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
(9-1)
motion for rockets
The Center of Mass:
Consider a system of two particles of masses m1 and m2
at positions x1 and x2 , respectively. We define the
position of the center of mass (com) as follows:
m1 x1 + m2 x2
xcom =
m1 + m2
We can generalize the above definition for a system of n particles as follows:
m x + m2 x2 + m3 x3 + ... + mn xn m1 x1 + m2 x2 + m3 x3 + ... + mn xn 1 n
xcom = 1 1
m1 + m2 + m3 + ... + mn
=
M
=
M
∑m x
i =1
i i
Here M is the total mass of all the particles M = m1 + m2 + m3 + ... + mn
We can further generalize the definition for the center of mass of a system of
particles in three dimensional space. We assume that the the i -th particle
r
( mass mi ) has position vector ri
n
r 1 r
rcom =
M
∑ mi ri
i =1
(9-2)
n
r 1 r
The position vector for the center of mass is given by the equation: rcom =
M
∑ mi ri
i =1
r ˆ
The position vector can be written as: rcom = xcomi + ycom ˆ + zcom k
ˆ j
r
The components of rcom are given by the equations:
n n n
1 1 1
xcom =
M
∑m x
i =1
i i ycom =
M
∑m y
i =1
i i zcom =
M
∑m z
i =1
i i
The center of mass has been defined using the quations
given above so that it has the following property:
The center of mass of a system of particles moves as though
all the system's mass were concetrated there, and that the
vector sum of all the external forces were applied there
The abolve statement will be proved later. An example is
given in the figure. A baseball bat is flipped into the air
and moves under the influence of the gravitation force. The
center of mass is indicated by the black dot. It follows a
parabolic path as discussed in Chapter 4 (projectile motion)
(9-3)
All the other points of the bat follow more complicated paths
The Center of Mass for Solid Bodies (9-4)
Solid bodies can be considered as systems with continuous distribution of matter
The sums that are used for the calculation of the center of mass of systems with
discrete distribution of mass become integrals:
1 1 1
xcom =
M ∫ xdm ycom =
M ∫ ydm zcom =
M ∫ zdm
The integrals above are rather complicated. A simpler special case is that of
dm M
uniform objects in which the mass density ρ = is constant and equal to
dV V
1 1 1
xcom = ∫ xdV ycom = ∫ ydV zcom = ∫ zdV
V V V
In objects with symetry elements (symmetry point, symmetry line, symmetry plane)
it is not necessary to eveluate the integrals. The center of mass lies on the symmetry
element. For example the com of a uniform sphere coincides with the sphere center
In a uniform rectanglular object the com lies at the intersection of the diagonals
C . C
z
Newton's Second Law for a System of Particles
Consider a system of n particles of masses m1 , m2 , m3, ..., mn
r r r r
O
and position vectors r1 , r2 , r3 ,..., rn , respectively.
x y
The position vector of the center of mass is given by:
r r r r r
Mrcom = m1r1 + m2 r2 + m3r3 + ... + mn rn We take the time derivative of both sides →
d r d r d r d r d r
M rcom = m1 r1 + m2 r2 + m3 r3 + ... + mn rn →
dt dt dt dt dt
r m r r r r r
Mvcom = m1v1 + m2 v2 + m3v3 + ... + mn vn Here vcom is the velocity of the com
F r F
m and v is the velocity of the i -th particle. We take the time derivative once more →
i m
Fd
r d r d r d r d r
M vcom = m1 v1 + m2 v2 + m3 v3 + ... + mn vn →
dt dt dt dt dt
r r r r r r
Macom = m1a1 + m2 a2 + m3a3 + ... + mn an Here acom is the acceleration of the com
r
and ai is the acceleration of the i -th particle
(9-5)
z r r r r r
Macom = m1a1 + m2 a2 + m3 a3 + ... + mn an
We apply Newton's second law for the i -th particle:
r r r
mi ai = Fi Here Fi is the net force on the i -th particle
O r r r r r
x y Macom = F1 + F2 + F3 + ... + Fn
r
The force Fi can be decomposed into two components: applied and internal
r rapp rint
Fi = Fi + Fi The above equation takes the form:
rapp rint rapp rint rapp rint rapp rint
r
( ) ( ) (
Macom = F1 + F1 + F2 + F2 + F3 + F3 + ... + Fn + Fn → ) ( )
r m rapp rapp rapp rapp rint rint rint rint
( ) (
Macom = F1 + F2 + F3 + ... + Fn + F1 + F2 + F3 + ... + Fn
F F
)
m r
The sum in the first parenthesis on the RHS of the equation above is just Fnet
m
F
The sum in the second parethesis on the RHS vanishes
by virtue of Newton's third law.
r r
The equation of motion for the center of mass becomes: Macom = Fnet
In terms of components we have:
Fnet , x = Macom , x Fnet , y = Macom, y Fnet , z = Macom, z (9-6)
r r
(9-7) Macom = Fnet Fnet , x = Macom, x
Fnet , y = Macom, y
Fnet , z = Macom, z
The equations above show that the center of mass of a system of particles
moves as though all the system's mass were concetrated there, and that the
vector sum of all the external forces were applied there. A dramatic example is
given in the figure. In a fireworks display a rocket is launched and moves under
the influence of gravity on a parabolic path (projectile motion). At a certain point
the rocket explodes into fragments. If the explosion had not occured, the rocket
would have continued to move on the parabolic trajectory (dashed line). The forces
of the explosion, even though large, are all internal and as such cancel out. The
only external force is that of gravity and this remains the same before and after the
explosion. This means that the center of mass of the fragments folows the same
parabolic trajectory that the rocket would have followed had it not exploded
v
m Linear Momentum
p r r
Linear momentum p of a particle of mass m and velocity v
r r
r r is defined as: p = mv
p = mv
The SI unit for lineal momentum is the kg.m/s
Below we will prove the following statement: The time rate of change of the linear
momentum of a particle is equal to the magnitude of net force acting on the
particle and has the direction of the force
r r
dp
In equation form: Fnet = We will prove that this equation using
dt
Newton's second law
r r
r r dp d r dv r r
p = mv → = ( mv ) = m = ma = Fnet
dt dt dt
This equation is stating that the linear momentum of a particle can be changed
only by an external force. If the net external force is zero, the linear momentum
cannot change r r
dp
Fnet = (9-8)
dt
z The Linear Momentum of a System of Particles
In this section we will extedend the definition of
linear momentum to a system of particles. The
r
O
y
i -th particle has mass mi , velocity vi , and linear
x r
momentum pi
We define the linear momentum of a system of n particles as follows:
r r r r r r r r r r
P = p1 + p2 + p3 + ... + pn = m1v1 + m2 v2 + m3v3 + ... + mn vn = Mvcom
The linear momentum of a system of particles is equal to the product of the
r
total mass M of the syetsm and the velocity vcom of the center of mass
m
r
p p r dP d r r r
= ( Mvcom ) = Macom = Fnet
m
The time ratemof change of P is:
dt dt
p r
The linear momentum P of a system of particles can be changed only
r r r
by a net external force Fnet . If the net external force Fnet is zero P cannot change
r
r r r r r r dP r
P = p1 + p2 + p3 + ... + pn = Mvcom = Fnet (9-9)
dt
Collision and Impulse
We have seen in the previous discussion that the momentum of an object can
change if there is a non-zero external force acting on the object. Such forces
exist during the collision of two objects. These forces act for a brief time
interval, they are large, and they are responsible for the changes in the linear
momentum of the colliding objects
Consider the collision of a baseball with a baseball bat
The collision starts at time ti when the ball touches the bat
and ends at t f when the two objects separate
r
The ball is acted upon by a force F (t ) during the collision
The magnitude F (t ) of the force is plotted versus t in fig.a
The force is non-zero only for the time interval ti < t < t f
r r
dp r
F (t ) = Here p is the linear momentum of the ball
dt
tf tf
r r r r
dp = F (t )dt → ∫ dp = ∫ F (t )dt (9-10)
ti ti
tf tf tf
r r r r r r
∫ dp = ∫ F (t )dt
ti ti
∫ dp = p f − pi = ∆p = change in momentum
ti
tf
r r
∫ F (t )dt is known as the impulse J of the collision
ti
r tf r r
J = ∫ F (t )dt The magnitude of J is equal to the area
ti
r r
under the F versus t plot of fig.a → ∆p = J
In many situations we do not know how the force changes
with time but we know the average magnitude Fave of the
collision force. The magnitude of the impulse is given by:
J = Fave ∆t where ∆t = t f − ti
Geometrically this means that the the area under the
∆p = J F versus t plot (fig.a) is equal to the area under the
J = Fave ∆t Fave versus t plot (fig.b)
(9-11)
Series of Collisions
Consider a target which collides with a steady stream of
r
identical particels of mass m and velocity v along the x-axis
A number n of the particles collides with the terget during a time interval ∆t.
Each particle undergoes a change ∆p in momentum due to the collision with
the target. During each collision a momentum change − ∆p is imparted on the
target. The Impulse on the target during the time interval ∆t is:
J = −n∆p The average force on the target is:
J −n∆p n
Fave = = = − m∆v Here ∆v is the change in the velocity
∆t ∆t ∆t
of each particle along the x-axis due to the collision with the target →
∆m ∆m
Fave = − ∆v Here is the rate at which mass collides with the target
∆t ∆t
If the particles stop after the collision then ∆v = 0 − v = −v
If the particles bounce backwards then ∆v = −v − v = −2v (9-12)
z Conservation of Linear Momentum
r
Consider a system of particles for which Fnet = 0
r
dP r r
= Fnet = 0 → P = Constant
O
y
dt
x r
If no net external force acts on a system of particles the total linear momentum P
cannot change
total linear momentum total linear momentum
at some initial time t = at some later time t
i f
The conservation of linear momentum is an importan principle in physics.
m
m It also provides a powerful rule we can use to solve problems in mechanics such as
p p
m
collisions.
p r
Note 1: In systems in which Fnet = 0 we can always apply conservation of linear
momentum even when the internal forces ared vely large as in the case of
colliding objects
Note 2: We will encounter problems (e.g. inelastic collisions) in which the energy
is not conserved but the linear momentum is (9-13)
Momentum and Kinetic Energy in Collisions
Consider two colliding objects with masses m1 and m2 ,
r r r r
initial velocities v1i and v2i and final velocities v1 f and v2 f ,
respectively
r
If the system is isolated i.e. the net force Fnet = 0 linear momentum is conserved
The conervation of linear momentum is true regardless of the the collision type
This is a powerful rule thet allows us to determine the results of a collision without
knowing the details. Collisions are divided into two broad classes: elastic and
inelastic.
A collision is elastic if there is no loss of kinetic energy i.e. K i = K f
A collision is inelastic if kinetic energy is lost during the collision due to conversion
into other forms of energy. In this case we have: K f < K i
A special case of inelastic collisions are known as
completely inlelastic. In these collisions the two colliding objects stick together
and they move as a single body. In these collisions the loss of kinetic energy
is maximum (9-14)
One Dimensional Inelastic Collisions
In these collisions the linear momentium of the colliding
r r r r
objects is comserved → p1i + p2i = p1 f + p2 f
r r r r
m1v1i + m1v2i = m1v1 f + m1v2 f
One Dimensional Completely Inelastic Collisions
In these collisions the two colliding objects stick together
and move as a single body. In the figure to the leftwe show
r
a special case in which v2i = 0. → m1v1i = m1V + m1V →
m1
V= v1i
m1 + m2
The velocity of the center of mass in this collision
r r r r
r P p1i + p2i m1v1i
is vcom = = =
m1 + m2 m1 + m2 m1 + m2
In the picture to the left we show some freeze-frames
of a totally inelastic collition
(9-15)
One-Dimensional Elastic Collisons
Consider two colliding objects with masses m1 and m2 ,
r r r r
initial velocities v1i and v2i and final velocities v1 f and v2 f ,
respectively
Both linear momentum and kinetic energy are conserved.
Linear momentum conservation: m1v1i + m1v2i = m1v1 f + m1v2 f (eqs.1)
2 2
m1v12i m1v2i m1v1 f m2 v2 f
2
Kinetic energy conservation: + = + (eqs.2)
2 2 2 2
We have two equations and two unknowns, v1 f and v1 f
If we solve equations 1 and 2 for v1 f and v1 f we get the following solutions:
m1 − m2 2m2
v1 f = v1i + v2i
m1 + m2 m1 + m2
2m1 m − m1
v2 f = v1i + 2 v2i
m1 + m2 m1 + m2
(9-16)
Special Case of elastic Collisions-Stationaly Target ( v2i = 0 )
The substitute v2i = 0 in the two solutions for v1 f and v1 f
m1 − m2 2m2 m − m2
v1 f = v1i + v2i → v1 f = 1 v1i
m1 + m2 m1 + m2 m1 + m2
2m1 m − m1 2m1
v2 f = v1i + 2 v2i → v2 f = v1i
m1 + m2 m1 + m2 m1 + m2
Below we examine several special cases for which we know the outcome
of the collision form experience
1. Equal masses m1 = m2 = m
x
m1 − m2 m−m m m
v1 f = v1i = v1i = 0
m1 + m2 m+m x
2m1 2m m m
v2 f = v1i = v1i = v1i
m1 + m2 m+m
The two colliding objects have exchanged velocities
(9-17)
m1
2. A massive target m2 ? m1 → =1
m2
x m1
m1 −1
m2 m − m2 m
v1 f = 1 v1i = 2 v1i ≈ −v1i
x m1 + m2 m1
m1 +1
m2 m2
m
2 1
v2 f =
2m1
v1i = m2 v ≈ 2 m1 v
m1 1i 1i
m1 + m2 +1 m2
m2
Body 1 (small mass) bounces back along the incoming path with its speed
v
v =0
practically unchanged.
Body 2 (large mass) moves forward with avery small
v
v m1
speed because =1
m2
(9-18)
m2
2. A massive projectile m1 ? m2 → =1
m1
x m2
m2 1−
m1 m1 − m2 m1
v = v1i = v ≈v
x 1f m1 + m2 m2 1i 1i
m2 1+
m1 m1
2m1 2
v2 f = v1i = v ≈ 2v1i
m1 + m2 m2 1i
1+
m1
v Body 1 (large mass) keeps on going scarcely slowed by the collision .
Body 2 (small mass) charges ahead at twice the speed of body 1
v =0
v
v
(9-19)
Collisions in Two Dimensions
(9-20)
In this section we will remove the restriction that the
colliding objects move along one axis. Instead we assume
that the two bodies that participate in the collision
move in the xy -plane. Their masses are m1 and m2
r r r r
The linear momentum of the sytem is conserved: p1i + p2i = p1 f + p2 f
If the system is elastic the kinetic energy is also conserved: K1i + K 2i = K1 f + K 2 f
We assume that m2 is stationary and that after the collision particle 1 and
particle 2 move at angles θ1 and θ 2 with the initial direction of motion of m1
In this case the conservation of momentum and kinetic energy take the form:
x − axis: m1v1i = m1v1 f cos θ1 + m2 v2 f cos θ 2 (eqs.1)
y − axis: 0 = − m1v1 f sin θ1 + m2 v2 f sin θ 2 (eqs.2)
1 1 1
m1v12i = m1v2 f + m2 v2 f (eqs.3) We have three equations and seven variables:
2 2
2 2 2
Two masses: m1 , m2 three speeds: v1i , v1 f , v2 f and two angles: θ1 , θ 2 . If we know
the values of four of these parameters we can calculate the remaining three