Linear Momentum - De Anza Collegenebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture29.pdf · Linear...

Linear Momentum

Lana Sheridan

De Anza College

Nov 6, 2017

Last time

• energy practice

Overview

• introducing momentum

• Newton’s Second Law: more general form

• relation to force

• relation to Newton’s third law

• conservation of momentum

• the rocket equation

Linear Momentum

Introduce a new quantity:

For a particle, linear momentum, p, is product of the particle’smass with its velocity.

p = mv

It is a vector.

Newton called it the “quantity of motion”.

Units: kg m/s

Linear Momentum



p = mv

It is a vector.


Units:

kg m/s

Linear Momentum



p = mv

It is a vector.


Units: kg m/s

Momentum vs. Kinetic energy

Momentum: p = mv

Kinetic energy: K = 12mv2 = p2

2m

• Both depend on m and v only.

• K is a scalar.

• p is a vector.

• In a system with many particles, can have∑

i pi = 0 even if∑i Ki 6= 0.

Question

Quick Quiz 9.11 Two objects have equal kinetic energies. How dothe magnitudes of their momenta compare?

(A) p1 < p2

(B) p1 = p2

(C) p1 > p2

(D) not enough information to tell

1From Serway & Jewett, page 250.

Question

Quick Quiz 9.11 Two objects have equal kinetic energies. How dothe magnitudes of their momenta compare?

(A) p1 < p2

(B) p1 = p2

(C) p1 > p2

(D) not enough information to tell ←


QuestionQuick Quiz 9.22 Your physical education teacher throws abaseball to you at a certain speed and you catch it. The teacher isnext going to throw you a medicine ball whose mass is ten timesthe mass of the baseball. You are given the following choices: Youcan have the medicine ball thrown with

(i) the same speed as the baseball(ii) the same momentum, or(iii) the same kinetic energy.

Rank these choices from easiest to hardest to catch, in terms ofwork done to catch the ball.

(A) (i), (ii), (iii)

(B) (iii), (ii), (i)

(C) (ii), (iii), (i)

(D) all are equivalent2From Serway & Jewett, page 250.

QuestionQuick Quiz 9.22 Your physical education teacher throws abaseball to you at a certain speed and you catch it. The teacher isnext going to throw you a medicine ball whose mass is ten timesthe mass of the baseball. You are given the following choices: Youcan have the medicine ball thrown with

(i) the same speed as the baseball(ii) the same momentum, or(iii) the same kinetic energy.

Rank these choices from easiest to hardest to catch, in terms ofwork done to catch the ball.

(A) (i), (ii), (iii)

(B) (iii), (ii), (i)

(C) (ii), (iii), (i) ←(D) all are equivalent


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 vS2f 5 85 i 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 pSi 5 0 S pSf 5 pSi S m1 vS1f 1 m2 vS2f 5 0

Solve this equation for vS1f and substitute numerical values:

vS1f 5 2m 2

m1 vS2f 5 2a0.030 kg

60 kgb 185 i m/s 2 5 20.042 i 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.

Conservation of Linear Momentum

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

d

dt

(∑i

pi

)= 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.)

Force and Momentum: Newton’s Second Law

Newton’s second law for a particle whose mass might change(more general!):

F =dp

dt

This is how Newton thought of his second law.

Also

Favg =∆p

∆t

Force and Momentum

This is the definition to use when mass is changing with time.

F =dp

dt

F =d(mv)

dt

F = mdv

dt+v

dm

dt

If m is constant dmdt = 0, and this reduces to:

F = m a

Generalized Newton’s second law example

#94, page 292

Sand from a stationary hopper falls onto a moving conveyor belt atthe rate of 5.00 kg/s as shown. The conveyor belt is supported byfrictionless rollers and moves at a constant speed of v = 0.750 m/sunder the action of a constant horizontal external force Fext

supplied by the motor that drives the belt. Find (a) the sand’s rateof change of momentum in the horizontal direction, (b) the forceof friction exerted by the belt on the sand, and (c) the externalforce Fext.

292 Chapter 9 Linear Momentum and Collisions

acceleration; the work–kinetic energy theorem, stating that the total work on the particle causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on the particle causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity 7.00 j m/s. Then, a constant net force 12.0 i N acts on the object for 5.00 s. (a) Calculate the object’s final velocity, using the impulse–momentum theorem. (b) Calculate its acceler-ation from aS 5 1 vSf 2 vSi 2/Dt. (c) Calculate its accel-eration from aS 5 g F

S/m . (d) Find the object’s vector

displacement from D rS 5 vSit 1 12 aSt 2. (e) Find the work

done on the object from W 5 FS

? D rS. (f) Find the final kinetic energy from 12mvf

2 5 12mvSf ? vSf . (g) Find the

final kinetic energy from 12mvi2 1 W. (h) State the result

of comparing the answers to parts (b) and (c), and the answers to parts (f) and (g).

91. A 2.00-g particle moving at 8.00 m/s makes a perfectly elastic head-on collision with a resting 1.00-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10.0 g. (c) Find the final kinetic energy of the incident 2.00-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

Challenge Problems 92. In the 1968 Olympic games, University of Oregon

jumper Dick Fosbury introduced a new technique of high jumping called the “Fosbury flop.” It contributed to raising the world record by about 30 cm and is cur-rently used by nearly every world-class jumper. In this technique, the jumper goes over the bar face-up while arching her back as much as possible as shown in Figure P9.92a. This action places her center of mass outside her body, below her back. As her body goes over the bar, her center of mass passes below the bar. Because a given energy input implies a certain elevation for her center of mass, the action of arching her back means that her body is higher than if her back were straight. As a model, consider the jumper as a thin uniform rod of length L. When the rod is straight, its center of mass is at its center. Now bend the rod in a circular arc so that it subtends an angle of 90.08 at the center of the arc as shown in Figure P9.92b. In this configuration, how far outside the rod is the center of mass?

u

b

Figure P9.92a

Mar

k Dad

swel

l/Get

ty Im

ages

93. Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds

M

vi. Particle m is traveling to the left, and particle 3m is traveling to the right. They undergo an elastic glanc-ing collision such that particle m is moving in the nega-tive y direction after the collision at a right angle from its initial direction. (a) Find the final speeds of the two particles in terms of vi. (b) What is the angle u at which the particle 3m is scattered?

94. Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.94. The conveyor belt is supported by fric-tionless rollers and moves at a constant speed of v 5 0.750 m/s under the action of a constant horizontal external force F

Sext supplied by the motor that drives

the belt. Find (a) the sand’s rate of change of momen-tum in the horizontal direction, (b) the force of fric-tion exerted by the belt on the sand, (c) the external force F

Sext , (d) the work done by F

Sext in 1 s, and (e) the

kinetic energy acquired by the falling sand each second due to the change in its horizontal motion. (f) Why are the answers to parts (d) and (e) different?

vFextS

Figure P9.94

95. On a horizontal air track, a glider of mass m carries a G-shaped post. The post supports a small dense sphere, also of mass m, hanging just above the top of the glider on a cord of length L. The glider and sphere are initially at rest with the cord vertical. (Fig-ure P9.57 shows a cart and a sphere similarly con-nected.) A constant horizontal force of magnitude F is applied to the glider, moving it through displace-ment x1; then the force is removed. During the time interval when the force is applied, the sphere moves through a displacement with horizontal component x2. (a) Find the horizontal component of the veloc-ity of the center of mass of the glider–sphere system when the force is removed. (b) After the force is removed, the glider continues to move on the track and the sphere swings back and forth, both without friction. Find an expression for the largest angle the cord makes with the vertical.

96. Review. A chain of length L and total mass M is released from rest with its lower end just touching the top of a table as shown in Figure P9.96a. Find the force exerted by the table on the chain after the chain has fallen through a distance x as shown in Figure P9.96b. (Assume each link comes to rest the instant it reaches the table.)

Q/C

S

S

LL ! x

a b

x

Figure P9.96

Generalized Newton’s second law example

(a) the sand’s rate of change of momentum in the horizontaldirection

dp

dt=

d(mv)

dt

= m��0

dv

dt+ v

dm

dt

= vdm

dt

= (0.750 m/s)(5.00 kg/s)

= 3.75 N

Newton’s Third Law and Conservation ofMomentum


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 vS1 and vS2 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 aS1 1 m2 aS2 5 0

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

m1 d vS1

dt1 m2

d vS2

dt5 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 vS1 1 m2 vS2 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 X 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



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 vS1 and vS2 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 aS1 1 m2 aS2 5 0

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

m1 d vS1

dt1 m2

d vS2

dt5 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 vS1 1 m2 vS2 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 X 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

dp1dt

= −dp2dt

d

dt(p1 + p2) = 0

Implies:ptotal = p1 + p2 does not change with time. Or, ∆ptotal = 0.


Newton’s third law ⇔ conservation of momentum

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

∆ptotal = 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:

Frocket = −Fexhaust

2Figure 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:

Fr =dp

dt= m

dv

dt+v

dm

dt

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

Newton’s second law for the exhaust:

Fe = medvedt

+ vedme

dt

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

The Rocket Equation



Fr =dp

dt= m

dv

dt+��

0

vdm

dt



Fe = medvedt

+ vedme

dt


The Rocket Equation



Fr =dp

dt= m

dv

dt+��

0

vdm

dt



Fe =��>

0

medvedt

+ vedme

dt


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

−Fe = Fr

−vedme

dt= m

dv

dt

but dmedt = − dm

dt since the rate mass in ejected is equal to the rateof loss of rocket mass:

vedm

dt= m

dv

dt

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

dt . 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

= −

∫mf

mi

vem

dm

= −ve[

ln(m)]mf

mi

∆v = ve ln

(mi

mf

)

The Rocket Equation

For a rocket burning fuel at a constant rate:

vf = vi + ve ln

(mi

mf

)

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 dm

dt

∣∣∣∣

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)

vf = vi − ve ln

(mi

mf

)

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

FThrust = vedm

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 dm

dt is negative.

Summary

• linear momentum

• momentum in an isolated system

• conservation of momentum

• rocket equation

(Uncollected) Homework Serway & Jewett,

• Read the first sections of Chapter 9, read ahead aboutnon-isolated systems.

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

Linear Momentum - De Anza Collegenebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture29.pdf · Linear...

Documents

Transcript of Linear Momentum - De Anza Collegenebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture29.pdf · Linear...