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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 17 "Conservation laws" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "We solve \+
a problem of ballistics using energy and momentum conservation. A bull
et of mass " }{TEXT 257 1 "m" }{TEXT -1 44 "  is shot into a suspended
 sand box of mass " }{TEXT 258 1 "M" }{TEXT -1 58 ". The aim is to det
ermine the initial speed of the bullet " }{TEXT 264 1 "v" }{TEXT -1 
89 "0 from the motion of the combined sandbox/bullet system that acqui
res the final velocity " }{TEXT 259 1 "V" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "The exercise will \+
help us to understand how the " }{TEXT 19 5 "solve" }{TEXT -1 256 " co
mmand works in Maple. We will also learn how to scale variables in suc
h a way that the results can be graphed without making assumptions abo
ut numerical values for the parameters in the problem. This represents
 a very useful trick for many applications." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 272 "The two equations required are
 the energy conservation and momentum conservation laws. We include th
e conversion of some mechancial energy into heat, i.e., we assume that
 a part of the bullet's kinetic energy is converted into translational
 (kinetic) energy of the system " }{TEXT 260 3 "M+m" }{TEXT -1 95 " , \+
and a fraction goes into heating the sand while the bullet is slowed d
own from the velocity " }{TEXT 263 1 "v" }{TEXT -1 5 "0 to " }{TEXT 
262 1 "V" }{TEXT -1 27 ". The latter is labeled as " }{TEXT 261 1 "Q" 
}{TEXT -1 92 ". The conditions are called inelastic, as some mechanica
l energy is removed from the system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 263 "The energy conservation law states tha
t the kinetic energy of the bullet before it hits the suspended sandbo
x equals the kinetic energy of the combined system after the hit plus \+
the amount of heat produced by the slow-down of the bullet inside the \+
box (friction)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "En:=m/2*
v0^2=(M+m)/2*V^2+Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EnG/,$*&%\"m
G\"\"\")%#v0G\"\"#\"\"\"#F)F,,&*&,&%\"MGF)F(F)F))%\"VGF,F-F.%\"QGF)" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Momentum conservation is stated \+
as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Mom:=m*v0=(M
+m)*V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$MomG/*&%\"mG\"\"\"%#v0GF(
*&,&%\"MGF(F'F(F(%\"VGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We so
lve the pair of equations for the final velocity and the unknown heat:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol:=solve(\{En,Mom\},
\{V,Q\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<$/%\"VG*&*&%\"mG
\"\"\"%#v0GF+\"\"\",&%\"MGF+F*F+!\"\"/%\"QG,$*&*(F*F-)F,\"\"#F-F/F+F-F
.F0#F+F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 216 "From this result it \+
is evident that the determination of the final velocity (e.g., from th
e amplitude of the suspended sand box motion), and knowledge of both m
asses m and M permits to determine the initial velocity " }{TEXT 265 
1 "v" }{TEXT -1 86 "0. This information is contained in the first part
 of the solution. Let us extract it:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "nops(sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "sol[1];" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/%\"VG*&*&%\"mG\"\"\"%#v0GF(\"\"\",&%\"MGF(F'F(!\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solve(%,v0);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#*&*&%\"VG\"\"\",&%\"MGF&%\"mGF&F&\"\"\"F)!\"
\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "We assume a mass of the sa
nd box 1000 times larger than for the bullet, e.g., 5 kg and 5 g respe
ctively:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(M=1000*m,%
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%\"VG\"%,5" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 178 "From the above we realize that we need to know t
he mass ratio of the sand box and bullet masses to obtain the relation
ship of initial bullet and final sand box/bullet velocities." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "An interesting
 conclusion concerns the fraction of  the bullet's initial kinetic ene
rgy converted to heat." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "so
l[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"QG,$*&*(%\"mG\"\"\")%#v0G
\"\"#\"\"\"%\"MGF)F-,&F.F)F(F)!\"\"#F)F," }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 75 "We follow the technique of assigning the solution. This f
ixes the variable " }{TEXT 266 1 "Q" }{TEXT -1 13 " in our case:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assign(sol[2]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#,$*&*(%\"mG\"\"\")%#v0G\"\"#\"\"\"%\"MGF'F+,&F,F'F&F'!\"\"#F'F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "fr:=subs(M=1000*m,Q)/(m/2*v0
^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#frG#\"%+5\"%,5" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 93 "We observe that the majority of the bulle
t's kinetic energy was used up to heat the sand box." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Let us look at the frac
tion in general:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "fr:=Q/(
m/2*v0^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#frG*&%\"MG\"\"\",&F&
\"\"\"%\"mGF)!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "How would w
e express this fraction as a function of the mass ratio " }{TEXT 293 
2 "r " }{TEXT -1 2 "= " }{TEXT 292 1 "M" }{TEXT -1 1 "/" }{TEXT 291 1 
"m" }{TEXT -1 2 " ?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "fr1:
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{TEXT -1 30 "The fact that for mass ratios " }{TEXT 294 1 "r" }{TEXT 
-1 3 " = " }{TEXT 277 1 "M" }{TEXT -1 1 "/" }{TEXT 278 1 "m" }{TEXT 
-1 242 " of the order of 100 or more the collision is entirely inelast
ic becomes obvious from the graph. One can make use of this fact and s
olve the problem approximately by momentum conservation alone (once on
 the RHS of the energy conservation law " }{TEXT 279 1 "Q" }{TEXT -1 
113 " dominates completely, the other term can be ignored, and the equ
ation becomes redundant in the determination of " }{TEXT 280 1 "V" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 161 "Now one can solve the \+
problem of the ballistic pendulum, where the kinetic energy of the bul
let/sand box system is converted into gravitational potential energy:
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "Mom:=m*v0=(M+m)*V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$MomG/*&%\"mG\"\"\"%#v0GF(*&,&%\"MGF(F'F(F(%\"VGF(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "V:=solve(Mom,V);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"VG*&*&%\"mG\"\"\"%#v0GF(\"\"\",&%\"MGF(F'F(!\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "En:=(M+m)/2*V^2=(M+m)*g
*h;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EnG/,$*&*&)%\"mG\"\"#\"\"\")
%#v0GF+F,F,,&%\"MG\"\"\"F*F1!\"\"#F1F+*(F/F1%\"gGF1%\"hGF1" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "height:=solve(En,h);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%'heightG,$*&*&)%\"mG\"\"#\"\"\")%#v0GF*F+F+*&)
,&%\"MG\"\"\"F)F2\"\"#F+%\"gG\"\"\"!\"\"#F2F*" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 93 "We can do variations on the theme, i.e., consider othe
r types of collisions in one dimension." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 162 "An elastic collision with different
 final velocities for the projectile and the target is an obvious case
. Let us restart the Maple engine to reset the variables:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 3 "En;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%#EnG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Energy conservation now reads:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "En:=m/2*v0^2=m/2*v^2+M/2*V^
2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EnG/,$*&%\"mG\"\"\")%#v0G\"\"
#\"\"\"#F)F,,&*&F(F-)%\"vGF,F-F.*&%\"MGF))%\"VGF,F-F." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 34 "Momentum conservation now becomes:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Mom:=m*v0=m*v+M*V;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$MomG/*&%\"mG\"\"\"%#v0GF(,&*&F'\"\"\"%\"v
GF(F(*&%\"MGF(%\"VGF(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "
sol:=solve(\{En,Mom\},\{V,v\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$
solG6$<$/%\"vG%#v0G/%\"VG\"\"!<$/F(*&*&F)\"\"\",&%\"mGF1%\"MG!\"\"F1\"
\"\",&F4F1F3F1!\"\"/F+,$*&*&F3F1F)F6F6F7F8\"\"#" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 80 "The solution contains two sets: the first describes \+
the initial condition (mass " }{TEXT 268 1 "M" }{TEXT -1 18 " at rest \+
and mass " }{TEXT 269 1 "m" }{TEXT -1 22 " moving with velocity " }
{TEXT 267 1 "v" }{TEXT -1 93 "0), i.e., the situation before the colli
sion. It is the second set that we are interested in." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assign(sol[2]);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 4 "v,V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$*&*&
%#v0G\"\"\",&%\"mGF&%\"MG!\"\"F&\"\"\",&F)F&F(F&!\"\",$*&*&F(F&F%F+F+F
,F-\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "We could be interest
ed in expressing the ratio of initial to final veocity of the light pa
rticle as a function of the mass ratio " }{TEXT 297 2 "r " }{TEXT -1 
4 " := " }{TEXT 296 1 "M" }{TEXT -1 1 "/" }{TEXT 295 1 "m" }{TEXT -1 
2 " :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "ratio:=simplify(su
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$\"+D*[>m#FQ$!1GweY``c**F/7$$\"+Rk:!o#FQ$!16'e]=;$e**F/7$$\"+,g(*)p#FQ
$!14'H/R!3g**F/7$$\"+ktp<FFQ$!1g!pFWh<'**F/7$$\"+e@OLFFQ$!1b5#Gh8J'**F
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y#FQ$!1lwPn\\&p'**F/7$$\"+t:_&z#FQ$!1H7_!yC!o**F/7$$\"+FH)3\"GFQ$!1YjB
(GM\"p**F/7$$\"+'4PU#GFQ$!1Sq8Qv1q**F/7$$\"+?j0QGFQ$!1H$QTj.5(**F/7$$
\"+2)4-&GFQ$!1WN=/F!=(**F/7$$\"+;)=J'GFQ$!15)f!ztis**F/7$$\"+>X#\\(GFQ
$!1>B;m/Ot**F/7$$\"+_6$p)GFQ$!1h$)*>#f3u**F/7$$\"+6uN)*GFQ$!1-6^zzvu**
F/7$$\"+Gg+5HFQ$!1NXI*>Da(**F/7$$\"+!=P4#HFQ$!1<&zLHNg(**F/7$$\"+%QQ=$
HFQ$!1t;H\\'Gm(**F/7$$\"+[cQUHFQ$!1fW%)o()=x**F/7$$\"+?n&=&HFQ$!1yBs+.
ox**F/7$$\"+WNYiHFQ$!1dE)zA=#y**F/7$$\"+7etrHFQ$!1V)pI&yny**F/7$$\"+r'
39)HFQ$!1%=\\I,Z\"z**F/7$$\"+9$p/*HFQ$!1D!y.7x&z**F/7$$\"+++++IFQ$!1!)
*>+)*>+)**F/-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"r6\"%!G
-%%VIEWG6$;F(Fi]l%(DEFAULTG-%*AXESTICKSG6$7F/F(Q#1.Fh^l/$\"+d**H5IF,Q!
6\"/$\"+\\D@rZF,Fh_l/$\"+8**f?gF,Fh_l/$\"+T+q*)pF,Fh_l/$\"+.D^\"y(F,Fh
_l/$\"++/)4X)F,Fh_l/$\"+r)**3.*F,Fh_l/$\"+$4DCa*F,Fh_l/\"\"\"Q%.1e2Fh^
l/$\"+'**H5I\"FQFh_l/$\"+b77x9FQFh_l/$\"+\"**f?g\"FQFh_l/$\"+/+(*)p\"F
QFh_l/$\"+]7:y<FQFh_l/$\"+S!)4X=FQFh_l/$\"+()**3.>FQFh_l/$\"+4DCa>FQFh
_l/$\"+++++?FQQ%.1e3Fh^l/$\"+'**H5I#FQFh_l/$\"+b77xCFQFh_l/$\"+\"**f?g
#FQFh_l/$\"+/+(*)p#FQFh_l/$\"+]7:yFFQFh_l/$\"+S!)4XGFQFh_l/$\"+()**3.H
FQFh_l/$\"+4DCaHFQFh_l/Fi]lQ%.1e4Fh^l/$\"+'**H5I$FQFh_l/$\"+b77xMFQFh_
l/$\"+\"**f?g$FQFh_l/$\"+/+(*)p$FQFh_l/$\"+]7:yPFQFh_l/$\"+S!)4XQFQFh_
l/$\"+()**3.RFQFh_l/$\"+4DCaRFQFh_lF^_l" 1 2 0 1 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "We see t
hat as the mass ratio " }{TEXT 273 1 "M" }{TEXT -1 1 "/" }{TEXT 274 1 
"m" }{TEXT -1 87 " exceeds about 100 the motion describes essentially \+
a reflection of the light particle " }{TEXT 272 1 "m" }{TEXT -1 43 " f
rom a wall, i.e., it approaches the limit" }}{PARA 0 "" 0 "" {TEXT -1 
1 " " }{TEXT 270 1 "v" }{TEXT -1 2 "=-" }{TEXT 271 1 "v" }{TEXT -1 2 "
0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 303 10 "Ex
ercise: " }}{PARA 0 "" 0 "" {TEXT -1 32 "Provide a graph of the ratio \+
of " }{TEXT 275 1 "V" }{TEXT -1 1 "/" }{TEXT 276 1 "v" }{TEXT -1 136 "
0 as a function of the mass ratio. Graph the ratio of final kinetic en
ergies of the collision partners as a function of  the mass ratio." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "A
 slightly more sophisticated problem is solved in R.L. Greene's book: \+
" }{TEXT 298 30 "Classical Mechanics with Maple" }{TEXT -1 17 " (Sprin
ger 1994)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
299 12 "Party trick:" }{TEXT -1 34 " A light and a heavy ball (masses \+
" }{TEXT 281 1 "m" }{TEXT -1 5 " and " }{TEXT 282 1 "M" }{TEXT -1 81 "
 respectively) are dropped simultaneously in a gravitational field fro
m a height " }{TEXT 285 2 "h0" }{TEXT -1 6 " with " }{TEXT 284 1 "m" }
{TEXT -1 19 " sitting on top of " }{TEXT 283 1 "M" }{TEXT -1 50 " init
ially. What height will the light ball reach?" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "We assume elastic collis
ions and ignore the acceleration by the gravitational field during the
 collision itself." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 163 "The first statement is energy conservation: when the two
-ball system hits the ground it has converted its gravitational potent
ial energy fully into kinetic energy:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 
"En1:=(m+M)*g*h0=(m+M)/2*v0^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E
n1G/*(,&%\"mG\"\"\"%\"MGF)F)%\"gGF)%#h0GF),$*&F'\"\"\")%#v0G\"\"#F/#F)
F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The large ball " }{TEXT 
286 1 "M" }{TEXT -1 70 " bounces elastically from the ground, i.e., re
verses its velocity to -" }{TEXT 287 1 "v" }{TEXT -1 40 "0. It then co
llides with the small ball " }{TEXT 288 1 "m" }{TEXT -1 36 " that is s
till moving downward with " }{TEXT 289 1 "v" }{TEXT -1 2 "0." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "En2:=M/2*v0^2+m/2*v0^2=M/2*V
^2+m/2*v^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$En2G/,&*&%\"MG\"\"\"
)%#v0G\"\"#\"\"\"#F)F,*&%\"mGF)F*F-F.,&*&F(F-)%\"VGF,F-F.*&F0F-)%\"vGF
,F-F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Mom2:=M*(-v0)+m*v0
=M*V+m*v;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Mom2G/,&*&%\"MG\"\"\"%
#v0GF)!\"\"*&%\"mGF)F*\"\"\"F),&*&F(F.%\"VGF)F)*&F-F.%\"vGF)F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "The final height of the small ball
 called " }{TEXT 290 1 "h" }{TEXT -1 38 " is obtained from energy cons
ervation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "En3:=m/2*v^2=m
*g*h;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$En3G/,$*&%\"mG\"\"\")%\"vG
\"\"#\"\"\"#F)F,*(F(F-%\"gGF)%\"hGF)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 90 "We can try the simple approach, i.e., we can let Maple grind on
 the set of four equations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "sol:=solve(\{En1,En2,Mom2,En3\},h);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$solG6\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Our intellige
nce is needed..." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "v0s:=so
lve(En1,v0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$v0sG6$*&-%%sqrtG6#
\"\"#\"\"\"-F(6#*&%\"gG\"\"\"%#h0GF0F+,$F&!\"\"" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 166 "The choice of root depends on the definition of the
 positive direction. If upwards is positive, we choose the negative ro
ot to describe downward motion at this point:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 11 "v0:=v0s[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#v0G,$*&-%%sqrtG6#\"\"#\"\"\"-F(6#*&%\"gG\"\"\"%#h0GF0F+!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sol:=solve(\{En2,Mom2,En3\},
h);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 76 "Our difficulties are not yet solved...  even thoug
h v0 has been substituted:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
4 "En2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&*(%\"MG\"\"\"%\"gGF'%#h0
GF'F'*(%\"mGF'F(\"\"\"F)F,F',&*&F&F,)%\"VG\"\"#F,#F'F1*&F+F,)%\"vGF1F,
F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "solve(Mom2,V);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(*(%\"MG\"\"\"-%%sqrtG6#\"\"#\"\"
\"-F*6#*&%\"gGF(%#h0GF(F-!\"\"*(%\"mGF(F)F-F.F-F(*&F5F-%\"vGF(F(F-F'!
\"\"F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "V:=simplify(%);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&,(*(%\"MG\"\"\"-%%sqrtG6#
\"\"#\"\"\"-F,6#*&%\"gGF*%#h0GF*F/!\"\"*(%\"mGF*F+F/F0F/F**&F7F/%\"vGF
*F*F/F)!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(En2
,v);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$*&-%%sqrtG6#\"\"#\"\"\"-F&6#
*&%\"gG\"\"\"%#h0GF.F)!\"\",$*&*(F%F)F*F),&%\"mGF.%\"MG!\"$F.F),&F5F.F
6F.!\"\"F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 234 "There are two root
s and we have to figure out which one is physically acceptable. The fi
rst describes the situation before the collision (downward motion with
 v0), i.e., we need the one after the collision which depends on the m
asses:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "v:=%[2];" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"vG,$*&*(-%%sqrtG6#\"\"#\"\"\"-F)6#*&%\"g
G\"\"\"%#h0GF1F,,&%\"mGF1%\"MG!\"$F1F,,&F4F1F5F1!\"\"!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(En3,h);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&*&%#h0G\"\"\"),&%\"mGF&%\"MG!\"$\"\"#\"\"\"F-*$),&F)F
&F*F&\"\"#F-!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "h:=%;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG*&*&%#h0G\"\"\"),&%\"mGF(%\"MG
!\"$\"\"#\"\"\"F/*$),&F+F(F,F(\"\"#F/!\"\"" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 113 "We are interested again in exploring the answer as a fun
ction of the mass ratio and eliminate the initial height:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "ratio:=simplify(subs(M=r*m,h)/h0);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&ratioG*&*$),&!\"\"\"\"\"%\"rG\"
\"$\"\"#\"\"\"F.*$),&F*F*F+F*\"\"#F.!\"\"" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 29 "semilogplot(ratio,r=1..1000);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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l/$\"++/)4X)F.Fi`l/$\"+r)**3.*F.Fi`l/$\"+$4DCa*F.Fi`l/F*Q%.1e2Fi_l/$\"
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$\"+()**3.HFgnFi`l/$\"+4DCaHFgnFi`l/Fj^lQ%.1e4Fi_l/$\"+'**H5I$FgnFi`l/
$\"+b77xMFgnFi`l/$\"+\"**f?g$FgnFi`l/$\"+/+(*)p$FgnFi`l/$\"+]7:yPFgnFi
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{TEXT -1 162 "The result is astonishing: the light ball can return to \+
a multiple of its original height! The curve indicates a limiting beha
viour. We use Maple to explore this:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "limit(ratio,r=infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "For mass \+
ratios of " }{TEXT 302 2 "r " }{TEXT -1 2 "= " }{TEXT 301 1 "M" }
{TEXT -1 1 "/" }{TEXT 300 4 "m > " }{TEXT -1 134 "100 the maximum poss
ible height ratio is reached. The lighter ball can reach nine times th
e original height from which it was dropped." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 236 "Practice the trick with variou
s balls (tennis, ping-pong, solid rubber, etc.) before trying to impre
ss your friends. It takes a little skill to drop the balls in such a w
ay that they remain on top of each other when they hit the ground." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 586 "This pro
blem has further-reaching applications in that it can explain the esse
nce of some rather complicated phenomena in physics. One example is th
e problem of supernova explosions following the collapse of a star. On
e can think of the imploding massive star as carrying out a bounce whi
ch drives the lighter outer parts of the system through a shockwave to
 reach very high speeds. This mechanism results in the spilling of mat
erial (including heavy, unstable elements [nuclei beyond iron, i.e., b
eyond the maximum in the binding energy/nucleon curve]) into outer spa
ce at high speeds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 284 "Of course, the conservation laws can't tell us much abou
t the detailed dynamics of the motion. It is important to realize, how
ever, that the differential equations which describe the detailed moti
on (in Mechanics it is Newton's law) have solutions which satisfy the \+
conservation laws." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "78" 0 }{VIEWOPTS 1 1 0 1 1 1803 }