6. the Impulse-Momentum Change Theorem

6. The Impulse-Momentum Change Theorem

6.1. Momentum

The sports announcer says "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter."

Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team which is on the move has the momentum. If an object is in motion (on the move) then it has momentum.

Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.

Momentum = mass • velocity

In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as

p = m • v where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.

1 The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units which are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s.

Momentum is a vector quantity. To fully describe the momentum of a 5- kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction.

From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object which is at rest is 0. Objects at rest do not have momentum - they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects.

The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object.

Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s.

2 If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.

Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving."

Check Your Understanding

Express your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers.

1. Determine the momentum of a ...

a. 60-kg halfback moving eastward at 9 m/s.

b. 1000-kg car moving northward at 20 m/s.

c. 40-kg freshman moving southward at 2 m/s.

2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...

a. its velocity were doubled.

b. its velocity were tripled.

c. its mass were doubled (by adding more passengers and a greater load)

d. both its velocity were doubled and its mass were doubled.

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3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field.

Compare the velocities of these three players.

How many times greater is the velocity of the halfback and the velocity of the tight end than the velocity of the lineman?

Which player has the greatest momentum? Explain.

A. The tight end travels twice the distance of the lineman in the same amount of time. Thus, the tight end is twice as fast (vtight end = 6 m/s). The halfback travels three times the distance of the lineman in the same amount of time. Thus, the halfback is three times as fast (vhalfback = 9 m/s).

B. Both the halfback and the tight end have the greatest momentum. The each have the same amount of momentum - 540 kg*m/s. The lineman only has 360 kg*m/s.

4 6.2. Momentum and Impulse Connection

Momentum is a commonly used term in sports. When a sports announcer says that a team has the momentum they mean that the team is really on the move and is going to be hard to stop.

The term momentum is a physics concept. Any object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum which an object has, the harder that it is to stop.

Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed.

The concepts should not seem like abstract information to you. You have observed this a number of times if you have watched the sport of football. In football, the defensive players apply a force for a given amount of time to stop the momentum of the offensive player who has the ball. You have also experienced this a multitude of times while driving. As you bring your car to a halt when approaching a stop sign or stoplight, the brakes serve to apply a force to the car for a given amount of time to change the car's momentum. An object with momentum can be stopped if a force is applied against it for a given amount of time.

A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down.

If the force acts opposite the object's motion, it slows the object down.

5 If a force acts in the same direction as the object's motion, then the force speeds the object up.

Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

These concepts are merely an outgrowth of Newton's second law.

Newton's second law (Fnet = m • a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.

If both sides of the above equation are multiplied by the quantity t, a new equation results.

This equation represents one of two primary principles to be used in the analysis of collisions during this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m• v must be the change in momentum. The equation really says that the

Impulse = Change in momentum

One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first

6 law which we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. The law can be expressed this way:

In a collision, an object experiences a force for a specific amount of time which results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form,

F • t = m •Δv.

In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed and thus his momentum. If the motion was represented by a ticker tape diagram, it might appear as follows:

At approximately the tenth dot on the diagram, the collision occurs and lasts for a certain amount of time; in terms of dots, the collision lasts for a time equivalent to approximately nine dots. In the halfback-defensive back collision, the halfback experiences a force which lasts for a certain amount of time to change his momentum. Since the collision causes the rightward-moving halfback to slow down, the force on the halfback must have been directed leftward. If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N•s. This impulse would cause a momentum change of 720 kg•m/s. In a collision, the impulse experienced by an object is always equal to the momentum change.

7 Now consider a collision of a tennis ball with a wall. Depending on the physical properties of the ball and wall, the speed at which the ball rebounds from the wall upon colliding with it will vary. The diagrams below depict the changes in velocity of the same ball. For each representation (vector diagram, velocity-time graph, and ticker tape pattern), indicate which case (A or B) has the greatest change in velocity, greatest acceleration, greatest momentum change, and greatest impulse. Support each answer.

Vector Diagram

Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse?

a. The velocity change is greatest in case B. The velocity changes from +30 m/s to -28 m/s. This is a change of 58 m/s (-) and is greater than in case A (-15 m/s). b. The acceleration is greatest in case B. Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above) c. The momentum change is greatest in case B. Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above). d. The impulse is greatest in case B. Impulse equals momentum change and the momentum change is greatest in case B (as stated above). 8 Velocity-Time Graph

Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse?

a. The velocity change is greatest in case A. The v changes from +5 m/s to -3 m/s. This is a change of 8 m/s (-) and is greater than in case B (-4 m/s). b. The acceleration is greatest in case A. Acceleration depends on velocity change and the velocity change is greatest in case A (as stated above). c. The momentum change is greatest in case A. Momentum change depends on velocity change and the velocity change is greatest in case A (as stated above). d. The impulse is greatest in case A. Impulse equals momentum change and the momentum change is greatest in case A (as stated above).

9 Ticker Tape Diagram

Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse?

a. The velocity change is greatest in case B. In each case the initial velocity is the same. In case B, the object rebounds in the opposite direction with a greater speed than in case A. This is equivalent to a change from +10 m/s to -5 m/s; whereas, case A has a change from +10 m/s to -2 m/s. b. The acceleration is greatest in case B. Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above) c. The momentum change is greatest in case B. Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above). d. The impulse is greatest in case B. Impulse equals momentum change and the momentum change is greatest in case B (as stated above)

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Observe that each of the collisions above involve the rebound of a ball off a wall. Observe that the greater the rebound effect, the greater the acceleration, momentum change, and impulse. A rebound is a special type of collision involving a direction change in addition to a speed change. The result of the direction change is a large velocity change. On occasions in a rebound collision, an object will maintain the same or nearly the same speed as it had before the collision. Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force.

Use the impulse-momentum change principle to fill in the blanks in the following rows of the table. As you do, keep these three major truths in mind:

• the impulse experienced by an object is the force•time • the momentum change of an object is the mass•velocity change • the impulse equals the momentum change •

Force time Impulse Mom. Change Mass Vel. Change

(N) (s) (N*s) (kg*m/s) (kg) (m/s) 1. 0.010 10 -4

2. 0.100 -40 10

3. 0.010 -200 50

4. -20 000 -200 -8

5. -200 1.0 50

11 There are a few observations which can be made in the above table which relate to the computational nature of the impulse-momentum change theorem.

First, observe that the answers in the table above reveal that the third and fourth columns are always equal; that is, the impulse is always equal to the momentum change. Observe also that the if any two of the first three columns are known, then the remaining column can be computed. This is true because the impulse=force • time. Knowing two of these three quantities allows us to compute the third quantity. And finally, observe that knowing any two of the last three columns allows us to compute the remaining column. This is true since momentum change = mass • velocity change.

There are also a few observations which can be made which relate to the qualitative nature of the impulse-momentum theorem. An examination of rows 1 and 2 show that force and time are inversely proportional; for the same mass and velocity change, a tenfold increase in the time of impact corresponds to a tenfold decrease in the force of impact. An examination of rows 1 and 3 show that mass and force are directly proportional; for the same time and velocity change, a fivefold increase in the mass corresponds to a fivefold increase in the force required to stop that mass. Finally, an examination of rows 3 and 4 illustrate that mass and velocity change are inversely proportional; for the same force and time, a twofold decrease in the mass corresponds to a twofold increase in the velocity change.

12 Check Your Understanding

Express your understanding of the impulse-momentum change theorem by answering the following questions. Click the button to view the answers.

1. A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds.

Which cart (#1 or #2) has the greatest acceleration? Explain.

Cart #2 has the greatest acceleration. Recall that acceleration depends on force and mass. They each have the same mass, yet cart #2 has the greater force.

Which cart (#1 or #2) has the greatest impulse? Explain.

The impulse is the same for each cart. Impulse is force*time and can be calculated to be 1.0 N*s for each cart.

Which cart (#1 or #2) has the greatest change in momentum? Explain.

The momentum change is the same for each cart. Momentum change equals the impulse; if each cart has the same impulse, then it would follow that they have the same momentum change.

13 2. In a physics demonstration, two identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below.

Which balloon (A or B) has the greatest acceleration? Explain.

Balloon B has the greatest acceleration. The rate at which the velocity changes is greatest for Balloon B; this is shown by the fact that the speed (distance/time) changes most rapidly.

Which balloon (A or B) has the greatest final velocity? Explain.

Balloon B has the greatest final velocity. At the end of the diagram, the distance traveled in the last interval is greatest for Balloon B.

Which balloon (A or B) has the greatest momentum change? Explain.

Balloon B has the greatest momentum change. Since the final velocity is greatest for Balloon B, its velocity change is also the greatest. Momentum change depends on velocity change. The balloon with the greatest velocity change will have the greatest momentum change.

14 Which balloon (A or B) experiences the greatest impulse? Explain.

Balloon B has the greatest impulse. Impulse is equal to momentum change. If balloon B has the greatest momentum change, then it must also have the greatest impulse.

3. Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below.

At what approximate location on the diagram (in terms of dots) does each car begin to experience the impulse?

The slowing down occurs at approximately the ninth dot (plus or minus a dot). The diagram shows that it is at that location that the cars begin to slow down.

Which car (A or B) experiences the greatest acceleration? Explain.

Car A has the greatest acceleration. The velocity change of each car is the same. (They start with the same velocity and each finish with zero velocity.) Yet car A accomplishes this change in less time. Car A accelerates "most rapidly." 15

Which car (A or B) experiences the greatest change in momentum? Explain.

The momentum change is the same for each car. The velocity change of each car is the same (they start with the same velocity and each finish with zero velocity), and the mass of each car is the same. Thus, the momentum change is the same for each car.

Which car (A or B) experiences the greatest impulse? Explain.

The impulse is the same for each car. The impulse equals the momentum change. If the momentum change is the same for each car, then so must be the impulse.

4. The diagram to the right depicts the before- and after-collision speeds of a car which undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car crumples up and sticks to the wall.

a. In which case (A or B) is the change in velocity the greatest? Explain.

Case A has the greatest velocity change. The velocity change is -9 m/s in case A and only -5 m/s in case B.

16 b. In which case (A or B) is the change in momentum the greatest? Explain.

Case A has the greatest momentum change. The momentum change is dependent upon the velocity change; the object with the greatest velocity change has the greatest momentum change.

c. In which case (A or B) is the impulse the greatest? Explain.

The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for Car A, then so must be the impulse.

d. In which case (A or B) is the force which acts upon the car the greatest (assume contact times are the same in both cases)? Explain.

The impulse is greatest for Car A. The force is related to the impulse (I=F*t). The bigger impulse for Car A is attributed to the greater force upon Car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds.

17 6.3. Real-World Applications

In a previous part it was said that

In a collision, an object experiences a force for a given amount of time which results in its mass undergoing a change in velocity (i.e., which results in a momentum change).

There are four physical quantities mentioned in the above statement - force, time, mass, and velocity change. The force multiplied by the time is known as the impulse and the mass multiplied by the velocity change is known as the change in momentum. The impulse experienced by an object is always equal to the change in its momentum.

In terms of equations, this was expressed as

This is known as the impulse-momentum change theorem.

In this part, we will examine some real-world applications of the impulse-momentum change theorem. We will examine some physics in action in the real world. In particular, we will focus upon

• the affect of collision time upon the amount of force an object experiences, and • the affect of rebounding upon the velocity change and hence the amount of force an object experiences.

As an effort is made to apply the impulse-momentum change theorem to a variety of real-world situations, keep in mind that the goal is to use the equation as a guide to thinking about how an alteration in the value of one variable might affect the value of another variable.

18 The Affect of Collision Time upon the Force

First we will examine the importance of the collision time in affecting the amount of force which an object experiences during a collision. In a previous part, it was mentioned that force and time are inversely proportional. An object with 100 units of momentum must experience 100 units of impulse in order to be brought to a stop. Any combination of force and time could be used to produce the 100 units of impulse necessary to stop an object with 100 units of momentum. This is depicted in the table below.

Combinations of Force and Time Required to Produce 100 units of Impulse

Force Time Impulse 100 1 100 50 2 100 25 4 100 10 10 100 4 25 100 2 50 100 1 100 100 0.1 1000 100

Observe that the greater the time over which the collision occurs, the smaller the force acting upon the object.

Thus, to minimize the affect of the force on an object involved in a collision, the time must be increased. And to maximize the affect of the force on an object involved in a collision, the time must be decreased.

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There are several real-world applications of this phenomena.

One example is the use of air bags in automobiles. Air bags are used in automobiles because they are able to minimize the affect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. When encountering a car collision, the driver and passenger tend to keep moving in accord with Newton's first law. Their motion carries them towards a windshield which results in a large force exerted over a short time in order to stop their momentum. If instead of hitting the windshield, the driver and passenger hit an air bag, then the time duration of the impact is increased. When hitting an object with some give1 such as an air bag, the time duration might be increased by a factor of 100. Increasing the time by a factor of 100 will result in a decrease in force by a factor of 100. Now that's physics in action.

The same principle explains why dashboards are padded. If the air bags do not deploy (or are not installed in a car), then the driver and passengers run the risk of stopping their momentum by means of a collision with the

1 the ability of a material or substance to bend or stretch when put under pressure 20 windshield or the dashboard. If the driver or passenger should hit the dashboard, then the force and time required to stop their momentum is exerted by the dashboard. Padded dashboards provide some give in such a collision and serve to extend the time duration of the impact, thus minimizing the affect of the force. This same principle of padding a potential impact area can be observed in gymnasiums (underneath the basketball hoops), in pole-vaulting pits, in baseball gloves and goalie mitts, on the fist of a boxer, inside the helmet of a football player, and on gymnastic mats. Now that's physics in action.

Fans of boxing frequently observe this same principle of minimizing the affect of a force by extending the time of collision. When a boxer recognizes that he will be hit in the head by his opponent, the boxer often relaxes his neck and allows his head to move backwards upon impact. In the boxing world, this is known as riding the punch. A boxer rides the punch in order to extend the time of impact of the glove with their head. Extending the time results in decreasing the force and thus minimizing the affect of the force in the collision. Merely increasing the collision time by a factor of ten would result in a tenfold decrease in the force. Now that's physics in action.

Nylon ropes are used in the sport of rock-climbing for the same reason. Rock climbers attach themselves to the steep cliffs by means of nylon ropes. If a rock climber should lose her grip on the rock, she will begin to fall. In such a situation, her momentum will ultimately be halted by means of the rope, thus preventing a disastrous fall to the ground below. The ropes are made of nylon or similar material because of its ability to stretch. If the rope is capable of stretching upon being pulled taut by the falling climber's mass, then it will apply a force upon the climber over a longer time period. Extending the time over which the climber's momentum is broken results in reducing the force exerted on the falling climber. For certain, the

21 rock climber can appreciate minimizing the affect of the force through the use of a longer time of impact. Now that's physics in action.

In racket and bat sports, hitters are often encouraged to follow-through when striking a ball. High speed films of the collisions between bats/rackets and balls have shown that the act of following through serves to increase the time over which a collision occurs. This increase in time must result in a change in some other variable in the impulse-momentum change theorem. Surprisingly, the variable which is dependent upon the time in such a situation is not the force. The force in hitting is dependent upon how hard the hitter swings the bat or racket, not the time of impact. Instead, the follow-through increases the time of collision and subsequently contributes to an increase in the velocity change of the ball. By following through, a hitter can hit the ball in such a way that it leaves the bat or racket with more velocity (i.e., the ball is moving faster). In tennis, baseball, racket ball, etc., giving the ball a high velocity often leads to greater success. Now that's physics in action.

You undoubtedly recall other illustrations of this principle. A common physics demonstration involves the catching of water balloons of varying speed and varying mass. A water balloon is thrown high into the air and successfully caught (i.e., caught without breaking). The key to the success of the demonstration is to contact the balloon with outstretched arms and carry the balloon for a meter or more before finally stopping its momentum. The effect of this strategy is to extend the time over which the collision occurred and so reduce the force. This same strategy is used by lacrosse players when catching the ball. The ball is "cradled" when caught; i.e., the lacrosse player reaches out for the ball and carries it inward toward her body as if

22 she were cradling a baby. The effect of this strategy is to lengthen the time over which the collision occurs and so reduce the force on the lacrosse ball. Now that's physics in action.

Another common physics demonstration involves throwing an egg into a bed sheet. The bed sheet is typically held by two trustworthy students and a volunteer is used to toss the egg at full speed into the bed sheet. The collision between the egg and the bed sheet lasts over an extended period of time since the bed sheet has some give in it. By extending the time of the collision, the affect of the force is minimized. In all my years, the egg has never broken when hitting the bed sheet. On occasion the volunteer has a wayward toss and is not as accurate as expected. The egg misses the bed sheet and collides with the wall. In these unexpected cases, the collision between wall and egg lasts for a short period of time, thus maximizing the affect of the force on the egg. The egg brakes and leaves the wall and floor in a considerable mess. And that's no yolk!

The Effect of Rebounding

Occasionally when objects collide, they bounce off each other as opposed to sticking to each other and traveling with the same speed after the collision. Bouncing off each other is known as rebounding. Rebounding involves a change in the direction of an object; the before- and after- collision direction is different. Rebounding was pictured and discussed earlier.

23 The importance of rebounding is critical to the outcome of automobile accidents.

In an automobile accident, two cars can either collide and bounce off each other or collide, crumple up and travel together with the same speed after the collision. But which would be more damaging to the occupants of the automobiles - the rebounding of the cars or the crumpling up of the cars?

Contrary to popular opinion, the crumpling up of cars is the safest type of automobile collision.

As mentioned above, if cars rebound upon collision, the momentum change will be larger and so will the impulse. A greater impulse will typically be associated with a bigger force.

Occupants of automobiles would certainly prefer small forces upon their bodies during collisions.

In fact, automobile designers and safety engineers have found ways to reduce the harm done to occupants of automobiles by designing cars which crumple upon impact. Automobiles are made with crumple zones.

Crumple zones are sections in cars which are designed to crumple up when the car encounters a collision. Crumple zones minimize the affect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse. Finally, the crumpling of the car lengthens the time over which the car's momentum is changed; by increasing the time of the collision, the force of the collision is greatly reduced.

24 6.4. The Law of Action-Reaction (Revisited)

A collision is an interaction between two objects which have made contact (usually) with each other. As in any interaction, a collision results in a force being applied to the two colliding objects. Such collisions are governed by Newton's laws of motion:

... in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

Newton's third law of motion is naturally applied to collisions between two objects.

In a collision between two objects, both objects experience forces which are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum).

According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the acceleration of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion. the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force which results during the collision.

25 Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club2 collides with a golf ball at rest upon a tee3, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects which causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.

Consider the collision between a moving seven-ball and an eight-ball that is at rest in the sport of table pool4. When the seven-ball collides with the eight-ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven- ball experiences a leftward force which causes it to slow down; the eight- ball experiences a rightward force which causes it to speed up. Since the two balls have equal masses, they will also experience equal

2 a long thin metal stick used in golf to hit the ball 3 a small object that you use in a game of golf to hold the ball above the ground before you hit it 4 a game in which you use a stick to hit numbered balls into holes around a table, which is often played in bars 26 accelerations. In a collision, there is a force on both objects which causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Consider the interaction between a male and female figure skater in pair figure skating. A woman (m = 45 kg) is kneeling on the shoulders of a man (m = 70 kg); the pair is moving along the ice at 1.5 m/s. The man gracefully tosses the woman forward through the air and onto the ice. The woman receives the forward force and the man receives a backward force. The force on the man is equal in magnitude and opposite in direction to the force on the woman. Yet the acceleration of the woman is greater than the acceleration of the man due to the smaller mass of the woman.

Many observers of this interaction have difficulty believing that the man experienced a backward force. "After all," they might argue, "the man did not move backward." Such observers are presuming that forces cause motion. In their minds, a backward force on the male skater would cause a backward motion. This is a common misconception. Forces cause acceleration, not motion. The male figure skater experiences a backwards force which causes his backwards acceleration. The male skater slows down while the woman skater speeds up. In every interaction (with no exception), there are forces acting upon the two interacting objects which are equal in magnitude and opposite in direction.

Collisions are governed by Newton's laws. The law of action-reaction (Newton's third law) explains the nature of the forces between the two interacting objects. According to the law, the force exerted by object 1 upon object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 upon object 1.

27 Check Your Understanding

Express your understanding of Newton's third law by answering the following questions. Click the button to check your answers.

1. While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the firefly or the force on the bus?

Trick Question! Each force is the same size. For every action, there is an equal ... (equal!). The fact that the firefly splatters only means that with its smaller mass, it is less able to withstand the larger acceleration resulting from the interaction. Besides, fireflies have guts and bug guts have a tendency to be splatterable. Windshields don't have guts. There you have it.

2. For years, space travel was believed to be impossible because there was nothing which rockets could push off of in space in order to provide the propulsion necessary to accelerate. This inability of a rocket to provide propulsion in space is because ...

a. space is void of air so the rockets have nothing to push off of.

b. gravity is absent in space.

c. space is void of air and so there is no air resistance in space.

28 d. ... nonsense! Rockets do accelerate in space and have been able to do so for a long time.

Answer: D

It is a common misconception that rockets are unable to accelerate in space. The fact is that rockets do accelerate. Rockets are able to accelerate due to the fact that they burn fuel and thrust the exhaust gases in a direction opposite the direction which they wish to accelerate.

3. Many people are familiar with the fact that a rifle recoils when fired. This recoil is the result of action-reaction force pairs. A gunpowder explosion creates hot gases which expand outward allowing the rifle to push forward on the bullet. Consistent with Newton's third law of motion, the bullet pushes backwards upon the rifle. The acceleration of the recoiling rifle is ...

a. greater than the acceleration of the bullet.

b. smaller than the acceleration of the bullet.

c. the same size as the acceleration of the bullet.

Answer: B

The force on the rifle equals the force on the bullet. Yet, acceleration depends on both force and mass. The bullet has a greater acceleration due to the fact that it has a smaller mass. Remember: acceleration and mass are inversely proportional.

29 Astronaut Catch

Imagine that you are hovering next to the space shuttle in earth-orbit and your buddy of equal mass who is moving 4 m/s (with respect to the ship) bumps into you. If she holds onto you, then how fast do the two of you move after the collision?

A question like this involves momentum principles. In any instance in which two objects collide and can be considered isolated from all other net forces, the conservation of momentum principle can be utilized to determine the post-collision velocities of the two objects. Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the two astronauts, the combined momentum of the two astronauts before the collision equals the combined momentum of the two astronauts after the collision.

The mathematics of this problem is simplified by the fact that before the collision, there is only one object in motion and after the collision both objects have the same velocity. That is to say, a momentum analysis would show that all the momentum was concentrated in the moving astronaut before the collision. And after the collision, all the momentum was the result of a single object (the combination of the two astronauts) moving at an easily predictable velocity. Since there is twice as much mass in motion after the collision, it must be moving with one-half the velocity. Thus, the two astronauts move together with a velocity of 2 m/s after the collision.

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30 6.5. Momentum Conservation Principle

One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value. To understand the basis of momentum conservation, let's begin with a short logical proof.

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as

31 Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. A common physics lab involves the dropping of a brick upon a cart in motion.

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The dropped brick is at rest and begins with zero momentum. The loaded cart (a cart with a brick on it) is in motion with considerable momentum. The actual momentum of the loaded cart can be determined using the velocity (often determined by a ticker tape analysis) and the mass. The total amount of momentum is the sum of the dropped brick's momentum (0 units) and the loaded cart's momentum. After the collision, the momenta of the two separate objects (dropped brick and loaded cart) can be determined from their measured mass and their velocity (often found from a ticker tape analysis). If momentum is conserved during the collision, then the sum of the dropped brick's and loaded cart's momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick. Momentum data for the interaction between the dropped brick and the loaded cart could be depicted in a table similar to the money table above.

Before After Change in Collision Collision Momentum Momentum Momentum Dropped Brick 0 units 14 units +14 units Loaded Cart 45 units 31 units -14 units Total 45 units 45 units

Note that the loaded cart lost 14 units of momentum and the dropped brick gained 14 units of momentum. Note also that the total momentum of the system (45 units) was the same before the collision as it was after the collision.

Carbrick.gif 33 Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand. The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West. Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.

Now suppose that a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The momentum of the medicine ball is 80 kg*m/s before the collision. The momentum of the clown is 0 m/s before the collision. The total momentum of the system before the collision is 80 kg*m/s. Therefore, the total momentum of the system after the collision must also be 80 kg*m/s. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg*m/s. Momentum is conserved in the collision.

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Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram.

Check Your Understanding

When fighting fires, a firefighter must use great caution to hold a hose which emits large amounts of water at high speeds. Why would such a task be difficult?

The hose is pushing lots of water (large mass) forward at a high speed. This means the water has a large forward momentum. In turn, the hose must have an equally large backwards momentum, making it difficult for the firefighters to manage.

35 6.6. Isolated Systems

Total system momentum is conserved for collisions occurring in isolated systems. But what makes a system of objects an isolated system? And is momentum conserved if the system is not isolated?

A system is a collection of two or more objects. An isolated system is a system which is free from the influence of a net external force which alters the momentum of the system. There are two criteria for the presence of a net external force; it must be...

• a force which originates from a source other than the two objects of the system • a force that is not balanced by other forces.

A system in which the only forces which contribute to the momentum change of an individual object are the forces acting between the objects themselves can be considered an isolated system.

Consider the collision of two balls on the billiards table. The collision occurs in an isolated system as long as friction is small enough that its influence upon the momentum of the billiard balls can be neglected. If so, then the only unbalanced forces acting upon the two balls are the contact forces which they apply to one another. These two forces are considered internal forces since they result from a source within the system - that source being the contact of the two balls. For such a collision, total system momentum is conserved.

If a system is not isolated, then the total system momentum is not conserved.

Because of the inevitability of friction and air resistance in any real collision, one might can conclude that no system is ever perfectly isolated. The reasoning would be that there will always be a resistance force of some kind robbing the system of its momentum. For this reason, the law of conservation of momentum must be some sort of pie-in-the-sky

36 idea which never has any applicability. Why does one ever need to learn a law that is always broken?

We would quickly agree that resistance forces such as friction and air resistance are inevitable.

However, one must be careful of overrating the impact of such forces on total system momentum. To illustrate, consider the collision between two 1000-kg cars - one which is initially at rest and one which is moving at 10 m/s (36 km/hr). Such a collision would last for a fraction of a second. Suppose the actual contact forces between cars occur for 0.01 seconds. Perhaps after this very short 0.01 seconds, the cars stick together and slide a short distance across the asphalt street. For such a realistic collision, a typical contact force might be close to 500 000 N. A typical friction force would be no larger than 10 000 N (and would probably be considerably less). Each force would deliver an impulse to the vehicles during the collision in order to change the momentum of the colliding vehicles. The 500 000 N contact force between vehicles is an internal force and does not serve to change the momentum of the system. The 10 000 N friction force is an external force and would alter the total system momentum by approximately 200 kg•m/s (100 kg•m/s for each car). This loss of system momentum must be compared to the total system momentum of 10 000 kg•m/s. In this very realistic case, friction would cause no more than a 2% loss of total system momentum. If the collision being analyzed involved objects with even greater speeds, the percent loss of total system momentum would be considerably less.

So one must again consider the charge that the law of conservation of momentum is a pie-in-the-sky idea. The best response is to say that it is a very accurate model approximating the exchange of momentum between colliding objects. In the collision just described, the worse case scenario is that the assumption of momentum conservation is 98% accurate. Because contact forces during collisions are so large compared to the inevitable resistance forces such as friction and air resistance, the law of momentum conservation is a great tool for analyzing collisions and providing an accurate estimate of a post-collision (or a pre-collision) speed.

37 6.7. Using Equations as a Recipe for Algebraic Problem-Solving

Consider the following problem:

A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.

If it can be assumed that the affect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system. Momentum should be conserved and the post-collision velocity (v) can be determined using a momentum table as shown below.

Before Collision After Collision Person 0 (60 kg) • v (15 kg) • (20 km/hr) Medicine ball (15 kg) • v = 300 kg • km/hr Total 300 kg • km/hr 300

38 Observe in the table above that the known information about the mass and velocity of the two objects was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball. To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:

60 • v + 15 • v = 300

75 • v = 300

v = 4 km/hr

Using algebra skills, it can be shown that v = 4 km/hr. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision.

Now consider a similar problem involving momentum conservation.

A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.

Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity (v) .

The collision between the ball and the catcher's mitt occurs in an isolated system, total system momentum is conserved. Thus, the total momentum before the collision (possessed solely by the baseball) equals the total 39 momentum after the collision (shared by the baseball and the catcher's mitt).

The table below depicts this principle of momentum conservation.

Before Collision After Collision Ball 0.15 kg • 45 m/s = 6.75 kg•m/s (0.15 kg) • v Catcher's Mitt 0 (0.25 kg) • v Total 6.75 kg•m/s 6.75 kg•m/s

Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expression 0.15 • v and 0.25 • v are used for the after-collision momentum of the baseball and catcher's mitt. To determine v (the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results:

0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s

0.40 kg • v = 6.75 kg•m/s

v = 16.9 m/s

Using algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.

40 6.8. Using Equations as a Guide to Thinking

A large fish is in motion at 2 m/s when it encounters a smaller fish which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish (and the smaller fish) after the collision?

The process of solving this problem involved using a conceptual understanding of the equation for momentum (p=m*v). This equation becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity.

The amount of mass in motion is increased from 3m to 4m (3m + m); that is mass is increased by a factor of 4/3 (1.33). To conserve momentum, an increase in mass by a factor of 1.33 must be accompanied by a decrease in velocity by a factor of 1.33. Thus divide the original velocity of 2 m/s by 4/3.

Answer: v = 1.5 m/s

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41 A railroad diesel engine has five times the mass of a boxcar. A diesel coasts backwards along the track at 4 m/s and couples together with the boxcar (initially at rest). How fast do the two trains cars coast after they have coupled together?

Answer: v = 3.3 m/s

The amount of mass in motion is increased from 5m to 6m (5m+m). That is, the total mass which is moving is increased by a factor of 6/5 (or 1.20). To conserve momentum, an increase in mass by a factor of 1.20 must be accompanied by a decrease in velocity by a factor of 1.20. Thus, divide the original velocity of 4 m/s by 1.2.

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42 6.9. Momentum Conservation in Explosions

Total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions.

In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the system (which is often a collection of fragments from the original object) have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved.

Momentum conservation is often demonstrated with a homemade cannon demonstration. A homemade cannon is placed upon a cart and loaded with a tennis ball. The cannon is equipped with a reaction chamber into which a small amount of fuel is inserted. The fuel is ignited, setting off an explosion which propels the tennis ball through the muzzle of the cannon. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball.

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In the exploding cannon demonstration, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved.

As another demonstration of momentum conservation, consider two low- friction carts at rest on a track. The system consists of the two individual carts initially at rest. The total momentum of the system is zero before the explosion. One of the carts is equipped with a spring loaded plunger which can be released by tapping on a small pin. The spring is compressed and the carts are placed next to each other. The pin is tapped, the plunger is released, and an explosion-like impulse sets both carts in motion along the track in opposite directions. One cart acquires a rightward momentum while the other cart acquires a leftward momentum. If 20 units of forward momentum are acquired by the rightward-moving cart, then 20 units of backwards momentum is acquired by the leftward-moving cart. The vector sum of the momentum of the individual carts is 0 units. Total system momentum is conserved.

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Equal and Opposite Momentum Changes

Just like in collisions, the two objects involved encounter the same force for the same amount of time directed in opposite directions. This results in impulses which are equal in magnitude and opposite in direction. And since an impulse causes and is equal to a change in momentum, both carts encounter momentum changes which are equal in magnitude and opposite in direction. If the exploding system includes two objects or two parts, this principle can be stated in the form of an equation as:

If the masses of the two objects are equal, then their post-explosion velocity will be equal in magnitude (assuming the system is initially at rest). If the masses of the two objects are unequal, then they will be set in motion by the explosion with different speeds. Yet even if the masses of the two objects are different, the momentum change of the two objects (mass • velocity change) will be equal in magnitude.

The diagram below depicts a variety of situations involving explosion-like impulses acting between two carts on a low-friction track. The mass of the carts is different in each situation. In each situation, total system momentum is conserved as the momentum change of one cart is equal and opposite the momentum change of the other cart.

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In each of the above situations, the impulse on the carts is the same - a value of 20 kg•cm/s (or cN•s). Since the same spring is used, the same impulse is delivered. Thus, each cart encounters the same momentum change in every situation - a value of 20 kg•cm/s. For the same momentum change, an object with twice the mass will encounter one-half the velocity change. And an object with four times the mass will encounter one-fourth the momentum change.

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