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04/19/23 PHY 113 -- Lecture 9 1
Announcements
1. Exam 1 will be returned at the end of class. Please rework the exam, to help “solidify” your knowledge of this material. (Up to 10 extra credit points granted for reworked exam – turn in old exam, corrections on a separate paper – due Tuesday, Oct. 7.)
2. Physics colloquium today – 4 PM Olin 101
SPS meeting at 11:30 AM (free pizza for physics majors and potential physics majors)
3. Today’s topic – Chapter 9 HRW
Center of mass
Definition of momentum
Conservation of momentum
04/19/23 PHY 113 -- Lecture 9 2
Exam 1 -- PHY 113 D
0
1
2
3
4
5
50 55 60 65 70 75 80 85 90 95 100
Grade
04/19/23 PHY 113 -- Lecture 9 3
of mass
rirj
x
y
ii
iii
m
m rrCOM
04/19/23 PHY 113 -- Lecture 9 4
m1=1.2kg
m2=2.5kg
m3=3.4kg
321
332211
321
332211
mmmymymym
y
mmmxmxmxm
x
com
com
Example:
04/19/23 PHY 113 -- Lecture 9 5
Another example of center of mass:
x
y
(1m,2m)
(2m,1m)M0
4M0
m
mMMM
MMM
com
ji
jir
2.18.1
5142
5241
0
00
0
00
04/19/23 PHY 113 -- Lecture 9 6
ii
iii
com m
m aa
Position of the center of mass:
Velocity of the center of mass:
Acceleration of the center of mass:
ii
iii
com m
m rr
ii
iii
com m
m vv
04/19/23 PHY 113 -- Lecture 9 7
i
i
i
ii
iii
ii dt
ddtdm
mpv
aF
Physics of composite systems:
Center-of-mass velocity:
Note that:
M
m
m
mi
ii
ii
iii
com
vv
v
dtd
Mtotali
icomv
FF
04/19/23 PHY 113 -- Lecture 9 8
04/19/23 PHY 113 -- Lecture 9 9
A new way to look at Newton’s second law:
dtd
dtmd
dtd
mmpvv
aF )(
Define linear momentum p = mv
Consequences:
1. If F = 0 p = constant
2. For system of particles:
0dtdp
i
i
ii dt
dpF
i
ii
i
ii dt
dconstant 0 0 If p
pF
04/19/23 PHY 113 -- Lecture 9 10
Conservation of (linear) momentum:
If , p = constant 0Fpdtd
Example:
Suppose a molecule of CO is initially at rest, and it suddenly decomposes into separate C and O atoms. In this process the chemical binding energy E0 is transformed into mechanical energy of the C and O atoms. What can you say about the motion of these atoms after the decomposition?
Before After
vC
vO
mCvC-mOvO=0
vC=vO mO/mC
04/19/23 PHY 113 -- Lecture 9 11
Further analysis:
E0 = ½ mCvC2 + ½ mOvO
2 = ½ mCvC2 (1 + mC/mO)
½ mCvC2 = E0/(1 + mC/mO) ~ E0/(1 + 12/16) = 4/7 E0
½ mOvO2 = 3/7 E0
Extra credit opportunity:
Work through the details of the above analysis and verify the results for yourself ( perhaps with a different diatomic molecule).
04/19/23 PHY 113 -- Lecture 9 12
Peer instruction question:
Suppose a nucleus which has an initial mass of Mi= 238 m0 (where m0 denotes a standard mass unit),suddenly decomposes into two smaller nuclei with M1 = 234 m0 and M2 = 4 m0.
If the velocity of nucleus #1 is V1 what is the velocity of nucleus #2?
(a) -0.017V1 (b) -V1 (c) -59.5V1
After the decomposition which nucleus has more energy?
(a) M1 (b) M2 (c) The nuclei have the same energy.
(d) Not enough information is given.
04/19/23 PHY 113 -- Lecture 9 13
Peer instruction question:
Romeo (60 kg) entertains Juliet (40 kg) by playing his guitar from the rear of their boat (100 kg) which is at rest in still water. Romeo is 2m away from Juliet who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo’s cheek. How does the boat move relative to the shore in this process? (Initially Juliet is closest to the shore.)
(a) 0.2 m away from shore (b) 0.4 m away from shore
(c) 0.2 m toward shore (d) 0.4 m toward shore
x
CM
x’ d
CM
x-x’= d mJ/Mtotal = 0.4 m
04/19/23 PHY 113 -- Lecture 9 14
Another example:
In this case, mechanical energy is not conserved. (Where does it go?) However, to our level of approximation, we will assume that momentum is conserved.
Before: After:
mcv0c + mTv0T = mcvc + mT v
04/19/23 PHY 113 -- Lecture 9 15
Note: In general, momentum is a vector quantity.
mcv0c + mTv0T = (mc+ mT) vf
jiv TTc
Tc
Tc
cf v
mmm
vmm
m00
04/19/23 PHY 113 -- Lecture 9 16
04/19/23 PHY 113 -- Lecture 9 17
Statement of conservation of momentum:
inφinθ0
φcosθcos
2211
221111
svmsvm
vmvmvm
ff
ffi
If mechanical (kinetic) energy is conserved, then:2222
12112
12112
1ffi vmvmvm
04/19/23 PHY 113 -- Lecture 9 18
Peer instruction question:
Given the previous example, summarized with these equations:
which of the following statements are true?
(a) It is in principle possible to solve the above equations uniquely.
(b) It is not possible to solve the above equations uniquely because the mathematics is too difficult.
(c) It is not possible to solve the above equations uniquely because there is missing physical information.
inφinθ0
φcosθcos
2211
221111
svmsvm
vmvmvm
ff
ffi
2222
12112
12112
1ffi vmvmvm
04/19/23 PHY 113 -- Lecture 9 19
One dimensional case:
Conservation of momentum: m1v1i + m2v2i = m1v1f+m2v2f
Conservation of energy:
½ m1v1i2 + ½ m2v2i
2 = ½ m1v1f2+ ½ m2v2f
2
Extra credit: Show that
iif
iif
vmmmm
vmmm
v
vmmm
vmmmm
v
221
121
21
12
221
21
21
211
2
2
04/19/23 PHY 113 -- Lecture 9 20
dtdp
F
Linear momentum: p = mv
Generalization for a composite system:
i
i
ii dt
dpF
)constant(;0;0If i
ii
i
ii dt
dp
pF
Conservation of momentum:
Energy may also be conserved ( for example, in an “elastic” collision)
Summary
04/19/23 PHY 113 -- Lecture 9 21
Another example
h1
1. m1 falls a distance h1 – energy conserved
2. m1 collides with m2 – momentum and energy conserved
3. m1 moves back up the incline to a height h1’
h1’
Extra credit:
Show that 1
2
21
211 h
mmmm
h
04/19/23 PHY 113 -- Lecture 9 22
Notion of impulse:
Ft = pdtdp
F
Example:
vi
vf
mi
f
p = mvf - mvi
p=(mvf sin f + mvf sin f) i
+(-mvf cos f + mvf cos f) j
pfpi
04/19/23 PHY 113 -- Lecture 9 23
Notion of impulse:
dtdp
F
Example:
vi
vf
mi
f
p = mvf - mvi
p=(mvf sin f + mvf sin f) i
+(-mvf cos f + mvf cos f) j
pfpi
p = pf - pi = F t
cause effect