CHAPTER 6Momentum andCollisionsPHYSICS IN ACTIONSoccer players must consider an enormousamount of information every time they setthe ball—or themselves—into motion.Once a player knows where the ball shouldgo, the player has to decide how to get itthere. The player also has to consider theball’s speed and direction in order to stopit or change its direction. The player in thephotograph must determine how muchforce to exert on the ball and how muchfollow-through is needed. To do this, hemust understand his own motion as well asthe motion of the ball.• International regulations specify the mass ofofficial soccer balls. How does the mass of aball affect the way it behaves when kicked?• How does the velocity of the player’s footaffect the final velocity of the ball?CONCEPT REVIEWNewton’s laws of motion(Sections 4-2 and 4-3)Kinetic energy (Section 5-2)Conservation of energy(Section 5-3)Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions 207

Imagine rolling a bowling ball down one lane at a bowling alley and rollinga playground ball down another lane at the same speed. The more-massivebowling ball, shown in Figure 6-1, will exert more force on the pins becausethe bowling ball has more momentum than the playground ball. When wethink of a massive object moving at a high velocity, we often say that theobject has a large momentum. A less massive object with the same velocity hasa smaller momentum.On the other hand, a small object moving with a very high velocity has alarge momentum. One example of this is hailstones falling from very highclouds. By the time they reach Earth, they can have enough momentum tohurt you or cause serious damage to cars and buildings.Momentum is so fundamental inNewton’s mechanics that Newtoncalled it simply “quantity of motion.”The symbol for momentum, p,comes from Leibniz’s use of the termprogress, defined as “the quantity ofmotion with which a body proceedsin a certain direction.”SAMPLE PROBLEM 6AMomentumPROBLEMA 2250 kg pickup truck has a velocity of 25 m/s to the east. What is themomentum of the truck?SOLUTIONGiven: m = 2250 kg v = 25 m/s to the eastUnknown: p = ?Use the momentum equation from page 208.p = mv = (2250 kg)(25 m/s)p = 5.6 × 10 4 kg •m/s to the eastCALCULATOR SOLUTIONYour calculator will give you theanswer 56250 for the momentum.The value for the velocity has only twosignificant figures, so the answermust be reported as 5.6 × 10 4 .MomentumPRACTICE 6A1. An ostrich with a mass of 146 kg is running to the right with a velocity of17 m/s. Find the momentum of the ostrich.2. A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.a. What is the total momentum of the child and the bike together?b. What is the momentum of the child?c. What is the momentum of the bike?3. What velocity must a car with a mass of 1210 kg have in order to havethe same momentum as the pickup truck in Sample Problem 6A?Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions209

Figure 6-2When the ball is moving very fast,the player must exert a large forceover a short time to change theball’s momentum and quickly bringthe ball to a stop.A change in momentum takes force and timeFigure 6-2 shows a player stopping a moving soccer ball. In a given time interval,he must exert more force to stop a fast ball than to stop a ball that is movingmore slowly. Now imagine a toy truck and a real dump truck starting fromrest and rolling down the same hill at the same time. They would accelerate atthe same rate, so their velocity at any instant would be the same, but it wouldtake much more force to stop the massive dump truck than to stop the toytruck in the same time interval. You have probably also noticed that a ballmoving very fast stings your hands when you catch it, while a slow-movingball causes no discomfort when you catch it.From examples like these, we see that momentum is closely related to force.In fact, when Newton first expressed his second law mathematically, he wroteit not as F = ma, but in the following form.F = ⎯ ∆ ⎯p∆tchange in momentumforce = ⎯⎯⎯time intervalWe can rearrange this equation to find the change in momentum in termsof the net external force and the time interval required to make this change.IMPULSE-MOMENTUM THEOREMF∆t =∆p or F∆t =∆p = mv f − mv iforce × time interval = change in momentumNSTATOPIC: MomentumGO TO: www.scilinks.orgsciLINKS CODE: HF2061impulsefor a constant external force, theproduct of the force and the timeover which it acts on an objectThis equation states that a net external force, F, applied to an object for acertain time interval, ∆t, will cause a change in the object’s momentum equalto the product of the force and the time interval. In simple terms, a small forceacting for a long time can produce the same change in momentum as a largeforce acting for a short time. In this book, all forces exerted on an object areassumed to be constant unless otherwise stated.The expression F∆t =∆p is called the impulse-momentum theorem. Theterm on the left side of the equation, F∆t, is called the impulse of the force Ffor the time interval ∆t.The equation F∆t =∆p explains why follow-through is important in somany sports, from karate and billiards to softball and croquet. For example,when a batter hits a ball, the ball will experience a greater change in momentumif the batter follows through and keeps the bat in contact with the ball fora longer time. Follow through is also important in many everyday activities,such as pushing a shopping cart or moving furniture. Extending the timeinterval over which a constant force is applied allows a smaller force to cause agreater change in momentum than would result if the force were applied for avery short time.210Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

SAMPLE PROBLEM 6BForce and impulsePROBLEMSOLUTIONA 1400 kg car moving westward with a velocity of 15 m/s collides with autility pole and is brought to rest in 0.30 s. Find the magnitude of the forceexerted on the car during the collision.Given: m = 1400 kg v i = 15 m/s to the west =−15 m/s∆t = 0.30 s v f = 0 m/sUnknown: F = ?Use the impulse-momentum theorem.F∆t =∆p = mv f − mv iF = ⎯ mv f − mv⎯ i∆t(1400 kg)(0 m/s) − (1400 kg)(−15 m/s)F = =⎯ 21 00 0 kg•m/s⎯⎯⎯⎯ ⎯0.30 s0.30sF = 7.0 × 10 4 N to the eastPRACTICE 6BForce and momentum1. A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationaryreceiver catches the ball and brings it to rest in 0.020 s. What isthe force exerted on the receiver?2. An 82 kg man drops from rest on a diving board 3.0 m above the surfaceof the water and comes to rest 0.55 s after reaching the water. What forcedoes the water exert on him?3. A 0.40 kg soccer ball approaches a player horizontally with a velocity of18 m/s to the north. The player strikes the ball and causes it to move inthe opposite direction with a velocity of 22 m/s. What impulse was deliveredto the ball by the player?4. A 0.50 kg object is at rest. A 3.00 N force to the right acts on the objectduring a time interval of 1.50 s.a. What is the velocity of the object at the end of this interval?b. At the end of this interval, a constant force of 4.00 N to the left isapplied for 3.00 s. What is the velocity at the end of the 3.00 s?Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions211

PRACTICE 6CStopping distance1. How long would it take the car in Sample Problem 6C to come to a stopfrom 20.0 m/s to the west? How far would the car move before stopping?Assume a constant acceleration.2. A 2500 kg car traveling to the north is slowed down uniformly from aninitial velocity of 20.0 m/s by a 6250 N braking force acting opposite thecar’s motion. Use the impulse-momentum theorem to answer the followingquestions:a. What is the car’s velocity after 2.50 s?b. How far does the car move during 2.50 s?c. How long does it take the car to come to a complete stop?3. Assume that the car in Sample Problem 6C has a mass of 3250 kg.a. How much force would be required to cause the same acceleration asin item 1? Use the impulse-momentum theorem.b. How far would the car move before stopping?A change in momentum over a longer timerequires less forceThe impulse-momentum theorem is used to design safetyequipment that reduces the force exerted on the human bodyduring collisions. Examples of this are the nets and giant airmattresses firefighters use to catch people who must jump outof tall burning buildings. The relationship is also used todesign sports equipment and games.Figure 6-4 shows an Inupiat family playing a traditionalgame. Common sense tells us that it is much better for thegirl to fall onto the outstretched blanket than onto the hardground. In both cases, however, the change in momentum ofthe falling girl is exactly the same. The difference is that theblanket “gives way” and extends the time of collision so thatthe change in the girl’s momentum occurs over a longer timeinterval. A longer time interval requires a smaller force toachieve the same change in the girl’s momentum. Therefore,the force exerted on the girl when she lands on the outstretchedblanket is less than the force would be if she were toland on the ground.Figure 6-4In this game, the girl is protected from injury because theblanket reduces the force of the collision by allowing it totake place over a longer time interval.Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions213

(a)(b)Now consider a falling egg. When the egghits a hard surface, like the plate in Figure6-5(a), the egg comes to rest in a very shorttime interval. The force the hard plate exertson the egg due to the collision is large. Whenthe egg hits a floor covered with a pillow, as inFigure 6-5(b), the egg undergoes the samechange in momentum, but over a muchlonger time interval. In this case, the forcerequired to accelerate the egg to rest is muchsmaller. By applying a small force to the eggover a longer time interval, the pillow causesthe same change in the egg’s momentum asthe hard plate, which applies a large force overa short time interval. Because the force in thesecond situation is smaller, the egg can withstandit without breaking.Figure 6-5A large force exerted over a short time (a) causes the samechange in the egg’s momentum as a small force exerted over alonger time (b).Section Review1. The speed of a particle is doubled.a. By what factor is its momentum changed?b. What happens to its kinetic energy?2. A pitcher claims he can throw a 0.145 kg baseball with as much momentumas a speeding bullet. Assume that a 3.00 g bullet moves at a speed of1.50 × 10 3 m/s.a. What must the baseball’s speed be if the pitcher’s claim is valid?b. Which has greater kinetic energy, the ball or the bullet?3. When a force is exerted on an object, does a large force always produce alarger change in the object’s momentum than a smaller force does? Explain.4. What is the relationship between impulse and momentum?5. Physics in Action A 0.42 kg soccer ball is moving downfield witha velocity of 12 m/s. A player kicks the ball so that it has a final velocity of18 m/s downfield.a. What is the change in the ball’s momentum?b. Find the constant force exerted by the player’s foot on the ball if thetwo are in contact for 0.020 s.214Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

6-2Conservation of momentumMOMENTUM IS CONSERVEDSo far in this chapter, we have considered the momentum of only one object ata time. Now we will consider the momentum of two or more objects interactingwith each other. Figure 6-6 shows a stationary soccer ball set into motionby a collision with a moving soccer ball. Assume that both balls are on asmooth gym floor and that neither ball rotates before or after the collision.Before the collision, the momentum of ball B is equal to zero because the ballis stationary. During the collision, ball B gains momentum while ball A losesmomentum. As it turns out, the momentum that ball A loses is exactly equalto the momentum that ball B gains.6-2 SECTION OBJECTIVES• Describe the interactionbetween two objects interms of the change inmomentum of each object.• Compare the total momentumof two objects beforeand after they interact.• State the law of conservationof momentum.• Predict the final velocities ofobjects after collisions, giventhe initial velocities.(a)(b)A B A BTable 6-1 shows the velocity and momentum of each soccer ball bothbefore and after the collision. The momentum of each ball changes due to thecollision, but the total momentum of the two balls together remains constant.Figure 6-6(a) Before the collision, ball A hasmomentum p A and ball B has nomomentum. (b) After the collision,ball B gains momentum p B .Table 6-1Momentum in a collisionBall ABall BMass Velocity Momentum Mass Velocity Momentumbeforecollisionaftercollision0.47 kg 0.84 m/s 0.40 kg•m/s 0.47 kg 0 m/s 0 kg•m/s0.47 kg 0.04 m/s 0.02 kg•m/s 0.47 kg 0.80 m/s 0.38 kg•m/sCopyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions215

NSTATOPIC: RocketryGO TO: www.scilinks.orgsciLINKS CODE: HF2062In other words, the momentum of ball A plus the momentum of ball B beforethe collision is equal to the momentum of ball A plus the momentum of ball Bafter the collision.p A,i + p B,i = p A,f + p B,fThis relationship is true for all interactions between isolated objects and isknown as the law of conservation of momentum.CONSERVATION OF MOMENTUMm 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,ftotal initial momentum = total final momentumIn its most general form, the law of conservation of momentum can bestated as follows:1. Ice skatingIf a reckless ice skater collideswith another skaterwho is standing on the ice, isit possible for both skaters tobe at rest after the collision?2. Space travelA spacecraft undergoes achange of velocity when itsrockets are fired. Howdoes the spacecraftchange velocity in emptyspace, where there isnothing for the gasesemitted by the rocketsto push against?The total momentum of all objects interacting with one another remains constantregardless of the nature of the forces between the objects.Momentum is conserved in collisionsIn the soccer-ball example, we found that the momentum of ball A does notremain constant and the momentum of ball B does not remain constant, butthe total momentum of ball A and ball B does remain constant. In general, thetotal momentum remains constant for a system of objects that interact withone another. In this case, in which the floor is assumed to be frictionless, thesoccer balls are the only two objects interacting. If a third object exerted aforce on either ball A or ball B during the collision, the total momentum ofball A, ball B, and the third object would remain constant.In this book, most conservation-of-momentum problems deal with onlytwo isolated objects. However, when you use conservation of momentum tosolve a problem or investigate a situation, it is important to include all objectsthat are involved in the interaction. Frictional forces—such as the frictionalforce between the soccer balls and the floor—will be disregarded in mostconservation-of-momentum problems in this book.Momentum is conserved for objects pushing away from each otherAnother example of conservation of momentum is when two or more interactingobjects that initially have no momentum begin moving away from eachother. Imagine that you initially stand at rest and then jump up, leaving theground with a velocity v. Obviously, your momentum is not conserved; beforethe jump, it was zero, and it became mv as you began to rise. However, thetotal momentum remains constant if you include Earth in your analysis. Thetotal momentum for you and Earth remains constant.If your momentum after you jump is 60 kg •m/s upward, then Earth musthave a corresponding momentum of 60 kg •m/s downward, because totalCopyright © by Holt, Rinehart and Winston. All rights reserved.

momentum is conserved. However,because Earth has an enormous mass(6 × 10 23 kg), that momentum correspondsto a tiny velocity (1 × 10 −23 m/s).Imagine two skaters pushing awayfrom each other, as shown in Figure 6-7.The skaters are both initially at rest witha momentum of p 1,i = p 2,i = 0. Whenthey push away from each other, theymove in opposite directions with equalbut opposite momentum so that thetotal final momentum is also zero (p 1,f +p 2,f = 0).(a)Figure 6-7(a) When the skaters stand facingeach other, both skaters have zeromomentum, so the total momentumof both skaters is zero.(b)(b) When the skaters push away fromeach other, their momentum is equalbut opposite, so the total momentumis still zero.Surviving a CollisionPucks and carts collide in physics labs all the timewith little damage. But when cars collide on a freeway,the resulting rapid change in speed can causeinjury or death to the drivers and any passengers.forces will break your hold on the baby. Because ofinertia, the baby will continue at the car’s originalvelocity and collide with the front windshield.Many types of collisions are dangerous, but head-oncollisions involve the greatest accelerations and thusthe greatest forces.When two cars going 100 km/h(62 mi/h) collide head-on, each car dissipates thesame amount of kinetic energy that it would dissipateif it hit the ground after being dropped from theroof of a 12-story building.The key to many automobile-safety features is theconcept of impulse. One way today’s cars make useof the concept of impulse is by crumpling duringimpact. Pliable sheet metal and frame structuresabsorb energy until the force reaches the passengercompartment, which is built of rigid metal for protection.Because the crumpling slows the car gradually, itis an important factor in keeping the driver alive.Even taking into account this built-in safety feature,the National Safety Council estimates that high-speedcollisions involve accelerations of 20 times the freefallacceleration. In other words, an 89 N (20 lb)infant could experience a force of 1780 N(400 lb) in a collision.If you are holding a baby in yourlap during a collision, it is very likely that these largeSeat belts are necessary to protect a body fromforces of such large magnitudes. They stretch andextend the time it takes a passenger’s body to stop,thereby reducing the force on the person. Seat beltsalso prevent passengers from hitting the insideframe of the car. During a collision, a person notwearing a seat belt is likely to hit the windshield, thesteering wheel, or the dashboard—often with traumaticresults.Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions217

SAMPLE PROBLEM 6DConservation of momentumPROBLEMSOLUTION1. DEFINEA 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of theboat and onto the dock. If the boater moves out of the boat with a velocityof 2.5 m/s to the right, what is the final velocity of the boat?Given: m 1 = 76 kg m 2 = 45 kg v 1,i = 0v 2,i = 0 v 1,f = 2.5 m/s to the rightUnknown: v 2,f = ?Diagram:m 1 = 76 kgv 1,f = 2.5 m/s2. PLANm 2 = 45 kgChoose an equation or situation: Because the total momentum of an isolatedsystem remains constant, the total initial momentum of the boater andthe boat will be equal to the total final momentum of the boater and the boat.m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,fBecause the boater and the boat are initially at rest, the total initial momentumof the system is equal to zero.m 1 v 1,i + m 2 v 2,i = 03. CALCULATETherefore, the final momentum of the system must also be equal to zero.m 1 v 1,f + m 2 v 2,f = 0Substitute the values into the equation(s) and solve:m 1 v 1,f + m 2 v 2,f = (76 kg × 2.5 m/s) + (45 kg × v 2,f )190 kg •m/s + 45 kg(v 2,f ) = 045 kg(v 2,f ) =−190 kg •m/sv 2,f = ⎯ −190 kg •m/s⎯45 kgv 2,f =−4.2 m/s4. EVALUATEThe negative sign for v 2,f indicates that the boat is moving to the left, in thedirection opposite the motion of the boater.v 2,f = 4.2 m/s to the left218Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

•INTERACTIVEPRACTICE 6DConservation of momentum1. A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttlebreaks. The astronaut is able to throw a 10.0 kg oxygen tank in a directionaway from the shuttle with a speed of 12.0 m/s, propelling the astronautback to the shuttle. Assuming that the astronaut starts from rest, find thefinal speed of the astronaut after throwing the tank.2. An 85.0 kg fisherman jumps from a dock into a 135.0 kg rowboat at reston the west side of the dock. If the velocity of the fisherman is 4.30 m/sto the west as he leaves the dock, what is the final velocity of the fishermanand the boat?3. Each croquet ball in a set has a mass of 0.50 kg. The green ball, travelingat 12.0 m/s, strikes the blue ball, which is at rest. Assuming that the ballsslide on a frictionless surface and all collisions are head-on, find the finalspeed of the blue ball in each of the following situations:a. The green ball stops moving after it strikes the blue ball.b. The green ball continues moving after the collision at 2.4 m/s in thesame direction.c. The green ball continues moving after the collision at 0.3 m/s in thesame direction.4. A boy on a 2.0 kg skateboard initially at rest tosses an 8.0 kg jug of waterin the forward direction. If the jug has a speed of 3.0 m/s relative to theground and the boy and skateboard move in the opposite direction at0.60 m/s, find the boy’s mass.Newton’s third law leads to conservation of momentumConsider two isolated bumper cars, m 1 and m 2 , before and after they collide.Before the collision, the velocities of the two bumper cars are v 1,i and v 2,i ,respectively. After the collision, their velocities are v 1,f and v 2,f , respectively. Theimpulse-momentum theorem, F∆t =∆p, describes the change in momentum ofone of the bumper cars. Applied to m 1 , the impulse-momentum theorem givesthe following:F 1 ∆t = m 1 v 1,f − m 1 v 1,iLikewise, for m 2 it gives the following:PHYSICST U T O RModule 7“Conservation of Momentum”provides an interactivelesson with guided problemsolvingpractice to teach youabout momentum and momentumconservation.F 2 ∆t = m 2 v 2,f − m 2 v 2,iCopyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions219

F 1 is the force that m 2 exerts on m 1 during the collision, and F 2 is the forcethat m 1 exerts on m 2 during the collision, as shown in Figure 6-8. Because theonly forces acting in the collision are the forces the two bumper cars exert oneach other, Newton’s third law tells us that the force on m 1 is equal to andopposite the force on m 2 (F 1 =−F 2 ). Additionally, the two forces act over thesame time interval, ∆t. Therefore, the force m 2 exerts on m 1 multiplied by thetime interval is equal to the force m 1 exerts on m 2 multiplied by the timeinterval, or F 1 ∆t =−F 2 ∆t. That is, the impulse on m 1 is equal to and oppositethe impulse on m 2 . This relationship is true in every collision or interactionbetween two isolated objects.Figure 6-8Because of the collision, the forceexerted on each bumper car causesa change in momentum for each car.The total momentum is the samebefore and after the collision.v1,i v2,i v 1,fv 2,fm 2 m 1 m2F 1 F 2m 1Before collision After collisionBecause impulse is equal to the change in momentum, and the impulse onm 1 is equal to and opposite the impulse on m 2 , the change in momentum ofm 1 is equal to and opposite the change in momentum of m 2 . This means thatin every interaction between two isolated objects, the change in momentumof the first object is equal to and opposite the change in momentum of thesecond object. In equation form, this is expressed by the following equation.m 1 v 1,f − m 1 v 1,i =−(m 2 v 2,f − m 2 v 2,i )This equation means that if the momentum of one object increases after a collision,then the momentum of the other object in the situation must decreaseby an equal amount. Rearranging this equation gives the following equationfor the conservation of momentum.m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,fFF 1F 2Figure 6-9This graph shows the force on eachbumper car during the collision.Although both forces vary withtime, F 1 and F 2 are always equal inmagnitude and opposite in direction.tForces in real collisions are not constantAs mentioned in Section 6-1, the forces involved in a collision are treated asthough they are constant. In a real collision, however, the forces may vary intime in a complicated way. Figure 6-9 shows the forces acting during the collisionof the two bumper cars. At all times during the collision, the forces onthe two cars are equal and opposite in direction. However, the magnitudes ofthe forces change throughout the collision—increasing, reaching a maximum,and then decreasing.When solving impulse problems, you should use the average force during thecollision as the value for force. In Chapter 2, you learned that the average velocityof an object undergoing a constant acceleration is equal to the constant velocityrequired for the object to travel the same displacement in the same time interval.Similarly, the average force during a collision is equal to the constant forcerequired to cause the same change in momentum as the real, changing force.220Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

Section Review1. A 44 kg student on in-line skates is playing with a 22 kg exercise ball. Disregardingfriction, explain what happens during the following situations.a. The student is holding the ball, and both are at rest. The student thenthrows the ball horizontally, causing the student to glide back at3.5 m/s.b. Explain what happens to the ball in part (a) in terms of the momentumof the student and the momentum of the ball.c. The student is initially at rest. The student then catches the ball, whichis initially moving to the right at 4.6 m/s.d. Explain what happens in part (c) in terms of the momentum of thestudent and the momentum of the ball.2. A boy stands at one end of a floating raft that is stationary relative to theshore. He then walks in a straight line to the opposite end of the raft,away from the shore.a. Does the raft move? Explain.b. What is the total momentum of the boy and the raft before the boywalks across the raft?c. What is the total momentum of the boy and the raft after the boywalks across the raft?3. High-speed stroboscopic photographs show the head of a 215 g golf clubtraveling at 55.0 m/s just before it strikes a 46 g golf ball at rest on a tee.After the collision, the club travels (in the same direction) at 42.0 m/s.Use the law of conservation of momentum to find the speed of the golfball just after impact.4. Two isolated objects have a head-on collision. For each of the followingquestions, explain your answer.a. If you know the change in momentum of one object, can you find thechange in momentum of the other object?b. If you know the initial and final velocity of one object and the mass ofthe other object, do you have enough information to find the finalvelocity of the second object?c. If you know the masses of both objects and the final velocities of bothobjects, do you have enough information to find the initial velocitiesof both objects?d. If you know the masses and initial velocities of both objects and thefinal velocity of one object, do you have enough information to findthe final velocity of the other object?e. If you know the change in momentum of one object and the initialand final velocities of the other object, do you have enough informationto find the mass of either object?Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions221

6-3Elastic and inelastic collisions6-3 SECTION OBJECTIVES• Identify different types ofcollisions.• Determine the changes inkinetic energy during perfectlyinelastic collisions.• Compare conservation ofmomentum and conservationof kinetic energy in perfectlyinelastic and elasticcollisions.• Find the final velocity ofan object in perfectlyinelastic and elastic collisions.COLLISIONSAs you go about your day-to-day activities, you probably witness many collisionswithout really thinking about them. In some collisions, two objects collideand stick together so that they travel together after the impact. An exampleof this is a collision between an arrow and a target, as shown in Figure 6-10.The arrow, sailing forward, collides with the target at rest. In an isolated system,the target and the arrow would both move together after the collision with amomentum equal to their combined momentum before the collision. In othercollisions, like a collision between a tennis racquet and a tennis ball, two objectscollide and bounce so that they move away with two different velocities.The total momentum remains constant in any type of collision. However,the total kinetic energy is generally not conserved in a collision because somekinetic energy is converted to internal energy when the objects deform. In thissection, we will examine different types of collisions and determine whetherkinetic energy is conserved in each type. We will primarily explore two extremetypes of collisions, elastic and perfectly inelastic.Perfectly inelastic collisionsperfectly inelastic collisiona collision in which two objectsstick together and move with acommon velocity after collidingWhen two objects collide and move together as one mass, like the arrow andthe target, the collision is called perfectly inelastic. Likewise, if a meteoritecollides head on with Earth, it becomes buried in Earth and the collision isnearly perfectly inelastic.Figure 6-10When an arrow pierces a target andremains stuck in the target, thearrow and target have undergone aperfectly inelastic collision (assumingno debris is thrown out).222Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

Perfectly inelastic collisions are easy to analyze in terms of momentumbecause the objects become essentially one object after the collision. The finalmass is equal to the combined mass of the two objects, and they move withthe same velocity after colliding.Consider two cars of masses m 1 and m 2 moving with initial velocities ofv 1,i and v 2,i along a straight line, as shown in Figure 6-11. The two cars sticktogether and move with some common velocity, v f , along the same line ofmotion after the collision. The total momentum of the two cars before thecollision is equal to the total momentum of the two cars after the collision.PERFECTLY INELASTIC COLLISIONm 1 m 2(b)m 1 v 1,i + m 2 v 2,i = (m 1 + m 2 ) v fv f(a)v 1,iv 2,iThis simplified version of the equation for conservation of momentum isuseful in analyzing perfectly inelastic collisions. When using this equation, it isimportant to pay attention to signs that indicate direction. In Figure 6-11, v 1,ihas a positive value (m 1 moving to the right), while v 2,i has a negative value(m 2 moving to the left).m 1 + m 2Figure 6-11The total momentum of the two carsbefore the collision (a) is the sameas the total momentum of the twocars after the inelastic collision (b).SAMPLE PROBLEM 6EPerfectly inelastic collisionsPROBLEMSOLUTIONA 1850 kg luxury sedan stopped at a traffic light is struck from the rearby a compact car with a mass of 975 kg. The two cars become entangledas a result of the collision. If the compact car was moving at a velocity of22.0 m/s to the north before the collision, what is the velocity of theentangled mass after the collision?Given: m 1 = 1850 kg m 2 = 975 kg v 1,i = 0 m/sv 2,i = 22.0 m/s to the northUnknown: v f = ?Use the equation for a perfectly inelastic collision.m 1 v 1,i + m 2 v 2,i = (m 1 + m 2 ) v fv f = ⎯ m 1v1,i + m⎯2v 2,im1+ m2v f = =⎯ 2.14 × 4(1850 kg)(0 m/s) + (975 kg)(22.0 m/s) 10kg •m/s⎯⎯⎯⎯ ⎯1850 kg + 975 kg2820kgv f = 7.59 m/s to the northCopyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions223

PRACTICE 6EPerfectly inelastic collisions1. A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kgtruck that is initially at rest at a stoplight. The car and truck stick togetherand move together after the collision. What is the final velocity of thetwo-vehicle mass?2. A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocerycart. The bag hits the cart with a horizontal speed of 5.5 m/s towardthe front of the cart. What is the final speed of the cart and bag?3. A 1.50 × 10 4 kg railroad car moving at 7.00 m/s to the north collides withand sticks to another railroad car of the same mass that is moving in thesame direction at 1.50 m/s. What is the velocity of the joined cars afterthe collision?4. A dry cleaner throws a 22 kg bag of laundry onto a stationary 9.0 kg cart.The cart and laundry bag begin moving at 3.0 m/s to the right. Find thevelocity of the laundry bag before the collision.5. A 47.4 kg student runs down the sidewalk and jumps with a horizontalspeed of 4.20 m/s onto a stationary skateboard. The student and skateboardmove down the sidewalk with a speed of 3.95 m/s. Find the following:a. the mass of the skateboardb. how fast the student would have to jump to have a final speed of 5.00 m/sCONCEPT PREVIEWInternal energy will be discussed inChapter 10.Kinetic energy is not constant in inelastic collisionsIn an inelastic collision, the total kinetic energy does not remain constant whenthe objects collide and stick together. Some of the kinetic energy is converted tosound energy and internal energy as the objects deform during the collision.This phenomenon helps make sense of the special use of the words elasticand inelastic in physics. We normally think of elastic as referring to somethingthat always returns to, or keeps, its original shape. In physics, the most importantcharacteristic of an elastic collision is that the objects maintain their originalshapes and are not deformed by the action of forces. Objects in aninelastic collision, on the other hand, are deformed during the collision andlose some kinetic energy.The decrease in the total kinetic energy during an inelastic collision can becalculated using the formula for kinetic energy from Chapter 5, as shown inSample Problem 6F. It is important to remember that not all of the initialkinetic energy is necessarily lost in a perfectly inelastic collision.224Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

SAMPLE PROBLEM 6FKinetic energy in perfectly inelastic collisionsPROBLEMSOLUTIONTwo clay balls collide head-on in a perfectly inelastic collision. The firstball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right.The mass of the second ball is 0.250 kg, and it has an initial velocity of3.00 m/s to the left. What is the final velocity of the composite ball of clay afterthe collision? What is the decrease in kinetic energy during the collision?Given: m 1 = 0.500 kg m 2 = 0.250 kg v 1,i = 4.00 m/s to the right =+4.00 m/sv 2,i = 3.00 m/s to the left =−3.00 m/sUnknown: v f = ? ∆KE = ?Use the equation for perfectly inelastic collisions from page 223.m 1 v 1,i + m 2 v 2,i = (m 1 + m 2 ) v fv f = ⎯ m 1v1,i + m⎯2v 2,im1+ m2v f =v f = ⎯ 1.2 5 kg• m/s⎯0.750kgUse the equation for kinetic energy from Chapter 5.Initial: KE i = KE 1,i + KE 2,iKE i = ⎯ 1 2 ⎯ m 1 v 1,i2 + ⎯ 1 2 ⎯ m 2 v 2,i2KE i = ⎯ 1 2 ⎯ (0.500 kg)(4.00 m/s) 2 + ⎯ 1 2 ⎯ (0.250 kg)(−3.00 m/s) 2KE i = 5.12 JFinal: KE f = KE 1,f + KE 2,f(0.500 kg)(4.00 m/s) + (0.250 kg)(−3.00 m/s)⎯⎯⎯⎯⎯0.500 kg + 0.250 kgv f = 1.67 m/s to the rightKE f = ⎯ 1 2 ⎯ (m 1 + m 2 )v f2KE f = ⎯ 1 2 ⎯ (0.750 kg)(1.67 m/s) 2KE f = 1.05 J∆KE = KE f − KE i = 1.05 J − 5.12 J∆KE =−4.07 JCopyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions225

PRACTICE 6FKinetic energy in perfectly inelastic collisions1. A 0.25 kg arrow with a velocity of 12 m/s to the west strikes and piercesthe center of a 6.8 kg target.a. What is the final velocity of the combined mass?b. What is the decrease in kinetic energy during the collision?2. During practice, a student kicks a 0.40 kg soccer ball with a velocity of8.5 m/s to the south into a 0.15 kg bucket lying on its side. The buckettravels with the ball after the collision.a. What is the final velocity of the combined mass?b. What is the decrease in kinetic energy during the collision?3. A 56 kg ice skater traveling at 4.0 m/s to the north suddenly grabs thehand of a 65 kg skater traveling at 12.0 m/s in the opposite direction asthey pass. Without rotating, the two skaters continue skating togetherwith joined hands.a. What is the final velocity of the two skaters?b. What is the decrease in kinetic energy during the collision?ELASTIC COLLISIONSelastic collisiona collision in which the totalmomentum and the total kineticenergy remain constantWhen a player kicks a soccer ball, the collision between the ball and the player’sfoot is much closer to elastic than the collisions we have studied so far. In this case,elastic means that the ball and the player’s foot remain separate after the collision.In an elastic collision, two objects collide and return to their originalshapes with no change in total kinetic energy. After the collision, the twoobjects move separately. In an elastic collision, both the total momentum andthe total kinetic energy remain constant.NSTATOPIC: CollisionsGO TO: www.scilinks.orgsciLINKS CODE: HF2064Most collisions are neither elastic nor perfectly inelasticIn the everyday world, most collisions are not perfectly inelastic. That is, collidingobjects do not usually stick together and continue to move as oneobject. However, most collisions are not elastic, either. Even nearly elastic collisions,such as those between billiard balls or between a football player’s footand the ball, result in some decrease in kinetic energy. For example, a footballdeforms when it is kicked. During this deformation, some of the kinetic energyis converted to internal elastic potential energy. In most collisions, someof the kinetic energy is also converted into sound, such as the click of billiardballs colliding. In fact, any collision that produces sound is not elastic; thesound represents a decrease in kinetic energy.226Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

Elastic and perfectly inelastic collisions are limiting cases; most collisionsactually fall into a category between these two extremes. In this third categoryof collisions, called inelastic collisions, the colliding objects bounce and moveseparately after the collision, but the total kinetic energy decreases in the collision.For the problems in this book, we will consider all collisions in which theobjects do not stick together to be elastic collisions. This means that we willassume that the total momentum and the total kinetic energy remain constantin all collisions that are not perfectly inelastic.Kinetic energy is conserved in elastic collisionsFigure 6-12 shows an elastic head-on collision between two soccer balls ofequal mass. Assume, as in earlier examples, that the balls are isolated on a frictionlesssurface and that they do not rotate. The first ball is moving to theright when it collides with the second ball, which is moving to the left. Whenconsidered as a whole, the entire system has momentum to the left.After the elastic collision, the first ball moves to the left and the secondball moves to the right. The magnitude of the momentum of the first ball,which is now moving to the left, is greater than the magnitude of themomentum of the second ball, which is now moving to the right. When consideredtogether, the entire system has momentum to the left, just as beforethe collision.Another example of a nearly elastic collision is the collision between a golfball and a club. After a golf club strikes a stationary golf ball, the golf ball movesat a very high speed in the same direction as the golf club. The golf club continuesto move in the same direction, but its velocity decreases so that themomentum lost by the golf club is equal to and opposite the momentumgained by the golf ball. If a collision is perfectly elastic, the total momentum andthe total kinetic energy remain constant throughout the collision.MOMENTUM AND KINETIC ENERGY REMAIN CONSTANTIN AN ELASTIC COLLISIONElastic and InelasticCollisionsMATERIALS LIST✔ 2 or 3 small balls ofdifferent typesSAFETY CAUTIONPerform this lab in an open space,preferably outdoors, away fromfurniture and other people.Drop one of the balls from shoulderheight onto a hard-surfacedfloor or sidewalk. Observe themotion of the ball before and after itcollides with the ground. Next,throw the ball down from the sameheight. Perform several trials, givingthe ball a different velocity eachtime. Repeat with the other balls.During each trial, observe theheight to which the ball bounces.Rate the collisions from most nearlyelastic to most inelastic. Describewhat evidence you have for oragainst conservation of kineticenergy and conservation ofmomentum for each collision.Based on your observations, do youthink the equation for elastic collisionsis useful to make predictions?m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f1⎯2 ⎯ 2m 1 v 1,i + ⎯ 1 2 ⎯ 2m 2 v 2,i = ⎯ 1 2 ⎯ 2m 1 v 1,f + ⎯ 1 2 ⎯ 2m 2 v 2,fRemember that v is positive if an object moves to the right and negative if itmoves to the left.(a) Initial(b) Impulse(c) Finalp A p B ∆p A = F ∆t∆p B = −F t p Ap B Figure 6-12In an elastic collision like thisone (b), both objects return totheir original shapes and move separatelyA BA BABafter the collision (c).∆Copyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions227

SAMPLE PROBLEM 6GElastic collisionsPROBLEMSOLUTION1. DEFINEA 0.015 kg marble moving to the right at 0.225 m/s makes an elastic headoncollision with a 0.030 kg shooter marble moving to the left at 0.180 m/s.After the collision, the smaller marble moves to the left at 0.315 m/s.Assume that neither marble rotates before or after the collision and thatboth marbles are moving on a frictionless surface. What is the velocity ofthe 0.030 kg marble after the collision?Given: m 1 = 0.015 kg m 2 = 0.030 kgv 1,i = 0.225 m/s to the right =+0.225 m/sv 2,i = 0.180 m/s to the left =−0.180 m/sv 1,f = 0.315 m/s to the left =−0.315 m/sUnknown: v 2,f = ?Diagram:0.225 m/s –0.180 m/sm 10.015 kgm 20.030 kg2. PLANChoose an equation or situation: Use the equation for the conservation ofmomentum to find the final velocity of m 2 , the 0.030 kg marble.m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,fRearrange the equation(s) to solve for the unknown(s): Rearrange the equationto isolate the final velocity of m 2 .m 2 v 2,f = m 1 v 1,i + m 2 v 2,i − m 1 v 1,fv 2,f =m 1 v 1,i + m 2 v 2,i − m 1 v 1,f⎯⎯⎯m23. CALCULATESubstitute the values into the equation(s) and solve: The rearranged conservation-ofmomentumequation will allow you to isolate and solve for the final velocity.v 2,f =v 2,f =v 2,f =(0.015 kg)(0.225 m/s) + (0.030 kg)(−0.180 m/s) − (0.015 kg)(−0.315 m/s)⎯⎯⎯⎯⎯⎯⎯⎯0.030 kg(3.4 × 10 −3 kg •m/s) + (−5.4 × 10 −3 kg •m/s) − (−4.7 × 10 −3 kg •m/s)⎯⎯⎯⎯⎯⎯⎯0.030 kg2.7 × 10 −3 kg •m/s⎯⎯3.0 × 10−2 kgv 2,f = 9.0 × 10 −2 m/s to the right228Chapter 6Copyright © by Holt, Rinehart and Winston. All rights reserved.

The Ongoing Quest for Truth and Justice:Enacting and Annulling Argentina’s Amnesty Laws2009transfer of power’. 142 The military tried to resist these demands, and expected to use theirremaining months in power to ‘negotiate with the politicians in order to ensure theacceptance of at least one of their conditions – namely, no investigation of human rightsviolations’. 143 However, a divided military was unable to negotiate from a unified position,and according to Vacs, ‘the hard-liners’ threats of implementing an internal coup andnullifying the elections if an agreement were not reached lacked substance because of theweakness of their position in the wake of the Malvinas defeat’. 144 Consequently, the militarybowed to pressure, and Alfonsín was inaugurated on 10 December 1983, without the newadministration offering ‘any explicit guarantees to the military’. 145The new president faced enormous difficulties as military rule had left Argentina in a deepeconomic crisis with a population divided in its views on the military’s actions, and withlarge numbers of individuals who could potentially be implicated in its crimes. 146 Alfonsínwas, however, fortunate that the military had been weakened by their defeat in theMalvinas/Falklands conflict and was therefore initially unable to oppose his policy of limitedaccountability. 1475. THEORETICAL UNDERPINNINGS OF ALFONSÍN’S HUMAN RIGHTS POLICY:DEVELOPING ‘THE THEORY OF THE TWO DEMONS’During his election campaign, Alfonsín had begun to outline the human rights policy that hisgovernment would pursue once in office. In designing this policy, he and his advisers soughtto address two key dilemmas. Firstly, whilst they were sympathetic to the demands fromcivil society for truth and justice, they also felt it was necessary for the newly democraticstate to have an effective military under civilian control. Consequently, they sought tobalance accountability for individual military officers with the preservation and reform ofthe armed forces as an institution. 148 Secondly, they were conscious that repudiating themilitary’s actions without condemning the left-wing violence of the 1970s could increase the‘chasms that separated Argentine society’. 149 Therefore, they sought ‘to condemn the roleof revolutionary armed organizations without adding to the suffering of many of thesurviving members who had endured loss of family, torture, and disappearance’. 150 In142 Vacs (n 83) 32.143 Ibid 32.144 Ibid 32.145 Ibid 32.146 Alfonsín (n 8) 16.147 Sriram (n4) 117.148 Carina Perelli, ‘Settling Accounts with Blood Memory: The Case of Argentina’ (1992) 59 Soc. Res 415, 431.149 Ibid 431.150 Ibid 431.22

CHAPTER 6SummaryKEY IDEASSection 6-1 Momentum and impulse• Momentum is a vector quantity defined as the product of an object’s massand velocity, p = mv.• A net external force applied constantly to an object for a certain timeinterval will cause a change in the object’s momentum equal to the productof the force and the time interval, F∆t =∆p.• The product of the constant applied force and the time interval during whichthe force is applied is called the impulse of the force for the time interval.Section 6-2 Conservation of momentum• In all interactions between isolated objects, momentum is conserved.• In every interaction between two isolated objects, the change in momentumof the first object is equal to and opposite the change in momentumof the second object.Section 6-3 Elastic and inelastic collisions• In a perfectly inelastic collision, two objects stick together and move asone mass after the collision.• Momentum is conserved but kinetic energy is not conserved in a perfectlyinelastic collision.• In an inelastic collision, kinetic energy is converted to internal elasticpotential energy when the objects deform. Some kinetic energy is alsoconverted to sound energy and internal energy.• In an elastic collision, two objects return to their original shapes and moveaway from the collision separately.• Both momentum and kinetic energy are conserved in an elastic collision.• Few collisions are elastic or perfectly inelastic.KEY TERMSelastic collision (p. 226)impulse (p. 210)momentum (p. 208)perfectly inelastic collision(p. 222)Variable symbolsQuantitiesUnitsp momentum kg • m/skilogram-meters per secondF∆t impulse N • s Newton-seconds =kilogram-meters per secondCopyright © by Holt, Rinehart and Winston. All rights reserved.Momentum and Collisions231

CHAPTER 6Laboratory ExerciseOBJECTIVES•Measure the mass andvelocity of two carts.•Calculate the momentumof each cart.•Verify the law of conservationof momentum.MATERIALS LIST✔ two carts, one with a springmechanism✔ balance✔ metric rulerPROCEDURECBL AND SENSORS✔ CBL✔ CBL motion detector✔ graphing calculator with linkcable✔ support stand with clamp✔ set of masses✔ tape✔ two 25 × 25 cm squares ofposter boardRECORDING TIMER✔ recording timer✔ paper tape✔ stopwatchCONSERVATION OF MOMENTUMThe product of the mass of a moving object and its velocity is called itsmomentum. Studying momentum helps physicists understand the relationshipbetween the motion of two interacting objects.According to the impulse-momentum theorem, the change in an object’smomentum is equal to the product of the force acting on the object and thetime interval during which the force acts. Newton’s third law states that everyaction is accompanied by an equal and opposite reaction. Thus, when a springloadedcart pushes off against another cart, the force on the first cart is accompaniedby an equal and opposite force on the second cart. Both of these forcesact for exactly the same time interval, so the change in momentum of the firstcart is equal and opposite to the change in momentum of the second cart, inthe absence of other forces.In this experiment you will study the momentum of two carts with unequalmasses. The carts will be placed together and will move apart when a compressedspring between them is released. You will find the mass and velocity of each cartin order to compare the momentum before and after the carts move apart.SAFETY• Tie back long hair, secure loose clothing, and remove loose jewelry toprevent their getting caught in moving or rotating parts.PREPARATION1. Determine whether you will be using the CBL and sensors or the recordingtimer to perform this experiment. Read the entire lab procedure forthe appropriate method. Plan the steps you will take.2. Prepare a data table in your lab notebook with seven columns and fourrows. In the first row, label the first through seventh columns Trial,m 1 (kg), m 2 (kg), Cart 1 Distance (m), Cart 2 Distance (m), Cart 1 Timeinterval (s), and Cart 2 Time interval (s). In the first column, label thesecond through fourth rows 1, 2, and 3.3. Choose a location where both carts will be able to move at least 1.0 mwithout any obstacles.238Chapter 6Recording timer procedure begins on page 240.Copyright © by Holt, Rinehart and Winston. All rights reserved.