linear momentum isolated systems: conservation of...

Linear MomentumIsolated Systems: Conservation of

Momentum

Lana Sheridan

De Anza College

Feb 26, 2019

Last time

• introduced momentum

• Newton’s Second Law: more general form

• relation to force

• momentum vs kinetic energy

Overview

• conservation of momentum

• relation to Newton’s third law

• the rocket equation

• applying the rocket equation

• conservation of momentum in isolated systems

Conservation of Linear Momentum

For an isolated system, ie. a system with no external forces, totallinear momentum is conserved.

9.2 Analysis Model: Isolated System (Momentum) 251

Example 9.1 The Archer

Let us consider the situation proposed at the beginning of Section 9.1. A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s (Fig. 9.2). With what velocity does the archer move across the ice after firing the arrow?

Conceptualize You may have conceptualized this problem already when it was introduced at the beginning of Section 9.1. Imagine the arrow being fired one way and the archer recoiling in the opposite direction.

Categorize As discussed in Section 9.1, we cannot solve this problem with models based on motion, force, or energy. Nonetheless, we can solve this problem very eas-ily with an approach involving momentum. Let us take the system to consist of the archer (including the bow) and the arrow. The system is not isolated because the gravitational force and the normal force from the ice act on the system. These forces, however, are vertical and perpendicular to the motion of the system. There are no external forces in the horizontal direction, and we can apply the isolated system (momentum) model in terms of momentum com-ponents in this direction.

Analyze The total horizontal momentum of the system before the arrow is fired is zero because nothing in the sys-tem is moving. Therefore, the total horizontal momentum of the system after the arrow is fired must also be zero. We choose the direction of firing of the arrow as the positive x direction. Identifying the archer as particle 1 and the arrow as particle 2, we have m1 5 60 kg, m2 5 0.030 kg, and v

S2f 5 85 î m/s.

AM

S O L U T I O N

Figure 9.2 (Example 9.1) An archer fires an arrow horizontally to the right. Because he is standing on frictionless ice, he will begin to slide to the left across the ice.

Using the isolated system (momentum) model, begin with Equation 9.5:

DpS 5 0 S pSf 2 pS

i 5 0 S pS

f 5 pS

i S m1 vS

1f 1 m2 vS

2f 5 0

Solve this equation for vS1f and substitute numerical values:

vS1f 5 2m 2m1

vS2f 5 2a0.030 kg60 kg b 185 î m/s 2 5 20.042 î m/s

Analysis Model Isolated System (Momentum)Imagine you have identified a system to be analyzed and have defined a system boundary. If there are no external forces on the system, the system is isolated. In that case, the total momentum of the system, which is the vector sum of the momenta of all members of the system, is conserved:

DpStot 5 0 (9.5)

Examples:

each other (Chapter 21)

Momentum

Systemboundary

If no external forces act on the system, the total momentum of the system is constant.

continued

Finalize The negative sign for vS1f indicates that the archer is moving to the left in Figure 9.2 after the arrow is fired, in the direction opposite the direction of motion of the arrow, in accordance with Newton’s third law. Because the archer

1Figures from Serway & Jewett.


For an isolated system, ie. a system with no external forces, totallinear momentum is conserved:

∆ #»p net = 0

This corresponds to a translational symmetry in the equations ofmotion.

(Note: before when speaking of energy “isolated” meant “not exchanging

energy” now, for momentum, it means, no external forces act on the

system.)

Newton’s Third Law and Conservation ofMomentum

248 Chapter 9 Linear Momentum and Collisions

applying a conservation principle, conservation of energy. Let us consider another situation and see if we can solve it with the models we have developed so far:

A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s. With what velocity does the archer move across the ice after firing the arrow?

From Newton’s third law, we know that the force that the bow exerts on the arrow is paired with a force in the opposite direction on the bow (and the archer). This force causes the archer to slide backward on the ice with the speed requested in the problem. We cannot determine this speed using motion models such as the particle under constant acceleration because we don’t have any information about the accel-eration of the archer. We cannot use force models such as the particle under a net force because we don’t know anything about forces in this situation. Energy models are of no help because we know nothing about the work done in pulling the bow-string back or the elastic potential energy in the system related to the taut bowstring. Despite our inability to solve the archer problem using models learned so far, this problem is very simple to solve if we introduce a new quantity that describes motion, linear momentum. To generate this new quantity, consider an isolated system of two particles (Fig. 9.1) with masses m1 and m2 moving with velocities v

S1 and v

S2 at

an instant of time. Because the system is isolated, the only force on one particle is that from the other particle. If a force from particle 1 (for example, a gravitational force) acts on particle 2, there must be a second force—equal in magnitude but opposite in direction—that particle 2 exerts on particle 1. That is, the forces on the particles form a Newton’s third law action–reaction pair, and F

S12 5 2 F

S21. We can

express this condition as

FS

21 1 FS

12 5 0

From a system point of view, this equation says that if we add up the forces on the particles in an isolated system, the sum is zero. Let us further analyze this situation by incorporating Newton’s second law. At the instant shown in Figure 9.1, the interacting particles in the system have accel-erations corresponding to the forces on them. Therefore, replacing the force on each particle with maS for the particle gives

m1 aS

1 1 m2 aS

2 5 0

Now we replace each acceleration with its definition from Equation 4.5:

m1 d vS1dt

1 m2 d vS2dt

5 0

If the masses m1 and m2 are constant, we can bring them inside the derivative oper-ation, which gives

d 1m1 vS1 2dt

1d 1m2 vS2 2

dt5 0

ddt1m1 vS1 1 m2 vS2 2 5 0 (9.1)

Notice that the derivative of the sum m1 vS

1 1 m2 vS

2 with respect to time is zero. Consequently, this sum must be constant. We learn from this discussion that the quantity mvS for a particle is important in that the sum of these quantities for an isolated system of particles is conserved. We call this quantity linear momentum:

The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity vS is defined to be the product of the mass and velocity of the particle:

pS ; mvS (9.2)

Definition of linear momentum of a particle

m2

m1F21S

F12S

v1S

v2S

Figure 9.1 Two particles inter-act with each other. According to Newton’s third law, we must have FS

12 5 2 FS

21.

Newton’s third law for two interacting particles:

#»

F21 = −#»

F12


248 Chapter 9 Linear Momentum and Collisions

applying a conservation principle, conservation of energy. Let us consider another situation and see if we can solve it with the models we have developed so far:

A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s. With what velocity does the archer move across the ice after firing the arrow?

From Newton’s third law, we know that the force that the bow exerts on the arrow is paired with a force in the opposite direction on the bow (and the archer). This force causes the archer to slide backward on the ice with the speed requested in the problem. We cannot determine this speed using motion models such as the particle under constant acceleration because we don’t have any information about the accel-eration of the archer. We cannot use force models such as the particle under a net force because we don’t know anything about forces in this situation. Energy models are of no help because we know nothing about the work done in pulling the bow-string back or the elastic potential energy in the system related to the taut bowstring. Despite our inability to solve the archer problem using models learned so far, this problem is very simple to solve if we introduce a new quantity that describes motion, linear momentum. To generate this new quantity, consider an isolated system of two particles (Fig. 9.1) with masses m1 and m2 moving with velocities v

S1 and v

S2 at

an instant of time. Because the system is isolated, the only force on one particle is that from the other particle. If a force from particle 1 (for example, a gravitational force) acts on particle 2, there must be a second force—equal in magnitude but opposite in direction—that particle 2 exerts on particle 1. That is, the forces on the particles form a Newton’s third law action–reaction pair, and F

S12 5 2 F

S21. We can

express this condition as

FS

21 1 FS

12 5 0

From a system point of view, this equation says that if we add up the forces on the particles in an isolated system, the sum is zero. Let us further analyze this situation by incorporating Newton’s second law. At the instant shown in Figure 9.1, the interacting particles in the system have accel-erations corresponding to the forces on them. Therefore, replacing the force on each particle with maS for the particle gives

m1 aS

1 1 m2 aS

2 5 0

Now we replace each acceleration with its definition from Equation 4.5:

m1 d vS1dt

1 m2 d vS2dt

5 0

If the masses m1 and m2 are constant, we can bring them inside the derivative oper-ation, which gives

d 1m1 vS1 2dt

1d 1m2 vS2 2

dt5 0

ddt1m1 vS1 1 m2 vS2 2 5 0 (9.1)

Notice that the derivative of the sum m1 vS

1 1 m2 vS

2 with respect to time is zero. Consequently, this sum must be constant. We learn from this discussion that the quantity mvS for a particle is important in that the sum of these quantities for an isolated system of particles is conserved. We call this quantity linear momentum:

The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity vS is defined to be the product of the mass and velocity of the particle:

pS ; mvS (9.2)

Definition of linear momentum of a particle

m2

m1F21S

F12S

v1S

v2S

Figure 9.1 Two particles inter-act with each other. According to Newton’s third law, we must have FS

12 5 2 FS

21.Newton’s third law for two interacting particles:

#»

F21 = −#»

F12d #»p 1dt

= −d #»p 2dt

d

dt( #»p 1 +

#»p 2) = 0

Implies:#»p net =

#»p 1 +#»p 2 does not change with time. Or, ∆

#»p net = 0.


Newton’s third law ⇔ conservation of momentum

No external forces (only internal action-reaction pairs):

∆ #»p net = 0

The Rocket EquationA case where the mass is changing.

A famous example: what happens to a rocket as it burns fuel.(“The rocket equation”)

The Rocket Equation

Main idea: conserve momentum between the rocket and theejected propellant.

Suppose that the rocket burns fuel at a steady rate with respect totime, and let the initial mass of the rocket and propellent be mi(at time ti ) and the final mass be mf (at tf ).

Newton’s third law:

#»

F rocket = −#»

F exhaust

0Figure by Wikipedia user Skorkmaz.

The Rocket Equation

Consider this interaction in the frame of the rocket and supposethe force Fr on the rocket is constant with time.

Newton’s Second Law for the rocket:

#»

F r =d #»p

dt= m

d #»v

dt+ #»v

dm

dt

In the rocket frame, the rocket is not moving ( #»v = 0).

Newton’s second law for the exhaust:

#»

F e = med #»vedt

+ #»vedmedt

Constant force ⇒ constant exhaust velocity in the frame of therocket. (dvedt = 0)

The Rocket Equation



#»

F r =d #»p

dt= m

d #»v

dt+��

0

#»vdm

dt



#»

F e = med #»vedt

+ #»vedmedt


The Rocket Equation



#»

F r =d #»p

dt= m

d #»v

dt+��

0

#»vdm

dt



#»

F e =��>

0

med #»vedt

+ #»vedmedt


The Rocket EquationSo, if #»v is the velocity of the rocket

−#»

F e =#»

F r

− #»vedmedt

= md #»v

dt

but dmedt = −dmdt since the rate mass in ejected is equal to the rate

of loss of rocket mass:

#»vedm

dt= m

d #»v

dt

Consider the x-axis to drop the vector notation, noticing #»ve pointsin the opposite direction of #»v and d

#»vdt . Therefore,

−vedm

dt= m

dv

dt

The Rocket Equation

Rocket’s change in velocity, ∆v?

dv

dt= −

vem

dm

dt

∆v = −

∫ tfti

vem(t)

dm

dtdt

= −

∫mfmi

vem

dm

= −ve[

ln(m)]mfmi

∆v = ve ln

(mimf

)(the “Tsiolkovsky rocket equation”)

The Rocket Equation

For a rocket burning fuel at a constant rate:

vf = vi + ve ln

(mimf

)

where mi is the initial mass of the rocket and mf is the final(smaller) mass.

The thrust on an object is the forward force on the objectgenerated by engines / a propulsion system.

For a contant velocity of the exhaust, ve :

Thrust =

∣∣∣∣ve dmdt∣∣∣∣

The Rocket EquationThe last slide gives the most commonly stated version of therocket equation. The vector form of these equations are:

(Both for a rocket burning fuel at a constant rate, constantexhaust velocity)

#»v f =#»v i −

#»v e ln

(mimf

)

The minus sign indicates that the exhaust velocity is opposite tothe direction of the acceleration of the rocket.

#»

FThrust =#»v e

dm

dt

The direction of the exhaust velocity is opposite to the direction ofthe acceleration of the rocket, but the rocket’s mass is decreasing,so dmdt is negative.

Example 9.17

A rocket moving in space, far from all other objects, has a speed of3.0× 103 m/s relative to the Earth. Its engines are turned on, andfuel is ejected in a direction opposite the rocket’s motion at aspeed of 5.0× 103 m/s relative to the rocket.1

(a) What is the speed of the rocket relative to the Earth once therocket’s mass is reduced to half its mass before ignition?

(b) What is the thrust on the rocket if it burns fuel at the rate of50 kg/s?

3Serway & Jewett, page 279.

Example 9.17

(a) Speed with half of mass gone? Let initial mass be mi

vf = vi + ve ln

(mimf

)

vf = (3.0× 103) + (5.0× 103) ln(

mi(1/2)mi

)vf = (3.0× 103) + (5.0× 103) ln (2)

vf = 6.5× 103 m/s

(b) Thrust if∣∣dmdt

∣∣ = 50 kg/s?FThrust =


FThrust = (5.0× 103)(50)FThrust = 2.5× 105 N

Example 9.17

(a) Speed with half of mass gone? Let initial mass be mi

vf = vi + ve ln

(mimf

)vf = (3.0× 103) + (5.0× 103) ln

(mi

(1/2)mi

)vf = (3.0× 103) + (5.0× 103) ln (2)

vf = 6.5× 103 m/s

(b) Thrust if∣∣dmdt

∣∣ = 50 kg/s?FThrust =


FThrust = (5.0× 103)(50)FThrust = 2.5× 105 N

Isolated Systems and Linear Momentum

If we have a system that interacts internally, but does notexperience external forces, momentum is conserved.

Example 9.2When discussing energy, we ignored the kinetic energy of the Earthwhen considering the energy of a system consisting of the Earthand a dropped ball. (∆Kball + ∆Ug = 0.) Verify this is areasonable thing to do.

Isolated Systems and Linear MomentumExample 9.2When discussing energy, we ignored the kinetic energy of the Earthwhen considering the energy of a system consisting of the Earthand a dropped ball. (∆Kball + ∆Ug = 0.) Verify this is areasonable thing to do.

Need to argue that ∆KEarth

Isolated Systems and Linear Momentum

Need to argue that ∆KEarth


So far we have been considering:

For an isolated system, ie. a system with no external forces, totallinear momentum is conserved.

9.2 Analysis Model: Isolated System (Momentum) 251

Example 9.1 The Archer

Let us consider the situation proposed at the beginning of Section 9.1. A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s (Fig. 9.2). With what velocity does the archer move across the ice after firing the arrow?

Conceptualize You may have conceptualized this problem already when it was introduced at the beginning of Section 9.1. Imagine the arrow being fired one way and the archer recoiling in the opposite direction.

Categorize As discussed in Section 9.1, we cannot solve this problem with models based on motion, force, or energy. Nonetheless, we can solve this problem very eas-ily with an approach involving momentum. Let us take the system to consist of the archer (including the bow) and the arrow. The system is not isolated because the gravitational force and the normal force from the ice act on the system. These forces, however, are vertical and perpendicular to the motion of the system. There are no external forces in the horizontal direction, and we can apply the isolated system (momentum) model in terms of momentum com-ponents in this direction.

Analyze The total horizontal momentum of the system before the arrow is fired is zero because nothing in the sys-tem is moving. Therefore, the total horizontal momentum of the system after the arrow is fired must also be zero. We choose the direction of firing of the arrow as the positive x direction. Identifying the archer as particle 1 and the arrow as particle 2, we have m1 5 60 kg, m2 5 0.030 kg, and v

S2f 5 85 î m/s.

AM

S O L U T I O N

Figure 9.2 (Example 9.1) An archer fires an arrow horizontally to the right. Because he is standing on frictionless ice, he will begin to slide to the left across the ice.

Using the isolated system (momentum) model, begin with Equation 9.5:

DpS 5 0 S pSf 2 pS

i 5 0 S pS

f 5 pS

i S m1 vS

1f 1 m2 vS

2f 5 0

Solve this equation for vS1f and substitute numerical values:

vS1f 5 2m 2m1

vS2f 5 2a0.030 kg60 kg b 185 î m/s 2 5 20.042 î m/s

Analysis Model Isolated System (Momentum)Imagine you have identified a system to be analyzed and have defined a system boundary. If there are no external forces on the system, the system is isolated. In that case, the total momentum of the system, which is the vector sum of the momenta of all members of the system, is conserved:

DpStot 5 0 (9.5)

Examples:

each other (Chapter 21)

Momentum

Systemboundary

If no external forces act on the system, the total momentum of the system is constant.

continued

Finalize The negative sign for vS1f indicates that the archer is moving to the left in Figure 9.2 after the arrow is fired, in the direction opposite the direction of motion of the arrow, in accordance with Newton’s third law. Because the archer

1Figures from Serway & Jewett.

Nonisolated Systems

What happens when the system is not isolated: when externalforces act?

The momentum of the system will change.

#»

F ext =d #»p sys

dt

How will it change?

Impulse & Nonisolated Systems

Generalized Newton’s second law:

#»

Fnet =d #»p

dt

where#»

Fnet =∑

i

#»

F i .How will it change?

#»p f =

∫#»

Fnet dt+#»p i

This change in momentum is called the impulse:

#»

I = ∆ #»p =

∫#»

Fnet dt

where#»

Fnet is the net external force

Nonisolated Systems

9.3 Analysis Model: Nonisolated System (Momentum) 255

continued

the ball and bat during the collision. When we use this approximation, it is impor-tant to remember that pSi and p

Sf represent the momenta immediately before and

after the collision, respectively. Therefore, in any situation in which it is proper to use the impulse approximation, the particle moves very little during the collision.

Q uick Quiz 9.3 Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2. (i) When a constant force is applied to object 1, it accelerates through a distance d in a straight line. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (a) p1 , p2 (b) p1 5 p2 (c) p1 . p2 (d) K1 , K2 (e) K1 5 K2 (f) K1 . K2 (ii) When a force is applied to object 1, it accelerates for a time interval Dt. The force is removed from object 1 and is applied to object 2. From the same list of choices, which statements are true after object 2 has accelerated for the same time interval Dt?

Q uick Quiz 9.4 Rank an automobile dashboard, seat belt, and air bag, each used alone in separate collisions from the same speed, in terms of (a) the impulse and (b) the average force each delivers to a front-seat passenger, from greatest to least.

Analysis Model Nonisolated System (Momentum)

Imagine you have identified a system to be analyzed and have defined a system boundary. If external forces are applied on the system, the system is nonisolated. In that case, the change in the total momentum of the system is equal to the impulse on the system, a statement known as the impulse–momentum theorem:

DpS 5 IS

(9.13)

Examples:

(Chapter 34)

Momentum

Systemboundary

Impulse

The change in the total momentum of the system is equal to the total impulse on the system.

Example 9.3 How Good Are the Bumpers?

In a particular crash test, a car of mass 1 500 kg col-lides with a wall as shown in Figure 9.4. The initial and final velocities of the car are vSi 5 215.0 î m/s and vSf 5 2.60 î m/s, respectively. If the collision lasts 0.150 s, find the impulse caused by the collision and the average net force exerted on the car.

Conceptualize The collision time is short, so we can imagine the car being brought to rest very rapidly and then moving back in the opposite direction with a reduced speed.

Categorize Let us assume the net force exerted on the car by the wall and friction from the ground is large compared with other forces on the car (such as

AM

S O L U T I O N+2.60 m/s

–15.0 m/sBefore

After

a b

Figure 9.4 (Example 9.3) (a) This car’s momentum changes as a result of its collision with the wall. (b) In a crash test, much of the car’s initial kinetic energy is transformed into energy associated with the damage to the car.

Hyun

dai M

otor

s/HO

/Lan

dov

Connection to Energy case

214 Chapter 8 Conservation of Energy

are familiar with work, we can simplify the appearance of equations by letting the simple symbol W represent the external work Wext on a system. For internal work, we will always use Wint to differentiate it from W.) The other four members of our list do not have established symbols, so we will call them TMW (mechanical waves), TMT (matter transfer), TET (electrical transmission), and TER (electromagnetic radiation). The full expansion of Equation 8.1 is

DK 1 DU 1 DEint 5 W 1 Q 1 TMW 1 TMT 1 TET 1 TER (8.2)

which is the primary mathematical representation of the energy version of the anal-ysis model of the nonisolated system. (We will see other versions of the nonisolated system model, involving linear momentum and angular momentum, in later chap-ters.) In most cases, Equation 8.2 reduces to a much simpler one because some of the terms are zero for the specific situation. If, for a given system, all terms on the right side of the conservation of energy equation are zero, the system is an isolated system, which we study in the next section. The conservation of energy equation is no more complicated in theory than the process of balancing your checking account statement. If your account is the sys-tem, the change in the account balance for a given month is the sum of all the transfers: deposits, withdrawals, fees, interest, and checks written. You may find it useful to think of energy as the currency of nature! Suppose a force is applied to a nonisolated system and the point of application of the force moves through a displacement. Then suppose the only effect on the system is to change its speed. In this case, the only transfer mechanism is work (so that the right side of Eq. 8.2 reduces to just W) and the only kind of energy in the system that changes is the kinetic energy (so that the left side of Eq. 8.2 reduces to just DK). Equation 8.2 then becomes

DK 5 W

which is the work–kinetic energy theorem. This theorem is a special case of the more general principle of conservation of energy. We shall see several more special cases in future chapters.

Q uick Quiz 8.1 By what transfer mechanisms does energy enter and leave (a) your television set? (b) Your gasoline-powered lawn mower? (c) Your hand-cranked pencil sharpener?

Q uick Quiz 8.2 Consider a block sliding over a horizontal surface with friction. Ignore any sound the sliding might make. (i) If the system is the block, this sys-tem is (a) isolated (b) nonisolated (c) impossible to determine (ii) If the system is the surface, describe the system from the same set of choices. (iii) If the system is the block and the surface, describe the system from the same set of choices.

Analysis Model Nonisolated System (Energy)Imagine you have identified a system to be analyzed and have defined a system boundary. Energy can exist in the system in three forms: kinetic, potential, and internal. The total of that energy can be changed when energy crosses the system boundary by any of six transfer methods shown in the diagram here. The total change in the energy in the system is equal to the total amount of energy that has crossed the system bound-ary. The mathematical statement of that concept is expressed in the conservation of energy equation:

DEsystem 5 o T (8.1)

Work Heat Mechanicalwaves

Mattertransfer

Electricaltransmission

Electromagneticradiation

Kinetic energyPotential energyInternal energy

Systemboundary

The change in the total amount of energy in the system is equal to the total amount of energy that crosses the boundary of the system.

Energy transfers cross theboundary

W =∫ #»F · d #»r

9.3 Analysis Model: Nonisolated System (Momentum) 255

continued

the ball and bat during the collision. When we use this approximation, it is impor-tant to remember that pSi and p

Sf represent the momenta immediately before and

after the collision, respectively. Therefore, in any situation in which it is proper to use the impulse approximation, the particle moves very little during the collision.

Q uick Quiz 9.3 Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2. (i) When a constant force is applied to object 1, it accelerates through a distance d in a straight line. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (a) p1 , p2 (b) p1 5 p2 (c) p1 . p2 (d) K1 , K2 (e) K1 5 K2 (f) K1 . K2 (ii) When a force is applied to object 1, it accelerates for a time interval Dt. The force is removed from object 1 and is applied to object 2. From the same list of choices, which statements are true after object 2 has accelerated for the same time interval Dt?

Q uick Quiz 9.4 Rank an automobile dashboard, seat belt, and air bag, each used alone in separate collisions from the same speed, in terms of (a) the impulse and (b) the average force each delivers to a front-seat passenger, from greatest to least.

Analysis Model Nonisolated System (Momentum)

Imagine you have identified a system to be analyzed and have defined a system boundary. If external forces are applied on the system, the system is nonisolated. In that case, the change in the total momentum of the system is equal to the impulse on the system, a statement known as the impulse–momentum theorem:

DpS 5 IS

(9.13)

Examples:

(Chapter 34)

Momentum

Systemboundary

Impulse

The change in the total momentum of the system is equal to the total impulse on the system.

Example 9.3 How Good Are the Bumpers?

In a particular crash test, a car of mass 1 500 kg col-lides with a wall as shown in Figure 9.4. The initial and final velocities of the car are vSi 5 215.0 î m/s and vSf 5 2.60 î m/s, respectively. If the collision lasts 0.150 s, find the impulse caused by the collision and the average net force exerted on the car.

Conceptualize The collision time is short, so we can imagine the car being brought to rest very rapidly and then moving back in the opposite direction with a reduced speed.

Categorize Let us assume the net force exerted on the car by the wall and friction from the ground is large compared with other forces on the car (such as

AM

S O L U T I O N+2.60 m/s

–15.0 m/sBefore

After

a b

Figure 9.4 (Example 9.3) (a) This car’s momentum changes as a result of its collision with the wall. (b) In a crash test, much of the car’s initial kinetic energy is transformed into energy associated with the damage to the car.

Hyun

dai M

otor

s/HO

/Lan

dov

Impulse crosses the boundary#»

I =∫ #»F dt = ∆ #»p

Summary

• conservation of momentum and Newton’s third law• the rocket equation• using the rocket equation

(Uncollected) Homework Serway & Jewett,

• Read along in Chapter 9.• Ch 9, onward from page 283. Probs: 1, 3, 5, 7, 61, 63

linear momentum isolated systems: conservation of...

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