can nd out more about the actual Ranger lunar missions at http nssdcgsfcnasagovplanetaIylunarrangerhtml r To send a spacecraft to the Moon we put it on top ofa large rocket contain ing lots of rocket fuel and re it upward At first the huge ship moves quite slowly but the speed increases rapidly When the rststage portion of the rocket has aghausted its fuel and is empty it is discarded and falls back to Earth By discarding an empty rocket smge we decrease the amount of mass that must be accelerated to even higher speeds There may be several stages that operate for a while and then are discarded before the spacecraft has ris en above most of Earth s atmosphere about 50 km say above the Earth and has acquired a high speed At that point all the fuel available for this mission has been used up and the spacecraft simply coasts toward the Moon through the vacuum of space We will model the Ranger 7 mission Smting 50 km above the Earth s sur face 5gtlt10i1 m a spacecraft coasts toward the Moon with an initial speed of about 104 ms Here are data we will need mass of spacecraft 17 kg mass ofEarthm 6gtltlOzi1 kg mass of Moon o 7x1022 kg radius of Moon 175x106 m dismnce from Earth to Moon m 4X108 m 400000 km center to center We re going to ignore the Sun in a simpli ed model even though it exerts a sizable gravitational force We re expecting the Moon mission to mke onl a few days during which time the Earth and Moon move in a nearly straight line with respect to the Sun because it mkes 365 days to go all the way around the Sun We take a reference frame xed to the Earth as repre 2 I 9 Problems senting approximately an inertial frame of reference with respect to which we can use the Momentum Principle For a simple model make the Earth and Moon be fixed in space during the mission Factors that would certainly influence the path of the space craft include the motion of the Moon around the Earth and the motion of the Earth around the Sun In addition the Sun and other planets exert gravitational forces on the spacecraft As a separate project you might like to include some of these additional factors a Compute the path of the spacecraft and display it either with a graph or with an animated image Here and in remaining parts 0 the problem report the step size At that gives accurate results that is cutting this step size has little effect on the results b By trying various initial speeds determine the approximate minimum launch speed needed to reach the Moon to two significant figures this is the speed that the spacecraft obtains from the multistage rocket at the time of release above the Earth s atmosphere What happens if the launch speed is less than this minimum value Be sure to check the step size issue c Use a launch speed 10 larger than the approximate minimum value found in part b How long does it take to go to the Moon in hours or days Be sure to check the step size issue e What is the impact speed of the spacecraft its speedjust before it hits the Moon s surface Make sure that your spacecraft crashes on the sur face of the Moon not at the Moon s center Be sure to check the step size issue You may have noticed that you don t actually need to know the mass m of the spacecraft in order to carry out the computation The gravitational force is proportional to m and the momentum is also proportional to m so m cancels However nongravitational forces such as electric forces are not proportional to the mass and there is no cancellation in that case We kept the mass m in the analysis in order to illustrate a general technique for pre dicting motion no matter what kind of force gravitational or not Problem 226 The effect of Venus on the Moon voyage In the Moon voyage analysis you used a simplified model in which you ne glected among other things the effect of Venus An important aspect of physical modeling is making estimates of how large the neglected effects might be lfwe take Venus into account make a rough estimate ofwhether the spacecraft will miss the Moon entirely How large a sideways deflection of the crash site will there be Explain your reasoning and approximations Venus and the Earth have similar size and mass At its closest approach to the Earth Venus is about 40 million kilometers away 4X1010 m In the real world Venus would attract the Earth and the Moon as well as the space craft but to get an idea of the size of the effects pretend that the Earth the Moon and Venus are all xed in position Figure 256 andjust investigate Venus s effect on the spacecraft 111 Mo on a 3 Venus Ea h Figure 256 The relative positions of Venus Earth and Moon in Problem 226 112 August 21 2005 847 pm Chapter 2 The Momentum Principle 220 Answers to exercises 21 page 55 22 page 55 23 page 55 24 page 56 25 page 58 26 page 58 27 page 58 28 page 59 29 page 63 210 page 63 211 page 63 212 page 63 213 page 63 214 page 67 215 page 67 216 page 68 217 page 72 218 page 82 219 page 82 220 page 82 221 page 86 222 page 90 223 page 90 225 page 91 226 page 92 0275 N 3571 NIn 714 N 760 724 96 Ns 7003 7005 002 N Earth water and air exert forces net force is zero zero glider track spring Earth air pf 10 011 kgIns A fe t 71500003000 kgIns impulse by ground a 7500001000 N a lt71x104 0 0 N pf 94 03 0 kgIns a 710 712 75 Ins b 3 6036 78 In c 86 In d 133 s e 266 s 1 7175 0 7183 In y h 1211 0yf 0At l2h 12M l2gh SaIne results if mass is changed time in air is 177 5 horizontal distance 88 In 38 In high y 0108 In 1 00365 kgIns 316 times as large 710 25gtlt10 710 N 25gtlt10 N 6x1024 kg a 28x108 0 728x108gt m b 396x108 m c 0707 070707 no units diInensionless d 7127x10200127gtlt1020gt N about a day or less a small portion of the circle about 002 s a small fraction of the roundtrip tiIne 32gtlt104 In note that Mt Everest is about 8gtlt103 In tall 28x10 8 N
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