Assignment 1 - Classical Mechanics 5103 8/23/11
Main Reading: In new text (Classical Mechanics with a BANG! (2008)) Unit 1 Chapter 1
thru Ch. 5.
Exercises Due Tue. Aug 30
Exercise 1.2.1 A correct SUVvsVW momentum conservation tensor relation for Fig. 2.6 is
as follows†:
 M
SUV
d[11] M V COM  d[11] M V IN  d[11] M V FIN where: d[11]  (1,1)d with d>0 and: M  

0
0
MVW



† This was given in class. A diagonal d[11]=(d, d) dot • product on left makes a scalar numerical equation.
Discuss and verify the following special case of this relation as it applies to Fig. 2.6 and
Fig. 3.1:
 M
SUV
V COM M V COM  V COM M V IN  V COM M V FIN where: M  

0
0
MVW
 

  4 0 
  0 1 
Show it is numerically correct for both elastic and inelastic cases in Fig. 3.1.
Discuss totally elastic condition V 2 (V V ) and its related symmetries.
Given elastic condition derive and check/evaluate numerically the following conservation
equation:
COM
1
IN
FIN
Elastic kinetic energy: KE 2 V M V 2 V M V
Derive totally inelastic kinetic energy. Show it is minimum kinetic energy achieved in an
elastic collision.
1
IN
Inelastic kinetic energy: IE 2 V
1
COM
IN
1
FIN
FIN
M VCOM
Express their difference E=KE-IE in terms of quadratic tensor forms
and V
IN
V IN M V IN
M V FIN .
Exercise 1.3.1 Plot a (VSUV-1,VSUV-2)=(60,10) collision diagram like Fig. 3.1 but with an
identical mass M=4 SUV replacing the VW. Draw energy ellipses as precisely as
possible.
Research Problem (Due Thur Sept. 1. This allows more than a week, but start now and do
it well!))
1.1.1 This is a dimensional analysis and power law problem. It involves Olympic weight
lifters but is a general piece of mechanics of everything. Perhaps you wonder why
toy cars can fall off cliffs with no damage while real ones can’t? Power laws like
Strength = const. (Area of crossection)~k·L2 are involved.
Olympic weight lifters are divided into classes according to their body weight.
Generally top performers are close to maximum allowed by their class (except for
“super-heavyweight" classes.)
(a) From dimensional arguments alone, you can predict that the Olympic records R in a
given event (say, the "clean and jerk" which is always the greatest record) should have
a definite functional relationship to the weight W=Mg of the performers: R= R(W).
Derive R(W) as a power law times an undetermined coefficient.
(b) Obtain a set of records from an almanac or book of records, and plot them against W
for a given event or events. See how well your theory and experiment jive. (Hint: It is
most convenient to plot on log-log graph paper. Why?)
(c) Use the results of (a) and (b) to answer: How many times his bodyweight could a man
lift if he was the size of an ant with a mass of M = 1 gm.? (A real ant is supposed to
lift five or ten times its body weight. How much better or worse is the ant doing than
v2
VFIN
1,-1
IN
VFIN
C0M= -VC0M
VIN
0,V VFIN
v,0
IN
VFIN
1,1 =V1,1
V IN
v,0
v1
VFIN
0,V
V IN
1,-1
V IN
C0M
"Antman"? )
Exercise 1.3.2. Ch. 3 thru Ch. 5 treat various 1D-2-body collisions. Some examples
V IN = (1,-1)
originate from initial velocity vectors 1,-1
for which m1 and m2 have equal speeds
(Here we study unit speed).
This exercise is intended to help match algebra and geometry by asking for the simplest
formulas for the various velocities in a figure above that are final elastic results of the
following initial velocity vectors.
V IN = (1,-1)
V IN = (v,0)
V IN
V IN = (0,V)
= (vCOM ,vCOM )
x
y
a. 1,-1
b. v,0
c. 0 ,V
d. COM
Derive the IN and FIN components of all vectors in terms of masses m1 and m2 only
V IN = (1,-1)
assuming the same total KE as 1,-1
has. (Check your results against the figure
above where ratio 2=m1/m2 holds.)
Indicate where the time reversed vector T·VIN of each VIN lies. (What does time reversal
do to velocity V?
Give a formula for the orange (dashed) and green (solid) tangent line slopes in terms of
m1 and m2.
…and compare to slope of the black line connecting major and minor radii in terms of m1
and m2.
Solutions to tensor momentum-energy-relation 1.2.1 exercises:

 



Given: V COM  21 (V IN V FIN )  21 ( 6    4  )10   5  10. This holds if there is time-reversal symmetry (T - symmetry).
 1   9 
 5 
T -symmetry demands V IN  V FIN in COM frame where V COM  0.
COM
COM
Momentum conservation axiom holds in all frames even if T -symmetry is broken. (Always true: VVW VVW
or
SUV
COM
( M SUV  MVW )VVW
or
SUV
IN
COM
or
SUV
FIN
COM
( M SUV  MVW )VVW
or
SUV
 

   4 0  to write momentum conservation axiom.
  0 1 
0
MVW
COM
 VVW
or
SUV
IN
IN
FIN

FIN
( M SUV VSUV  MVW VVW )
or
SUV
 V COM M V FIN






100 5 5  4 0   5   100 5 5  4 0   6 
 0 1  5 
 0 1  1 

COM
( M SUV VSUV  MVW VVW )  VVW
 V COM M V IN
V COM M V COM

FIN
 ( M SUV VSUV  MVW VVW )  ( M SUV VSUV  MVW VVW )
 M
SUV
Use mass tensor: M  

0
VVW
IN
COM
 VSUV .)



 100 5 5  4 0   4 
 0 1  9 



100α125

100α125

100α125
With T -symmetry this becomes the following elastic kinetic energy (KE) conservation theorem :
 21 (V IN V FIN ) M V IN
V COM M V COM
V COM M V COM  21 V IN M V FIN 
Mass matrix M-symmetry is used: V
IN
1 IN
2V
FIN
MV

M V IN
V



100 6 1  4 0   4   100
 0 1  9 

 21 (V IN V FIN ) M V FIN

FIN
4
MV
1 FIN
2V
M V FIN

KE  7, 250 E-units
IN
 4 0  6 
9 

  100 105  10, 250 E-units
 0 1  1 

Inelastic kinetic energy IE= 21 V COM M V COM is less than elastic KE= 21 V IN M V IN as derived here:
IE  21 V COM M V COM 
KE

6, 250

IE

1 IN
4V
M V IN 

3,625
1 IN
4V
MV
IN

1 IN
4V
M V FIN

1
2 KE

1 IN
4V
M V FIN
2,625
1 IN
4V
M V FIN
 1,000 E-units
Solution Exercise 1.3.1 … (VSUV-1,VSUV-2)=(60,10) collision like Fig. 3.1 with M=4 SUV
replacing VW.
Exercise 1.3.1 Equal mass M=4 SUV collision geometry for elastic and inelastic cases.
Research Problem
1.1.2 This is a dimensional analysis and power law problem. It involves Olympic weight
lifters but is a general piece of mechanics that applies to everything. (Have you
wondered why toy cars can fall off cliffs without damage while yours cannot?)
Olympic weight lifters are divided into classes according to their body weight.
Generally top performers are close to maximum allowed by their class (except for
“super-heavyweight" classes.)
(a) From dimensional arguments alone, you can predict that the Olympic records R in a
given event (say, the "clean and jerk" which is always the greatest record) should have
a definite functional relationship to the weight W=Mg of the performers: R= R(W).
Derive R(W) as a power law times an undetermined coefficient.
(b) Obtain a set of records from an almanac or book of records, and plot them against W
for a given event or events. See how well your theory and experiment jive. (Hint: It is
most convenient to plot on log-log graph paper. Why?)
(c) Use the results of (a) and (b) to answer: How many times his bodyweight could a man
lift if he was the size of an ant with a mass of M = 1 gm.? (A real ant is supposed to
lift five or ten times its body weight. How much better or worse is the ant doing than
"Antman"? )
Solutions.
Strength is proportional to crossection-area : S=kSA= kSL2.
Weight is proportional to volume : W=kWV= kWL3.
Strength is proportional to cross-section-area : S= kSL2= kS[(W/kW)1/3]2= (kS/kW1/3) W2/3.
If dimensional theory is correct then log-log graph would give straight line of slope 2/3:
log S=log (kS/kW1/3)+(2/3)log W.
If W=125kg man could lift S=400kg (that’s pushing it!) then: S=400= (kS/kW1/3) 1252/3=
(kS/kW1/3)25.
This gives a coefficient of : (kS/kW1/3)=8. (That would be 3.2 times mass-weight.)
Given the same structure, a W=1kg man could lift S=8kg. (8 times weight.)
A W=10-3kg=1gm “ant” man could lift S=8·10-2kg=80gm. (80 times weight.)
A W=10+3kg “Elephant” man could lift S=8·102kg=800kg. (0.8 times weight.)
A W=10+6kg “KingKong” man could lift S=8·104kg=80,000kg. (only 8% of weight.)
Solutions for Exercise 1.3.2
v2
VFIN
1,-1
IN
VFIN
C0M= -VC0M
VIN
0,V VFIN
v,0
IN
VFIN
1,1 =V1,1
V IN
v,0
v1
VFIN
0,V
V IN
1,-1
V IN
C0M
Exercise 1.3.2 Solutions. .. m1 and m2 formulas for… final elastic velocities … for initial
velocity vectors.
IN
V1,-1
= (1,-1)
IN
Vv,0
= (v,0)
V0IN,V = (0,V)
IN
VCOM
= (vCOM
,vCOM
)
x
y
a.
b.
c.
d.
Indicate T·VIN of each VIN … slope of orange (dashed) and green (solid) black line.
v2
slope
V
IN
1,-1
TV
FIN
1,-1
VIN
-(m /m )
0,V VFIN
v,0 1 2
slope
-(m1/m2 )
slope
-Γ(m1/m2 )
=VIN
-1,1
slope
+(1/1)
Time
reversal
T
V IN
v,0
slope
+(m1/m2 )
VFIN
0,V
V IN
1,-1
IN
slope
V C0M
+(m1/m2 )
v1
Exercise 1.3.2 Solutions(contd.)
IN
V1,1
KE determined by
1
2 1
2 1
1
2 1
2
=(1,-1) is : KE  2 m11  2 m21  2 (m1  m2 )  2 m1v  2 m2V .
2
2
All others on the KE-ellipse have same KE so: v  (m1  m2 ) / m1 and: V  (m1  m2 ) / m2 .
Lines of slope=+1 include dashed tangents to ellipse at VCOM and main COM line
(That is due to Galileo’s translation symmetry a.k.a Newton momentum conservation.)
Lines with negative slope are slope=-m1/m2 except for the black line that has slope=IN
m1 /m2
=-V/v.
(That is due to Galileo-Newton momentum conservation.)
is m1/m2=2.)
(In this graph the mass ratio
IN
COM COM
The COM collision frame starts with VCOM = (v1 ,v2 ) and has the same KE and slope
-m1/m2 , too.
mv
  + m v 
m v
v
 + m v m / m 
m v
 + m / m v   m v  (1+ m / m )
( m  m )·m / m v 
So:
KE 12 (m1  m2 ) 12 m1v 2 12 m1 v1COM
12
12
12
1
COM
1
1
COM
1
1
2
2
2
v1COM  m2 /m1 ,
1
2
2
1
2
1
1
COM
1
2
2
2
COM
1
2
1
2
2
COM
2
2
COM
1 1
2
1
COM
1
2
COM
1
2
m2 v2COM  0
1
2
1
COM
1
m1 / m2  v2COM
2
1
2
2
v2COM   m1 /m2
.
But, for ELASTIC (“ka-bong”) cases only the linear momentum conservation (Newton
I.) is needed. The more complicated quadratic formula is a corollary of Newton axiom I.
and T-symmetry.
v

v
FIN
1
FIN
2
 m1  m2
2m2 

  2m1
m2  m1   v1IN 

 IN 
m1  m2

 v2 
General vFIN related to vIN by vFIN=2VCOM-vIN or:
.
This is in Ch. 4 (Eqs. (4.1) and (4.3)) and used later in Ch. 5 Eq. (5.1) to give all v’s.


 v1COM FIN  
 COM FIN   
 v2
 



m2 
m2 
 m1  3m2 




IN
FIN
m1 
m1   v1   3m1  m2 
 v1  
 v1IN   1 
for
:

,

for
:
 IN  
 IN    

  FIN 
m1  m2
 v2  
 v2   1
m1 
m1   v2 


m2 
m2 
(See figures.)
Optional Practice Exercises with answers. Peek only after trying. Some of these may
show up on quizzes.
Exercise 1.1.1 Be sure you know how to ruler & compass construct arctan(y/x) and
arcsec(r/x) and their complimentary arcot(x/y) and arcsec(r/y) as well as the geometric mean
a  b in
Fig. 1.8 (3rd frame).
Use this to construct √5 and the Golden Means
resonance.


G 
1 5
2 that

will be used later to study

(G± are important because they satisfy G  G  1 and G  G  1 . G± are the “most
irrational” numbers.)
Use the preceding to find Golden angle artan(G+) and its corresponding Golden Rectangle.
This should resemble the rectangle(s) in Fig. 1.5 with an interesting twist: the famous
whirling square construction.
(Do it!)
Exercise 1.2.1 Redraw Fig. 2.5 for initial speeds (VSUV=40,VVW=10) with the SUV only
twice the mass of the VW. (Hummer-Lite) Include also a line describing the frame in
which the SUV is initially stationary and another for which the SUV is finally stationary.
Practice exercise solutions:
Exercise 1.1.1 … constructing √5 and the Golden Means
G 
1 5
2
Exercise 1.2.1 Redraw Fig. 2.5 for initial speeds (VSUV=40,VVW=10) with the SUV only
twice the mass of the VW. (Hummer-Lite) Include also a line describing the frame in
which the SUV is initially stationary and another for which the SUV is finally stationary.
Solutions to 1.2.1 Intercepts of 45° lines with v1 or v2 axes indicate the IN or FIN frames
for VW or SUV.
v2
vCOM=0
frame
v1
v1=0
frames
v2=0
frames