Impulse and Momentum in Physics PDF

MOMENTUM, IMPULSE, AND COLLISIONS 8 LEARNING GOALS...

MOMENTUM, IMPULSE, AND COLLISIONS 8 LEARNING GOALS By studying this chapter, you will learn: The meaning of the momentum of a particle, and how the impulse of the net force acting on a particle causes its momentum to change. The conditions under which the total momentum of a system of particles is constant (conserved). Which could potentially do greater damage to this carrot: a.22-caliber bullet ? moving at 220 m> s as shown here, or a lightweight bullet of the same length How to solve problems in which two bodies collide with each other. and diameter but half the mass moving at twice the speed? The important distinction among elastic, inelastic, and completely here are many questions involving forcesS that cannot be answered by T inelastic collisions. S directly applying Newton’s second law, gF ⴝ ma. For example, when a The definition of the center of mass moving van collides head-on with a compact car, what determines which of a system, and what determines way the wreckage moves after the collision? In playing pool, how do you decide how the center of mass moves. how to aim the cue ball in order to knock the eight ball into the pocket? And How to analyze situations such as when a meteorite collides with the earth, how much of the meteorite’s kinetic rocket propulsion in which the mass energy is released in the impact? of a body changes as it moves. A common theme of all these questions is that they involve forces about which we know very little: the forces between the car and the moving van, between the two pool balls, or between the meteorite and the earth. Remarkably, we will find in this chapter that we don’t have to know anything about these forces to answer questions of this kind! Our approach uses two new concepts, momentum and impulse, and a new con- servation law, conservation of momentum. This conservation law is every bit as important as the law of conservation of energy. The law of conservation of momentum is valid even in situations in which Newton’s laws are inadequate, such as bodies moving at very high speeds (near the speed of light) or objects on a very small scale (such as the constituents of atoms). Within the domain of Newtonian mechanics, conservation of momentum enables us to analyze many situations that would be very difficult if we tried to use Newton’s laws directly. Among these are collision problems, in which two bodies collide and can exert very large forces on each other for a short time. 8.1 Momentum and Impulse S S In Chapter 6 we re-expressed Newton’s second law for a particle, gF ⴝ ma , in terms of the work–energy theorem. This theorem helped us tackle a great number of physics problems S and led us to the law of conservation of energy. Let’s now S return to gF ⴝ ma and see yet another useful way to restate this fundamental law. 241 242 CHAPTER 8 Momentum, Impulse, and Collisions Newton’s Second Law in Terms of Momentum Consider a particle of constant mass m. (Later in this chapter we’ll see how to S S deal with situations in which the mass of a body changes.) Because a ⴝ d v>dt, we can write Newton’s second law for this particle as S S dv d S gF ⴝ m ⴝ 1mv2 (8.1) dt dt We can move the mass m inside the derivative S because it is constant. Thus New- ton’s second law says that the net force gF acting on a particle equals the time S rate of change of the combination mv, the product of the particle’s mass and velocity. We’ll call this combination the momentum, or linear momentum, of S the particle. Using the symbol p for momentum, we have S S p ⴝ mv (definition of momentum) (8.2) 8.1 The velocity and momentum vectors The greater the mass m and speed v of a particle, the greater is its magnitude of of a particle. momentum mv. Keep in mind, however, that momentum is a vector quantity with y the same direction as the particle’s velocity (Fig. 8.1). Hence a car driving north at 20 m>s and an identical car driving east at 20 m>s have the same magnitude of S S v momentum 1mv2 but different momentum vectors 1mv2 because their directions m S S are different. p 5 mv We often express the momentum of a particle in terms of its components. x If the particle has velocity components vx , vy , and vz , then its momentum compo- O nents px , py , and pz (which we also call the x-momentum, y-momentum, and S Momentum p is a vector quantity; z-momentum) are given by a particle’s momentum has the same S direction as its velocity v. px = mvx py = mvy pz = mvz (8.3) These three component equations are equivalent to Eq. (8.2). The units of the magnitude of momentum are units of mass times speed; the SI units of momentum are kg # m>s. The plural of momentum is “momenta.” If we now substitute the definition of momentum, Eq. (8.2), into Eq. (8.1), we get S S dp gF ⴝ (Newton’s second law in terms of momentum) (8.4) dt 8.2 If a fast-moving automobile stops suddenly in a collision, the driver’s momentum (mass times velocity) changes The net force (vector sum of all forces) acting on a particle equals S the time S from a large value to zero in a short time. rate of change of momentum of the particle. This, not gF ⴝ ma , is the An air bag causes the driver to lose form in which Newton originally stated his second law (although he called momentum more gradually than would an momentum the “quantity of motion”). This law is valid only in inertial frames abrupt collision with the steering wheel, reducing the force exerted on the driver as of reference. well as the possibility of injury. According to Eq. (8.4), a rapid change in momentum requires a large net force, while a gradual change in momentum requires less net force. This principle is used in the design of automobile safety devices such as air bags (Fig. 8.2). The Impulse–Momentum Theorem S S A particle’s momentum p ⴝ mv and its kinetic energy K = 12 mv2 both depend on the mass and velocity of the particle. What is the fundamental difference between these two quantities? A purely mathematical answer is that momentum is a vector whose magnitude is proportional to speed, while kinetic energy is a scalar proportional to the speed squared. But to see the physical difference between momentum and kinetic energy, we must first define a quantity closely related to momentum called impulse.  8.1 Momentum and Impulse 243 S Let’s first consider a particle acted on by a constant net force gF during a Application Woodpecker Impulse The pileated woodpecker (Dryocopus pileatus) time interval ¢t from t 1 to t 2. (We’ll look at the case of varying forces shortly.) has been known to strike its beak against a S The impulse of the net force, denoted by J , is defined to be the product of the net tree up to 20 times a second and up to 12,000 times a day. The impact force can be force and the time interval: as much as 1200 times the weight of the bird’s head. Because the impact lasts such a S S S J ⴝ gF1t 2 - t 12 ⴝ gF ¢t (assuming constant net force) (8.5) short time, the impulse—the product of the net force during the impact multiplied by the duration of the impact—is relatively small. (The S Impulse is a vector quantity; its direction is the same as the net force gF. Its woodpecker has a thick skull of spongy bone as well as shock-absorbing cartilage at the magnitude is the product of the magnitude of the net force and the length of time that the net force acts. The SI unit of impulse is the newton-second 1N # s2. base of the lower jaw, and so avoids injury.) Because 1 N = 1 kg # m>s2, an alternative set of units for impulse is kg # m>s, the same as the units of momentum. To see what impulse is good for, let’s go back to Newton’s S second law as restated in terms of momentum, Eq. (8.4). If the net force gF is constant, then S S d p >dt is also constant. In that case, d p >dt is equal to the total change in momen- S S tum p 2 ⴚ p 1 during the time interval t 2 - t 1 , divided by the interval: S S S p2 ⴚ p1 gF ⴝ t2 - t1 Multiplying this equation by 1t 2 - t 12, we have S S S gF1t 2 - t 12 ⴝ p 2 ⴚ p 1 Comparing with Eq. (8.5), we end up with a result called the impulse–momentum theorem: S S S J ⴝ p2 ⴚ p1 (impulse–momentum theorem) (8.6) 8.3 The meaning of the area under a The change in momentum of a particle during a time interval equals the impulse graph of gFx versus t. of the net force that acts on the particle during that interval. (a) The impulse–momentum theorem also holds when forcesS are not constant. To The area under the curve of net force versus S see this, we integrate both sides of Newton’s second law gF ⴝ d p >dt over time time equals the impulse of the net force: t2 between the limits t 1 and t 2 : ΣFx Area 5 Jx 5 1oFx dt t1 S t2 p2 t2 S We can also calculate the dpS S S S gF dt ⴝ dt ⴝ S d p ⴝ p 2 ⴚ p 1 impulse by replacing the Lt1 Lt1 dt Lp1 varying net force with an S S average net force: The integral on the left is defined to be the impulse J of the net force gF during (Fav)x Area 5 Jx this interval: 5 (Fav)x (t2 2 t1) t2 S S t Jⴝ gF dt (general definition of impulse) (8.7) t1 t2 Lt1 t2 ⫺ t1 S S S With this definition, the impulse–momentum S theorem J ⴝ p 2 ⴚ p 1 , Eq. (8.6), is (b) valid even when the net force gF variesSwith time. S We can define an S average net force Fav such that even when gF is not con- ΣFx Large force that acts stant, the impulse J is given by for a short time S S J ⴝ Fav1t 2 - t 12 (8.8) The area under both curves is the same, so both forces S S S deliver the same impulse. When gF is constant, gF ⴝ Fav and Eq. (8.8) reduces to Eq. (8.5). Figure 8.3a shows the x-component of net force gFx as a function of time Smaller force that during a collision. This might represent the force on a soccer ball that is in con- acts for a longer time tact with a player’s foot from time t 1 to t 2. The x-component of impulse during this interval is represented by the red area under the curve between t 1 and t 2. This t 244 CHAPTER 8 Momentum, Impulse, and Collisions area is equal to the green rectangular area bounded by t 1 , t 2 , and 1Fav2x , so 1Fav2x 1t 2 - t 12 is equal to the impulse of the actual time-varying force during the same interval. Note that a large force acting for a short time can have the same impulse as a smaller force acting for a longer time if the areas under the force–time curves are the same (Fig. 8.3b). In this language, an automobile airbag (see Fig. 8.2) provides the same impulse to the driver as would the steering wheel or the dash- board by applying a weaker and less injurious force for a longer time. Impulse and momentum are both vector quantities, and Eqs. (8.5)–(8.8) are all vector equations. In specific problems, it is often easiest to use them in compo- nent form: t2 Jx = gFx dt = 1Fav2x 1t 2 - t 12 = p2x - p1x = mv2x - mv1x Lt1 (8.9) t2 Jy = gFy dt = 1Fav2y 1t 2 - t 12 = p2y - p1y = mv2y - mv1y Lt1 and similarly for the z-component. Momentum and Kinetic Energy Compared We can now see the fundamental difference S between momentum and kinetic S S energy. The impulse–momentum theorem J ⴝ p 2 ⴚ p 1 says that changes in a particle’s momentum are due to impulse, which depends on the time over which the net force acts. By contrast, the work–energy theorem Wtot = K 2 - K 1 tells us that kinetic energy changes when work is done on a particle; the total work depends on the distance over which the net force acts. Consider a particle that S S S starts from rest at t 1 so that v1 ⴝ 0. Its initial momentum is p 1 ⴝ mv1 ⴝ 0, and 1 its initial kinetic energy is K 1 = 2 mv1 = 0. Now let a constant net force equal to 2 S F act on that particle from time t 1 until time t 2. During this interval, the particle ActivPhysics 6.1: Momentum and Energy moves a distance s in the direction of the force. From Eq. (8.6), the particle’s Change momentum at time t 2 is S S S S p2 ⴝ p1 ⴙ J ⴝ J S S where J ⴝ F1t 2 - t 12 is the impulse that acts on the particle. So the momentum of a particle equals the impulse that accelerated it from rest to its present speed; impulse is the product of the net force that accelerated the particle and the time required for the acceleration. By comparison, the kinetic energy of the particle at 8.4 The kinetic energy of a pitched base- ball is equal to the work the pitcher does t 2 is K 2 = Wtot = Fs, the total work done on the particle to accelerate it from on it (force multiplied by the distance rest. The total work is the product of the net force and the distance required to the ball moves during the throw). The accelerate the particle (Fig. 8.4). momentum of the ball is equal to the impulse the pitcher imparts to it (force multiplied by the time it took to bring Here’s an application of the distinction between momentum and kinetic energy. Suppose you have a choice between catching a 0.50-kg ball moving at 4.0 m>s or a 0.10-kg ball moving at 20 m>s. Which will be easier to catch? Both ? the ball up to speed). balls have the same magnitude of momentum, p = mv = 10.50 kg214.0 m>s2 = Net force ΣF S 10.10 kg2120 m>s2 = 2.0 kg # m>s. However, the two balls have different values of kinetic energy K = 12 mv2; the large, slow-moving ball has K = 4.0 J, while the small, fast-moving ball has K = 20 J. Since the momentum is the same for S Displacement s in time Dt both balls, both require the same impulse to be brought to rest. But stopping the 0.10-kg ball with your hand requires five times more work than stopping the 0.50-kg Kinetic energy gained ball because the smaller ball has five times more kinetic energy. For a given force S S by ball 5 ΣF · s that you exert with your hand, it takes the same amount of time (the duration of S the catch) to stop either ball, but your hand and arm will be pushed back five times Momentum gained by ball 5 ΣF Dt farther if you choose to catch the small, fast-moving ball. To minimize arm strain, you should choose to catch the 0.50-kg ball with its lower kinetic energy. Both the impulse–momentum and work–energy theorems are relationships between force and motion, and both rest on the foundation of Newton’s laws. They are integral principles, relating the motion at two different times separated  8.1 Momentum and Impulse 245 bySa finite interval. S By contrast, Newton’s second law itself (in either of the forms S S gF ⴝ ma or gF ⴝ d p >dt) is a differential principle, relating the forces to the rate of change of velocity or momentum at each instant. Conceptual Example 8.1 Momentum versus kinetic energy Consider again the race described in Conceptual Example 6.5 it from rest. As in Conceptual Example 6.5, the net force on each S (Section 6.2) between two iceboats on a frictionless frozen lake. boat equals the constant horizontal wind force F. Let ¢t be the The boats have masses mS and 2m, and the wind exerts the same time a boat takes to reachS the Sfinish line, so that the impulse on the constant horizontal force F on each boat (see Fig. 6.14). The boats boat during that time is J ⴝ F ¢ t. Since the boat starts from rest, S start from rest and cross the finish line a distance s away. Which this equals the boat’s momentum p at the finish line: boat crosses the finish line with greater momentum? S S p ⴝ F ¢t SOLUTION S In Conceptual Example 6.5 we asked how the kinetic energies of Both boats are subjected to the same force F, but they take dif- the boats compare when they cross the finish line. We answered ferent times ¢t to reach the finish line. The boat of mass 2m accel- this by remembering that a body’s kinetic energy equals the total erates more slowly and takes a longer time to travel the distance s; work done to accelerate it from rest. Both boats started from rest, thus there is a greater impulse on this boat between the starting and and the total work done was the same for both boats (because the finish lines. So the boat of mass 2m crosses the finish line with a net force and the displacement were the same for both). Hence greater magnitude of momentum than the boat of mass m (but with both boats had the same kinetic energy at the finish line. the same kinetic energy). Can you show that the boat of mass 2m Similarly, to compare the momenta of the boats we use the idea has 12 times as much momentum at the finish line as the boat of that the momentum of each boat equals the impulse that accelerated mass m? Example 8.2 A ball hits a wall You throw a ball with a mass of 0.40 kg against a brick wall. It EXECUTE: (a) With our choice of x-axis, the initial and final hits the wall moving horizontally to the left at 30 m>s and x-components of momentum of the ball are p1x = mv1x = 10.40 kg21-30 m>s2 = -12 kg # m>s rebounds horizontally to the right at 20 m>s. (a) Find the p2x = mv2x = 10.40 kg21+20 m>s2 = +8.0 kg # m>s impulse of the net force on the ball during its collision with the wall. (b) If the ball is in contact with the wall for 0.010 s, find the average horizontal force that the wall exerts on the ball dur- From the x-equation in Eqs. (8.9), the x-component of impulse ing the impact. equals the change in the x-momentum: Jx = p2x - p1x SOLUTION = 8.0 kg # m>s - 1-12 kg # m>s2 = 20 kg # m>s = 20 N # s IDENTIFY and SET UP: We’re given enough information to deter- mine the initial and final values of the ball’s momentum, so we can (b) The collision time is t 2 - t 1 = ¢t = 0.010 s. From the use the impulse–momentum theorem to find the impulse. We’ll x-equation in Eqs. (8.9), Jx = 1Fav2x1t 2 - t 12 = 1Fav2x ¢t, so 20 N # s then use the definition of impulse to determine the average force. Jx Figure 8.5 shows our sketch. We need only a single axis because 1Fav2x = = = 2000 N ¢t 0.010 s the motion is purely horizontal. We’ll take the positive x-direction to be to the right. In part (a) our target variable is the EVALUATE: The x-component of impulse Jx is positive—that is, to x-component of impulse, Jx, which we’ll find from the x-components the right in Fig. 8.5. This is as it should be: The impulse represents of momentum before and after the impact, using Eqs. (8.9). In part (b), the “kick” that the wall imparts to the ball, and this “kick” is cer- our target variable is the average x-component of force 1Fav2x; once tainly to the right. we know Jx, we can also find this force by using Eqs. (8.9). CAUTION Momentum is a vector Because momentum is a vector, we had to include the negative sign in writing p1x = -12 kg # m>s. Had we carelessly omitted it, we would have calcu- 8.5 Our sketch for this problem. lated the impulse to be 8.0 kg # m>s - 112 kg # m>s2 = -4 kg # m>s. This would say that the wall had somehow given the ball a kick to the left! Make sure that you account for the direction of momentum in your calculations. ❙ The force that the wall exerts on the ball must have such a large magnitude (2000 N, equal to the weight of a 200-kg object) to Continued 246 CHAPTER 8 Momentum, Impulse, and Collisions change the ball’s momentum in such a short time. Other forces that 8.6 Typically, a tennis ball is in contact with the racket for act on the ball during the collision are comparatively weak; for approximately 0.01 s. The ball flattens noticeably due to the instance, the gravitational force is only 3.9 N. Thus, during the tremendous force exerted by the racket. short time that the collision lasts, we can ignore all other forces on the ball. Figure 8.6 shows the impact of a tennis ball and racket. Note that the 2000-N value we calculated is the average horizon- tal force that the wall exerts on the ball during the impact. It corre- sponds to the horizontal line (Fav)x in Fig. 8.3a. The horizontal force is zero before impact, rises to a maximum, and then decreases to zero when the ball loses contact with the wall. If the ball is relatively rigid, like a baseball or golf ball, the collision lasts a short time and the maximum force is large, as in the blue curve in Fig. 8.3b. If the ball is softer, like a tennis ball, the collision time is longer and the maximum force is less, as in the orange curve in Fig. 8.3b. Example 8.3 Kicking a soccer ball A soccer ball has a mass of 0.40 kg. Initially it is moving to the left impulse on the ball, Jx and Jy, and the components of the average at 20 m>s, but then it is kicked. After the kick it is moving at 45° net force on the ball, 1Fav2x and 1Fav2y. We’ll find them using the upward and to the right with speed 30 m>s (Fig. 8.7a). Find the impulse–momentum theorem in its component form, Eqs. (8.9). impulse of the net force and the average net force, assuming a col- EXECUTE: Using cos 45° ⫽ sin 45° ⫽ 0.707, we find the ball’s lision time ¢t = 0.010 s. velocity components before and after the kick: SOLUTION v1x = -20 m>s v1y = 0 IDENTIFY and SET UP: The ball moves in two dimensions, so we v2x = v2y = 130 m>s210.7072 = 21.2 m>s must treat momentum and impulse as vector quantities. We take From Eqs. (8.9), the impulse components are the x-axis to be horizontally to the right and the y-axis to be verti- Jx = p2x - p1x = m1v2x - v1x2 cally upward. Our target variables are the components of the net = 10.40 kg2321.2 m>s - 1-20 m>s24 = 16.5 kg # m>s 8.7 (a) Kicking a soccer ball. (b) Finding the average force on the ball from its components. Jy = p2y - p1y = m1v2y - v1y2 = 10.40 kg2121.2 m>s - 02 = 8.5 kg # m>s (a) Before-and-after diagram From Eq. (8.8), the average net force components are y v2 5 30 m s / Jx Jy 1Fav2x = = 1650 N 1Fav2y = = 850 N ¢t ¢t S AFTER The magnitude and direction of the average net force Fav are 45° m 5 0.40 kg Fav = 211650 N22 + 1850 N22 = 1.9 * 10 3 N x 850 N O u = arctan = 27° v1 5 20 m s / BEFORE 1650 N (b) Average force on the ball The ball was not initially at rest, so its final velocity does not have the same direction as the average force that acted on it. S (Fav)y Fav S EVALUATE: Fav includes the force of gravity, which is very small; u the weight of the ball is only 3.9 N. As in Example 8.2, the aver- age force acting during the collision is exerted almost entirely by (Fav)x the object that the ball hit (in this case, the soccer player’s foot). Test Your Understanding of Section 8.1 Rank the following situations according to the magnitude of the impulse of the net force, from largest value to small- est value. In each situation a 1000-kg automobile is moving along a straight east–west road. (i) The automobile is initially moving east at 25 m>s and comes to a stop in 10 s. (ii) The automobile is initially moving east at 25 m>s and comes to a stop in 5 s. (iii) The automobile is initially at rest, and a 2000-N net force toward the east is applied to it for 10 s. (iv) The automobile is initially moving east at 25 m>s, and a 2000-N net force toward the west is applied to it for 10 s. (v) The automobile is initially moving east at 25 m>s. Over a 30-s period, the automobile reverses direction and ends up moving west at 25 m>s. ❙  8.2 Conservation of Momentum 247 8.2 Conservation of Momentum The concept of momentum is particularly important in situations in which we 8.8 Two astronauts push each other as have two or more bodies that interact. To see why, let’s consider first an idealized they float freely in the zero-gravity system of two bodies that interact with each other but not with anything else—for environment of space. example, two astronauts who touch each other as they float freely in the zero- gravity environment of outer space (Fig. 8.8). Think of the astronauts as particles. Each particle exerts a force on the other; according to Newton’s third law, the two forces are always equal in magnitude and opposite in direction. Hence, the A B impulses that act on the two particles are equal and opposite, and the changes in momentum of the two particles are equal and opposite. Let’s go over that again with some new terminology. For any system, the forces that the particles of the system exert on each other are called internal forces. Forces No external forces act on the two-astronaut exerted on any part of the system by some object outside it are called S external forces. system, so its total momentum is conserved. For the system shown in Fig. 8.8, the internal forces are FB on A , exerted by y y S particle B on particle A, and FA on B, exerted by particle A on particle B. There are no external forces; when this is the case, we have an isolated system. S x S x S S FB on A FA on B The net force on particle A is FB on A and the net force on particle B is FA on B , so from Eq. (8.4) the rates of change of the momenta of the two particles are The forces the astronauts exert on each other form an action–reaction pair. S S S d pA S d pB FB on A ⴝ FA on B ⴝ (8.10) dt dt The momentum of each particle changes, but S these changes S are related to each other by Newton’s third law: The two forces FB on A and FA on B are always equal S S S in magnitude and opposite in direction. That is, FB on A ⴝ ⴚFA on B , so FB on A ⴙ S FA on B ⴝ 0. Adding together the two equations in Eq. (8.10), we have S S S S S S d pA d pB d1p A ⴙ p B2 FB on A ⴙ FA on B ⴝ ⴙ ⴝ ⴝ 0 (8.11) dt dt dt The rates of change of the two momenta are equal and opposite, so the rate of S S change of the vector sum p A ⴙ p B is zero. We now define the total momentum S P of the system of two particles as the vector sum of the momenta of the individual 8.9 Two ice skaters push each other as particles; that is, they skate on a frictionless, horizontal surface. (Compare to Fig. 8.8.) S S S P ⴝ pA ⴙ pB (8.12) Then Eq. (8.11) becomes, finally, S S S dP FB on A ⴙ FA on B ⴝ ⴝ0 (8.13) dt S The time rate of change of the total momentum P is zero. Hence the total The forces the skaters exert on each momentum of the system is constant, even though the individual momenta of the other form an action–reaction pair. particles that make up the system can change. y If external forces are also present, they must be included on the left side of y Eq. (8.13) along with the internal forces. Then the total momentum is, in general, S S nA nB not constant. But if the vector sum of the external forces is zero, S as in Fig. 8.9, these forces have no effect on the left side of Eq. (8.13), and dP>dt is again zero. Thus we have the following general result: S x S x FB on A FA on B If the vector sum of the external forces on a system is zero, the total momentum of the system is constant. S S wA wB This is the simplest form of the principle of conservation of momentum. This Although the normal and gravitational principle is a direct consequence of Newton’s third law. What makes this principle forces are external, their vector sum is zero, useful is that it doesn’t depend on the detailed nature of the internal forces that so the total momentum is conserved. 248 CHAPTER 8 Momentum, Impulse, and Collisions act between members of the system. This means that we can apply conservation of momentum even if (as is often the case) we know very little about the internal forces. We have used Newton’s second law to derive this principle, so we have to be careful to use it only in inertial frames of reference. We can generalize this principle for a system that contains any number of par- ticles A, B, C,... interacting only with one another. The total momentum of such a system is S S P ⴝ pA ⴙ pB ⴙ S Á S ⴝ m AvA ⴙ m BvB ⴙ S Á (total momentum of a system of particles) (8.14) We make the same argument as before: The total rate of change of momentum of the system due to each action–reaction pair of internal forces is zero. Thus the ActivPhysics 6.3: Momentum Conservation total rate of change of momentum of the entire system is zero whenever the vector and Collisions ActivPhysics 6.7: Explosion Problems sum of the external forces acting on it is zero. The internal forces can change the ActivPhysics 6.10: Pendulum Person- momenta of individual particles in the system but not the total momentum of the Projectile Bowling system. 8.10 When applying conservation of CAUTION Conservation of momentum means conservation of its components When you momentum, remember that momentum is apply the conservation of momentum to a system, remember that momentum is a vector a vector quantity! quantity. Hence you must use vector addition to compute the total momentum of a system S (Fig. 8.10). Using components is usually the simplest method. If pAx , pAy , and pAz are the S pB A system of two components of momentum of particle A, and similarly for the other particles, then pA B particles with Eq. (8.14) is equivalent to the component equations pA 5 18 kg · m s / momenta in A pB 5 24 kg · m s / different directions Px = pAx + pBx + Á Py = pAy + pBy + Á (8.15) You CANNOT find the magnitude of the total momentum by adding the magnitudes of the Pz = pAz + pBz + Á individual momenta! If the vector sum of the external forces on the system is zero, then Px , Py , and Pz are all P 5 pA 1 pB ⫽ 42 kg · m s / constant. ❙ Instead, use vector addition: In some ways the principle of conservation of momentum is more general than S pB the principle of conservation of mechanical energy. For example, mechanical energy is conserved only when the internal forces are conservative—that is, when S S S S the forces allow two-way conversion between kinetic and potential energy—but pA P ⴝ pA 1 pB conservation of momentum is valid even when the internal forces are not conser- u vative. In this chapter we will analyze situations in which both momentum and S S P 5 0 pA 1 pB 0 mechanical energy are conserved, and others in which only momentum is con- 5 30 kg · m s at u 5 37° / served. These two principles play a fundamental role in all areas of physics, and we will encounter them throughout our study of physics. Problem-Solving Strategy 8.1 Conservation of Momentum IDENTIFY the relevant concepts: Confirm that the vector sum of EXECUTE the solution: the external forces acting on the system of particles is zero. If it 1. Write an equation in symbols equating the total initial and final isn’t zero, you can’t use conservation of momentum. x-components of momentum, using px = mvx for each particle. Write a corresponding equation for the y-components. Velocity SET UP the problem using the following steps: components can be positive or negative, so be careful with 1. Treat each body as a particle. Draw “before” and “after” signs! sketches, including velocity vectors. Assign algebraic symbols 2. In some problems, energy considerations (discussed in Sec- to each magnitude, angle, and component. Use letters to label tion 8.4) give additional equations relating the velocities. each particle and subscripts 1 and 2 for “before” and “after” 3. Solve your equations to find the target variables. quantities. Include any given values such as magnitudes, angles, or components. EVALUATE your answer: Does your answer make physical sense? 2. Define a coordinate system and show it in your sketches; define If your target variable is a certain body’s momentum, check that the positive direction for each axis. the direction of the momentum is reasonable. 3. Identify the target variables.  8.2 Conservation of Momentum 249 Example 8.4 Recoil of a rifle A marksman holds a rifle of mass m R = 3.00 kg loosely, so it can is pRx = m RvRx. Our target variables are vRx, pBx, pRx, and the recoil freely. He fires a bullet of mass m B = 5.00 g horizontally final kinetic energies K B = 12 m BvBx2 and K R = 12 m RvRx2. with a velocity relative to the ground of vBx = 300 m>s. What is the recoil velocity vRx of the rifle? What are the final momentum EXECUTE: Conservation of the x-component of total momentum and kinetic energy of the bullet and rifle? gives Px = 0 = m BvBx + m RvRx SOLUTION mB 0.00500 kg IDENTIFY and SET UP: If the marksman exerts negligible hori- vRx = - vBx = - ¢ ≤ 1300 m>s2 = - 0.500 m>s mR 3.00 kg zontal forces on the rifle, then there is no net horizontal force on the system (the bullet and rifle) during the firing, and the total The negative sign means that the recoil is in the direction opposite horizontal momentum of the system is conserved. Figure 8.11 to that of the bullet. shows our sketch. We take the positive x-axis in the direction of The final momenta and kinetic energies are aim. The rifle and the bullet are initially at rest, so the initial pBx = m BvBx = 10.00500 kg21300 m>s2 = 1.50 kg # m>s x-component of total momentum is zero. After the shot is fired, the bullet’s x-momentum is pBx = m BvBx and the rifle’s x-momentum K B = 12 m BvBx2 = 12 10.00500 kg21300 m>s22 = 225 J pRx = m RvRx = 13.00 kg21-0.500 m>s2 = -1.50 kg # m>s 8.11 Our sketch for this problem. K R = 12 m RvRx2 = 1213.00 kg21-0.500 m>s22 = 0.375 J EVALUATE: The bullet and rifle have equal and opposite final momenta thanks to Newton’s third law: They experience equal and opposite interaction forces that act for the same time, so the impulses are equal and opposite. But the bullet travels a much greater distance than the rifle during the interaction. Hence the force on the bullet does more work than the force on the rifle, giving the bullet much greater kinetic energy than the rifle. The 600:1 ratio of the two kinetic energies is the inverse of the ratio of the masses; in fact, you can show that this always happens in recoil situations (see Exercise 8.26). Example 8.5 Collision along a straight line Two gliders with different masses move toward each other on a 8.12 Two gliders colliding on an air track. frictionless air track (Fig. 8.12a). After they collide (Fig. 8.12b), glider B has a final velocity of +2.0 m>s (Fig. 8.12c). What is the vA1x 5 2.0 m s / vB 1x 5 22.0 m s / final velocity of glider A? How do the changes in momentum and in velocity compare? (a) Before collision x A B SOLUTION mA 5 0.50 kg mB 5 0.30 kg IDENTIFY and SET UP: As for the skaters in Fig. 8.9, the total ver- tical force on each glider is zero, and the net force on each individ- (b) Collision A B x ual glider is the horizontal force exerted on it by the other glider. The net external force on the system of two gliders is zero, so their total momentum is conserved. We take the positive x-axis to be to vA2x vB 2x 5 2.0 m s / the right. We are given the masses and initial velocities of both gliders and the final velocity of glider B. Our target variables are (c) After collision x A B vA2x, the final x-component of velocity of glider A, and the changes in momentum and in velocity of the two gliders (the value after the collision minus the value before the collision). This is positive (to the right in Fig. 8.12) because A has a greater EXECUTE: The x-component of total momentum before the collision is magnitude of momentum than B. The x-component of total momen- Px = m AvA1x + m BvB1x tum has the same value after the collision, so = 10.50 kg212.0 m>s2 + 10.30 kg21-2.0 m>s2 Px = m AvA2x + m BvB2x = 0.40 kg # m>s Continued 250 CHAPTER 8 Momentum, Impulse, and Collisions We solve for vA2x: The changes in x-velocities are Px - m BvB2x 0.40 kg # m>s - 10.30 kg212.0 m>s2 vA2x - vA1x = 1-0.40 m>s2 - 2.0 m>s = -2.4 m>s vA2x = = mA 0.50 kg vB2x - vB1x = 2.0 m>s - 1-2.0 m>s) = + 4.0 m>s = -0.40 m>s EVALUATE: The gliders were subjected to equal and opposite The changes in the x-momenta are interaction forces for the same time during their collision. By m AvA2x - m AvA1x = 10.50 kg21-0.40 m>s2 the impulse–momentum theorem, they experienced equal and - 10.50 kg212.0 m>s2 = - 1.2 kg # m>s opposite impulses and therefore equal and opposite changes in momentum. But by Newton’s second law, the less massive glider m BvB2x - m BvB1x = 10.30 kg212.0 m>s2 1B2 had a greater magnitude of acceleration and hence a greater - 10.30 kg21-2.0 m>s2 = + 1.2 kg # m>s velocity change. Example 8.6 Collision in a horizontal plane Figure 8.13a shows two battling robots on a frictionless surface. EXECUTE: The momentum-conservation equations and their solu- Robot A, with mass 20 kg, initially moves at 2.0 m>s parallel to the tions for vB2x and vB2y are x-axis. It collides with robot B, which has mass 12 kg and is ini- m AvA1x + m BvB1x = m AvA2x + m BvB2x tially at rest. After the collision, robot A moves at 1.0 m>s in a direction that makes an angle a = 30° with its initial direction m AvA1x + m BvB1x - m AvA2x vB2x = (Fig. 8.13b). What is the final velocity of robot B? mB 120 kg212.0 m>s2 + 112 kg2102 B R SOLUTION - 120 kg211.0 m>s21cos 30°2 = IDENTIFY and SET UP: There are no horizontal external forces, so 12 kg the x- and y-components of the total momentum of the system are = 1.89 m>s both conserved. Momentum conservation requires that the sum of the x-components of momentum before the collision (subscript 1) m AvA1y + m BvB1y = m AvA2y + m BvB2y must equal the sum after the collision (subscript 2), and similarly m AvA1y + m BvB1y - m AvA2y S for the sums of the y-components. Our target variable is vB2, the vB2y = mB final velocity of robot B. 120 kg2102 + 112 kg2102 B R - 120 kg211.0 m>s21sin 30°2 = 8.13 Views from above of the velocities (a) before and 12 kg (b) after the collision. = -0.83 m>s (a) Before collision Figure 8.13b shows the motion of robot B after the collision. The S magnitude of vB2 is y S vB2 = 211.89 m>s22 + 1-0.83 m>s22 = 2.1 m>s A vA1 B x and the angle of its direction from the positive x-axis is O -0.83 m>s b = arctan = -24° 1.89 m>s (b) After collision EVALUATE: We can check our answer by confirming that the components of total momentum before and after the collision are y S vA2 equal. Initially robot A has x-momentum m AvA1x = 120 kg2 vA2y 12.0 m>s2 = 40 kg # m>s and zero y-momentum; robot B has a zero momentum. After the collision, the momentum compo- nents are m AvA2x = 120 kg211.0 m>s21cos 30°2 = 17 kg # m>s A vA2x a and m BvB2x = 112 kg211.89 m>s2 = 23 kg # m>s; the total x- momentum is 40 kg # m>s, the same as before the collision. The x O b vB2x B final y-components are m AvA2y = 120 kg211.0 m>s21sin 30°2 = vB2y b 10 kg # m>s and m BvB2y = 112 kg21-0.83 m>s2 = - 10 kg # m>s; S the total y-component of momentum is zero, the same as before the vB2 collision.  8.3 Momentum Conservation and Collisions 251 Test Your Understanding of Section 8.2 A spring-loaded toy sits at 8.14 Two gliders undergoing an elastic rest on a horizontal, frictionless surface. When the spring releases, the toy breaks collision on a frictionless surface. Each into three equal-mass pieces, A, B, and C, which slide along the surface. Piece A glider has a steel spring bumper that exerts moves off in the negative x-direction, while piece B moves off in the negative y-direction. a conservative force on the other glider. (a) What are the signs of the velocity components of piece C? (b) Which of the three (a) Before collision pieces is moving the fastest? ❙ S Springs S vA1 vB1 8.3 Momentum Conservation and Collisions A B To most people the term collision is likely to mean some sort of automotive dis- aster. We’ll use it in that sense, but we’ll also broaden the meaning to include any strong interaction between bodies that lasts a relatively short time. So we include (b) Elastic collision not only car accidents but also balls colliding on a billiard table, neutrons hitting atomic nuclei in a nuclear reactor, the impact of a meteor on the Arizona desert, A B and a close encounter of a spacecraft with the planet Saturn. If the forces between the bodies are much larger than any external forces, as is Kinetic energy is stored as potential the case in most collisions, we can neglect the external forces entirely and treat the energy in compressed springs. bodies as an isolated system. Then momentum is conserved and the total momen- tum of the system has the same value before and after the collision. Two cars col- (c) After collision liding at an icy intersection provide a good example. Even two cars colliding on S vA2 S vB2 dry pavement can be treated as an isolated system during the collision if the forces between the cars are much larger than the friction forces of pavement against tires. A B Elastic and Inelastic Collisions The system of the two gliders has the same If the forces between the bodies are also conservative, so that no mechanical kinetic energy after the collision as before it. energy is lost or gained in the collision, the total kinetic energy of the system is the same after the collision as before. Such a collision is called an elastic collision. A collision between two marbles or two billiard balls is almost completely elastic. Figure 8.14 shows a model for an elastic collision. When the gliders collide, their springs are momentarily compressed and some of the original kinetic energy is momentarily converted to elastic potential energy. Then the gliders bounce apart, 8.15 Two gliders undergoing a com- the springs expand, and this potential energy is converted back to kinetic energy. pletely inelastic collision. The spring bumpers on the gliders are replaced by A collision in which the total kinetic energy after the collision is less than before Velcro®, so the gliders stick together after the collision is called an inelastic collision. A meatball landing on a plate of collision. spaghetti and a bullet embedding itself in a block of wood are examples of inelastic (a) Before collision collisions. An inelastic collision in which the colliding bodies stick together and move as one body after the collision is often called a completely inelastic collision. Velcro® S S Figure 8.15 shows an example; we have replaced the spring bumpers in Fig. 8.14 vA1 vB1 with Velcro®, which sticks the two bodies together. A B CAUTION An inelastic collision doesn’t have to be completely inelastic It’s a common misconception that the only inelastic collisions are those in which the colliding bodies stick together. In fact, inelastic collisions include many situations in which the bodies do not stick. If two cars bounce off each other in a “fender bender,” the work done to deform (b) Completely inelastic collision the fenders cannot be recovered as kinetic energy of the cars, so the collision is inelastic (Fig. 8.16). ❙ A B Remember this rule: In any collision in which external forces can be neglected, momentum is conserved and the total momentum before equals the total momentum The gliders stick together. after; in elastic collisions only, the total kinetic energy before equals the total kinetic energy after. (c) After collision Completely Inelastic Collisions S v2 Let’s look at what happens to momentum and kinetic energy in a completely inelastic collision of two bodies (A and B), as in Fig. 8.15. Because the two bod- S A B ies stick together after the collision, they have the same final velocity v2: S S S The system of the two gliders has less kinetic vA2 ⴝ vB2 ⴝ v2 energy after the collision than before it. 252 CHAPTER 8 Momentum, Impulse, and Collisions 8.16 Automobile collisions are intended Conservation of momentum gives the relationship to be inelastic, so that the structure of the S S S car absorbs as much of the energy of the m A vA1 ⴙ m B vB1 ⴝ 1m A + m B2v2 (completely inelastic collision) (8.16) collision as possible. This absorbed energy cannot be recovered, since it goes into a If we know the masses and initial velocities, we can compute the common final S permanent deformation of the car. velocity v2. Suppose, for example, that a body with mass m A and initial x-component of velocity vA1x collides inelastically with a body with mass m B that is initially at rest 1vB1x = 02. From Eq. (8.16) the common x-component of velocity v2x of both bodies after the collision is mA (completely inelastic collision, v2x = v (8.17) m A + m B A1x B initially at rest) Let’s verify that the total kinetic energy after this completely inelastic colli- sion is less than before the collision. The motion is purely along the x-axis, so the kinetic energies K 1 and K 2 before and after the collision, respectively, are K1 = 12 mA vA1x2 mA 2 K 2 = 12 1m A + m B2v2x2 = 12 1m A + m B2a b vA1x2 mA + mB The ratio of final to initial kinetic energy is K2 mA (completely inelastic collision, = (8.18) K1 mA + mB B initially at rest) The right side is always less than unity because the denominator is always greater than the numerator. Even when the initial velocity of m B is not zero, it is not hard to verify that the kinetic energy after a completely inelastic collision is always less than before. Please note: We don’t recommend memorizing Eq. (8.17) or (8.18). We derived them only to prove that kinetic energy is always lost in a completely inelastic collision. Example 8.7 A completely inelastic collision We repeat the collision described in Example 8.5 (Section 8.2), but EXECUTE: From conservation of momentum, this time equip the gliders so that they stick together when they collide. Find the common final x-velocity, and compare the initial m AvA1x + m BvB1x = 1m A + m B2v2x and final kinetic energies of the system. m AvA1x + m BvB1x v2x = mA + mB SOLUTION 10.50 kg212.0 m>s2 + 10.30 kg21-2.0 m>s2 IDENTIFY and SET UP: There are no external forces in the x-direction, = 0.50 kg + 0.30 kg so the x-component of momentum is conserved. Figure 8.17 shows our sketch. Our target variables are the final x-velocity v2x and the = 0.50 m>s initial and final kinetic energies K 1 and K 2. Because v2x is positive, the gliders move together to the right after the collision. Before the collision, the kinetic energies are 8.17 Our sketch for this problem. K A = 12 m AvA1x2 = 1210.50 kg212.0 m>s22 = 1.0 J K B = 12 m BvB1x2 = 1210.30 kg21-2.0 m>s22 = 0.60 J The total kinetic energy before the collision is K 1 = K A + KB = 1.6 J. The kinetic energy after the collision is K 2 = 121m A + m B2v2 x 2 = 1210.50 kg + 0.30 kg210.50 m>s22 = 0.10 J  8.3 Momentum Conservation and Collisions 253 1 EVALUATE: The final kinetic energy is only 16 of the original; 15 16 is together, the energy is stored as potential energy of the spring. In both converted from mechanical energy to other forms. If there is a wad of cases the total energy of the system is conserved, although kinetic chewing gum between the gliders, it squashes and becomes warmer. energy is not. In an isolated system, however, momentum is always If there is a spring between the gliders that is compressed as they lock conserved whether the collision is elastic or not. Example 8.8 The ballistic pendulum Figure 8.18 shows a ballistic pendulum, a simple system for meas- conserved during this stage, however, because there is a net exter- uring the speed of a bullet. A bullet of mass m B makes a com- nal force (the forces of gravity and string tension don’t cancel pletely inelastic collision with a block of wood of mass m W, which when the strings are inclined). is suspended like a pendulum. After the impact, the block swings SET UP: We take the positive x-axis to the right and the positive up to a maximum height y. In terms of y, m B, and m W, what is the y-axis upward. Our target variable is v1. Another unknown quan- initial speed v1 of the bullet? tity is the speed v2 of the system just after the collision. We’ll use momentum conservation in the first stage to relate v1 to v2, SOLUTION and we’ll use energy conservation in the second stage to relate v2 to y. IDENTIFY: We’ll analyze this event in two stages: (1) the embed- ding of the bullet in the block and (2) the pendulum swing of the EXECUTE: In the first stage, all velocities are in the ⫹x-direction. block. During the first stage, the bullet embeds itself in the Momentum conservation gives block so quickly that the block does not move appreciably. The mBv1 = 1mB + mW2v2 supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet–block system, and the hori- mB + mW v1 = v2 zontal component of momentum is conserved. Mechanical mB energy is not conserved during this stage, however, because a At the beginning of the second stage, the system has kinetic energy nonconservative force does work (the force of friction between K = 121m B + m W2v22. The system swings up and comes to rest for bullet and block). an instant at a height y, where its kinetic energy is zero and the In the second stage, the block and bullet move together. The potential energy is 1m B + m W2gy; it then swings back down. only forces acting on this system are gravity (a conservative Energy conservation gives force) and the string tensions (which do no work). Thus, as the 1 2 1mB + mW2v22 = 1mB + mW2gy block swings, mechanical energy is conserved. Momentum is not v2 = 22gy 8.18 A ballistic pendulum. We substitute this expression for v2 into the momentum equation: mB + mW v1 = 22gy mB EVALUATE: Let’s plug in the realistic numbers m B = 5.00 g = 0.00500 kg, m W = 2.00 kg, and y = 3.00 cm = 0.0300 m. We then have Before collision v1 0.00500 kg + 2.00 kg v1 = 2219.80 m>s2210.0300 m2 mB 0.00500 kg mW = 307 m>s The speed v2 of the block just after impact is v2 = 22gy = 2219.80 m>s2210.0300 m2 = 0.767 m>s v2 Top of swing The speeds v1 and v2 seem realistic. The kinetic energy of the Immediately bullet before impact is 1210.00500 kg21307 m>s22 = 236 J. Just y after collision after impact the kinetic energy of the system is 1212.005 kg2 mB 1 mW 10.767 m>s22 = 0.590 J. Nearly all the kinetic energy disap- pears as the wood splinters and the bullet and block become warmer. 254 CHAPTER 8 Momentum, Impulse, and Collisions Example 8.9 An automobile collision A 1000-kg car traveling north at 15 m>s collides with a 2000-kg 8.19 Our sketch for this problem. truck traveling east at 10 m>s. The occupants, wearing seat belts, are uninjured, but the two vehicles move away from the impact point as one. The insurance adjustor asks you to find the velocity of the wreckage just after impact. What is your answer? SOLUTION IDENTIFY and SET UP: We’ll treat the cars as an isolated system, so that the momentum of the system is conserved. We can do so because (as we show below) the magnitudes of the horizontal forces that the cars exert on each other during the collision are much larger than any external forces such as friction. Figure 8.19 shows our sketch S and the coordinate axes. We can find the total momentum P before the collision using Eqs. (8.15). The momen- tum has the same S value just after the collision; hence we can find the S velocity S V just after the collision (our target variable) using P ⴝ MV, where M = m C + m T = 3000 kg is the mass of the S S S From P ⴝ MV, the direction of the velocity V just after the collision wreckage. is also u = 37°. The velocity magnitude is S 2.5 * 10 4 kg # m>s EXECUTE: From Eqs. (8.15), the components of P are P Px = pCx + pTx = m CvCx + m TvTx V = = = 8.3 m>s M 3000 kg = 11000 kg2102 + 12000 kg2110 m>s2 = 2.0 * 10 4 kg # m>s EVALUATE: This is an inelastic collision, so we expect the total kinetic energy to be less after the collision than before. As you can Py = pCy + pTy = m CvCy + m TvTy show, the initial kinetic energy is 2.1 * 10 5 J and the final value is = 11000 kg2115 m>s2 + 12000 kg2102 1.0 * 10 5 J. = 1.5 * 10 4 kg # m>s We can now justify our neglect of the external forces on the vehi- cles during the collision. The car’s weight is about 10,000 N; if the S The magnitude of P is coefficient of kinetic friction is 0.5, the friction force on the car P = 212.0 * 10 4 kg # m>s22 + 11.5 * 10 4 kg # m>s22 during the impact is about 5000 N. The car’s initial kinetic energy is 1 2 11000 kg2115 m>s2 = 1.1 * 10 J, so -1.1 * 10 J of work 2 5 5 = 2.5 * 10 4 kg # m>s must be done to stop it. If the car crumples by 0.20 m and its direction is given by the angle u shown in Fig. 8.19: in stopping, a force of magnitude 11.1 * 105 J2>10.20 m2 = 5.5 * 105 N would be needed; that’s 110 times the friction force. Py 1.5 * 10 4 kg # m>s So it’s reasonable to treat the external force of friction as negligible tan u = = = 0.75 u = 37° Px 2.0 * 10 4 kg # m>s compared with the internal forces the vehicles exert on each other. Classifying Collisions It’s important to remember that we can classify collisions according to energy considerations (Fig. 8.20). A collision in which kinetic energy is conserved is called elastic. (We’ll explore these in more depth in the next section.) A collision in which the total kinetic energy decreases is called inelastic. When the two bodies have a common final velocity, we say that the collision is completely inelastic. There are also cases in which the final kinetic energy is greater than the initial value. Rifle recoil, discussed in Example 8.4 (Section 8.2), is an example. 8.20 Collisions are classified according to energy considerations. S S S S S S vA1 vB1 vA1 vB1 vA1 vB1 A B A B