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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;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Momentum c
onservation is stated as follows:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "Mom:=m*v0=(M+m)*V;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 75 "We solve the pair of equations for the final velocity and the u
nknown heat:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol:=solve(
\{En,Mom\},\{V,Q\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 216 "From thi
s result it is evident that the determination of the final velocity (e
.g., from the amplitude of the suspended sand box motion), and knowled
ge of both masses 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);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "sol[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "
solve(%,v0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "We assume a mass
 of the sand box 1000 times larger than for the bullet, e.g., 5 kg and
 5 g respectively:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(
M=1000*m,%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "From the above w
e realize that we need to know the mass ratio of the sand box and bull
et masses to obtain the relationship 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 energy converted to heat." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "sol[2];" }}}{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;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "fr:=subs(M=1000*m,Q)/(m/2*v0^2);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 93 "We observe that the majority of the bullet's kinet
ic 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 fraction in \+
general:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "fr:=Q/(m/2*v0^2
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "How would we express this f
raction 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:=subs(M=m*r,fr);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fr1:=simplify(fr1);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "semilogplot(fr1,r=1..1000);" }}}
{EXCHG {PARA 0 "" 0 "" {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 collis
ion is entirely inelastic becomes obvious from the graph. One can make
 use of this fact and solve the problem approximately by momentum cons
ervation alone (once on the RHS of the energy conservation law " }
{TEXT 279 1 "Q" }{TEXT -1 113 " dominates completely, the other term c
an be ignored, and the equation 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 th
e kinetic energy of the bullet/sand box system is converted into gravi
tational potential energy:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Mom:=m*v0=(M+m)*V;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "V:=solve(Mom,V);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 26 "En:=(M+m)/2*V^2=(M+m)*g*h;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "height:=solve(En,h);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 93 "We can do variations on the theme, i.e., \+
consider other types of collisions in one dimension." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "An elastic collision w
ith 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;" }}}{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;" }}}{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;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol:=solve(\{En,Mom\},\{V,v\});" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "The solution contains two sets: t
he first describes the initial condition (mass " }{TEXT 268 1 "M" }
{TEXT -1 18 " at rest and mass " }{TEXT 269 1 "m" }{TEXT -1 22 " movin
g with velocity " }{TEXT 267 1 "v" }{TEXT -1 93 "0), i.e., the situati
on before the collision. It is the second set that we are interested i
n." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assign(sol[2]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "v,V;" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 129 "We could be interested in expressing the ratio of ini
tial to final veocity of the light particle 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(subs(M=r*m,v)/v0);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "semilogplot(ratio,r=1..1000);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 30 "We see that as the mass ratio " }{TEXT 
273 1 "M" }{TEXT -1 1 "/" }{TEXT 274 1 "m" }{TEXT -1 87 " exceeds abou
t 100 the motion describes essentially a reflection of the light parti
cle " }{TEXT 272 1 "m" }{TEXT -1 43 " from 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 -1 10 "Exercise: " }}{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 energies 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 sophist
icated problem is solved in R.L. Greene's book: " }{TEXT 298 30 "Class
ical Mechanics with Maple" }{TEXT -1 17 " (Springer 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 d
ropped simultaneously in a gravitational field from a height " }{TEXT 
285 2 "h0" }{TEXT -1 6 " with " }{TEXT 284 1 "m" }{TEXT -1 19 " sittin
g on top of " }{TEXT 283 1 "M" }{TEXT -1 50 " initially. What height w
ill the light ball reach?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 112 "We assume elastic collisions and ignore the accel
eration by the gravitational field during the collision itself." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "The firs
t statement is energy conservation: when the two-ball system hits the \+
ground it has converted its gravitational potential 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;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The large ball " }
{TEXT 286 1 "M" }{TEXT -1 70 " bounces elastically from the ground, i.
e., reverses its velocity to -" }{TEXT 287 1 "v" }{TEXT -1 40 "0. It t
hen collides with the small ball " }{TEXT 288 1 "m" }{TEXT -1 36 " tha
t is still 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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Mom2:=M*(-v
0)+m*v0=M*V+m*v;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "The final hei
ght of the small ball called " }{TEXT 290 1 "h" }{TEXT -1 38 " is obta
ined from energy conservation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "En3:=m/2*v^2=m*g*h;" }}}{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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 
"Our intelligence is needed..." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "v0s:=solve(En1,v0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "T
he choice of root depends on the definition of the positive direction.
 If upwards is positive, we choose the negative root to describe downw
ard motion at this point:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "v0:=v0s[2];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sol:=sol
ve(\{En2,Mom2,En3\},h);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Our di
fficulties are not yet solved...  even though v0 has been substituted:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "En2;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 14 "solve(Mom2,V);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 15 "V:=simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "solve(En2,v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
234 "There are two roots and we have to figure out which one is physic
ally acceptable. The first describes the situation before the collisio
n (downward motion with v0), i.e., we need the one after the collision
 which depends on the masses:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 8 "v:=%[2];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(En3,
h);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "h:=%;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 113 "We are interested again in exploring the
 answer as a function of the mass ratio and eliminate the initial heig
ht:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "ratio:=simplify(subs
(M=r*m,h)/h0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plot
s):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "semilogplot(ratio,r=
1..1000);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "The result is aston
ishing: the light ball can return to a multiple of its original height
! The curve indicates a limiting behaviour. We use Maple to explore th
is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "limit(ratio,r=infini
ty);" }}}{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 possible height rati
o is reached. The lighter ball can reach nine times the original heigh
t from which it was dropped." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 236 "Practice the trick with various balls (tennis,
 ping-pong, solid rubber, etc.) before trying to impress your friends.
 It takes a little skill to drop the balls in such a way that they rem
ain on top of each other when they hit the ground." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 586 "This problem has further
-reaching applications in that it can explain the essence of some rath
er complicated phenomena in physics. One example is the problem of sup
ernova explosions following the collapse of a star. One can think of t
he imploding massive star as carrying out a bounce which drives the li
ghter outer parts of the system through a shockwave to reach very high
 speeds. This mechanism results in the spilling of material (including
 heavy, unstable elements [nuclei beyond iron, i.e., beyond the maximu
m in the binding energy/nucleon curve]) into outer space at high speed
s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 284 "Of
 course, the conservation laws can't tell us much about the detailed d
ynamics of the motion. It is important to realize, however, that the d
ifferential equations which describe the detailed motion (in Mechanics
 it is Newton's law) have solutions which satisfy the conservation law
s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 
17 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }