chapter 7 linear momentum. mfmcgraw-phy 1401chap07b- linear momentum: revised 6/28/2010 2 linear...

Chapter 7

Linear Momentum

MFMcGraw-PHY 1401 Chap07b- Linear Momentum: Revised 6/28/2010

2

Linear Momentum

• Definition of Momentum

• Impulse

• Conservation of Momentum

• Center of Mass

• Motion of the Center of Mass

• Collisions (1d, 2d; elastic, inelastic)


3

Momentum From Newton’s Laws

Consider two interacting bodies with m2 > m1:

If we know the net force on each body then tm

t netFav

The velocity change for each mass will be different if the masses are different.

F21 F12m1 m2


4

tm 2111 Fv

Rewrite the previous result for each body as:

ttm 211222 FFv

Combine the two results:

ifif mm

mm

222111

2211

vvvv

vv

Momentum

From the 3rd Law


5

The quantity (mv) is called momentum (p).

p = mv and is a vector.

The unit of momentum is kg m/s; there is no derived unit for momentum.

Define Momentum


6

From slide (4):

21

2211

pp

vv

mm

The change in momentum of the two bodies is “equal and opposite”. Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other.

Δp1 +Δp2 = 0

Momentum


7

Impulse

Definition of impulse:

t Fp

One can also define an average impulse when the force is variable.

We use impulses to change the momentum of the system.


8

Example (text conceptual question 7.2): A force of 30 N is applied for 5 sec to each of two bodies of different masses.

30 N

m1 or m2

Take m1 < m2

(a) Which mass has the greater momentum change?

t FpSince the same force is applied to each mass for the same interval, p is the same for both masses.

Impulse Example


9

(b) Which mass has the greatest velocity change?

Example continued:

m

pv

Since both masses have the same p, the smaller mass (mass 1) will have the larger change in velocity.

(c) Which mass has the greatest acceleration?

t

v

aSince av the mass with the greater velocity change will have the greatest acceleration (mass 1).


10

An object of mass 3.0 kg is allowed to fall from rest under the force of gravity for 3.4 seconds. Ignore air resistance.

What is the change in momentum?

Want p = mv.

(downward) m/s kg 100

m/sec 3.33

vp

v

av

m

tg

t

Change in Momentum Example


11

What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds? Assume frictionless ice.

N 5.7

s 0.20

m/s 0.3kg 0.50av

av

av

F

t

m

t

t

vpF

Fp

The force will be in the direction of motion.

Change in Momentum Example


12

Impulse

iV

fV

Since we are applying the change of momentum over a time interval consisting of the instant before the impact to the instant after the impact, we ignore the effect of gravity over that time period

f i f i AvgΔP = P - P = M V - V = F Δt

i f

f i

V = 20 m s ; V = -15 m s ; M = 3kg

ΔP = M V - V = 3(-15 - 20)

ΔP = 3 -35 = -105 kg m s


13

Impulse and Momentum

The force in impulse problems changes rapidly so we usually deal with the average force


14

Work and Impulse Comparison

Work changes ET

Work = 0 ET is conserved

Work = Force x Distance

Impulse changes

Impulse = 0 is conserved

Impulse = Force x Time

P

P


15

Conservation of Momentum

v1i

v2i

m1 m2

m1>m2

m1 m2

A short time later the masses collide.

What happens?


16

During the interaction:

N1

w1

F21x

y

1121

11 0

amFF

wNF

x

y

2212

22 0

amFF

wNF

x

y

There is no net external force on either mass.

N2

w2

F12


17

The forces F12 and F21 are internal forces. This means that:

ffii

ifif

2121

2211

21

pppp

pppp

pp

In other words, pi = pf. That is, momentum is conserved. This statement is valid during the interaction (collision) only.


18

Momentum Examples


19

Generic Zero Initial Momentum Problem

These problems occur whenever all the objects in the system are initially at rest and then separate.

P Ai

p = p + p = 0 P Ai

p' = p' + p' = 0


20

Changing the Mass While Conserving the Momentum


21



22



23

Astronaut TossingHow Long Will the Game Last?


24

How Long Will the Game Last?

0 00

V 3VV

2 2

0-V 0V

0V

2

0-V

0 00

V VV

2 2

0 00

V 3VV

2 2


25

The Ballistic Pendulum

Conservation of Momentum

Conservation of Energy


26

The High Velocity Ballistic Pendulum


27

Relative Speed Relationship

• The relative speed relationship applies to elastic collisions.

• In an elastic collision the total kinetic energy is also conserved.

• If a second equation is needed to solve a momentum problem, the relative speed relationship is easier to deal with than the KE equation.


28

Relative Speed Relationship


29

Relative Speed in Problem Solving

Need a second equation when solving for two variables in an elastic collision problem?

The relative speed relationship is preferred over the conservation ofkinetic energy relationship


30

A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s.

What is the recoil speed of the rifle as the bullet leaves the barrel?

As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved.

m/s 1.82m/s 820kg 4.5

kg 01.0

0

br

br

rrbb

fi

vm

mv

vmvm

pp

Rifle Example


31

Center of Mass


32

Center of Mass

The center of mass (CM) is the point representing the mean (average) position of the matter in a body.

This point need not be located within the body.


33

The center of mass (of a two body system) is found from:

21

2211

mm

xmxmxcm

This is a “weighted” average of the positions of the particles that compose a body. (A larger mass is more important.)

Center of Mass


34

In 3-dimensions write

ii

iii

cm m

m rr where

iimmtotal

The components of rcm are:

ii

iii

cm m

xmx

ii

iii

cm m

ymy

ii

iii

cm m

zmz

Center of Mass


35

Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of 10.0 grams. Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.0 cm, 5.0 cm)?

x

y

A

x

ba

bb

ba

bbaacm mm

xm

mm

xmxmx

ba

bb

ba

bbaacm mm

ym

mm

ymymy

Center of Mass Example


36

cm 8

cm 2g30g 10

g 10

b

bcm

x

xx

cm 20

cm 5g 10g 30

g 10

b

bcm

y

yy

Center of Mass Example


37

The positions of three particles are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (1.0 m, 2.0 m). The masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively.

What is the location of the center of mass?

x

y

3

2

1

Find the Center of Mass


38

m 92.1

kg 364

m 1kg 3m 2kg 6m 4kg 4321

332211cm

mmm

xmxmxmx

m 38.1

kg 364

m 2kg 3m 4kg 6m 0kg 4321

332211cm

mmm

ymymymy

Example continued:


39

Motion of the Center of Mass

For an extended body, it can be shown that p = mvcm.

From this it follows that Fext = macm.


40

For the center of mass:

ii

iii

m

mt

rvr cmcm

Solving for the velocity of the center of mass:

ii

iii

ii

i

ii

m

m

mt

m vr

vcm

Or in component form:

ii

ixii

m

vm ,

xcm,v

ii

iyii

m

vm ,

ycm,v

Center of Mass Velocity


41

Center of Mass Reference System


42

Center of Mass Treatment


43



44



45



46

Center of Mass Reference System

r

1r

2r

R

1m

2m

1 1 2 2

1 2

2 1

m r + m rR =

m + m

r = r - r

21

1 2

12

1 2

mr = R - r

m + m

mr = R + r

m + m

Center of Mass System

Laboratory System


47

Collisions in One Dimension

When there are no external forces present, the momentum of a system will remain unchanged. (pi = pf)

If the kinetic energy before and after an interaction is the same, the “collision” is said to be perfectly elastic. If the kinetic energy changes, the collision is inelastic.


48

If, after a collision, the bodies remain stuck together, the loss of kinetic energy is a maximum, but not necessarily a 100% loss of kinetic energy. This type of collision is called perfectly inelastic.

Collisions in One Dimension


49

Elastic Collision - Unequal Masses

In the later slides, the initially moving mass, mn , will called be m1 while the initially stationary mass, mc , will be m2.


50

Elastic Velocity Ratios

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

0.0 5.0 10.0 15.0 20.0 25.0

Mass Ratio

Ve

loc

ity

Ra

tio

U1

U2

Elastic Collision - KE Transfer

2

1 1

20

1

m- 1

u m= -

mv 1+m

2

20

1

u 2=

mv 1+m


51

Elastic Collision - Fractional KE “Loss”

For equal masses all of the KE is transferred from m1 to m2

The curve is like an energy transfer function from m1 to m2. For small m2 the mass m1 just rolls on through. For large m2 the smaller m1 just bounces off without transferring much energy. Efficient energy transfer requires equal masses.


52

Fractional KE Loss

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0 5 10 15 20 25

Mass Ratio

Fra

cti

on

al K

E L

os

s

Fractional KE Loss

U2 Velocity Ratio

Elastic Collision - KE Transfer

For m2 > m1 the velocity of the 2nd mass gets a smaller velocity and even the larger mass can’t overcome the shrinking v2 factor in KE2

2

12

2

1

m4

mf =

m1+

m

2

20

1

u 2=

mv 1+m


53

Inelastic Collision - Equal Masses

v1 v2 = 0

u

m1 m2

m1 + m2

11 1 1 2 1

1 2

2 21 1i 1 1 f 1 22 2

2

211f 1 2 12

1 2

21 11f 1 1 i2

1 2 1 2

f 1

i 1 2

mm v = m + m u; u = v

m + m

KE = m v ; KE = m + m u

mKE = m + m v

m + m

m mKE = m v = KE

m + m m + m

KE m=

KE m + m

Before

After

Momentum

Kinetic Energy


54

In a railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.0m. The two cars couple together upon collision.

(a) What is the speed of the two cars after the collision?

m/s 2.0

0

121

1

212111

2121

vmm

mv

vmmvmvmvm

pppp

pp

ffii

fi

Inelastic Collisions in One Dimension


55

(b) Suppose instead that both cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.0 m/s.

m/s 25.0

00

12

12

2211

2121

ii

ii

ffii

fi

vm

mv

vmvm

pppp

pp

Inelastic Collisions in One Dimension


56

A projectile of 1.0 kg mass approaches a stationary body of 5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse direction along the same line with a speed of 5.0 m/s.

What is the speed of the 5.0 kg mass after the collision?

m/s0.3m/s 5.0m/s 10kg 0.5

kg 0.1

0

112

12

221111

2121

fif

ffi

ffii

fi

vvm

mv

vmvmvm

pppp

pp

One Dimension Projectile


57

Body A of mass M has an original velocity of 6.0 m/s in the +x-direction toward a stationary body (B) of the same mass. After the collision, body A has vx = +1.0 m/s and vy = +2.0 m/s.

What is the magnitude of body B’s velocity after the collision?

A

B

vAi

Initial

B

AFinal

Collision in Two Dimensions

Angle is 90o


58

fxfxix

fxfxixix

fxix

vmvmvm

pppp

pp

221111

2121

0

fyfy

fyfyiyiy

fyiy

vmvm

pppp

pp

2211

2121

00

x momentum:

Solve for v2fx: Solve for v2fy:

m/s 00.5

11

2

11112

fxix

fxixfx

vv

m

vmvmv

y momentum:

m/s 00.2

12

112

fyfy

fy vm

vmv

m/s 40.522

22

2 fxfyf vvvThe mag. of v2 is

Collision in Two Dimensions


59

Collisions in Two and Three Dimensions


60

Totally Inelastic Collision


61

Totally Inelastic Collision

If the objects stick together after the collision, then the collision is INELASTIC.

Just because the objects DON’T stick together after the collision, doesn’t mean the collision is ELASTIC.


62

General Glancing CollisionInitially only one object is moving


63

General Glancing CollisionInitially only one object is moving


64

Another 3-Vector Problem

Momentum problems where one of the objects is initially at rest yield 3-vector problems that can be solved using triangles, instead of simultaneous equations.


65

Combining Conservation of Energy, Momentum, and KE Transfer in Equal Mass Inelastic Collisions


66

Δh1Δh2

211 2

1

mg h = mv

v = 2g h 12

mv = (m + m)u

u = v

2122

22

(2m)u = 2mgΔh

u = 2gΔh

2 21 1

2

u v 2gΔh ΔhΔh = = = =

2g 8g 8g 4

Conservation of Energy Conservation of EnergyConservation of Momentum


67

Summary

• Definition of Momentum

• Impulse

• Center of Mass

• Conservation of Momentum


68

Extra Slides


69

Conservation of MomentumTwo objects of identical mass have a collision. Initially object 1 is traveling to the right with velocity v1 = v0. Initially object 2 is at rest v2 = 0.

After the collision: Case 1: v1’ = 0 , v2’ = v0 (Elastic)

Case 2: v1’ = ½ v0 , v2’ = ½ v0 (Inelastic)In both cases momentum is conserved.

Case 1 Case 2


70

Conservation of Kinetic Energy?

KE = Kinetic Energy = ½mvo

2

Case 1 Case 2

KE After = KE Before KE After = ½ KE Before

(Conserved) (Not Conserved)


71

Where Does the KE Go?

Case 1 Case 2

KE After = KE Before KE After = ½ KE Before

(Conserved) (Not Conserved)

In Case 2 each object shares the Total KE equally.

Therefore each object has KE = 25% of the original KE Before


72

50% of the KE is Missing

Each rectangle represents KE = ¼ KE

Before


73

Turn Case 1 into Case 2Each rectangle represents KE = ¼ KEBefore

Take away 50% of KE. Now the total system KE is correct

But object 2 has all the KE and object 1 has none

Use one half of the 50% taken to speed up object 1. Now it has 25% of the initial KE Use the other half of the 50% taken to slow down object 2. Now it has only 25% of the the initial KE

Now they share the KE equally and we see where the missing 50% was spent.

chapter 7 linear momentum. mfmcgraw-phy 1401chap07b- linear momentum: revised 6/28/2010 2 linear...

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