Fundamentals of Physics
Chapter 9a Systems of Particles
1. A Special Point
2. The Center of Mass
3. Newton’s 2nd Law for a System of Particles
4. Linear Momentum
5. Linear Momentum of a System of Particles
6. Conservation of Linear Momentum
7. A Rocket
8. External Forces & Internal Energy Changes
Review & Summary
Questions
Exercises & Problems
2008
Physics 2111 Fundamentals of Physics
Chapter 9
1
The Center of Mass
•
2008
The center of mass of a body or system of bodies is the point that moves
as though all of the mass were concentrated there and all external forces
were applied there.
Physics 2111 Fundamentals of Physics
Chapter 9
2
The Center of Mass
Consider 2 particles separated by distance d:
The center of mass, xcom, of the system is the point that moves as though all of
the mass were concentrated there and all external forces were applied there.
xcom
2008
m1 x1  m2 x2

m1  m2
Physics 2111 Fundamentals of Physics
Chapter 9
3
Center of Mass (cm)
xcom
xcom is closer to the larger mass.
xcom
xcom 
2008
m2
d
m1  m2
Physics 2111 Fundamentals of Physics
Chapter 9
4
The Center of Mass
The center of mass, xcom, of the system is the point that moves as though all of the
mass were concentrated there and all external forces were applied there.
Many particles distributed in 3 dimensions:
2008
xcom
1

M
ycom
1

M
zcom
1

M
m x

1
rcom 
M

m
r
 ii
xcom
Many particles distributed along the x-axis:
n
1

M
i 1
i
i
n
i 1
n
m x
i
i
m
yi
i 1
n
i 1
i
n
m z
i 1
i
i
Physics 2111 Fundamentals of Physics
Chapter 9
5
Center of Mass?

M 
m
i
i
i
 30  81  42  16
i
i
 30  82  41  20
m x
m y
xcom 
m x
m
i
i
i
2008
 3  8  4  15

16
 1.1
15
ycom 
m y
m
i
i
i

20
 1.3
15
Physics 2111 Fundamentals of Physics
Chapter 9
6
CM of Solid Bodies

1
rcom 
M
xcom 
 x dm
ycom
 y dm
zcom
Uniform density case:
2008
1
M
1

M
1

M


r dm
dm   dV 
 z dm
M
dV
V
Physics 2111 Fundamentals of Physics
Chapter 9
7
Newton’s 2nd Law for a System of Particles

rcom  1 M

M rcom 

M vcom 

M acom 
n

m r
i 1
n
i i

m r
i 1
i i
n

m v
i 1
i
i
n

 mi ai 
i 1


Fnet  M acom


 Fi  Fnet
n
i 1
Net Force of all external forces
Internal forces cancel (Newton’s 3rd Law)
Total mass of the system
Acceleration of the center of mass
2008
Physics 2111 Fundamentals of Physics
Chapter 9
8
Newton’s 2nd Law for a System of Particles
The center of mass of a system moves like a particle of mass M under the influence of
the net external force acting on the system.


Fnet  M acom
• Net of all external forces
Internal forces cancel (Newton’s 3rd Law)
• Total mass of the system
• Acceleration of the center of mass
2008
Physics 2111 Fundamentals of Physics
Chapter 9
9
Linear Momentum


p  mv
Linear momentum of a particle:

dp
dt

d
dt
 

mv
 m

dv
dt

 ma
The net force acting on a particle equals the time rate of change of the particle’s linear
momentum:

Fnet
2008


dp
dt
Physics 2111 Fundamentals of Physics
Newton’s 2nd Law
Chapter 9
10
Linear Momentum of a System of Particles
Total linear momentum of a system of particles:

P 
n

 pi 
i 1
But

M vcom 
n

m v
i 1
i
i
n

m v
i 1
i
i
The linear momentum of a system of particles equals the product of the total mass and
the velocity of the center of mass:


P  M vcom
Differentiating:
2008




dvcom
dP
d
M vcom   M

 M acom
dt
dt
dt


dP
 Fnet
dt
Physics 2111 Fundamentals of Physics
Chapter 9
11
Conservation of Momentum


dP
 Fnet
dt
Law of conservation of momentum: If no net external force acts on a system of
particles, the total linear momentum of the system cannot change:

P  constant


Pf  Pi
If the component of the net external force on a closed system is zero along an axis,
then the component of linear momentum of the system along that axis cannot
change.
2008
Physics 2111 Fundamentals of Physics
Chapter 9
12
A shell explodes at its highest point.
v0 = 20 m/s
 = 600
M
M
2
2
M
xmax ?
xE  R 
1
2
Coordinates of the explosion:
Momentum in the x-direction at the explosion:
Projectile motion from the explosion:
2008
1
2
 v02 

 sin 2 0
g


yE 
v0 sin  0 2
2g
M 
M v0 cos  0    V0
 2 
xmax  xE  V0 t   xE  V0
Physics 2111 Fundamentals of Physics
2 yE
g
Chapter 9
13
Method 2: the com continues on its way!
2008
Physics 2111 Fundamentals of Physics
Chapter 9
14
The com continues on its way!
2008
Physics 2111 Fundamentals of Physics
Chapter 9
15
When the person walks to the front of the boat, how far is the boat from the pier?
d ?
no friction
“the system” = person + boat

Fnet  0  motion of the com does not change
the system is initially at rest
 vcom  0
xcom before  xcom after
606.5  1203.5
60  120

60d   120d
 3
60  120
d  2 .5 m
2008
Physics 2111 Fundamentals of Physics
Chapter 9
16
Conservation of ??
Initially both are moving with velocity v0; then the man starts to run with velocity
vrel with respect to the car.
m
M 
momentum before:
momentum after:
2008
w
g
W
g
M
 m v0
M v  m v  vrel 
v  v0 
Physics 2111 Fundamentals of Physics
m vrel
M m
Chapter 9
17
ROCKETS
2008
Physics 2111 Fundamentals of Physics
Chapter 9
18
Rockets
NYT, January 13, 1920
•
for after the rocket quits our air and really starts on its longer journey,
its flight would be neither accelerated nor maintained by the
explosion of the charges it then might have left. To claim that it would
be is to deny a fundamental law of dynamics, and only Dr. Einstein
and his chosen dozen, so few and fit, are licensed to do that.
…..That Professor Goddard, with his "chair" in Clark College and the
countenancing of the Smithsonian Institution, does not know the
relation of action to reaction, and of the need to have something
better than a vacuum against which to react--to say that would be
absurd. Of course he only seems to lack the knowledge ladled out
daily in high schools.
NYT, July, 1969
Further investigation and experimentation have confirmed the findings of
Isaac Newton in the 17th century and it is now definitely established that a
rocket can function in a vacuum as well as in an atmosphere. The Times
regrets the error
2008
Physics 2111 Fundamentals of Physics
Chapter 9
19
Rockets
“The System”
Rocket
Burning Fuel
Exhaust Gas
The rocket + exhaust gas has constant momentum in the absence of
external forces (e.g. in outer space).
Constant Total Energy?
Constant Mechanical Energy?
2008
Physics 2111 Fundamentals of Physics
Chapter 9
20
Fundamentals of Physics
Chapter 9b Collisions
1. Collisions
2. Impulse & Linear Momentum
3. Momentum & Kinetic Energy
4. Inelastic Collisions in 1-Dimension
5. Elastic Collisions in 1-Dimension
6. Collisions in 2-Dimensions
Review & Summary
Questions
Exercises & Problems
2008
Physics 2111 Fundamentals of Physics
Chapter 9
21
Collisions
A collision is an isolated event in which two or more bodies exert relatively strong forces
on each other for a relatively short time.
Moving freely
“Isolated” - no significant external forces during
the collision.
Interacting
A collision does not require physical contact.
Moving freely
2008
Physics 2111 Fundamentals of Physics
Chapter 9
22
Impulse During a Collision
Newton’s 2nd Law:


dP
Fnet 
dt
(eq. 9-23)
Newton’s 3rd Law
During a collision, the left & right bodies briefly exert a force on each other:
F(t) acts for time Dt
An Impulse
2008
Physics 2111 Fundamentals of Physics
Chapter 9
23
Impulse – Linear Momentum Theorem



p f  pi  Dp  J

J 
(vectors!)

 F t  dt


J  Favg Dt
equal areas
2008
Physics 2111 Fundamentals of Physics
Chapter 9
24
A Series of Collisions
A stream of objects (e.g. bullets, a stream of water, …) hitting a target:

J

Dp
What is the force on the target?

J   n Dp   n m Dv   Dm Dv
Favg  


J   Favg Dt
Dm
Dv
Dt
Dv depends on what happens to the objects in the stream? Stop, bounce back,…
2008
Physics 2111 Fundamentals of Physics
Chapter 9
25
Throw a ball against the wall
J=?
Favg = ?
m  300 g  0.3 kg
v  6.0 m s
Dt  10 ms  0.01 s
  300
a) Impulse on the ball?
vxf  vxi  v cos
Dvx  0
v yf   v yi  v sin
Dv y  2 v sin
J  D p  m Dv  2 mv sin
J  18
. Ns
b) Average force on the wall?
Favg 
2008
J
 18
. Ns

  18
. x 102 N
Dt
0.01 s
Physics 2111 Fundamentals of Physics
Chapter 9
26
Momentum & Kinetic Energy in Collisions
Collisions in a closed, isolated system:
– “Closed” - no mass enters or leaves
– “Isolated” - no external forces
– Elastic Collision
• Total Kinetic Energy is conserved.
• Total Linear momentum is conserved.
– Inelastic Collision
• Total Linear momentum is conserved.
• Completely Inelastic Collision
– The bodies stick together after they collide.
The linear momentum of each colliding body may change but the total momentum
of the system cannot change, whether the collision is elastic or inelastic.
2008
Physics 2111 Fundamentals of Physics
Chapter 9
27
Inelastic Collisions in One Dimension
momentum vectors:


pbefore  pafter




p1i  p2i  p1 f  p2 f




m1 v1i  m2 v2i  m1 v1 f  m2 v2 f
2008
Physics 2111 Fundamentals of Physics
Chapter 9
28
Completely Inelastic Collisions
stuck together


pbefore  pafter



p1i  p2i  p12 f


m1 v1i  m2 0  m1  m2  V
V 
2008
m1
v1i
m1  m2
Physics 2111 Fundamentals of Physics
Chapter 9
29
Velocity of the Center of mass
Completely Inelastic Case:
For an isolated system, the velocity of the center of
mass cannot be changed by a collision:


P  M vcom
M 
m
i

M vcom 

m
v
 i i



P  p1i  p2i
vcom 
2008
p1i  p2i
 constant
m1  m2
Physics 2111 Fundamentals of Physics
Chapter 9
30
The Ballistic Pendulum
Velocity of a speeding bullet?
•Step 1: Bullet quickly stops in block
Completely inelastic collision.
Little time for block to move.


m
V 
v
mM
•Step 2: Bullet & block swing upward
Mechanical Energy Conserved.
1
2
m  M  V 2
 m  M  g h
(Tension in cords approximately perpendicular to block’s direction of travel; no work done.)
2008
Physics 2111 Fundamentals of Physics
Chapter 9
31
Colliding Blocks
m1 = 2.0 kg, 10 m/s
m2 = 5.0 kg, 3 m/s
k = 1120 N/m
Conservation of momentum:
m1 v1i  m2 v2i  m1 v1 f  m2 v2 f
At maximum compression:
Conservation of total mechanical energy:
v1 f  v2 f  v
Ki  K f  U spring
Ki 
2
Kf 
1
U 
1
algebra  Dx 
2008
1
m1 v1i 2 
2
m
2
k x2
1
1
2
m2 v2i 2
 m2  v 2
m1 m2
v1i  v2i 

k m1  m2 
Physics 2111 Fundamentals of Physics
Chapter 9
32
Elastic Collisions in One Dimension
In an elastic collision, the kinetic energy of each colliding body may change,
but the total kinetic energy of the system does not change.
m1 v1i  m1 v1 f  m2 v2 f
1
2
m1 v1i
v1 f
2008
2

1
2
m1 v1 f
m  m2
 1
v
m1  m2 1i
2

1
2
v2 f
m2 v2 f
2
2 m1

v
m1  m2 1i
Physics 2111 Fundamentals of Physics
Chapter 9
33
Elastic Collisions with a Moving Target
m1 v1i  m2 v2i  m1 v1 f  m2 v2 f
1
2
m1 v1i
2

v1 f
1
m2 v2i 2 
1
2
m1 v1 f
2

1
2
m2 v2 f
2
m1  m2
2 m2

v 
v
m1  m2 1i
m1  m2 2i
v2 f 
2008
2
2 m1
m  m1
v1i  2
v2 i
m1  m2
m1  m2
Physics 2111 Fundamentals of Physics
Chapter 9
34
Planetary Assist (“slingshot”)
v = 12 km/s
vJ= 13 km/s
v1 f
v1 f 
m1  m2
2 m2
v1i 
v
m1  m2
m1  m2 2i
m2  m1
m1  m2
 1
m1  m2
2 m2
 2
m1  m2
v1 f   v1i  2 v2i
v1 f     12  2   13
v1 f   38 km s
2008
Physics 2111 Fundamentals of Physics
Chapter 9
35
Collisions in Two Dimensions


pbefore  pafter




p1i  p2i  p1 f  p2 f

For p21  0:
m1 v1i 
m1 v1 f cos1  m2 v2 f cos2
0   m1 v1 f sin1  m2 v2 f sin2
2008
Physics 2111 Fundamentals of Physics
Chapter 9
36
Billiard Ball Collision



m v1i  m v1 f  m v2 f



v1i 
v1 f 
v2 f
1
2
m v1i
2
v1i
2


1
2
m v1 f
2
v1 f
2

1
2
m v2 f
2
v2 f
2

Pythagorean Theorem!
   900
Equal mass elastic scattering.
e.g. billiards
2008
Physics 2111 Fundamentals of Physics
Chapter 9
37
Equal Mass Elastic Collision
proton + proton  proton + proton
2008
Physics 2111 Fundamentals of Physics
Chapter 9
38