67
Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum conservation axiom Totally Inelastic “ka-runch”collisions* Perfectly Elastic “ka-bong” and Center Of Momentum (COM) symmetry* Comments on idealization in classical models Geometry of Galilean translation symmetry 45° shift in (V 1 ,V 2 )-space Time reversal symmetry ...of COM collisions Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisions Matrix or tensor algebra of collisions Deriving Energy Conservation Theorem Numerical details of collision tensor algebra *Download Superball Collision Simulator http://www.uark.edu/ua/modphys/testing/markup/BounceItWeb.html * Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html 1 Wednesday, August 26, 2015

Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

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Page 1: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Lecture 1 Tue. 8.25.2015

Axiomatic development of classical mechanics(Ch. 1 and Ch. 2 of Unit 1)

Geometry of momentum conservation axiomTotally Inelastic “ka-runch”collisions*Perfectly Elastic “ka-bong” and Center Of Momentum (COM) symmetry*Comments on idealization in classical models

Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions

Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Deriving Energy Conservation Theorem

Numerical details of collision tensor algebra

*Download Superball Collision Simulator http://www.uark.edu/ua/modphys/testing/markup/BounceItWeb.html

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html

1Wednesday, August 26, 2015

Page 2: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelastic

case

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

6600mph1100mph

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html

2Wednesday, August 26, 2015

Page 3: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelasticcase

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

6600mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)and solve...

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html

3Wednesday, August 26, 2015

Page 4: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelasticcase

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

6600mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)and solve...

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html

4Wednesday, August 26, 2015

Page 5: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelasticcase

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)and solve...

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html5Wednesday, August 26, 2015

Page 6: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelasticcase

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.MSUVVVSUV+MVWVVVW =constant is AAxxiioomm ##11

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html6Wednesday, August 26, 2015

Page 7: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelasticcase

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV6600mph

1100mph

= ΔVSUV

ΔVVWMSUV-mVW

4-1 =

MSUVVVSUV+MVWVVVW =constant is AAxxiioomm ##11

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

*Car Collision simulator http://www.uark.edu/ua/modphys/testing/markup/CMMotionWeb.html7Wednesday, August 26, 2015

Page 8: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of momentum conservation axiomTotally Inelastic “ka-runch”collisions*Perfectly Elastic “ka-bong” and Center Of Momentum (COM) symmetry*Comments on idealization in classical models

8Wednesday, August 26, 2015

Page 9: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!slope=+1

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Totally Inelastic (Ka − Runch!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

9Wednesday, August 26, 2015

Page 10: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of momentum conservation axiomTotally Inelastic “ka-runch”collisions*Perfectly Elastic “ka-bong” and Center Of Momentum (COM) symmetry*Comments on idealization in classical models

10Wednesday, August 26, 2015

Page 11: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!slope=+1

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Totally Inelastic (Ka − Runch!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

11Wednesday, August 26, 2015

Page 12: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!) ... (and a circle...)

Totally Elastic (Ka − Bong!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

12Wednesday, August 26, 2015

Page 13: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

Superball Collision Simulator

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

*Download Superball Collision Simulator http://www.uark.edu/ua/modphys/testing/markup/BounceItWeb.html13Wednesday, August 26, 2015

Page 14: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

( )VSUVIN=60mphVVWIN =10mph

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVW

VSUV

( )VSUVFIN=50mphVVWFIN =50mph

VVW=VSUVline

0

INITIAL

10 30 70 80

90

10

30

70

F FINAL

45°45° “K

a-Runch!”

I

“Ka-Runch!”(Extreme inelastic collision)4

-1

= ΔVSUV

ΔVVWMSUV-mVW

4-1 =slope

Fig. 2.1in Unit 1

...But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines!)

Conventional solution: Look up the usualmomentum and energy formulas/axioms:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final)

14Wednesday, August 26, 2015

Page 15: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

( )VSUVIN=60mphVVWIN =10mph

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVW

VSUV

( )VSUVFIN=50mphVVWFIN =50mph

VVW=VSUVline

0

INITIAL

10 30 70 80

90

10

30

70

F FINAL

45°45° “K

a-Runch!”

I

“Ka-Runch!”(Extreme inelastic collision)4

-1

= ΔVSUV

ΔVVWMSUV-mVW

4-1 =slope

Fig. 2.1in Unit 1

Fig. 2.2in Unit 1

Notice “Ka-Bong”Figure 2.2 scaling(ft./min. is more realistic)

Conventional solution: momentum and energy formulas:ΣimVi(initial) = ΣimVi(final)ΣimV2i(initial) = ΣimV2i(final) ( )V

SUV

IN=60ft per min

VVW

IN =10ft per min

50ft per min20 40 60 90

20

40

60

80

50ft per min

100ft per min

100ft per min

VVW

VSUV

( )VSUV

COM=50ft per min

VVW

COM =50ft per minVVW=V

SUV

line

0

“Ka-Bong!” (Ideal elastic collision)

INITIAL

10 30 70 80

90

10

30

70

45°

45°

“Ka-Runch!”

( )VSUV

FIN=40ft per min

VVW

FIN =90ft per min

F FINAL

Center

of

Momentum

“!hnuR-aK”

I

15Wednesday, August 26, 2015

Page 16: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of momentum conservation axiomTotally Inelastic “ka-runch”collisions*Perfectly Elastic “ka-bong” and Center Of Momentum (COM) symmetry*Comments on idealization in classical models

16Wednesday, August 26, 2015

Page 17: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

The SUV and VW Idealized thought experiments

Idealization 1. Ignore background.(No rolling friction, air resistance, etc.)

Idealization 2.Make each 1-dimensional.(Cars “constrained” to ride on frictionless rail)

System now hasjust two “dimensions”or “degrees-of-freedom”

Landscape 1.1 Idealized model for collision model and thought experiments

17Wednesday, August 26, 2015

Page 18: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

3 translationdimensions

3 translationdimensions

6 translationaldegrees of freedomfor SUV and VW.

Translation (Each body has 3 translational degrees of freedom) (Intoduced in Units 1 and 2)

3 rotationaldimensions

yaw-pitch-rollEuler angles

3 rotationaldimensions

yaw-pitch-rollEuler angles

6 rotationaldegrees of freedomfor SUV and VW.

Rotation (Each body has 3 rotationaldegrees of freedom) (Intoduced in Units 3 and 7)

SUV and VW system involves12 rigid-body degrees of freedom

Summary of Classical Mechanical Degrees of Freedom

Vibration (Each body has many vibrational degrees of freedom) (Intoduced in Units 3-8)

An N-atom molecule has3N-6 vibrational degrees of freedom

18Wednesday, August 26, 2015

Page 19: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

3 translationdimensions

3 translationdimensions

6 translationaldegrees of freedomfor SUV and VW.

Translation (Each body has 3 translational degrees of freedom) (Intoduced in Units 1 and 2)

3 rotationaldimensions

yaw-pitch-rollEuler angles

3 rotationaldimensions

yaw-pitch-rollEuler angles

6 rotationaldegrees of freedomfor SUV and VW.

Rotation (Each body has 3 rotationaldegrees of freedom) (Intoduced in Units 3 and 7)

SUV and VW system involves12 rigid-body degrees of freedom

Summary of Classical Mechanical Degrees of Freedom

Vibration (Each body has many vibrational degrees of freedom) (Intoduced in Units 3-8)

An N-atom molecule has3N-6 vibrational degrees of freedom

19Wednesday, August 26, 2015

Page 20: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

3 translationdimensions

3 translationdimensions

6 translationaldegrees of freedomfor SUV and VW.

Translation (Each body has 3 translational degrees of freedom) (Intoduced in Units 1 and 2)

3 rotationaldimensions

yaw-pitch-rollEuler angles

3 rotationaldimensions

yaw-pitch-rollEuler angles

6 rotationaldegrees of freedomfor SUV and VW.

Rotation (Each body has 3 rotationaldegrees of freedom) (Intoduced in Units 3 and 7)

SUV and VW system involves12 rigid-body degrees of freedom

Summary of Classical Mechanical Degrees of Freedom

Vibration (Each body has many vibrational degrees of freedom) (Intoduced in Units 3-8)

An N-atom molecule has3N-6 vibrational degrees of freedom

Landscape 1.2 Some idealized classical model degrees of freedom

Generalized Curvilinear Coordinates (GCC)introduced in Unit 1 Chapters 10 -12

20Wednesday, August 26, 2015

Page 21: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions

21Wednesday, August 26, 2015

Page 22: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

Fig. 2.5ain Unit 1

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

(In some direction x,y, or z…)(In that direction…)

22Wednesday, August 26, 2015

Page 23: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

FSUV(F)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUV

(b) ... and to five or six other reference frames

Fig. 2.5ain Unit 1

Fig. 2.5bin Unit 1

(In some direction x,y, or z…)(In that direction…)

23Wednesday, August 26, 2015

Page 24: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions

24Wednesday, August 26, 2015

Page 25: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

Fig. 2.5ain Unit 1

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

FSUV(F)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUV

(b) ... and to five or six other reference frames

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUVFig. 2.5b

in Unit 1

TTiimmee--rreevveerrssaall ((FF--II))ssyymmmmeettrryy ppaaiirrss((FFoouurr eexxaammpplleess))

FSUV(F)

Time-reversal means flip t to -t…(Run a movie backwards)

Final F and Initial Itrade places…

25Wednesday, August 26, 2015

Page 26: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

Fig. 2.5ain Unit 1

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

FSUV(F)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUV

(b) ... and to five or six other reference frames

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUVFig. 2.5b

in Unit 1

TTiimmee--rreevveerrssaall ((FF--II))ssyymmmmeettrryy ppaaiirrss((FFoouurr eexxaammpplleess))

FSUV(F)

Time-reversal means flip t to -t…(Run a movie backwards)

That means you flip Velocity V to -V…(Everything goes backwards)

Final F and Initial Itrade places…

26Wednesday, August 26, 2015

Page 27: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions

27Wednesday, August 26, 2015

Page 28: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

Fig. 2.5ain Unit 1

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

FSUV(F)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUV

(b) ... and to five or six other reference frames

Fig. 2.5bin Unit 1

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FVW(I)

FSUV(F)

ISUV(F)

FCOM

ICOM

FSUV(I)

ISUV(I)

FVW(F)

IVW(F)

F

I

IEarth

IVW(I)

VVW

VSUV

TTHHEECCOOMM TTiimmee--rreevveerrssaallssyymmmmeettrryy ppaaiirr((JJuusstt 11 ccaassee))

Time-reversal means flip t to -t…(Run a movie backwards)

That means you flip Velocity V to -V…(Everything goes backwards)

There is just one velocity framein which the time-reversed collisionlooks just like the original collision

That is the Center-of-Momentum

(COM)-frame

28Wednesday, August 26, 2015

Page 29: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Deriving Energy Conservation Theorem Energy Ellipse geometry

29Wednesday, August 26, 2015

Page 30: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

MSUV=4

mVW=1

MSUV

-mVW

slope =

Developing Conservation-of-Momentum The key axiom of mechanics

Wow! This is constant! (Says the axiom)

30Wednesday, August 26, 2015

Page 31: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

MSUV=4

mVW=1

MSUV

-mVW

slope =

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

Wow! This is constant! (Says the axiom)

31Wednesday, August 26, 2015

Page 32: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

MSUV=4

mVW=1

MSUV

-mVW

slope =

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

Wow! This is constant! (Says the axiom)

32Wednesday, August 26, 2015

Page 33: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

Wow! This is constant! (Says the axiom)

33Wednesday, August 26, 2015

Page 34: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector: u= 1

1 10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector:

u=1

1

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

34Wednesday, August 26, 2015

Page 35: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector: u= 1

1 10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector:

u=1

1

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Define funny-unit vector : u= 11

⎛⎝⎜

⎞⎠⎟

Developing Conservation-of-Momentum The key axiom of mechanics

35Wednesday, August 26, 2015

Page 36: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Deriving Energy Conservation Theorem Energy Ellipse geometry

36Wednesday, August 26, 2015

Page 37: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector: u= 1

1 10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

Define funny-unit vector:

u=1

1

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Define funny-unit vector : u= 11

⎛⎝⎜

⎞⎠⎟

Developing Conservation-of-Momentum The key axiom of mechanics

37Wednesday, August 26, 2015

Page 38: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

Define funny-unit vector:

u=1

1

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Define funny-unit vector : u= 11

⎛⎝⎜

⎞⎠⎟

Developing Conservation-of-Momentum The key axiom of mechanics

38Wednesday, August 26, 2015

Page 39: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

Define funny-unit vector:

u=1

1

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Define funny-unit vector : u= 11

⎛⎝⎜

⎞⎠⎟

Developing Conservation-of-Momentum The key axiom of mechanics

39Wednesday, August 26, 2015

Page 40: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

Denote Center of Momentum VCOMwith engineer’s centering symbol

Define funny-unit vector:

u=1

1

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

40Wednesday, August 26, 2015

Page 41: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

TTiimmee--RReevveerrssaallSSyymmmmeettrryy AAxxiioomm

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Developing Conservation-of-Momentum The key axiom of mechanics

41Wednesday, August 26, 2015

Page 42: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Deriving Energy Conservation Theorem Energy Ellipse geometry

42Wednesday, August 26, 2015

Page 43: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M=

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Developing Conservation-of-Momentum The key axiom of mechanics

43Wednesday, August 26, 2015

Page 44: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Developing Conservation-of-Momentum The key axiom of mechanics

44Wednesday, August 26, 2015

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10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V=u•M•V IN

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

=u•M•V FIN

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

= 4090

⎛⎝⎜

⎞⎠⎟

V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

= 6010

⎛⎝⎜

⎞⎠⎟

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Points of vectorsVFIN and VCOM and V IN all lie on momentum - conservation line Vector VCOM is along 45° line

Developing Conservation-of-Momentum The key axiom of mechanics

45Wednesday, August 26, 2015

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10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

V COM•M•V COM - 1/2V FIN•M•V IN

=1/2V IN•M•V IN=1/2V FIN•M•V FIN

ByMM--ssyymmmmeettrryyM=MTT : V FIN•M•V IN=V IN•M•V FIN

this becomes:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

(by u• product)

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

46Wednesday, August 26, 2015

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Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Completing derivation of Energy Conservation Theorem Energy Ellipse geometry

47Wednesday, August 26, 2015

Page 48: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

VSUV

VVW

(MSUV+M

VW)V COM=M

SUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

Conservation of momentum line:

const.=V COM=MSUVVSUV

IN+MVWVVW

IN=MSUVVSUV

FIN+MVWVVW

FIN

(MSUV+M

VW) (M

SUV+M

VW)

Mass weighted average velocity at anytime is Center of Mass velocity V COM:

MSUV0

0 MVW

M= =MTranspose

V FIN=VSUV

FIN

VVW

FIN

V IN=VSUV

IN

VVW

IN

V COM= =V COM

=V COMu

V COM

V COM

1

1

Express this using velocity vectors:

...and matrix operators:

...that give momentum vector:

whose sum of components is constant.

MSUV0

0 MVW

P=M•V= = =VSUV

VVW

PSUV

PVW

MSUVVSUV

MVWVVW

const.=u•P=PSUV+P

VW=M

SUVVSUV+M

VWVVW=u•M•V

V COM•M•V COM=V COM•M•V IN=V COM•M•V FIN

Then: V COM=V COMu gives:

V COM•M•V COM - 1/2V FIN•M•V IN

=1/2V IN•M•V IN=1/2V FIN•M•V FIN

ByMM--ssyymmmmeettrryyM=MTT : V FIN•M•V IN=V IN•M•V FIN

this becomes:

These are equations for energy conservation ellipse:

By TT--SSyymmmmeettrryy AAxxiioomm: V COM=1/2(V IN+V FIN). Substituting:

V COM•M•V COM=1/2(V IN+V FIN)•M•V IN=1/2(V IN+V FIN)•M•V FIN

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

MM--ssyymmmmeettrryyM=MTT

Algebra, Geometry, and Physics of Momentum Conservation Axiom

These are equations for energy conservation ellipse:

const. = 1/2MSUVVSUV

2 + 1/2MVWVVW

2

(by u• product)

= 1/2MSUVVSUV

2 + 1/2MVWVVW

2

=Kinetic Energy=KE is now defined

and proved a constant under T-Symmetry

Define velocity vector points

VFIN=VSUV

FIN

VVWFIN

⎝⎜⎜

⎠⎟⎟

, VCOM =VSUV

COM

VVWCOM

⎝⎜⎜

⎠⎟⎟

, V IN =VSUV

IN

VVWIN

⎝⎜⎜

⎠⎟⎟

Developing Conservation-of-Momentum The key axiom of mechanics

leading toConservation-of-Energy Theorem

48Wednesday, August 26, 2015

Page 49: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Algebra,Geometry, and Physics of momentum conservation axiom Vector algebra of collisionsMatrix or tensor algebra of collisions Deriving Energy Conservation Theorem Energy Ellipse geometry

49Wednesday, August 26, 2015

Page 50: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

VSUV

VVW

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

Conservation of momentum line: (...one of ∞-many...)

Algebra, Geometry, and Physics of MMoommeennttuumm CCoonnsseerrvvaattiioonn AAxxiioomm

V COM•M•V COM-1/2V FIN•M•V IN

=1/2V IN•M•V IN=1/2V FIN•M•V FIN

MSUV=4

mVW=1

MSUV

-mVW

slope =

All lines of slope -MSUV/mVW

...are bisected by the

(slope=1)-COM line

MMoommeennttuummCCoonnsseerrvvaattiioonnAAxxiioomm

plus

TT--SSyymmmmeettrryyAAxxiioomm(M=MTT implied)

gives

KKiinneettiicc EEnneerrggyyCCoonnsseerrvvaattiioonnTThheeoorreemm

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

These are equations for energy conservation ellipse:These are equations for energy conservation ellipse:

KE = 1/2MSUVVSUV

2 + 1/2MVWVVW

2

1 = VSUV

2 + VVW

2 = x2 + y2

2·KE 2·KE a2 + b2

MSUV

MVW

Developing Conservation-of-Momentum The key axiom of mechanics

leading toConservation-of-Energy Theorem

50Wednesday, August 26, 2015

Page 51: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

VSUV

VVW

10 20 30 40 50 60 70 80 90 100

10

20

30405060

708090

100

V IN

V FIN

Conservation of momentum line: (...one of ∞-many...)

Algebra, Geometry, and Physics of MMoommeennttuumm CCoonnsseerrvvaattiioonn AAxxiioomm

V COM•M•V COM-1/2V FIN•M•V IN

=1/2V IN•M•V IN=1/2V FIN•M•V FIN

MSUV=4

mVW=1

MSUV

-mVW

slope =

All lines of slope -MSUV/mVW

...are bisected by the

(slope=1)-COM line

MMoommeennttuummCCoonnsseerrvvaattiioonnAAxxiioomm

plus

TT--SSyymmmmeettrryyAAxxiioomm(M=MTT implied)

gives

KKiinneettiicc EEnneerrggyyCCoonnsseerrvvaattiioonnTThheeoorreemm

TT--SSyymmmmeettrryy AAxxiioommV COM=(V IN+V FIN)/2

These are equations for energy conservation ellipse:These are equations for energy conservation ellipse:

KE = 1/2MSUVVSUV

2 + 1/2MVWVVW

2

1 = VSUV

2 + VVW

2 = x2 + y2

2·KE 2·KE a2 + b2

MSUV

MVW

Developing Conservation-of-Momentum The key axiom of mechanics

leading toConservation-of-Energy Theorem

If and only if…there is T-Symmetry

51Wednesday, August 26, 2015

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VVW100110120

-120

(60,10)Initial-point

Final“Ka-Bong”-point

a=√=60.21

2·KEMSUV

b=√=120.42

2·KEmVW

ElasticKineticEnergyellipse(KE=7,250)

MomentumPTotal=250line

(50,50)

(40,90)

MSUV=4MSUV=4

mVW=1mVW=1

VSUV

Fig. 3.1 ain Unit 1

Fig. 3.1

Developing Conservation-of-Momentum The key axiom of mechanics

leading toConservation-of-Energy Theorem

If and only if…there is T-Symmetry

52Wednesday, August 26, 2015

Page 53: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

VVW

100

110

120

-120

(60,10)

Initial-point

Final“Ka-Bong”-point

a=√=60.21

2·KE

MSUV

b=√=120.42

2·KEmVW

Elastic

Kinetic

Energy

ellipse

(KE=7,250)

Momentum

PTotal=250

line

(50,50)

(40,90)

MSUV=4M

SUV=4

mVW=1m

VW=1

VSUV

Fig. 3.1 ain Unit 1

a=√=55.9

2·IE

MSUV

b=√=111.8

2·IEmVW

COM

Energy

ellipse

(ECOM=1,000)

elastic

FIN

IN

inlastic

FIN is

VCOM

point

Inelastic

Kinetic

Energy

ellipse

(IE=6,250)

Fig. 3.1 bin Unit 1

Fig. 3.1

53Wednesday, August 26, 2015

Page 54: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

II

Felastic

I

(a) IdealElastic(Ka-Bong!)

(b) Genericinelastic(Ka-whump!)

(c) Totallyinelastic(Ka-Runch!)

aelastic=2EM2√

belastic=2EM1

(1,8)

(2,2)

(3,-4)

Eelastic=35

IECOM=14

Ewhump=23.33

Fwhump(4/3,6)

FCOM

M1=6M2=1

FPass-thru

bCOM=√2·IEM1

bwhump=2EwhumpM1√

aCOM=√2·IEM2

2EwhumpM2√

√awhump=

As usual in physics, opposite extremes are easier to analyzethan the generic “real(er) world” in between!

Fig. 3.2in Unit 1

(This case has Bush era requisiteSUV mass of the 6 ton “Hummer”)

Next: The X-2pen-

launcher

The X-2pen-

launcher

Here T-Symmetryis best

Here T-Symmetryis less

Here T-Symmetryis least

54Wednesday, August 26, 2015

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Numerical details of collision tensor algebra

55Wednesday, August 26, 2015

Page 56: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

56Wednesday, August 26, 2015

Page 57: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

A product of total momentum PTotal and is expressed by tensor quadratic forms v•M•u

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

VCOM

VCOMPTotal =

VCOM i

M iV IN =

VCOM i

M iVFIN =

VCOM i

M iVCOM =VCOMMTotalV

COM

57Wednesday, August 26, 2015

Page 58: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Write this out with the numbers used in Fig. 3.1 where .

A product of total momentum PTotal and is expressed by tensor quadratic forms v•M•u

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

VCOM

VCOMPTotal =

VCOM i

M iV IN =

VCOM i

M iVFIN =

VCOM i

M iVCOM =VCOMMTotalV

COM

VCOM = 50

50PTotal = 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 50 50( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟= 50MTotal50 = 12,500

= 100 ⋅125 = 100 ⋅125 = 50 ⋅250

(2 pages back)

58Wednesday, August 26, 2015

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PTotal =250 is the same at IN, FIN, and COM. Now use T-symmetry:

A product of total momentum PTotal and is expressed by tensor quadratic forms v•M•u

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

VCOM

VCOMPTotal =

VCOM i

M iV IN =

VCOM i

M iVFIN =

VCOM i

M iVCOM =VCOMMTotalV

COM

50PTotal = 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 50 50( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟= 50MTotal50 = 12,500

= 100 ⋅125 = 100 ⋅125 = 50 ⋅250

VCOM =(

V FIN+

V IN )/2

V COM PTotal =V FIN+

V IN

2i

M iV IN =

V FIN+

V IN

2i

M iV FIN =

V FIN+

V IN

2i

M i

V FIN+

V IN

2

Write this out with the numbers used in Fig. 3.1 where . VCOM = 50 (2 pages back)

59Wednesday, August 26, 2015

Page 60: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

PTotal =250 is the same at IN, FIN, and COM. Now use T-symmetry:

A product of total momentum PTotal and is expressed by tensor quadratic forms v•M•u

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

VCOM

VCOMPTotal =

VCOM i

M iV IN =

VCOM i

M iVFIN =

VCOM i

M iVCOM =VCOMMTotalV

COM

50PTotal = 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 50 50( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟= 50MTotal50 = 12,500

= 100 ⋅125 = 100 ⋅125 = 50 ⋅250

VCOM =(

V FIN+

V IN )/2

V COM PTotal =V FIN+

V IN

2i

M iV IN =

V FIN+

V IN

2i

M iV FIN =

V FIN+

V IN

2i

M i

V FIN+

V IN

2

V COM PTotal −V FIN i

M iV IN

2=V IN i

M iV IN

2=V FIN i

M iV FIN

2=V IN i

M iV IN

4+V FIN i

M iV FIN

4

Write this out with the numbers used in Fig. 3.1 where . VCOM = 50 (2 pages back)

Subtracting:V FIN i

MiV IN

2

=V IN i

MiV FIN

2

⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟

60Wednesday, August 26, 2015

Page 61: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

PTotal =250 is the same at IN, FIN, and COM. Now use T-symmetry:

A product of total momentum PTotal and is expressed by tensor quadratic forms v•M•u

General Inertia Tensor M or inertia matrix of n2 coefficients Mjk for dimension n=2, 3,....

P1 = M11V1 + M12V2

P2 = M21V1 + M22V2

⎫⎬⎪

⎭⎪ denoted :

P =

M iV or :

P1

P2

⎝⎜⎜

⎠⎟⎟=

M11 M12

M21 M22

⎝⎜⎜

⎠⎟⎟

V1

V2

⎝⎜⎜

⎠⎟⎟

V1COM =V2

COM ≡VCOMWith 45° diagonal VCOM so: .

PTotal = M1V1IN +M2V2

IN = M1V1FIN +M2V2

FIN = M1VCOM +M2V

COM = MTotalVCOM

VCOM

VCOMPTotal =

VCOM i

M iV IN =

VCOM i

M iVFIN =

VCOM i

M iVCOM =VCOMMTotalV

COM

50PTotal = 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 50 50( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟= 50MTotal50 = 12,500

= 100 ⋅125 = 100 ⋅125 = 50 ⋅250

VCOM =(

V FIN+

V IN )/2

V COM PTotal =V FIN+

V IN

2i

M iV IN =

V FIN+

V IN

2i

M iV FIN =

V FIN+

V IN

2i

M i

V FIN+

V IN

2

V COM PTotal −V FIN i

M iV IN

2=V IN i

M iV IN

2=V FIN i

M iV FIN

2=V IN i

M iV IN

4+V FIN i

M iV FIN

4

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

Write this out with the numbers used in Fig. 3.1 where . VCOM = 50 (2 pages back)

Subtracting:V FIN i

MiV IN

2

=V IN i

MiV FIN

2

⎜⎜⎜⎜⎜⎜⎜

⎟⎟⎟⎟⎟⎟⎟

61Wednesday, August 26, 2015

Page 62: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

With (M12 =0=M21) kinetic energy is the same at V=VIN and V=VFIN.

V COM PTotal −V FIN i

M iV IN

2=

V IN i

M iV IN

2=

V FIN i

M iV FIN

2= KEElastic

12,500 − 10,5002

=60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟

2=

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

2= KEElastic

12,500 − 5,250 = 7,250 = 7,250

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

KEElastic= 21 Vi

MiV

62Wednesday, August 26, 2015

Page 63: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

With (M12 =0=M21) kinetic energy is the same at V=VIN and V=VFIN.

V COM PTotal −V FIN i

M iV IN

2=

V IN i

M iV IN

2=

V FIN i

M iV FIN

2= KEElastic

12,500 − 10,5002

=60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟

2=

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

2= KEElastic

12,500 − 5,250 = 7,250 = 7,250

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

Consider kinetic energy KEinelastic=IE when V=VCOM. It is reduced by 1,000 from 7,250 to 6,250.

KEInelastic=V COM PTotal−VCOM i

MiVCOM

2= 1

2V COM PTotal=

VCOM i

MiVCOM

2=V IN i

MiV IN

4+V FIN i

MiV IN

4= 1

2KEElastic+

V FIN i

MiV IN

4

12,500 − 12,5002

= 6,250 = 6,250 = 3,625 + 2,625 = IE

KEElastic= 21 Vi

MiV

63Wednesday, August 26, 2015

Page 64: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

With (M12 =0=M21) kinetic energy is the same at V=VIN and V=VFIN. KEElastic= 21 Vi

MiV

V COM PTotal −V FIN i

M iV IN

2=

V IN i

M iV IN

2=

V FIN i

M iV FIN

2= KEElastic

12,500 − 10,5002

=60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟

2=

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

2= KEElastic

12,500 − 5,250 = 7,250 = 7,250

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

Consider kinetic energy KEinelastic=IE when V=VCOM. It is reduced by 1,000 from 7,250 to 6,250.

KEInelastic=V COM PTotal−VCOM i

MiVCOM

2= 1

2V COM PTotal=

VCOM i

MiVCOM

2=V IN i

MiV IN

4+V FIN i

MiV IN

4= 1

2KEElastic+

V FIN i

MiV IN

4

12,500 − 12,5002

= 6,250 = 6,250 = 3,625 + 2,625 = IE

KEInelastic=2

1 VCOM i

MiVCOM =2

1 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 50

50⎛

⎝⎜⎞

⎠⎟= 6,250

64Wednesday, August 26, 2015

Page 65: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

With (M12 =0=M21) kinetic energy is the same at V=VIN and V=VFIN. KEElastic= 21 Vi

MiV

V COM PTotal −V FIN i

M iV IN

2=

V IN i

M iV IN

2=

V FIN i

M iV FIN

2= KEElastic

12,500 − 10,5002

=60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟

2=

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

2= KEElastic

12,500 − 5,250 = 7,250 = 7,250

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

Consider kinetic energy KEinelastic=IE when V=VCOM. It is reduced by 1,000 from 7,250 to 6,250.

KEInelastic=V COM PTotal−VCOM i

MiVCOM

2= 1

2V COM PTotal=

VCOM i

MiVCOM

2=V IN i

MiV IN

4+V FIN i

MiV IN

4= 1

2KEElastic+

V FIN i

MiV IN

4

12,500 − 12,5002

= 6,250 = 6,250 = 3,625 + 2,625 = IE

KEElastic − KEInelastic =V IN i

M iV IN

4−V FIN i

M iV IN

41,000 = 3,625 − 2,625 = KE − IE

The difference is inelastic “crunch” energy KE-IE or, for elastic cases, potential energy of compression. KEInelastic=2

1 VCOM i

MiVCOM =2

1 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 50

50⎛

⎝⎜⎞

⎠⎟= 6,250

65Wednesday, August 26, 2015

Page 66: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

With (M12 =0=M21) kinetic energy is the same at V=VIN and V=VFIN. KEElastic= 21 Vi

MiV

V COM PTotal −V FIN i

M iV IN

2=

V IN i

M iV IN

2=

V FIN i

M iV FIN

2= KEElastic

12,500 − 10,5002

=60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟

2=

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

2= KEElastic

12,500 − 5,250 = 7,250 = 7,250

Transpose symmetry (Mjk =Mkj) of the M-matrix implies:

V FIN i

M iV IN =

V IN i

M iV FIN

40 90( ) i 4 00 1

⎝⎜⎞

⎠⎟i 60

10⎛

⎝⎜⎞

⎠⎟= 60 10( ) i 4 0

0 1⎛

⎝⎜⎞

⎠⎟i 40

90⎛

⎝⎜⎞

⎠⎟

= 100 ⋅105 = 100 ⋅105 = 10,500

Consider kinetic energy KEinelastic=IE when V=VCOM. It is reduced by 1,000 from 7,250 to 6,250.

KEInelastic=V COM PTotal−VCOM i

MiVCOM

2= 1

2V COM PTotal=

VCOM i

MiVCOM

2=V IN i

MiV IN

4+V FIN i

MiV IN

4= 1

2KEElastic+

V FIN i

MiV IN

4

12,500 − 12,5002

= 6,250 = 6,250 = 3,625 + 2,625 = IE

KEElastic − KEInelastic =V IN i

M iV IN

4−V FIN i

M iV IN

41,000 = 3,625 − 2,625 = KE − IE

The difference is inelastic “crunch” energy KE-IE or, for elastic cases, potential energy of compression. KEInelastic=2

1 VCOM i

MiVCOM =2

1 50 50( ) i 4 00 1

⎝⎜⎞

⎠⎟i 50

50⎛

⎝⎜⎞

⎠⎟= 6,250

This difference is the same in all reference frames including COM frame where KEinelastic=IE is zero. 66Wednesday, August 26, 2015

Page 67: Axiomatic development of classical mechanics...2015/08/25  · Lecture 1 Tue. 8.25.2015 Axiomatic development of classical mechanics (Ch. 1 and Ch. 2 of Unit 1) Geometry of momentum

Fig. 3.4 Galilean Frame Views of collision like Fig. 2.5 or Fig. 3.1 with Bush (6:1) SUV. (a) Earth frame view (b) Initial VW frame (VW initially fixed) (c) COM frame view (d) Final VW frame (VW ends up fixed) Fig. 3.5 Momentum (P=const.)-lines and energy (KE=const.)-ellipses

appropriate for Fig. 3.4.

Note “crunch” energy ElasticKE-inelasticIE = 0.21 is the same in all frames including COM-frame.

COM-frame

67Wednesday, August 26, 2015