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Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Physics 111 Lecture 08
Linear Momentum, Collisions, Systems of Particles
SJ 8th Ed.: Chap 9.1 –9.7
•New Concepts -Overview
•Linear Momentum, Isolated and Non-Isolated
systems
•Newton’s Second Law in terms of the
Momentum
•Conservation of Linear Momentum
•Impulse-M
omentum Theorem
•What is a Collision?
•Momentum and Kinetic Energy in Collisions
–Inelastic Collisions in One Dimension
–Elastic Collisions in One Dimension
–Collisions in Two Dimensions
•Center of Mass
–Systems of Particles & Solid Bodies
•Linear Momentum for a System of Particles
–Newton’s Second Law for a System
–Momentum Conservation
9.1
Lin
ear
Mo
men
tum
9.2
Mo
men
tum
in
Iso
late
d S
yste
ms
9.3
Mo
men
tum
in
No
n-I
so
late
d S
yste
ms
9.4
Co
llis
ion
s in
On
e D
imen
sio
n
9.5
Co
llis
ion
s in
Tw
o D
imen
sio
ns
9.6
Th
e C
en
ter
of
Mass
9.7
Syste
ms o
f M
an
y P
art
icle
s
9.8
Defo
rmab
le S
yste
ms
9.9
Ro
cket
Pro
pu
lsio
n
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
New Concepts
Mom
ent
um:
•fo
r a p
art
icle
: m
ass
x v
elo
city
•co
nserv
ed in
isol
ate
d s
yst
em
sd
tPd
F ne
t
rr
= ===
Im
puls
e:
•m
oment
um c
hang
e f
or a
non
-is
olate
d s
yst
em
tPF a
ve
∆ ∆∆∆∆ ∆∆∆= ===
rr
tF
P
I
ave
∆ ∆∆∆= ===
∆ ∆∆∆≡ ≡≡≡
rr
r
Col
lisi
ons:
•m
any
part
icle
s in
tera
ctin
g in
a s
yst
em
•
collis
ions
may c
onse
rve m
oment
um a
nd e
nerg
y•
bro
adly
app
lica
ble
mod
el (e
.g,
pool
balls
, ato
ms,
gala
xie
s…)
Mass
cent
er:
O
ne p
oint
in
a b
ody o
r sy
stem
of
part
icle
s m
oves
as
if a
ll th
e m
ass
and
exte
rnal fo
rces
are
loc
ate
d t
here
•CM
is
an
ave
rage
pos
itio
n wit
h w
eig
hti
ng b
y p
art
icle
s’m
ass
es
•CM
is
rela
ted t
o ce
nter
of g
ravi
ty,
but
not
qui
te t
he s
am
e
App
lica
bilit
y:SINGLE
PARTICLES
MASS
ZERO SIZE
SYSTEMS OF
PARTICLES
INTERACTING
RIGID BODIES
FIXED SHAPE
CAN ROTATE &
VIBRATE
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Linear Momentum of a Single Particle
How
do
you
measu
re “
quant
ity o
f m
otio
n”fo
r a b
ody
•Pr
opor
tion
al to
mass
and
velo
city
vm
pr
r≡ ≡≡≡
v
to
pa
ralle
l
Ve
cto
rr
mvr
•U
nits
: s
lug
.ft/
s
gm
.cm
/s,
k
g.m
/s,
] p [
= ===
Newto
n’s
Seco
nd L
aw f
or 1
pa
rtic
le,
in t
erm
s of
mom
ent
um
dtp
dF
ne
t
rr
= ===
Th
e n
et
forc
e o
n a
part
icle
eq
uals
th
e r
ate
of
ch
an
ge o
f it
s m
om
en
tum
Abov
e h
olds
als
o fo
r sy
stem
s of
part
icle
s & r
igid
bod
ies
Deri
vati
on:
dtv
dm
vd
tm d
dt
]v
[m d
dtp
d
rr
rr
+ +++= ===
≡ ≡≡≡
usually zero, as
no mass is lost
ne
tF
a
m
d
tpd
rr
r
= ==== ===
gd
x
df
dx
dg
f
g
(x)]
[f
(x)
dxd
+ +++≡ ≡≡≡R
ule
C
ha
in
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Momentum Conservation –An isolated pair of particles
m1
m2
1vr
2vr
12
Fr2
1Fr
Two
part
icle
s in
tera
ctin
g vi
a f
orce
F(a
ny k
ind).
•F
21
and
F12
are
IN
TERN
AL “
thir
d law”
act
ion-
react
ion
forc
es
02
11
2
F
F
= ===+ +++
rr
Seco
nd L
aw:
] p
p [
dtd
dtp
d
d
tpd
02
1
rr
rr
+ +++= ===
+ +++= ===
21
dtp
dF
dtp
dF
12
21
21
rr
rr
= ==== ===
Tot
al m
oment
um o
f th
e s
yst
em
:
p
p
p2
1
tot
rr
v+ +++
≡ ≡≡≡
dt
pd
tot
0= ===
∴ ∴∴∴
rT
he t
ota
l li
near
mo
men
tum
of
an
iso
late
d
sys
tem
of
two
or
mo
re p
art
icle
s is c
on
sta
nt
When
the s
yst
em
chang
es
state
:f
fi
ip
p
p
p2
12
1
rr
rr
+ +++= ===
+ +++
Mom
ent
um a
long
each
Cart
esi
an
axis
is
cons
erv
ed
sepa
rate
ly
fxfx
xixi
p
p
p
p
21
21
+ +++= ===
+ +++
p
p
p
pfy
fyyi
yi2
12
1+ +++
= ===+ +++
p
p
p
pfz
fzzi
zi2
12
1+ +++
= ===+ +++
“Isolated”means momentum
transfer across system
boundary = 0.
Energy may be flowing by
non-m
echanical means
The internal forces alwayscancel in pairs
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Newton’s Second Law for a System of Particles
NET E
XTERN
AL f
orce
act
s on
each
part
icle
. Syst
em
is
NO
T I
SO
LATED n
ow
F21
and
F12
are
sti
ll in
tern
al, a
ctio
n-re
act
ion
forc
es,
so
02
11
2
F
F= ===
+ +++r
r
App
ly S
eco
nd L
aw t
o in
div
idua
l pa
rtic
les
dtp
dF
F
F
dtp
dF
F
F
(2)
12
ext,
2n
et,
2
(1)
21
ext,
1n
et,
1
21
rr
rr
rr
rr
= ===+ +++
= ===
= ===+ +++
= ===
d
t
pd
F to
t
ext
net,
rv
= ===T
he n
et
ex
tern
al
forc
e o
n a
sys
tem
eq
uals
th
e r
ate
of
ch
an
ge o
f to
tal
lin
ear
mo
me
ntu
m (
mass c
en
ter
mo
men
tum
)
c
on
sta
nt
is
tot
the
ne
xt
,n
et
If
p
,
F
rr
0= ===
MOMENTUM CONSERVATION:
m1
m2
1vr
2vr
12
Fr2
1Fr
ext,
1Fr
ext,
2Fr
] p
p [
dtd
dtp
d
d
tpd
F
F
F
F2
1ext,
2ext,
1n
et,
2n
et,
1
rr
rr
rr
rr
+ +++= ===
+ +++= ===
+ +++= ===
+ +++2
1
p
p
p
to
t2
1
vr
r= ===
+ +++
Repl
ace
:
F F
F
ext
net,
ext,
2ext,
1
rr
r≡ ≡≡≡
+ +++
Add (
1)
to (
2),
not
ing
that
inte
rnal fo
rces
canc
el
d
t
pd
F
tot
ext
net,
rv
= ===∴ ∴∴∴
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple -
Line
ar
Mom
ent
um in
an
Isol
ate
d S
yst
em
An
ast
rona
ut p
ushes
a b
ox h
ard
eno
ugh t
ogi
ve it
a s
peed v
2f=
0.3
m/s
rela
tive
to
a t
hir
dob
serv
er.
Bot
h a
stro
naut
and
box
are
in
itally
at
rest
rela
tive
to
the s
am
e o
bse
rver.
What
is t
he a
stro
naut
’s f
inal sp
eed
v1f?
m1
= 6
0 k
g
m2
= 8
0 k
g
•Syst
em
(ast
rona
ut +
box
) is
iso
late
d.
•The p
ush o
n th
e b
ox is
an
inte
rnal fo
rce.
•
Tot
al line
ar
mom
ent
um is
cons
tant
acti
on
-reacti
on
inte
rnal
forc
es a
ct
2f
1f
2i
1i
to
tp
p
0
p
p
pr
rr
rv
+ +++= ===
= ===+ +++
= ===
Ch
oo
se +
x a
lon
g v
22
f2
1f
1v
m
v
m
0
+ +++
= ===
s/m
.
.
v
mm
v
60
80
2f
121f
40
30
− −−−= ===
× ×××− −−−
= ===− −−−
= ===
min
us s
ign
mean
s o
pp
os
ite
to
v2f
Is
the k
ineti
c ene
rgy o
f th
e s
yst
em
als
o co
nserv
ed?
0
K
K
K
2i
1i
i= ===
+ +++= ===
23
080
24
060
2 22
2 11
.21
.21
f21
f21
2f
1f
f
v
mv
m
K
K
K
× ×××× ×××
+ +++= ===
+ +++= ===
+ +++= ===
Jo
ule
s
8.4
K
f= ===
Where
did
the e
nerg
y c
ome f
rom
?
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple:
Ball h
itti
ng a
wall
A 2
kg
ball m
ovin
g hor
izon
tally
hit
s a v
ert
ical wall a
nd r
ebou
nds.
W
hat
is t
he b
all’
s m
oment
um c
hang
e∆ ∆∆∆
palo
ng
x
Wh
at
frac
tio
n f
o
f th
e i
nit
ial
kin
eti
c e
nerg
y i
s lo
st?
mv
0
m
vf
Can
rega
rd s
yst
em
as
eit
her:
•is
olate
d (
ball +
wall)
� ���co
llis
ion
forc
e is
inte
rnal
•no
n-is
olate
d (
ball
only
) � ���
collis
ion
forc
e is
exte
rnal
a)
Let
v0
= +
5.0
m/s
an
d
vf=
-2
.5 m
/s
5
.0)
-2
(-2
.5
)v
v(m
p
p
p
f0
f= ===
− −−−= ===
− −−−= ===
∆ ∆∆∆0
kg
.m/s
15
p
− −−−
= ===∆ ∆∆∆
decreased
Where
did
the m
oment
um g
o?
Wall g
ain
ed it
75
00
5
52
1
2
1
2
02
22
0
.
..
vv
v
vv
K
KK
f f
0
f0
f0
los
t= ===
− −−−= ===
− −−−
= ===− −−−
= ===− −−−
≡ ≡≡≡KE n
ot c
onse
rved.
75%
con
vert
ed t
o heat
(int
ern
al ene
rgy)
b)
Let
v0
= +
5.0
m/s
an
d
vf=
-5
.0 m
/s
5
.0)
-2
(-5
.0
)v
v(m
p
p
p
f0
f= ===
− −−−= ===
− −−−= ===
∆ ∆∆∆0
kg
.m/s
20
p
− −−−
= ===∆ ∆∆∆
0.0
vv
f
flo
st
= ===
− −−−
= ===
2
1
0
Ball r
efl
ect
ed p
erf
ect
ly e
last
ically
c)
Let
v0
= +
5.0
m/s
an
d
vf=
0.0
m/s
kg
m/s
1
0-
)
v(
m
p
= ===
− −−−= ===
∆ ∆∆∆0
0.0
f lo
st
1= ===
Ball r
efl
ect
ed c
ompl
ete
ly in-
ela
stic
ally
100%
of
KE
conv
ert
ed t
o heat
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Impulse: Momentum transfer in a non-isolated system
Impulse is the effect of collision forces acting over time, transferring momentum
Re-arr
ang
e t
he S
eco
nd L
aw:
d
tF
pd
ext
net,
tot
vr
= ===
FOR A SYSTEM:
FOR A PARTICLE:
d
tF
pd
n
et
vr
= ===
Int
egr
ate
ove
r a f
init
e t
ime int
erv
al
∆ ∆∆∆t:
d
tF
p
t
n
et
∫ ∫∫∫ ∆ ∆∆∆= ===
∆ ∆∆∆v
r
dt
F
p
t
ext
net,
tot
∫ ∫∫∫ ∆ ∆∆∆= ===
∆ ∆∆∆v
r
The r
ight
sides
of t
he a
bov
e a
re c
alled t
he “
Im
puls
e”
Th
e I
mp
uls
e-M
om
en
tum
Th
eo
rem
:
Th
e c
han
ge i
n m
om
en
tum
of
a p
art
icle
or
syste
m e
qu
als
th
e i
mp
uls
e o
f th
e
net
forc
e a
cti
ng
on
it.
•Im
pu
lse i
s a
ve
cto
r•
If a
syste
m is
iso
late
d,
the I
mp
uls
e t
ran
sfe
rred
to
it
via
fo
rces =
0•
If t
he
ne
t ex
tern
al
forc
e a
cti
ng
on
a s
ys
tem
is
no
n-z
ero
, th
e I
mp
uls
e i
sn
on
-zero
an
d t
he s
yste
m is
no
tis
ola
ted
.
p
rr
∆ ∆∆∆= ===
Ι ΙΙΙ
p
to
t
rr
∆ ∆∆∆= ===
Ι ΙΙΙ
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
The “Impulse Approximation”
FOR A SYSTEM:
FOR A PARTICLE:
Wor
k wit
h t
ime a
vera
ge f
orce
. fo
rce(t
ime)
deta
ils
may n
ot b
e k
nown.
d
tF
t1
F t
n
et
avg
net,
∫ ∫∫∫ ∆ ∆∆∆∆ ∆∆∆
≡ ≡≡≡v
r
dt
F t1
F
t
ext
net,
avg
ext,
net,
∫ ∫∫∫ ∆ ∆∆∆∆ ∆∆∆
≡ ≡≡≡v
r
p
t
F
avg
net,
rr
r∆ ∆∆∆
= ===∆ ∆∆∆
= ===Ι ΙΙΙ
tot
avg
ext,
net,
p
t
F
rr
r∆ ∆∆∆
= ===∆ ∆∆∆
= ===Ι ΙΙΙ
THE TIME AVERAGED FORCE PRODUCES THE SAME IMPULSE AS THE TIME DEPENDENT FORCE
Ass
ume o
ne larg
e f
orce
act
s fo
r a s
hor
t ti
me,
as
in a
col
lisi
on.
Oth
er
forc
es
(lik
e g
ravi
ty)
have
negl
igib
le e
ffect
dur
ing
the c
ollisi
on.
vm
p
t
F
avg
rr
rr
∆ ∆∆∆= ===
∆ ∆∆∆= ===
∆ ∆∆∆= ===
Ι ΙΙΙ
net
Fr
are
a
= ===Ι ΙΙΙr
t∆ ∆∆∆
Ac
tual
Imp
uls
e
net
Fr
t∆ ∆∆∆
t
F
avg
∆ ∆∆∆= ===
Ι ΙΙΙr
r
Imp
uls
e A
pp
roxim
ati
on
tF
p
p
tF
p
p
tm
a
v
m
mv
ta
v
v
avg
if
avg
if
avg
if
avg
if
∆ ∆∆∆= ===
− −−−= ===
Ι ΙΙΙ
∆ ∆∆∆+ +++
= ===
∆ ∆∆∆+ +++
= ===
∆ ∆∆∆+ +++
= ===Can
deri
ve a
lso
usin
g k
inem
ati
cs
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ples
app
lyin
g th
e I
mpu
lse A
ppro
xim
ati
on
Ball h
itti
ng a
wall
(see a
n earl
ier
slid
e)
In parts a,b,& c the impulse delivered to the ball was ∆ ∆∆∆p(negative)
The impulse delivered to the wall would be –
∆ ∆∆∆p(positive)
Bum
per
Test
A c
ar
of m
ass
1500 k
g co
llides
wit
h a
wall a
ndre
bou
nds
wit
h v
elo
city
show
n. f
ind t
he im
puls
e o
f th
e c
ollisi
on,
whic
h last
s 0.1
50 s
and
the a
vera
ge
net
forc
e e
xert
ed o
n th
e c
ar
Ign
ore f
orce
s ot
her
than
the w
all’s.
Nand
Fghave
no e
ffect
–ig
nore
the y
-dir
ect
ion.
Let
the s
yst
em
(no
n-is
olate
d)
be j
ust
the c
ar.
)v
v(m
pp
p
xi
xf
xi
xf
xx
− −−−= ===
− −−−= ===
∆ ∆∆∆= ===
Ι ΙΙΙr
kg
.m/s
10
2.6
4
))
1(
.2(
4x
× ×××= ===
− −−−− −−−
= ===Ι ΙΙΙ
560
1500
t
F
avg
∆ ∆∆∆= ===
Ι ΙΙΙr
r
Use
:
N 10
1.7
6
.
10
2.6
4 t
F
64
xavg
net,
x,
× ×××= ===
× ×××= ===
∆ ∆∆∆Ι ΙΙΙ= ===
15
00
If the car crumples up and does not bounce back, is the force larger or smaller?
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple:
Mach
ine g
un b
ulle
ts h
itti
ng a
targ
et
A s
tream
of
ident
ical bul
lets
hit
s th
e t
arg
et
At
a s
teady r
ate
. W
hat
is t
he a
vera
ge f
orce
on t
he t
arg
et,
ass
umin
g it
is
very
mass
ive a
nddoe
s no
t m
ove t
o th
e r
ight.
Bul
lets
+ t
arg
et
form
an
isol
ate
d s
yst
em
Ign
ore y
-dir
ect
ion
forc
es
as
they d
o no
t aff
ect
mot
ion
alo
ng x
Bu
lle
t m
ass =
mB
ulle
t s
peed
= v
i =
+v
Bu
lle
t ra
te =
ρ
ρ
ρ
ρ =
N/∆ ∆∆∆
t
Fav,t
arg
Two
case
s:a)
Bu
lle
ts b
ou
nce
back w
ith
sp
ee
d
vf
=
-v
(perf
ectl
y e
las
tic
)b
)B
ulle
ts s
tick t
o t
he t
arg
et:
vf
=
0
(perf
ec
tly i
ne
lasti
c)
vi
vf ∆ ∆∆∆p
on
eb
ull
et
Usi
ng I
mpu
lse A
ppro
xim
ati
on:
t
p
t
F
arg
tta
rga
vg
targ
,∆ ∆∆∆
∆ ∆∆∆= ===
∆ ∆∆∆
Ι ΙΙΙ= ===
bu
ll
on
eb
ull
o
ne
bu
lla
vg
targ
,p
t
pN
t
p
F
∆ ∆∆∆ρ ρρρ
− −−−= ===
∆ ∆∆∆
∆ ∆∆∆− −−−
= ===∆ ∆∆∆
∆ ∆∆∆− −−−
= ===)
vm
(v
v
m
p
if
bu
ll
on
e− −−−
= ===∆ ∆∆∆
= ===∆ ∆∆∆
Use
3rd
Law:
0
p
pb
ull
arg
t= ===
∆ ∆∆∆+ +++
∆ ∆∆∆O
r:0
bu
lla
rgt
= ===Ι ΙΙΙ
+ +++Ι ΙΙΙ
Case a
:
Case b
:
ρ ρρρ= ===
ρ ρρρ= ===
2m
v
v)
-m
(-v
-
Fa
vg
targ
,
ρ ρρρ= ===
ρ ρρρ= ===
m
v
v)
-m
(0
-
F
avg
targ
,
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
What is a collision?
An event:several bodies exert forces on each other.
There is a clear-cut “before”and “after”the event.
Dir
ect
co
nta
ct
No
n-c
on
tact
The system is isolated (for duration of collision).
Total momentum is conserved.
Interactions are 3rd Law internal forces.
BEFORE
DURING
AFTER
De
tails
ma
y
no
t b
e k
no
wn
Ca
n k
no
w
mo
me
nta
an
de
nerg
ies
Ca
n k
no
w
mo
me
nta
an
de
nerg
ies
The “impulse approximation”can be applied:
•The collision duration is usually “short”.
•The collision forces are assumed to be overwhelmingly
larger than any external forces present during the collision.
We ask only about initial and final state information.
Ass
umpt
ions
valid f
or:
Sub
ato
mic
part
icle
s -
10
-18
sec
10
-27
kgCol
lidin
g ga
laxie
s -
10
+8
yr.
10
+40
kg
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Types of collisions (2 particle examples)
p
p in
itia
l,
tot
fin
al
,to
t
rr
= ===
Use
kin
eti
c ene
rgy c
onse
rvati
on a
lso
if c
ollisi
on is
“perf
ect
ly e
last
ic”
When
collis
ion
forc
es
are
kno
wn,
can
use s
eco
nd law t
o extr
act
deta
il.
Oth
er
tim
es
“sca
tteri
ng e
xpe
rim
ent
s”are
don
e t
o pr
obe a
nd inf
er
forc
es
from
co
llis
ion
resu
lts
Partially Inelastic Collision: Some KE lost. Particles reboundpart way
Perfectly Elastic Collision: Particles rebound fully. KE conserved.
p
p p
p i
if
f2
12
1
rr
rr
+ +++= ===
+ +++
Perfectly Inelastic Collision: Particles stick together. Some KE lost.
ff
fv
v
v
21
rr
r= ===
= ===
i2
i1
ff
1v
m v
m
vm
vm
21
22
1
rr
rr
+ +++= ===
+ +++
Use
mom
ent
um c
onse
rvati
on.
Defi
ne c
ollidin
g sy
stem
to
be iso
late
d f
or p r
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Perfectly Inelastic Collisions
fin
al
,to
tin
itia
l,
tot
p p
rr
= ===
Iso
late
d s
yst
em
. M
oment
um is
cons
erv
ed
ff
fv
v
v
rr
r≡ ≡≡≡
= ===2
1Pa
rtic
les
stic
k to
geth
er � ���
one u
nkno
wn � ���
fewer
equ
ati
ons
Kin
eti
c Ene
rgy is
NO
T c
onse
rved,
ene
rgy los
s is
not
spe
cifi
ed
f2
1i
i1
v )m
m(
vm
vm
rr
r+ +++
= ===+ +++
22
1
Sp
ecia
l cases:
a)
m2
is s
tati
on
ary
targ
et � ���
v2
i =
0
b)
m1
= m
2, v
1i=
-v
2i.
� ���
Pto
t=
0, v
f=
0
mm
vm
v
21
i1
f+ +++
= ===1
rr
mm
vm
vm
v
21
ii
1f
+ ++++ +++= ===
22
1
rr
r
1,2
, o
r 3
D
v
mp
mp
v f
totf
tot,
toti
tot,
cm
r
rr
r= ===
≡ ≡≡≡≡ ≡≡≡
The m
ass
cent
er
of a
syst
em
is
the p
oint
whic
h a
cts
as
if a
ll t
he m
ass
is
conc
ent
rate
d t
here
. In
this
pro
ble
m v
f= v
cm:
Gene
rally:
Mass
cent
er
velo
city
is
cons
tant
in
an
isol
ate
d s
yst
em
Fra
ctio
n of
kin
eti
c ene
rgy r
em
ain
ing
2 22
2 11
2
21
iv)
m(21
iv)
m(21
fv)
mm(
21
ifre
mKK
f
+ +++= ===
≡ ≡≡≡+ +++
Sp
ecia
l cases:
a)
b)
21
1 mm
m
rem
f + +++
= ===
0
f re
m= ===
What if the direction of time is reversed? Explosion!
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple:
Cra
shin
g air
-tr
ack
cars
An
air
cart
of
mass
m1and
spe
ed v
1i
mov
es
alo
ng t
he x
-axis
tow
ard
and
co
llid
es
wit
h a
seco
nd a
ir c
art
of
mass
m
2th
at
is a
t re
st.
The c
art
s st
ick
toge
ther.
Fin
d t
he v
elo
city
of
the
cent
er
of m
ass
of
this
syst
em
(a)
befo
re a
nd (
b)
aft
er
the c
art
s co
llid
e.
m1
m1
m2
m2
v1
iv
2i=
0
vf
•Ine
last
ic c
ollisi
on,
isol
ate
d s
yst
em
•M
oment
um is
cons
erv
ed,
1 d
imens
iona
l i2
1
1f
21
ii
1f
v m
m
m
v
mm
vm
vm
v
1
22
1
+ +++= ===
⇒ ⇒⇒⇒+ ++++ +++
= ===
rr
r
v
mm
p
v
;v
mm
pv
f,cm
1
ito
t,i,
cm
f1
fto
t,f
cm
,= ===
+ +++≡ ≡≡≡
= ===+ +++
≡ ≡≡≡
22
a)
& b
) M
ass
cent
er
velo
city
is
the s
am
e b
efo
re a
nd
aft
er
the c
ollisi
on,
beca
use m
oment
um is
cons
erv
ed.
vf=
½v
1iif
m1
= m
2
∑ ∑∑∑∑ ∑∑∑≡ ≡≡≡
iii
cm
m
vm
v
For many
particles:
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Example: Ballistic Pendulum
The b
ulle
t (m
ass
m1,
velo
city
v1
A)
collid
es
wit
h t
he h
ang
ing
blo
ck a
nd b
eco
mes
em
bedded in
it.
The b
lock
+ b
ulle
t sw
ings
up
to
a m
axim
um h
eig
ht
h a
bov
e it’s
init
ial
posi
tion
. G
iven
h,
find
the ini
tial sp
eed o
f th
e b
ulle
t.
Ste
p 1:
Dur
ing
the c
ollisi
on (
A t
o B)
•Syst
em
= b
ulle
t + b
lock
is
isol
ate
d•
Perf
ect
ly ine
last
ic c
ollisi
on,
1 d
imens
ion
•Lin
ear
mom
ent
um is
cons
erv
ed,
not
KE
•Im
puls
e A
ppro
xim
ati
on h
olds
A2
1
1B
v m
m
m
v
1+ +++
= ===
m1
= 9
.5 g
mm
2=
5.4
kg
h =
6.3
cm
C
g
hm
mm
v
A2
1
21
1
+ +++= ===
Eva
luate
:
s/
m
v
A633
1= ===
gh
v
B2
2= ===
)(
UK
)(
UK
E
CC
BB
mech
+ +++= ===
+ +++= ===
gh
)m
m(
v)m
m(
B21
21
2
21
00
+ ++++ +++
= ===+ +++
+ +++
Ste
p 2:
Aft
er
collis
ion
end
s (a
t B)
•Syst
em
= b
ulle
t + b
lock
+ E
art
h•
Tens
ion
doe
s no
wor
k –
norm
al to
mot
ion
•M
ech
ani
cal ene
rgy is
cons
erv
ed
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Perfectly Elastic Collisions
p
p in
itia
l,
tot
fin
al
,to
t
rr
= ===
Iso
late
d s
yst
em
. M
oment
um is
cons
erv
ed
Part
icle
s m
ove s
epa
rate
ly a
fter
collis
ion
–do
not
stic
k to
geth
er
Abov
e p
rovi
des
only
1 e
quati
on (
in 1
D b
elo
w a
long
x-axis
)
i2
i1
ff
1v
m v
m
vm
vm
21
22
1
rr
rr
+ +++= ===
+ +++
(1)
i2
i1
ff
1
vm
vm
vm
vm
21
22
1+ +++
= ===+ +++
1 equation, 2 unknowns if initial v’sgiven.
Need 1 more equation to solve
Perf
ect
ly E
last
ic � ���
Tot
al Kin
eti
c Ene
rgy is
cons
erv
ed,
part
icle
KE’s
may c
hang
e
if
K
K = ===
(2)
f21
f1
21i
21i
121
v
m
vm
vm
v
m2 2
2
2 1
2 22
2 1+ +++
= ===+ +++
Fra
ctio
n of
kin
eti
c ene
rgy r
em
ain
ing
ela
sti
c
perf
ectl
y
if
ifre
m
1
KK
f
= ===≡ ≡≡≡
Sol
ve E
quati
ons
(1)
and
(2)
sim
ulta
neou
sly (
see t
ext,
page
244)
(9
.21)
ii
1f
v Mm
v
M
mm
v 2
2
1
2
1
2
+ +++
− −−−
= ===
(9.2
2)
i1
i1
f
v M
mm
v
M
2m
v 2
2
12
− −−−+ +++
= ===
2m
m
M
1+ +++
≡ ≡≡≡
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ples
of E
last
ic C
ollisi
ons
1)
Balls
have
equ
al m
ass
:m
1=
m2
= m
Eq
uati
on
s9.2
1/9
.22 i
mp
ly:
v
v
if
21
0+ +++
= ===
0v
v i
f+ +++
= ===1
2
BALLS EXCHANGE MOMENTA
v1
iv
2i
= 0
2)
Targ
et
ball
m2
is a
t re
st b
efo
re c
ollisi
onEqu
ati
ons
9.2
1/9
.22 im
ply:
i1
fv
M
mm
v 1
2
1
− −−−
= ===i
1f
vM
2m
v 1
2
= ===
2.1
) Cue
ball
stri
king
ano
ther
pool
ball
at
rest
. m
1=
m2
= m
Abov
e im
ply:
v
f
01
= ===
if
v v
1
2= ===
Cu
e b
all
sto
ps
[n
o d
raw
or
foll
ow
]
Str
uck b
all
carr
ies o
ff a
ll m
om
en
tum
2.2
) M
ass
ive t
arg
et
ball.
m2>
> m
1
Abov
e im
ply:
v
v
if
11
− −−−= ===
02
v
f
≈ ≈≈≈
Refl
ec
ted
off
targ
et]
Targ
et
mo
ves o
ff w
ith
neg
lig
ible
sp
eed
2.3
) M
ass
ive p
roje
ctile b
all.
m1>
> m
2
Abov
e im
ply:
v
v
if
11
= ===
1i
fv
v
2
2≈ ≈≈≈
Mo
me
ntu
m u
nc
han
ged
by c
oll
isio
n w
ith
targ
et]
Refl
ec
ts o
ff m
1as i
t sw
eep
s t
hro
ug
h
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Ela
stic
Col
lisi
on E
xam
ple:
Two
meta
l sp
here
s su
spend
ed b
y c
ords
just
to
uch w
hen
they a
re a
t re
st.
Sph
ere
1 is
rele
ase
d f
rom
rest
at
heig
ht
h1.
It
collid
es
ela
stic
ally
wit
h S
phere
2 w
hic
h is
at
rest
in
itally
and
ris
es
to h
eig
ht
h2 a
fter
the
collis
ion.
Sph
ere
1 r
ises
to h
eig
ht
h1’aft
er
the c
ollisi
on.
h2
T
path
to
te
nsio
n
mech
E
)(
⊥ ⊥⊥⊥
= ===∆ ∆∆∆
0
1
path
to
te
nsio
n
mech
E
)(
⊥ ⊥⊥⊥
= ===∆ ∆∆∆
0
3
co
llis
ion
ela
sti
c
(2)
(1)
v
m21
gh
m 2 i
11
11
= ===AT BOTTOM
BEFORE
COLLISION
g
h
v i
11
2= ===
(3)
v
m21
gh
m 2
f2
22
2= ===
v
m21
gh
m 2 f
1'
11
1= ===
v
2g1
h
2f
22
= ===
v2
g1
h 2 f
'1
1= ===
(2)
i11
fv
mm
mm
v 1
22
1
+ +++− −−−= ===
TO THE LEFT
IF
m2> m
1
i1
fv
mm
2m
v 1
21
2
+ +++= ===
ela
sti
cc
oll
isio
ne
qu
ati
on
sw
ith
v2i=
0
Eva
lua
tio
n
m1 =
30 g
m
m2
= 7
5 g
m h
1=
8.0
cm
v1i =
1.2
5 m
/sv
1f=
-0
.54 m
/sv
2f =
0.7
15 m
/s
h2 =
2.6
cm
h
1’
= 1
.5 c
m
Sp
ecia
l C
ase
m1
= m
2
01
= ===f
v
vv
if
12
= ===
h h
1= ===
2
Sys
tem
co
nti
nu
es in
de
fin
ite
ly
0= ===
1h
'
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Elastic Collisions in Two Dimensions
Iso
late
d s
yst
em
of
2 p
art
icle
s.
Mom
ent
um is
cons
erv
ed s
epa
rate
ly f
or x
and
y
Spe
cial ca
se:
Mot
ion
is in
x-y
plane
, m
2= t
arg
et
at
rest
,“g
lanc
ing”
impa
ct,
perf
ect
ly e
last
ic.
θ θθθ φ φφφm
1
x
y m2
iv
vi
i1
1= ===
r
02
= ===i
vr
fv
2
r
fv
1
r
m2
“im
pa
ct
pa
ram
ete
r” BEFORE
AFTERm
1
i1
ff
1v
m )
co
s(
vm
)co
s(
vm
12
21
= ===φ φφφ
+ +++θ θθθ
(1) For x:
(2) For y:
p
vm
py,
init
,to
ti
1x,
init
,to
t0
1= ===
= ===
02
21
)
sin
(v
m)
sin
(v
m f
f1
= ===φ φφφ
− −−−θ θθθ
Conserve momentum separately for x and y:
Elastic collision � ���
Kinetic energy conserved
2 1
2 22
2 1i
121
f21
f1
21v
m
v
m
v
m
r= ===
+ +++(3):
K
K in
itia
l,
tot
fin
al
,to
t= ===
Thre
e E
quati
ons
foun
d
-Seve
n Pa
ram
ete
rs in
them
v
v
v
m
m 2f
1f
1i
21
φ φφφθ θθθ
If
any
4 o
f th
em
are
kno
wn,
the p
roble
m c
an
be s
olve
d n
umeri
cally
p
p in
itia
l,
tot
fin
al
,to
t
rr
= ===
p p
p p
ii
ff
21
21
rr
rr
+ +++= ===
+ +++i
2i
1f
f1
vm
vm
v
mv
m 2
12
21
rr
rr
+ +++= ===
+ +++
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple:
Inela
stic
col
lisi
on in
2 d
imens
ions
CA
R:
m1
= 1
200 k
g, v
1i
= 60 k
m/h
r
A c
ar
dri
ving
East
col
lides
wit
h a
tru
ck d
rivi
ngN
orth
. T
he t
wo
vehic
les
stic
k t
ogeth
er.
Fin
dth
eir
fin
al co
mm
on s
peed a
nd d
irect
ion
x
y
TR
UC
K
CA
R
CA
R +
TR
UC
K
m1,
v1i
m2,
v2i
vf
θ θθθ
TR
UC
K:
m2
= 3
000 k
g, v
2i
= 40 k
m/h
r
Ine
last
ic c
ollisi
on,
isol
ate
d s
yst
em
mom
ent
um is
cons
erv
ed
p
p in
itia
l,
tot
fin
al
,to
t
rr
= ===i
2i
1f
1v
m v
m v ]
m[m
21
2
rr
r+ +++
= ===+ +++
Wor
k x &
y d
irect
ions
sepa
rate
ly
i1
fx1
vm
v ]
m[m
12
= ===+ +++
i2
fy1
vm
v ]m
[m 2
2= ===
+ +++
iv
mm
m
v
1
1fx
12
+ +++= ===
v
m
m
m
v i
1
2fy
2
2+ +++
= ===
CO
NV
ER
T:
v1
i=
60 k
m/h
r x 1
/3600 h
r/s x
1000 m
/km
v1
i =
16.6
7 m
/sV
2i
= 11.1
1 m
/sm
/s
4.7
6
v fx
= ===m
/s
7.9
4
v fy
= ===
km
/s
33
.3
m
/s
9.2
5
]v
[v v
1/2
2 fy2 fx
f= ===
= ===+ +++
= ===
5
9
]/v
[vta
n
o
fxfy
-1= ===
= ===θ θθθ
Num
eri
cal Eva
luati
on:
Fra
ctio
n of
kin
eti
c ene
rgy r
em
ain
ing
47
.4%
KK
f
iv)
m(21
iv)
m(21
fv)
mm(
21
ifre
m= ===
+ +++= ===
≡ ≡≡≡+ +++
2 22
2 11
2
21
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Systems -Mass Center
The c
ent
er
of m
ass
of a
bod
y o
r sy
stem
of
bod
ies
is t
he p
oint
that
mov
es
as
if a
ll of
the m
ass
were
con
cent
rate
d t
here
and
all
exte
rnal
forc
es
were
app
lied t
here
The hammer spins about its CM as
CM follows projectile trajectory
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Mass Center motion for an exploding sky-rocket
(Inelastic collision run in reverse)
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Mass Center: Average Position W
eighted by Mass
Con
cept
: Ave
rage
sco
re o
n a c
lass
exam
50 s
tud
en
ts4 q
uesti
on
s o
nth
e t
est
Sco
re%
#
Stu
den
ts0 0
25 5
50 15
75 20
100 1
0
stu
den
ts#
T
ota
l
stu
den
ts]
#x
[s
co
re ∑ ∑∑∑
= ===A
VE
RA
GE
SC
OR
E
67.5
50
xx
xx
= ==== ===
+ ++++ +++
+ +++1
01
00
20
75
15
50
52
5
Exam
ple:
Two
part
icle
s on
x-axis
x =
0m
1m
2
x1
x2
∑ ∑∑∑∑ ∑∑∑= ===
+ ++++ +++= ===
iii
21
cm
m
xm
m
m
xm
xm
x
21
21
Beco
mes s
imp
le a
vera
ge if
m1
= m
2
Mass
cent
er
of a
syst
em
of
N p
art
icle
s
kz
jy
ix
rcm
cm
cm
cm
+ ++++ +++
= ===v
tot
N jj
j
cm
M
x m
x∑ ∑∑∑
= ===≡ ≡≡≡
1
m
MN j
jto
t∑ ∑∑∑
= ===≡ ≡≡≡
1
M
rm
r
tot
N jj
j
cm
∑ ∑∑∑= ===
≡ ≡≡≡1
r
rIn
vect
or
form
(3D):
Alo
ng x
:
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Mass
Cent
er
Exam
ples
Equ
al m
ass
es
alo
ng a
lin
e
m1 =
m2
= m
3=
m4
= m
5 =
m
x1 =
1, x
2=
2,
x3
= 3
, x
4=
-1,
x5 =
-2
53
m
mm
mm
m
m
xm
x
iii
cm
= ===× ×××
− −−−× ×××
− −−−× ×××
+ +++× ×××
+ +++× ×××
= ==== ===
∑ ∑∑∑∑ ∑∑∑5
21
32
1
Une
qual m
ass
es
at
vert
ices
of e
quilate
ral tr
iang
le
m1
m2
m3
aa
a
m1 =
1.2
kg
, m
2=
2.5
kg
, m
3=
3.4
kg
2 dimensional problem
r 1 =
(0,0
), r 2
= (
a,0
),
r 3=
(a/2
,asin
(60
o))
,
sin
(60
) =
.866
m 0.8
3
.
..
..
..
.
m
xm
x
iii
cm
= ===+ +++
+ +++
× ×××+ +++
× ×××+ +++
× ×××= ===
= ===∑ ∑∑∑∑ ∑∑∑
43
52
21
70
43
41
52
02
1
Sid
e a
= 1
.4 m
m 0.5
8
.
..
)sin
(.
..
.
m
ym
y
o
iii
cm
= ===+ +++
+ +++
× ×××+ +++
× ×××+ +++
× ×××= ===
= ===∑ ∑∑∑∑ ∑∑∑
43
52
21
60
41
43
05
20
21
Une
qual m
ass
es,
sam
e loc
ati
ons
as
abov
e
m1 =
m,
m2
= 2
m,
m3
= 4
m,
m4
= 5
m,
m5 =
6m
0
m
mm
mm
m
m
xm
x
iii
cm
= ===× ×××
− −−−× ×××
− −−−× ×××
+ +++× ×××
+ +++× ×××
= ==== ===
∑ ∑∑∑∑ ∑∑∑1
8
26
15
34
22
1!!
CM
at
orig
in
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Mass Center of a Solid Body
App
roach
:•
Div
ide c
onti
nuou
s bod
y int
o N
chun
ks,
each
wit
h m
ass
∆ ∆∆∆m
k.
•W
rite
∆ ∆∆∆m
kas
a m
ath
fun
ctio
n•
App
ly t
he m
ass
cent
er
form
ulas
for
a s
yst
em
of
N p
art
icle
s (s
umm
ati
ons)
•Take
the lim
it a
s th
e m
ass
chun
ks b
eco
me
infi
nite
sim
ally
sm
all.
•Pe
rfor
m t
he r
esu
ltin
g in
tegr
ati
on(s
)
x
y
z
V
ol
x
en
sit
yd
m
k∆ ∆∆∆
= ===∆ ∆∆∆
len
gth
)
(ma
ss
/un
it
de
ns
ity
m
as
s
lin
ea
r
(x
)d
x
)
x(
d
m≡ ≡≡≡
λ λλλλ λλλ
= ===For
1D b
ody:
For
2D b
ody:
For
3D b
ody:
)
y)
(x,
d
yd
x
)y,
x(
d
ma
rea
(m
as
s/u
nit
d
en
sit
y
ma
ss
s
urf
ac
e≡ ≡≡≡
σ σσσσ σσσ
= ===
vo
lum
e)
(m
as
s/u
nit
d
en
sit
y
ma
ss
vo
lum
ez
y,
x,
)
(
d
z
dy
dx
)z,
y,x(
dm
≡ ≡≡≡ρ ρρρ
ρ ρρρ= ===
3D OBJECT
tot
N jj
j
cm
M
xm
x∑ ∑∑∑
= ===≡ ≡≡≡
1∫ ∫∫∫
≡ ≡≡≡
bo
dy
tot
cm
xd
mM
1
x
∫ ∫∫∫∫ ∫∫∫
ρ ρρρ= ===
≡ ≡≡≡
bo
dy
bo
dy
tot
dV
dm
Mwhere
d
m r
M
1
r
vo
lto
tcm
∫ ∫∫∫≡ ≡≡≡
rv
•OFTEN CAN USE SYMMETRY TO AVOID
ONE OR MORE INTEGRATIONS
•OFTEN CAN ASSUME DENSITY IS UNIFORM
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Exam
ple:
Mass
Cent
er
of a
Rod
•O
ne D
imens
iona
l bod
y•
By s
ym
metr
y,
ycm
and
zcm
equ
al zero
•Lin
ear
mass
dens
ity is
cons
tant
Mass M
, L
en
gth
L
λ
λ
λ
λ =
M
/L
dx
)x(
dm
λ λλλ= ===
∫ ∫∫∫∫ ∫∫∫
λ λλλ= ===
≡ ≡≡≡L
bo
dy
tot
cm
x
dx
M1
x
dm
M
1
x0
20
2
0
Lx
L1
xd
xM
x
L
21L
cm
= ==== ===
λ λλλ= ===
∫ ∫∫∫As
expe
cted b
y s
ym
metr
y
•W
hat
if t
he lin
ear
mass
dens
ity is
NO
T U
NIFO
RM
?
•For
exam
ple c
hoo
se:
2w
ith
M/L
x
)x(
≡ ≡≡≡
α αααα ααα
= ===λ λλλ
33
313
0
3
0
2
0
L
L
M
x
M
d
xx
M
x
dx
)x(
M1
x
LL
L
cm
= ===α ααα
= ===α ααα
= ===α ααα
= ===λ λλλ
= ===∫ ∫∫∫
∫ ∫∫∫Shif
ted
dx
)x(
dm
λ λλλ= ===
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Systems of Particles
System
boundary
Now generalize the momentum principles to N particles
•Int
ern
al co
llisi
on f
orce
s•
Exte
rnal fo
rces
can
act
dir
ect
lyon
part
icle
s in
the s
yst
em
•
Rig
id b
odie
s add f
ixed p
osit
ions
due
to
inte
rnal fo
rces
Exte
rnal
forc
e e
xam
ple
s:
•G
ravit
y,
Ele
ctr
osta
tic,
Mag
neti
c f
orc
es
•C
on
tact
or
co
llis
ion
fo
rces w
ith
exte
rnal ag
en
ts
Inte
rnal
forc
e E
xam
ple
s:
•C
oll
isio
ns b
etw
een
part
icle
s w
ith
in t
he s
yste
m•
Mo
lecu
lar
forc
es g
ivin
g t
he t
he
syste
m s
hap
e a
nd
fo
rm•
New
ton
’s 3
rdL
aw
im
plies t
hese c
an
cel
an
d h
ave n
o n
et
eff
ect
on
mo
tio
n o
f th
e s
yste
m a
s a
wh
ole
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Mass Center Velocity and Acceleration for a System
of Particles
M
r m
r
tot
N jj
j
cm
∑ ∑∑∑= ===
≡ ≡≡≡1
r
rM
ass
Cent
er
Defi
niti
onAss
ume t
otal m
ass
Mto
tis
con
stant
and
each
mass
mjre
main
s in
the s
yst
em
Dif
fere
ntia
te b
oth s
ides
–te
rm b
y t
erm
for
the s
um o
n th
e r
ight
tot
N jj
jc
mto
tc
mp
v
m
vM
p
rr
vv
= ==== ===
= ===∑ ∑∑∑
= ===1
d
t
rd
v c
mve
loc
ity
c
en
ter
m
as
sc
m
rr
≡ ≡≡≡= ===
d
trd
v
jj
part
icle
o
f
velo
cit
yj
r
r≡ ≡≡≡
= ===
MASS CENTER
MOMENTUM
v
m1
v
N jj
jM
cm
tot
∑ ∑∑∑= ===
≡ ≡≡≡∴ ∴∴∴
1
rv
a
m
a
MN j
jj
cm
tot
∑ ∑∑∑= ===
≡ ≡≡≡1
vv
EACHTERM EQUALS
F
j
rIS THE NET FORCE ON PARTICLE j INCLUDING EXTERNAL AND INTERNAL FORCES
F
j
r
INTERNAL FORCES ALWAYS CANCEL IN 3rd LAW PAIRS
dt
pd
F
aM
c
me
xt
ne
t,c
mto
t
rv
v= ===
≡ ≡≡≡∴ ∴∴∴
Dif
fere
ntia
te b
oth s
ides
aga
in –
term
by t
erm
for
the s
um o
n th
e r
ight
d
t
vd
a cm
on
accele
rati
cen
ter
m
ass
cm
rv
= ==== ===
jjj
j
pa
rtic
le
of
o
na
cc
ele
rati
jmF
d
tvd
a
rr
r= ===
= ==== ===
a
m
a
N jj
jM
cm
tot
∑ ∑∑∑= ===
≡ ≡≡≡∴ ∴∴∴
1
1v
v
Co
pyri
gh
t R
. J
an
ow
–F
all
2011
Newton’s Second Law for the Mass Center
F
aM
ext
net,
cm
tot
vv
≡ ≡≡≡
v
Mp
p cm
tot
cm
tot
rr
r= ===
= ===
IF MASS is CONSTANT
For
a b
ody o
r sy
stem
of
part
icle
s:The m
ass
cent
er
mov
es
as
if it
is a
part
icle
wit
h a
ll th
e m
ass
co
ncent
rate
d t
here
, act
ed o
n by t
he n
et
exte
rnal fo
rce.
•Int
ern
al fo
rces
canc
el due
to
the 3
rdLaw
•Int
ern
al co
llisi
ons,
expl
osio
ns,
etc
do
not
aff
ect
CM
mot
ion
•Sin
gle p
art
icle
s re
spon
d t
o bot
h int
ern
al and
exte
rnal fo
rces.
The m
ass
cent
er
obeys
the S
eco
nd L
aw a
s st
ate
d a
bov
e
Resu
lts
of p
revi
ous
slid
e:
d
t
pd
F
cm
ext
net,
rv
= ===
EVEN IF MASS is NOT CONSTANT
THE NET EXTERNAL FORCE EQUALS THE RATE OF CHANGE OF MASS CENTER MOMENTUM
