Chapter 9
Center of Mass and Linear Momentum
In this chapter we will introduce the following new
concepts:
-Center of mass (com) for a system of particles
-The velocity and acceleration of the center of mass
-Linear momentum for a single particle and a system of particles
We will derive the equation of motion for the center of mass, and
discuss the principle of conservation of linear momentum.
Finally, we will use the conservation of linear momentum to study
collisions in one and two dimensions and derive the equation of
motion for rockets.
What is the centre of mass of an
object?
Consider a bat thrown into the air. If you
look closely, you will find that one special
point of the bat moves in a simple
parabolic path, just as a particle would if
tossed into the air.
That special point moves as though (1) the bat's total mass were
concentrated there and (2) the gravitational force on the bat acted
only there.
That special point is said to be the center of mass of the bat.
In general: DEFINITION of CENTRE OF MASS
The center of mass of a body or a system of bodies is the point that
moves as though all of the mass were concentrated there and all
external forces were applied there.
Centre of Mass (com)
• Fig. 9-2a shows two particles
with masses m1 and m2 that
are a distance d apart.
• The origin of the x-axis is
chosen as the position of m1.
• The position of the centre of
mass (com) of this twosystem is then defined as:
xcom
m2

d
m1  m2
(9-1)
Fig.9-2b
• Two particles with
masses m1 and m2 ,
have positions x1 and x2
respectively , and are a
distance d apart.
• The centre of mass is
now defined as:
xcom
m1 x1  m2 x2

m1  m2
Substitute m1 + m2 = M (sum total of masses of two
particles)
xcom
m1 x1  m2 x2

M
(9-3)
(9-2)
Centre of Mass of a system of
particles
• If there are n particles in a system, stretched
out along the x-axis, then:
• M = m1 + m2 + .... + mn, and
• The position of the centre of mass is:
1 n
m1 x1  m2 x2  .... mn xn
  mi xi (9-4)
xcom 
M i 1
M
• If the particles are spread out in three dimensions, the
centre of mass must be defined by three position
coordinates:
1 n
1 n
1 n
xcom 
mi xi ; ycom 
mi yi ; zcom 
mi zi



(9-5)
M i 1
M i 1
M i 1
Solid uniform objects
• An ordinary object like a baseball bat, consist
of so many particles (atoms) that it can be
considered as a continuous distribution of
matter.
• Each particle is then a differential mass
element dm,
• And the sums of eq. 9-5 become integrals and
the coordinates of the centre of mass are
defined as:
xcom
1

M
 xdm ;
ycom
1

M
 ydm ;
with M the mass of the object.
zcom
1

M
 zdm;
(9-9)
• It is very difficult to determine these integrals,
but uniform objects have uniform densities,
mass per volume, thus
dm M


(9-10)
• Density = mass / volume:
• Then :
M 
dm   dV
V 
• And substitute into eq. 9-9 then:
xcom
Also
And
1
1
M 

x dV 

M V 
V
 xdV
1
ycom   ydV
V
1
zcom   zdV
V
dV
V
• If an object has a point, a line or plane of
symmetry, then the centre of mass of the
object lies at that point, on the line or in the
plane of symmetry.
• It can also happen that the centre of mass lies
outside of the object, e.g. a horseshoe and a
doughnut.
Centre of Mass
Centre of Mass
“ Sample Problem” 9-1
on p. 204(204)(171) in H&R.
• Fig. 9-3 Three particles
form an equilateral triangle
with each side = a.
• The Centre of Mass is found
by the position vector rcom.
Particle
1
2
3
Mass
(kg)
1.2
2.5
3.4
x (cm) y (cm)
0
140
70
0
0
120
• M = 1.2 + 2.5 + 3.4 = 7.1 kg
• x-coordinate:
xcom
m1 x1  m2 x2  m3 x3

M
(1.2kg)(0)  (2.5kg)(140cm)  (3.4kg)(70cm)
xcom 
= 83 cm
7.1kg
And also ycom = 58 cm
Thus position vector: r com = xcom î + ycomĵ = (83 î + 58 ĵ ) cm
Calculate the value and position of rcom like any other
vector. rcom= 101.26 cm and 35˚ with the positive x-axis
Newton’s Second Law for a System of Particles
• If you roll a cue ball at a second billiard ball that is at rest, you expect
that the two-ball system will continue to have some forward motion
after impact.
• What continues to move forward, in a steady motion completely
unaffected by the collision, is the centre of mass of the two-ball
system.
• We replace the pair of billiard balls with an assemblage of n particles
of (possibly) different masses. We are interested not in the
individual motions of these particles but only in the motion of their
center of mass.
• Although the center of mass is just a point, it moves like a particle
whose mass is equal to the total mass of the system;
• we can assign a position, a velocity, and an acceleration to it.
• The vector equation that describes the motion of
the centre of mass of such a system of particles, is
then: Fnet = Manet
(9-14)
• This equation is Newton’s second law for the
motion of the centre of mass of a system of
particles.
• It looks like the equation applicable to the motion of
a single particle F = ma.
NB:
1. Fnet is the net force of all external forces that
act on the system.
2. M is the total mass of the system. We assume
that no mass enters or leaves the system as it
moves, so that M remains constant. The system is
said to be closed.
3. acom is the acceleration of the centre of mass of the
system.
4. Because acom can consists of x-, y- and z-components,
we thus have:
(9-15)
• Now we again examine the behaviour of the billiard balls.
• Once the cue ball has begun to roll, no net external force
acts on the (two-ball) system.
• Thus Fnet = 0, and acom = 0 .
• Acceleration = the rate of change in velocity,
• thus the velocity of the centre of mass does not change.
• When the two balls collide, the forces that come into play
are internal forces so they do not contribute to the net
force Fnet, which remains zero.
• Thus, the center of mass of the system, which was
moving forward before the collision, must continue to
move forward after the collision, with the same speed
and in the same direction.
Proof of
eq.9-14
n
• Eq.9-8 : rcom
1

M

 mi ri ,

 i 1 

• Thus Mrcom  m1r1  m2 r2  .... mn rn ,
(9-16)
• Differentiate
eq.9-16 with respect to time:

drcom





M
 Mvcom  m1v1  m2 v2  m3v3  ...  mn vn
dt
(9-17)
• Differentiate
eq.9-17 with respect to time:

dvcom





M
 Macom  m1a1  m2 a2  m3 a3  ...  mn an
dt
(9-18)
But from Newton II, Fnet = ma or Fi = miai,
Thus Ma  F  F  F  .... F
1

 Macom  Fnet
com
2
3
n
(9-19)
(9-14)
Sample Problem 9-3, p.208
Linear Momentum
• The linear momentum of a particle with mass m and
velocity v is defined as a vector p,
• p = mv
• With p and v in the same direction.
• SI unit: kilogram-meter per second or kg.m/s.
• Newton’ s Second Law of motion in terms of
momentum:
• The time rate of change of the momentum of a particle
is equal to the net force acting on the particle and is in
the direction of that force.

• In equation form:

Fnet
dp

dt
(9-23)
• The linear momentum of an object can only be
changed by a net external force!




dp
Substitute p  mv
into Fnet 
dt



Then F  dp  d (mv )  m dv  ma
net
dt dt
dt
The Linear Momentum of a System of Particles
• For a system of n particles, each with its own
mass, velocity, and linear momentum
interacting with each other, the system as a
whole has a total linear momentum P, which is
defined to be the vector sum of the individual
particles' linear momentum.
(9-24)
If we compare this equation with Eq. 9-17 , we see that
(9-25)
• This 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.
• Differentiating eq.9-25 with respect to time


gives:
dvcom
dP

M
 Macom
(9-26)
dt
dt
• Comparing Fnet
=
Ma
,
eq.9-14,
with
eq.9-26,
net

then: 
dP
(system of particles)
Fnet 
(9-27)
dt
Collision and Impulse
• The momentum, p, of any particle cannot change
unless a net external force, Fnet acts on it.
• Two ways to change p:
- push or throw object;
- collision with another
object
• In a collision (crash), - the external force is brief
- has large magnitude
- suddenly changes object’s p.
• In a collision where one object moves and the other
is stationary, the moving object is called the
projectile and the stationary object the target.
Single Collision
• Let projectile = a ball and target = a bat
• Collision is brief, but force on ball big enough to
slow, stop or even reverse its motion.
• So from eq.9-27,
dp = F(t)dt
(9-28)
• Find the net change in ball’s momentum by
integrating both sides from time ti just before
collision to time tf just after collision:
•
t
t


 dp   F (t )dt (9-29)
• left side:
• right side:
f
f
ti
ti
pf – pi = Δp
impulse = J
tf
tf


 dp   F (t )dt
ti
(9-29)
ti
tf


J   F (t ) dt Impulse defined
ti


p  J
Linear momentum-Impulse theorem
(9-30)
(9-31)
In vector form:

 
p f  pi  J
(9-32)
In component form:
 px  J x
(9-33)
p fx  pix   Fx dt
(9-34)
tf
And
ti
• Often we do not know how the force varies with
time, but we know the magnitude of the
average force, Favg, and the duration Δt (= tf – ti)
of the collision.
• Then the magnitude of the impulse is: J = FavgΔt
(9-35)
•
•
•
•
•
What if instead we focused on the bat?
According to Newton III: Fball = -Fbat
Same magnitude, but opposite directions
Thus from eq.9-35: Jball = -Jbat
Do Sample Problems 9-4 & 9-5, p213-214
A series of Collisions
• Now let's consider the force on a body when it
undergoes a series of identical, repeated collisions.
• For example, as a prank, we might adjust one of
those machines that fires tennis balls to fire them at
a rapid rate directly at a wall.
• Each collision would produce a force on the wall, but
that is not the force we are seeking. We want the
average force Favg on the wall during the
bombardment—that is, the average force during a
large number of collisions.
Fig.9-10 A steady stream of projectiles,
with identical linear momenta, collides
with a target, which is fixed in place. The
average force Favg on the target is to the
right and has a magnitude that depends
on the rate at which the projectiles collide
or, equivalently, the rate at which mass
collides.
• Each projectile has initial momentum mv and
undergoes a change p in linear momentum
because of the collision. The total change in
linear momentum for n projectiles during
interval t is n p. The resulting impulse J on the
target during t is along the x axis and has the
same magnitude of n p but is in the opposite
direction.
• We can write this relation in component form as:
J = -np
(9-36)
where the (–) sign indicates that J and p have
opposite directions.
• By rearranging, the average force Favg , acting on
the target during the collisions is:
J
n
n
Favg 
  p   mv (9-37)
t
t
t
• This equation gives Favg i.t.o. n/t, the rate at
which the projectiles collide against the target, and
v the change in the velocity of the projectiles.
Conservation of Linear Momentum
• Assume that the net external force on a system of
particles is zero (an isolated system) and that no
particles leave or enter the system (the system is
closed).
• If Fnet = 0 is substituted into Eq. 9-23, then dP/ dt = 0
• P = constant (closed, isolated system)
(9-42)
• In words: If no external force act on a system of
particles, the total linear momentum of the system
does not change.
• Known as: Law of Conservation of Linear Momentum:
– Pi = Pf (closed, isolated system)
(9-43)
– [Tot. Lin. momentum at a initial time ti ] = [Tot. Lin.
momentum at a later time tf ]
• Eq’s. 9-42 and 9-43 are vector equations and thus
equivalent to three equations that correspond with the
conservation of linear momentum in the three
directions of the xyz –axis system.
• Depending on the forces that act on the system, linear
momentum can be conserved in one or two directions,
but not necessarily in all directions.
• If the component of the net external force on a closed
system is zero along an axis, then the component of
linear momentum on the system along the axis cannot
change.
• Do Sample Problems 9-6 to 9-8, p215-217
Momentum vs Kinetic Energy in
Collisions
• Consider a closed, isolated system of 2 particles:
• P = constant, if Fnet = 0
• Now consider the kinetic energy of the 2 particles that
collide:
• If the total kinetic energy, Ktot , before and after a
collision remains the same, then it is a elastic collision.
• Often K is lost and transformed into other forms of
energy during a collision, ex. thermal energy or sound
energy, then it is an inelastic collision.
• The biggest loss in K takes place during a collision where
after the objects stick together, and is called a
completely inelastic collision.
• During a collision in a closed, isolated system,
the momentum of individual bodies can
change, but the total momentum P of the
system cannot change, regardless if the
collision was elastic or inelastic.
• This is another way to express the Law of
Conservation of Linear Momentum
(discussed in Section 9-6).
Inelastic Collision in One-Dimention
Fig. 10-8 shows two bodies just before
and just after they have a onedimensional collision (meaning that the
motions before and after the collision
are along a single axis). The velocities
before the collision (subscript i) and
after the collision (subscript f) are
indicated.
• The two bodies form our system, which is closed and
isolated. We can write the law of conservation of linear
momentum for this two-body system as:
[total momentum Pi before collision] = [total
momentum Pf after collision]
• in symbol form:
p1 i + p2 i = p1 f + p2 f (conservation of momentum) (9-50)
Inelastic Collision in One-Dimention
• Because the motion is one-dimensional, we can
drop the overhead arrows for vectors and use
only components along the axis. Thus, from p =
mv, we can rewrite Eq. 9-50 as
• m1v1 i + m2v2 i = m1v1 f + m2v2 f
(9-51)
• With this we can calculate one of the unknown
values if the others were given.
Completely Inelastic Collision
Fig. 9-15 Before the collision m2
(target) is in rest and m1
(projectile) moves to the right.
After the collision both the twee
saam met dieselfde snelheid V.
• For this case Eq. 9-51 becomes
m1v1 i = (m1 + m2)V
m1
• Or
V 
v1i
m1  m2
• V can thus be calculated from this.
• Note that V must be smaller than v1i .
(9-52)
(9-53)
Velocity of Centre of Mass
• In a closed, isolated system, the velocity vcom of the
centre of mass of the system cannot be changed by
a collision because, with the system isolated, there
is no net external force to change it.
• To get an expression for vcom, let us return to the
two-body system and one-dimensional collision of
Fig. 9-15 . From Eq. 9-25 (P = Mvcom), we can relate
vcom to the total linear momentum P of that twobody system by writing



P  Mvcom  (m1  m2 )vcom


P  p1i  p2i
(9-54)
(9-55)
Velocity of Centre of Mass
• If P is replaced by Eq.9-55 in Eq.9-54 and then
solved for vcom , we get:




P
p1i  p2i
vcom 

(9-56)
m1  m2 m1  m2
• The right side of this equation is a constant, and
vcom has that same constant value before and after
the collision.
NB: Sample Problem 9-9, p219-220
Principle of function of ballistic pendulum
Vbullet&block
 vbullet
mbullet

vbullet
mbullet  mblock
mbullet&block

Vbullet&block
mbullet
From the conservation of mechanical
energy:
1
2
mbullet&blockV  mbullet&block gh
2
V  2gh
 vbullet
mbullet&block

mbullet
2 gh
(0.0095 5.4)kg
2
 vbullet 
2(9.8m / s )(0.063m)  630m / s
0.0095kg
Elastic Collision in One-dimention
• Ktot before collision = Ktot after collision
• In an elastic collision, the kinetic energy of each colliding particle
can change, but the total kinetic energy of the system does not
change.
• Thus ex. For a stationary target:
m1v1i = m1v1f + m2v2f (lin.momentum)
(9-63)
• and ½m1v1i2 = ½m1v1f2 + ½m2v2f2 (kin.energy) (9-64)
• Thus from eq.9-63: m1(v1i - v1f )= m2v2f
(9-65)
• And from eq.9-64: m1v1i2 - m1v1f2 = m2v2f2
m1(v1i + v1f)(v1i - v1f) = m2v2f2 (9-66)
• Divide eq.9-66 by 9-65 and with further algebra it follows that:
and
m1  m2
v1 f 
v1i
m1  m2
2m1
v2 f 
v1i
m1  m2
(9-67)
(9-68)
m1  m2
v1 f 
v1i
m1  m2
2m1
v2 f 
v1i
m1  m2
(9-67)
(9-68)
• Equal masses: m1 = m2
then v1f = 0 and v2f = v1i
two bodies exchange velocities
• Massive target: m2 >> m1
then v1f ≈ -v1i en v2f ≈ (2m1/m2)v1i
(9-69)
body 1 bounces back with almost the same
speed, body 2 moves forward very slowly.
m1  m2
v1 f 
v1i
m1  m2
2m1
v2 f 
v1i
m1  m2
(9-67)
(9-68)
• Massive projectile: m1 >> m2
then v1f ≈ v1i en v2f ≈ 2v1i
(9-70)
body 1 moves forward with almost the same
speed, body 2 moves forward at double the
speed of body 1.
• Do Sample Problem 9-11, p223