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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 30 "Classical Mechanics with \+
Maple" }}{PARA 256 "" 0 "" {TEXT -1 28 "Section 3.3: Linear Momentum" 
}}{PARA 19 "" 0 "" {TEXT -1 41 "Dr. Harald Kammerer\nmaple@jademountai
n.de" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Initialisation" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 57 "libname:=\"C:/mylib/m6dynlib\",\"C:/mylib/m6
dynfig\",libname:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "with(l
inalg):with(plots):with(plottools):with(dynamics);with(figures_chapter
_3);" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 19 "3.3 Linear Momentum" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 194 "Above we considered the integral \+
of a force along a path between two positions. Now we consider the int
egral over time. In general the force acting on a mass particle is var
iable over the time, " }{XPPEDIT 18 0 "F=F(t)" "6#/%\"FG-F$6#%\"tG" }
{TEXT -1 6 ". The " }{TEXT 256 15 "linear momentum" }{TEXT -1 27 " of \+
the force is defined as" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "L = int(F(
t),t = t1 .. t2):" "6#/%\"LG-%$intG6$-%\"FG6#%\"tG/F+;%#t1G%#t2G" }}
{PARA 0 "" 0 "" {TEXT -1 45 "We apply this definition on the law of mo
tion" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "int(F(t),t=t1..t2)=int(m*a,t \+
= t1 .. t2):" "6#/-%$intG6$-%\"FG6#%\"tG/F*;%#t1G%#t2G-F%6$*&%\"mG\"\"
\"%\"aGF3/F*;F-F." }}{PARA 0 "" 0 "" {TEXT -1 11 "This yields" }}
{PARA 257 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "int(m*a,t = t1 .. t2) \+
= m*int(1,v = v1 .. v2):" "6#/-%$intG6$*&%\"mG\"\"\"%\"aGF)/%\"tG;%#t1
G%#t2G*&F(F)-F%6$F)/%\"vG;%#v1G%#v2GF)" }}{PARA 0 "" 0 "" {TEXT -1 14 
"Finally we get" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "L = m*v2-m*v1;" "6
#/%\"LG,&*&%\"mG\"\"\"%#v2GF(F(*&F'F(%#v1GF(!\"\"" }}{PARA 0 "" 0 "" 
{TEXT -1 12 "Notice that " }{XPPEDIT 18 0 "F, v1,  v2" "6%%\"FG%#v1G%#
v2G" }{TEXT -1 23 " and consequently also " }{XPPEDIT 18 0 "L;" "6#%\"
LG" }{TEXT -1 13 " are vectors." }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 
32 "Example: Straight Central Impact" }}{PARA 0 "" 0 "" {TEXT -1 375 "
The linear momentum is very useful to describe what happens in the cas
e of the impact of two mass particles. Here we consider the straight c
entral impact. To imagine the situation we consider two balls that are
 moving translatoric on an straight trajectory. Both centers of gravit
y shall be before and after the impact on this trajectory (see Fig. 10
). We use the coordinate " }{TEXT 257 1 "x" }{TEXT -1 80 " to describe
 the motion. Before the impact the velocities of the mass particles " 
}{TEXT 260 2 "m1" }{TEXT -1 4 " and" }{TEXT 261 0 "" }{TEXT -1 1 " " }
{TEXT 262 2 "m2" }{TEXT -1 5 " are " }{TEXT 258 2 "v1" }{TEXT -1 5 " a
nd " }{TEXT 259 2 "v2" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 67 "display(Fig_3_7(),scaling=constrained,axes=none,title
=\"Figure 10\");" }}{PARA 0 "" 0 "" {TEXT -1 42 "We divide the process
 into three sections:" }}{PARA 0 "" 0 "" {TEXT -1 47 "1. The motion be
fore the impact, described by  " }{TEXT 263 2 "v1" }{TEXT -1 5 " and \+
" }{TEXT 264 3 "v2," }}{PARA 0 "" 0 "" {TEXT -1 17 "2. the impact and
" }}{PARA 0 "" 0 "" {TEXT -1 46 "3. the motion after the impact, descr
ibed by  " }{TEXT 265 3 "v1q" }{TEXT -1 5 " and " }{TEXT 266 4 "v2q." 
}}{PARA 0 "" 0 "" {TEXT -1 40 "The impact happens in the time interval
 " }{TEXT 267 2 "t1" }{TEXT -1 4 " to " }{TEXT 268 2 "t2" }{TEXT -1 
34 ". During this time an inner force " }{XPPEDIT 18 0 "F(t)" "6#-%\"F
G6#%\"tG" }{TEXT -1 197 " acts between both particles. When we cut bot
h particles free the force acts on every particle in the oppostite dir
ection with the same absolute value. For both particles we get the lin
ear momentum" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ1a := m1*(v
1q-v1) = -L:" "6#>%%EQ1aG/*&%#m1G\"\"\",&%$v1qGF(%#v1G!\"\"F(,$%\"LGF,
" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ1b := m2*(v2q-v2) = L:
" "6#>%%EQ1bG/*&%#m2G\"\"\",&%$v2qGF(%#v2G!\"\"F(%\"LG" }}}{PARA 0 "" 
0 "" {TEXT -1 5 "with " }{XPPEDIT 18 0 "L = int(F(t),t = t1 .. t2);" "
6#/%\"LG-%$intG6$-%\"FG6#%\"tG/F+;%#t1G%#t2G" }{TEXT -1 51 " the total
 linear momentum during the time periode " }{XPPEDIT 18 0 "Delta*t=t2-
t1" "6#/*&%&DeltaG\"\"\"%\"tGF&,&%#t2GF&%#t1G!\"\"" }{TEXT -1 268 ". T
his linear momentum has the same absolute value for both particles but
 points in the opposite direction, same as the passive force between b
oth particles. For further calculation we need some information about \+
the force F(t). To get this we divide the time periode " }{XPPEDIT 18 
0 "Delta*t;" "6#*&%&DeltaG\"\"\"%\"tGF%" }{TEXT -1 16 " into two parts
 " }{XPPEDIT 18 0 "Delta*t1;" "6#*&%&DeltaG\"\"\"%#t1GF%" }{TEXT -1 5 
" and " }{XPPEDIT 18 0 "Delta*t2;" "6#*&%&DeltaG\"\"\"%#t2GF%" }{TEXT 
-1 24 ". During the first part " }{XPPEDIT 18 0 "Delta*t1" "6#*&%&Delt
aG\"\"\"%#t1GF%" }{TEXT -1 57 " both particles are compressed, during \+
the second period " }{XPPEDIT 18 0 "Delta*t2" "6#*&%&DeltaG\"\"\"%#t2G
F%" }{TEXT -1 88 " both particles are decompressed. Consequently we se
parate two partial linear momentums " }{TEXT 269 2 "L1" }{TEXT -1 5 " \+
and " }{TEXT 270 3 "L2 " }{TEXT -1 69 "for the process of compression \+
and the decompression. Further we set " }{XPPEDIT 18 0 "t1=0" "6#/%#t1
G\"\"!" }{TEXT -1 51 " for the following consideration. Then we can wr
ite" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "L1 = int(F(t),t = 0 .. Delta*t
1):" "6#/%#L1G-%$intG6$-%\"FG6#%\"tG/F+;\"\"!*&%&DeltaG\"\"\"%#t1GF1" 
}}{PARA 0 "" 0 "" {TEXT -1 75 "At the end of the compression period bo
th particles have the same velocity " }{TEXT 271 2 "vm" }{TEXT -1 46 "
. This yields for the partial linear momentum " }{TEXT 272 2 "L1" }
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ2a := m1*(vm
-v1) = -L1:" "6#>%%EQ2aG/*&%#m1G\"\"\",&%#vmGF(%#v1G!\"\"F(,$%#L1GF," 
}}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "EQ2b := m2*(vm-v2) = L1:" "
6#>%%EQ2bG/*&%#m2G\"\"\",&%#vmGF(%#v2G!\"\"F(%#L1G" }}}{PARA 0 "" 0 "
" {TEXT -1 60 "Next we need a relation between the partial linear mome
ntum " }{TEXT 273 2 "L2" }{TEXT -1 58 " during the decompression and t
he partial linear momentum " }{TEXT 274 2 "L1" }{TEXT -1 2 ". " }
{TEXT 275 0 "" }{TEXT -1 30 "Here we use a hypothesis from " }{TEXT 
276 6 "Newton" }{TEXT -1 0 "" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "L2 = \+
e*L1:" "6#/%#L2G*&%\"eG\"\"\"%#L1GF'" }}{PARA 0 "" 0 "" {TEXT -1 19 "w
ith the parameter " }{TEXT 277 1 "e" }{TEXT -1 109 " which depends mai
nly on the characteristics of the material of both particles. This par
ameter is called the " }{TEXT 281 25 "coefficient of elasticity" }
{TEXT -1 26 " (don't confound with the " }{TEXT 282 23 "modulus of ela
sticity E" }{TEXT -1 72 "). One limit case is given by totatally elast
ic material. Then there is " }{XPPEDIT 18 0 "L2 = L1;" "6#/%#L2G%#L1G
" }{TEXT -1 8 " and so " }{XPPEDIT 18 0 "e=1" "6#/%\"eG\"\"\"" }{TEXT 
-1 214 ". This is nearly valid whe one drops a rubber ball on the hard
 ground. There is no enduring deformation at the end of the impact pro
cess. The second limit case is given by totally inelastic material. Th
en there is " }{XPPEDIT 18 0 "L2 = 0;" "6#/%#L2G\"\"!" }{TEXT -1 8 " a
nd so " }{XPPEDIT 18 0 "e=0" "6#/%\"eG\"\"!" }{TEXT -1 194 ". This is \+
valid when plasticine drops to the ground. It stays there after the im
pact without further motion. Real materials are always between these t
wo limit cases and so we can say in general " }{XPPEDIT 18 0 "0<=e" "6
#1\"\"!%\"eG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "e<=1" "6#1%\"eG\"\"
\"" }{TEXT -1 77 ". The total linear momentum during the total impact \+
process is then given by " }{XPPEDIT 18 0 "L = L1+L2;" "6#/%\"LG,&%#L1
G\"\"\"%#L2GF'" }{TEXT -1 7 " and so" }}{EXCHG {PARA 0 "> " 0 "" 
{XPPEDIT 19 1 "EQ3 := L = (1+e)*L1:" "6#>%$EQ3G/%\"LG*&,&\"\"\"F)%\"eG
F)F)%#L1GF)" }}}{PARA 0 "" 0 "" {TEXT -1 10 "For given " }{TEXT 278 1 
"e" }{TEXT -1 28 " we can solve the equations " }{TEXT 279 22 "EQ1a, E
Q1b, EQ2a, EQ2b" }{TEXT -1 5 " and " }{TEXT 280 3 "EQ3" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "sol:=solve(\{EQ1a, EQ1b, EQ2a, EQ2b
, EQ3\},\{vm,L,L1,v1q,v2q\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 12 "assign(sol):" }}}{PARA 0 "" 0 "" {TEXT -1 50 "At last we get for
 the velocities after the impact" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "v1q:=collect(v1q,\{v1,v2\});" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "v2q:=collect(v2q,\{v1,v2\});" }}}{PARA 0 "" 0 "" 
{TEXT -1 55 "The relations for this calculation are included in the " 
}{HYPERLNK 17 "dynamics" 2 "dynamics" "" }{TEXT -1 14 " package. See \+
" }{HYPERLNK 17 "impact" 2 "impact" "" }{TEXT -1 37 " to learn how to \+
use these functions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 65 "Now we consider a simple way to get the coefficient of \+
elasicity " }{TEXT 283 1 "e" }{TEXT -1 135 " for a ball falling on the
 ground. We assume that the ground doesn't move before and after the i
mpact.  We drop the ball with the mass " }{TEXT 287 1 "m" }{TEXT -1 
17 " from the height " }{TEXT 285 5 "h1=1m" }{TEXT -1 123 ". Then we m
easure the maximum height that it reaches after the impact on the grou
nd. We assume the ball reaches the height " }{TEXT 286 7 "h2=0.3m" }
{TEXT -1 18 " after the impact." }}{PARA 0 "" 0 "" {TEXT -1 66 "(See F
ig. 11. Click the picture and press \"play\" in the menu bar.)" }}
{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h1example := 1:" "6#>%*h1examp
leG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "h2example := .3:
" "6#>%*h2exampleG-%&FloatG6$\"\"$!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 104 "display(Fig_3_8(h1example,h2example),insequence=true
,scaling=constrained,axes=framed,title=\"Figure 11\");" }}}{PARA 0 "" 
0 "" {TEXT -1 37 "To get the coefficient of elasticity " }{TEXT 284 1 
"e" }{TEXT -1 202 " we consider the situation immediately before and i
mmediately after the impact. To calculate the velocities we consider t
he conservation of energy. We choose the ground as the reference horiz
ontal with " }{XPPEDIT 18 0 "z=0" "6#/%\"zG\"\"!" }{TEXT -1 26 ". At t
he initial position " }{XPPEDIT 18 0 "z=h0" "6#/%\"zG%#h0G" }{TEXT -1 
77 " and at the end of the considered motion, when the ball reaches th
e position " }{XPPEDIT 18 0 "z=h1" "6#/%\"zG%#h1G" }{TEXT -1 29 " the \+
velocity of the ball is " }{XPPEDIT 18 0 "v_start=0" "6#/%(v_startG\"
\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "v_end=0" "6#/%&v_endG\"\"!" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 59 "So we get the kinetic en
ergy at the beginning of the motion" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "Tstart:=kinetic_energy(m,0);" }}}{PARA 0 "" 0 "" 
{TEXT -1 38 "and also at the end of the time period" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 "Tend:=kinetic_energy(m,0);" }}}{PARA 0 "" 
0 "" {TEXT -1 50 "The kinetic energy imediately before the impact is" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "T1:=kinetic_energy(m,v1);
" }}}{PARA 0 "" 0 "" {TEXT -1 31 "and imediately after the impact" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "T2:=kinetic_energy(m,v2);" }
}}{PARA 0 "" 0 "" {TEXT -1 57 "The gravitational potential is at the b
egin of the motion" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Vstart
:=gravitational_energy(m,g,h1);" }}}{PARA 0 "" 0 "" {TEXT -1 46 "At th
e end we have the gravitational potential" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "Vend:=gravitational_energy(m,g,h2);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }{TEXT -1 35 "The potential before impact is z
ero" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "V1:=0;" }}}{PARA 0 "" 0 "" 
{TEXT -1 51 "The conservation of energy before the impact yields" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "Energy_conservation_1:=Tstar
t+Vstart=T1+V1;" }}}{PARA 0 "" 0 "" {TEXT -1 45 "and we get for the ve
locity before the impact" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "
v1:=solve(Energy_conservation_1,v1);" }}}{PARA 0 "" 0 "" {TEXT -1 54 "
We need the negative solution, because the coordinate " }{TEXT 288 1 "
z" }{TEXT -1 108 " and also the velocity was choosen upwards positive \+
but the velocity points downwards in reality. That means" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "v1:=-sqrt(v1[1]**2);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "After impact, the potential is also 0." }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "V2:=0;" }}}{PARA 0 "" 0 "" {TEXT 
-1 50 "After the impact the conservation of energy yields" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "Energy_conservation_2:=T2+V2=Tend+V
end;" }}}{PARA 0 "" 0 "" {TEXT -1 45 "and we get for the velocity befo
re the impact" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "v2:=solve(E
nergy_conservation_2,v2);" }}}{PARA 0 "" 0 "" {TEXT -1 45 "Now we need
 the positive solution, that means" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "v2:=sqrt(v2[1]**2);" }}}{PARA 0 "" 0 "" {TEXT -1 14 "
The procedure " }{HYPERLNK 17 "impact" 2 "impact" "" }{TEXT -1 43 " yi
elds for the velocities after the impact" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 79 "v[ball_after_impact]:=expand(impact(m[ball],m[ground]
,v[ball],v[ground],e)[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
81 "v[ground_after_impact]:=expand(impact(m[ball],m[ground],v[ball],v[
ground],e)[2]);" }}}{PARA 0 "" 0 "" {TEXT -1 28 "with the mass of the \+
ground " }{XPPEDIT 18 0 "m[ground]" "6#&%\"mG6#%'groundG" }{TEXT -1 
19 " and the velocitiy " }{XPPEDIT 18 0 "v[ground]" "6#&%\"vG6#%'groun
dG" }{TEXT -1 58 " of the ground before the impact. The mass of the ba
ll is " }{XPPEDIT 18 0 "m[ball]" "6#&%\"mG6#%%ballG" }{TEXT -1 39 " an
d its velocity before the impact is " }{XPPEDIT 18 0 "v[ball]" "6#&%\"
vG6#%%ballG" }{TEXT -1 35 ". It is evident that we can assume " }
{XPPEDIT 18 0 "m[ground] = infinity;" "6#/&%\"mG6#%'groundG%)infinityG
" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "v[ground] = 0;" "6#/&%\"vG6#%'gr
oundG\"\"!" }{TEXT -1 45 " without making significant errors. So we ge
t" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "v[ball_after_impact]:=l
imit(v[ball_after_impact],\{v[ground]=0,m[ground]=infinity\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "v[ground_after_impact]:=limi
t(v[ground_after_impact],\{v[ground]=0,m[ground]=infinity\});" }}}
{PARA 0 "" 0 "" {TEXT -1 33 "Applying this result we get with " }
{XPPEDIT 18 0 "v1=v[ball]" "6#/%#v1G&%\"vG6#%%ballG" }{TEXT -1 5 " and
 " }{XPPEDIT 18 0 "v2=v[ball_after_impact]" "6#/%#v2G&%\"vG6#%2ball_af
ter_impactG" }}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "eq:=v2=-e*v1;" 
"6#>%#eqG/%#v2G,$*&%\"eG\"\"\"%#v1GF*!\"\"" }}}{PARA 0 "" 0 "" {TEXT 
-1 49 "At last we get for the cooefficient of elasticity" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "e:=simplify(solve(eq,e),symbolic);
" }}}{PARA 0 "" 0 "" {TEXT -1 53 "For the concrete values from the exa
mple above we get" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "subs(\{
h1=h1example,h2=h2example\},e);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}}}{MARK "0 0 0" 16 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }