Axiomatic development of classical mechanics...2015/08/25 · Lecture 1 Tue. 8.25.2015 Axiomatic...
Transcript of Axiomatic development of classical mechanics...2015/08/25 · Lecture 1 Tue. 8.25.2015 Axiomatic...
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions
21Wednesday, August 26, 2015
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
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
Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions
24Wednesday, August 26, 2015
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
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
Geometry of Galilean translation symmetry45° shift in (V1,V2)-spaceTime reversal symmetry...of COM collisions
27Wednesday, August 26, 2015
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Numerical details of collision tensor algebra
55Wednesday, August 26, 2015
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
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
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
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
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
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
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
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
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
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
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
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