Bilingual Mechanics
Chapter 5
Center of Mass and
Linear Momentum
制作 张昆实
制作 张昆实 赵 俊
Yangtze University
Yangtze University
Chapter 5 Center of Mass and
Linear Momentum
5-1 What Is Physics?
5-2 The Center of Mass
5-3 Newton’s Second Law for a System
of Particles
5-4 Linear Momentum
5-5 The Linear Momentum of a System
of Particles
5-6 Collision and Impulse
Chapter 5 Center of Mass and
Linear Momentum
5-7 Conservation of Linear Momentum
5-8 Momentum and Kinetic Energy in
Collisions
5-9 Inelastic Collisions in One Dimension
5-10 Elastic Collisions in One Dimension
5-11 Collisions in Two Dimensions
5-12 Systems with Varying Mass: A
Rocket
5-1 What Is Physics
★ Analyzing complicated motion requires
simplification via an understanding of
physics.
★ We discuss how the complicated motion
of a system of objects can be simplified if
we determine a special point of the system
the center of mass of that system.
★ In this chapter we learn what is physics
through studying linear momentum and
collisions.
5-2 The Center of Mass
If we toss a small ball into the air, it
will follow a parabolic path. What about
a grenade?
5-2 The Center of Mass
Every part of the grenade moves in a
different way except one special point,
the center of mass, of the grenade,which
still moves in a parabolic path .
The center of mass of a system of particles
is the point that moves as though
(1) all the system’s mass were concentrated
there and
(2) all external forces were applied there.
5-2 The Center of Mass
Systems of Particles
The position of the center of mass (com) of this
two-particle system can be difined as:
y
m2
xcom 
Discuss
(1) if m2
m1  m2
0 ,
d
xcom  0
(5-1)
xcom
m2
m1
d
(Fig.5-2a)
the center of mass lies at the position of
(2) If m1  0 , xcom  d
m1
the center of mass lies at the position of
m2
(3) If m1  m2 , xcom  1 2 d
the center of mass shuld be halfway between them.
x
5-2 The Center of Mass
For a more generalized situation, in which the coordinate
system has been shifted leftward.
y
xcom
xcom
x1m1  x2 m2

m1  m2
(5-2)
o
m1
x1
x2
d
m2
x
(Fig. 5-2b)
Note:
in spite of the shift of the coordinate system, the
center of mass is still the same distance from each particle.
xcom
x1m1  x2 m2

M
(5-3)
( M  m1  m2 )
For a system of n particles
xcom
x1m1  x2 m2 

M
 xn mn
1

M
n
m x
i 1
i i
(5-4)
5-2 The Center of Mass
If particles are distributed in three dimensions, the center
of mass must be identified by three coordinates:
xcom
1

M
n
 mi xi
i 1
ycom 
1
M
n
m y
i 1
i
i
zcom
1

M
n
m z
i 1
i i
(5-5)
Define the center of mass with the language of vectors.
The position of a particle is given by a position vector:
ri  xi i  yi j  zi k
(5-6)
The position of the center of mass of a system of particles
is given by a position vector:
rcom  xcom i  ycom j  zcom k
(5-7)
5-2 The Center of Mass
The three scalar equation of
1
Eq.5-5 can be replaced by a rcom 
M
single vector equation
n
m r
i 1
i i
(5-8)
Solid Bodies
An ordinary solid object contains many particles,
can be treated as a continuous distribution of
matter. m  dm ( differential mass elements )
sum→integral
1
1

m
x
y


i i
com
M i 1
M
1 n
xcom
zcom 
mi zi (5-5)
mi yi


M i 1
i 1
the coordinates of the center of mass will be
1
1
1
zcom 
zdm (5-9)
ycom 
ydm
xcom 
xdm



M
M
M
n
n
5-2 The Center of Mass
We consider only uniform objects (uniform density  )
dm M


dV
V
Substitute dm  (M V )dV
xcom
1
  xdV
V
ycom
(5-10)
into Eq.5-9 gives
1
  ydV
V
zcom
1
  zdV
V
(5-11)
We can bypass one or more of these integrals if an object
has a point, a line, or a plane of symmetry. The center of
mass of such an object then lies at that point, or that line, or
in that plane.
Note:
the center of mass of an object need
not lie winthin the object.
5-3 Newton’s Second Law for a System of Particles
If we roll a cue ball at a second billiard ball
that is at rest, How do they move after inpact?
the second billiard ball
In fact, what continues to move forward is
the center of mass of the two-ball system.
The center of mass moves like a particle
whose mass is equal to the total mass of
the system. Then,the motion of it will be
governed by Eq.(5-14)
Note:
Fnet  Macom
cue ball
(Newton’s Second Law ) (5-14)
(1) Fnet is the net force of all external forces
acted on the system.
(2)
M is the total mass of the system.
5-3 Newton’s Second Law for a System of Particles
(3) acom is the acceleration of the center of mass of the
system.
Eq.5-14 is equivalent to three equations along the three
coordinate axes.
(5-14)
F  Ma
net
Fnet , x  Macom, x
com
Fnet , y  Macom, y Fnet , z  Macom, z
(5-15)
Go back and examine the behavior of the billiard balls.
The cue ball has
begun to roll
no net external force
acts on the system
Fnet  0
acom  0
Fnet  0
acom  0
internal forces
Two balls collide
don’t contribute to
the net force
Thus, the center of mass must still move forward after the
collision with the same speed and in the same direction.
5-3 Newton’s Second Law for a System of Particles
Newton’s Second Law applies not only to a system of
particles but also to a solid body.
Proof of Equation 5-14
From
For a system of n particles,
rcom
1

M
Mrcom  m1r1  m2 r2  m3r3 
n
m r
i 1
i i
 mn rn
(5-8)
(5-16)
Differentiating Eq.5-16 with respect to time gives
Mvcom  m1v1  m2v2  m3v3 
 mnvn
(5-17)
Differentiating Eq.5-17 with respect to time leads to
Macom  m1a1  m2 a2  m3a3 
 mn an
(5-18)
We can rewrite Eq.5-18 as
Macom  F1  F2  F3 
 Fn  Fnet
(5-19)
5-4 Linear Momentum
The linear momentum of a particle
is a vector quantity p , defined as
p
p  mv
(5-22)
v
and
have the same direction, the SI unit for momentum is
kilogram-meter per second ( kg  m / s ) .
Express Newton’s Second Law of Motion in terms of
momentum:
The time rate of change of the momentum of particle
is equal to the net force acting on the particle and is in
the direction of that force.
Fnet
Substituting for
p
dp

dt
(5-23)
from Eq.5-22 gives
dp d
dv
Fnet 
 (mv )  m
 ma
dt dt
dt
Equivalent
expressions
5-5 The Linear Momentum of a System of Particles
Now consider a system of n particles. The system as a
whole has a total linear momentum p , which is defined
to be the vector sum of the individual particles’ linear
momenta.
p  p1  p2  p3   pn
(5-24)
 m1v1  m2v2  m3v3 
 mn v n
Compare it with Eq.5-17, we see the linear momentum of
the system Mvcom  m1v1  m2v2  m3v3   mnvn (5-17)
p  Mvcom
(linear momentum, system of particles) (5-25)
The equation above gives us another way to define the
linear momentum of a system of particles:
 The
linear momentum of a system of particles is equal to
the product of the total mass M of the system and the
velocity of the center of mass.
5-5 The Linear Momentum of a System of Particles
Take the time derivative of Eq.5-25, we find
dvcom
dp
M
 Macom
dt
dt
(5-26)
Comparing Eq.5-14 and 5-26 allows us to write
Newton’s second law for a system of particles in
the equivalent form
Fnet  Macom
Fnet
dp

dt
(Newton’s Second Law ) (5-14)
(system of particles) (5-27)
5-6 Collision and Impulse
★ In everyday language, a collision occurs
when an objects crash into each other.
Example: tennis ball contacts with racket
the collision ofa ball with a bat
★ A collision is an isolated event in which two
or more bodies (the colliding bodies) exert
relatively strong forces on each other for a
relatively short time.
definition
5-6 Collision and Impulse
★ Single collision
L
R
During a head-on collision:
A third law force pair F (t ) ,  F (t )
 F (t )
x
F (t )
The change of the linear momentum depends on:
The forces and the action time t
Apply Newton’s second law F  dP dt to ball R
dP  F (t )dt
Integrating Eq. 5-28 from ti

Pf
Pi
tf
dP   F (t )dt
ti
(5-28)
to t f :
(5-29)
5-6 Collision and Impulse

Pf
Pi
tf
dP   F (t )dt
(5-29)
L
ti
J 

tf
F (t ) dt (5-30)
ti
F (t )
R
x
 F (t )
★ J is the impulse, which is a measure of both the
magnitude and the duration of the collision force.
tf
J   F (t )dt
ti
F
J  Favg t
F
(5-35)
F (t )
J
ti
rectangle
Favg
t
tf
t
J
ti
t
tf
t
5-6 Collision and Impulse

Pf
Pi
tf
dP   F (t )dt
(5-29)
L
ti
J 

tf
ti
F (t ) dt (5-30)
F (t )
R
 F (t )
From Eqs. 5-29 and 5-30
P  Pf  Pi  J
(5-31) (5-32)
( linear momentum-Impulse theorem )
The change in an object’s momentum is equal to
the impulse on the object.
Component
form
Px  J x
tf
Pfx  Pix   Fx dt  J x
ti
(5-33)
(5-34)
x
5-6 Collision and Impulse
Series of collisions
m
A steady stream of identical
projectiles collides with a
fixed target. Find the average
v
Favg
Target
Projectiles
force Favg on the target during the bombardment.
The total change in momentum for n
projectiles in t is : np
The impulse J on the target in t : J   np (5-36)
Combining Eq. 5-35
Favg
J  Favg t and 5-36:
J
n
n

  p   mv
t
t
t
(5-37)
x
5-6 Collision and Impulse
Series of collisions
The rate at which the
projectiles collide with
the target
v
Favg
Target
Projectiles
J
n
n
(5-37)
Favg 
  p   mv
t
t
t
In t , an amount of mass m  nm collide with
the target, so
The rate at which the mass
m
(5-40)
Favg  
v collides with the target
t
m
v
If vi  v, v f  0 v  0  v  v Favg 
t m
v
If vi  v, v f  v v  v  v  2v Favg  2
t
x
5-7 Conservation of Linear Momentum
Suppose that the net external force acting on a system of
particles is zero (isolated), and that no particles leave or
enter the system(closed)，Then F  0
dp
0
dt
p =constant
net
(closed,isolated system) (5-42)
 If no net external force acts on a system of particles,
the total linear momentum p of the system cannot
change.
Law of conservation of linear momentum
Eq.5-42 can also be written as
pi  p f
(closed,isolated system)
(5-43)
For a closed, isolated system
total linear momentum
at some initial time ti
=
total linear momentum
at some later time t f
5-7 Conservation of Linear Momentum
 If
the component of the net external force
on a closed system is zero along an axis
then the component of the linear momentum
of the system along that axis cannot change.
5-8 Momentum and Kinetic Energy in Collisions
Discussion collisions in closed (no mass enters
or leaves them) and isolated (no net external
forces act on the bodies within them ) systems.
Kinatic Energy（in collisions）
1. Elastic collision: is a special type of collision
in which the kinetic energe of the system of
colliding bodies is conserved.
2. Inelastic collision: after the collision the kinetic
energe of the system is not conserved.
3. Completely inelastic collision: after the collision
the bodies stick together and have the same final
velocity ( the greatest loss of kinetic energe ).
5-8 Momentum and Kinetic Energy in Collisions
Linear Momentum（in collisions）
Regardless of the details of the impulses in a
collision; Regardless of what happens to the
kinetic energy of the system, The total linear
momentum P of a closed, isolated systems
cannot change. (no external force !)
The law of conservation of linear momentum
In a closed , isolated system containing a collision,
the linear momentum of each colliding body may
change but the total linear momentum P of the
system cannot change, whether the collision is
elastic or inelastic.
5-9 Inelastic Collisions in One Dimension
One Dimensional Inelastic
collision
Two colliding bodies form
a closed , isolated system
along the x axis.
v2i
v1i
befor
x
v1 f
v2 f
after
x
m1
m2
The law of conservation of linear momentum
Pi  Pf
P1i  P2i  P1 f  P2 f
(5-50)
before、 after collision
The motion is one-dimensional, component form:
m1v1i  m2 v2i  m1v1 f  m2 v2 f
(5-51)
5-9 Inelastic Collisions in One Dimension
One-Dimensional Completely Inelastic Collision
Before the collision body 2
is at rest, body 1 moves
directly toward it. After the
collision, the stuck-together
bodies move with the same
velocity V .
v1i
befor
Projectile
m1
after collision
v2i  0
Target
m2
V
m1  m2
The law of conservation of linear momentum
Pix  Pfx
m1v1i  (m1  m2 )V
(5-52)
before、 after collision
m1
V 
v1i
m1  m2
(5-53)
x
x
5-9 Inelastic Collisions in One Dimension
Velocity of Center of Mass
In a closed, isolated system,
the velosity vcom of the center
of mass of the system cannot
be changed by a collision for
there is no net external force
to change it.
Find out
vcom
m1
(left side=constant)
p1i  p2i
P
vcom 

m1  m2 m1  m2
(constant)
v2i  0
Collision !
P  Mvcom  (m1  m2 )vcom
P  p1i  p2i
m2
v1i vcom
x
m1  m2
V  vcom
(5-54)
(5-55)
(5-56)
Completely
inelastic
collision
5-10 Elastic Collisions In One Dimension
We can approximate some
of the everyday collisions
as being elastic;
we can
approximate that the total
kinetic energy of the cilliding bodies is conserved.
Total kinetic energy
before the collision
☞
=
v1i
befor
Projectile
after
m1
v2i  0
Target
m2
v1 f
m1
v2 f
m2
Total kinetic energy
after the collision
In an elastic collision, the kinetic energy of
each colliding body may change, but the total
kinetic energy of the system does not change.
x
x
5-10 Elastic Collisions In One Dimension
v1i
Stationary Target
Before the collision body 2
is at rest, body 1 moves
toward it (head on collision).
befor
Projectile
after
m1
v2i  0
Target
m2
v1 f
v2 f
m2
m1
Assume this two body system
is closed and Isolated. Then the
net linear momentum of the system is conserved.
m1v1i  m1v1 f  m2 v2 f
( linear momentum ) (5-63)
If the collision is also elastic, the total kinetic
☞energy of the system is also conserved.
1
2
mv  mv  m v
2
1 1i
1
2
2
1 1f
1
2
2
2 2f
( kinetic energy )
(5-64)
x
x
5-10 Elastic Collisions In One Dimension
Stationary Target
m1v1i  m1v1 f  m2 v2 f
1
2
mv  mv  m v
2
1 1i
1
2
2
1 1f
1
2
2
2 2f
( linear momentum ) (5-63)
( kinetic energy )
(5-64)
If know m1、m2、v1i , Then can find out v1 f 、v2 f
Rewrite Eq.5-63 m (v  v )  m v
(5-65)
1
1i
1f
2 2f
Rewrite Eq.5-64
m1 (v1i  v1 f )(v1i  v1 f )  m v
2
2 2f
v1 f
m1  m2

v1i
m1  m2
v2 f
2m1

v1i
m1  m2
(5-66)
(5-67)
(5-68)
5-10 Elastic Collisions In One Dimension
Stationary Target
v1 f
m1  m2

v1i
m1  m2
v2 f
(5-67)
2m1

v1i(5-68)
m1  m2
Discussion: A few special situations
m1  m2 , Eqs. (5-67) and (5-68)
v2 f  v1i Changing vilocities !
reduce to v1 f  0 and
2. A massive target: If m2
m1 , Eqs. (5-67) and (5-68)
1. Equal masses: If
reduce to
v1 f  v1i
and
bounds back
3. A massive projectile: If
reduce to
v1 f  v1i
m1
and
v2 f 
 v
2 m1
m2
1i
(5-69)
m2 ,Eqs. (5-67) and (5-68)
v2 f  2v1i
(5-70)
5-10 Elastic Collisions In One Dimension
1.Conservation of the linear momentum
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Moving Target
v1i
2.Conservation of the kinetic energy
1
2
mv  m v  mv  m v
2
1 1i
1
2
2
2 2i
1
2
2
1 1f
1
2
2
2 2f
m1
v2i
m2
x
head on elastic collision
To work out v1 f and v2 f , rewrite these conservation equations:
m1 (v1i  v1 f )  m2 (v2i  v2 f )
(5-73)
(5-74)
m1 (v1i  v1 f )(v1i  v1 f )  m2 (v2i  v2 f )(v2i  v2 f )
m1  m2
2m2
v1 f 
v1i 
v2i
(5-75)
m1  m2
m1  m2
2m1
m2  m1
v2 f 
v1i 
v2i (5-76)
m1  m2
m1  m2
5-11 Collisions in Two Dimensions
Two dimension (not head-on)
elastic collision
System: closed + isolated
Conservation equations:
P1i  P2i  P1 f  P2 f
glancing
collision
(5-77)
y
v2 f
m2 
2
m1 v1i
x
1
v1 f
K1i  K 2i  K1 f  K 2 f (5-78)
Rewrite Eq.5-77 for components along the x、y axis
m1v1i  m1v1 f cos 1  m2v2 f cos  2
(x axis) (5-79)
0  m1v1 f sin 1  m2 v2 f sin  2
(y axis) (5-80)
1
2
mv  mv  m v
2
1 1i
1
2
2
1 1f
1
2
2
2 2f
(5-81)
5-12 Systems with Varying Mass: A Rocket
Now let us study a special system: a rocket
5-12 Systems with Varying Mass: A Rocket
Most of the mass of a
rocket on its launching
pad is fuel, all of which
will eventually be burned
and ejected from the
nozzle of the rocket
engine
Consider the rocket and
its ejected combustion
products as a system, the
total mass is still constant.
China´s manned spacecraft Shenzhou-7 blasts off
5-12 Systems with Varying Mass: A Rocket
Finding the Acceleration
t
M
v
Watching in a inertial reference
frame; In deep space, no gra(a)
t  dt
vitional or atmospheric drag
M  dM
dM
forces
At an arbitrary time
t:
p  Mv
x
v  dv
U
x
(b)
dt later
p   dMU  ( M  dM )(v  dv )
An time interval
For a closed 、isolated system, the linear momentum
must be conserved.
Pi  Pf
(5-82)
Mv  dMU  ( M  dM )(v  dv)
(5-83)
5-12 Systems with Varying Mass: A Rocket
t  dt
inertial
Earth reference frame
dM
M  dM
vrel
v  dv
U
x
Eq.5-83 can be simplified by using the relative speed
velocity of rocket velocity of rocket
relative to frame = relative to products
absolute velocity
relative velocity
+
velocity of products
relative to frame
convected velocity
vabs  vrel  vcon
For the onedimensional
motion
(v  dv)  vrel  U
U  v  dv  vrel
(5-84)
5-12 Systems with Varying Mass: A Rocket
U  v  dv  vrel
Substituting it into
yields:
dM
Replace
dt
comsuption
Mv  dMU  (M  dM )(v  dv)
dMvrel  Mdv
(5-85)
dM
dv

vrel  M
dt
dt
(5-86)
by  R , where
Rvrel  Ma
Rvrel
(5-84)
R
(5-83)
is the mass rate of fuel
(first rocket equation)
the thrust of the rocket engine
(5-87)
T  Ma
5-12 Systems with Varying Mass: A Rocket
Finding the Velocity
Now, let us study how the velocity of a rocket changes as it
consums its fuel.
From dMv  Mdv (5-85)
rel
dM
dv  vrel
M
Integrating leads to

vf
vi
dv  vrel 
Mi
v f  vi  vrel ln
Mf
Mf
Mi
dM
M
(second rocket equation)
(5-88)
The advantage of multistage rockets !