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
10/2/2003
PHY 113 -- Lecture 9
1
Exam 1 -- PHY 113 D
5
4
3
2
1
0
50
55
60
65
70
75
80
85
90
95
100
Grade
10/2/2003
PHY 113 -- Lecture 9
2
ŷ
of mass
ri
rj
r COM ≡
∑mr
∑m
i
i
i
i
i
x̂
10/2/2003
PHY 113 -- Lecture 9
3
Example:
m1=1.2kg
m2=2.5kg
m3=3.4kg
xcom
m1 x1 + m2 x2 + m3 x3
=
m1 + m2 + m3
ycom =
10/2/2003
PHY 113 -- Lecture 9
m1 y1 + m2 y2 + m3 y3
m1 + m2 + m3
4
Another example of center of mass:
y
(1m,2m)
M0
4M0
rcom
10/2/2003
(2m,1m)
x
⎛ M 0 ⋅1 + 4 M 0 ⋅ 2 M 0 ⋅ 2 + 4 M 0 ⋅1 ⎞
= ⎜⎜
i+
j ⎟⎟m
5M 0
5M 0
⎝
⎠
= (1.8i + 1.2 j)m
PHY 113 -- Lecture 9
5
Position of the center of mass:
r com ≡
∑mr
∑m
i
i
i
i
i
Velocity of the center of mass:
v com ≡
∑mv
∑m
i
i
i
i
i
Acceleration of the center of mass: a com ≡
∑ma
∑m
i
i
i
i
i
10/2/2003
PHY 113 -- Lecture 9
6
Physics of composite systems:
dmi v i
dp i
∑i Fi = ∑i miai =∑i dt = ∑i dt
Center-of-mass velocity:
∑m v ∑m v
≡
≡
M
∑m
i
v com
i
i
i
i
i
i
i
Note that:
∑ Fi ≡ Ftotal = M
i
10/2/2003
dv com
dt
PHY 113 -- Lecture 9
7
10/2/2003
PHY 113 -- Lecture 9
8
A new way to look at Newton’s second law:
F = ma = m
dv d (mv ) dp
=
≡
dt
dt
dt
Define linear momentum p = mv
Consequences:
1. If F = 0
Î
dp
=0
dt
2. For system of particles:
Î p = constant
∑ Fi = ∑
i
If
∑F = 0
i
i
10/2/2003
dp i
⇒∑
=0
dt
i
PHY 113 -- Lecture 9
i
dp i
dt
⇒ ∑ p i = constant
i
9
Conservation of (linear) momentum:
dp
=F=0
dt
Example:
If
, p = constant
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
10/2/2003
PHY 113 -- Lecture 9
vO
mCvC-mOvO=0
vC=vO mO/mC10
Further analysis:
E0 = ½ mCvC2 + ½ mOvO2 = ½ mCvC2 (1 + mC/mO)
Î ½ mCvC2 = E0/(1 + mC/mO) ~ E0/(1 + 12/16) = 4/7 E0
Î 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).
10/2/2003
PHY 113 -- Lecture 9
11
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.
10/2/2003
PHY 113 -- Lecture 9
12
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
(c) 0.2 m toward shore
(b) 0.4 m away from shore
(d) 0.4 m toward shore
x-x’= d mJ/Mtotal = 0.4 m
x’
x
10/2/2003
CM
d
CM
PHY 113 -- Lecture 9
13
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
10/2/2003
PHY 113 -- Lecture 9
14
Note: In general, momentum is a vector quantity.
mcv0c + mTv0T
vf =
10/2/2003
= (mc+ mT) vf
mc
mT
v0c i +
v0T j
mc + mT
mc + mT
PHY 113 -- Lecture 9
15
10/2/2003
PHY 113 -- Lecture 9
16
Statement of conservation of momentum:
m1v1i = m1v1 f cos θ + m2 v2 f cos φ
0 = m1v1 f sinθ − m2 v2 f sinφ
If mechanical (kinetic) energy is conserved, then:
1
2
m
v
2 1 1i =
10/2/2003
1
2
1 m v2
m
v
+
2 1 1f
2 2 2f
PHY 113 -- Lecture 9
17
Peer instruction question:
Given the previous example, summarized with these equations:
m1v1i = m1v1 f cos θ + m2 v2 f cos φ
0 = m1v1 f sinθ − m2 v2 f sinφ
1
2
m
v
2 1 1i =
1
2
1 m v2
m
v
+
2 1 1f
2 2 2f
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.
10/2/2003
PHY 113 -- Lecture 9
18
One dimensional case:
Conservation of momentum: m1v1i + m2v2i = m1v1f+m2v2f
Conservation of energy:
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2+ ½ m2v2f2
Extra credit: Show that
10/2/2003
⎛ m1 − m2 ⎞
⎛ 2m2 ⎞
v1 f = ⎜
⎟v1i + ⎜
⎟ v2 i
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
⎛ 2m1 ⎞
⎛ m2 − m1 ⎞
PHY
113
-Lecture
9
v2 f = ⎜
⎟ v2 i
⎟v1i + ⎜
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
19
Summary
Linear momentum: p = mv
F=
dp
dt
Generalization for a composite system:
dp i
∑ Fi = ∑
i
i dt
Conservation of momentum:
If
dp i
= 0; ⇒ ∑ p i = (constant )
∑ Fi = 0; ⇒ ∑
i
i dt
i
Energy may also be conserved ( for example, in an
“elastic” collision)
10/2/2003
PHY 113 -- Lecture 9
20
Another example
h1
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’
Extra credit:
10/2/2003
Show that
2
⎛ m − m2 ⎞
h1′ = ⎜ 1
⎟ h1
113 -- Lecture 9
⎝ m1 + mPHY
2⎠
21
Notion of impulse:
F=
dp
dt
ÎF∆t = ∆p
Example:
∆p = mvf - mvi
m
10/2/2003
vi
θi
vf
θf
∆p=(mvf sin θf + mvf sin θf) i
+(-mvf cos θf + mvf cos θf) j
pi
PHY 113 -- Lecture 9
pf
22
Notion of impulse:
dp
F=
dt
cause
effect
∆p = mvf - mvi
∆p=(mvf sin θf + mvf sin θf) i
Example:
+(-mvf cos θf + mvf cos θf) j
m
10/2/2003
vi
θi
vf
θf
∆p = pf - pi = F ∆t
pi
PHY 113 -- Lecture 9
pf
23