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Chapter 9
Linear Momentum and Collisions
Consider an isolated system consisting of just two
particles. If a force from particle 1 acts on
particle 2, then a force of the same magnitude
from particle 2 acts on particle 1 (by Newton’s
third law).
v1
m1
F21
F12
m2
v2
Two isolated particles interacting with each other. Newton’s third law ensures
that F12= -F21.
Some math...
F21  F12  0
m1a1  m2a2  0
dv1
dv 2
m1
 m2
0
dt
dt
d ( m1v1 ) d ( m2 v 2 )

0
dt
dt
d
( m1v1  m2 v 2 )  0
dt
What can we say about m1v1 + m2v2 if its derivative is zero?
m1v1 + m2v2 is a constant as long as the
system remains isolated.
Definition: The linear momentum of a particle or an object
that can be modeled as a particle of mass m moving with
a velocity v is defined to be the product of the mass and
velocity.
p  mv
We can write Newton’s second law in terms of momentum.
dv d
dp
 F  ma  m dt  dt (mv)  dt
The time rate of change of the linear momentum of a particle is
equal to the net force acting on the particle.
Conservation of momentum
d
( m1v1  m2 v 2 )  0
dt
Therefore, if two (or more) particles in an isolated system
interact, the total momentum of the system remains
constant. In other words, the total momentum of an
isolated system at all times equals its initial momentum.
p1i + p2i = p1f + p2f
pix = pfx
piy = pfy
piz = pfz
Practice Problem: Archery on ice
A 60.0 kg archer stands at rest on frictionless ice
and fires a 0.50 kg arrow horizontally at 50 m/s.
With what velocity does the archer move across
the ice after firing the arrow?
Impulse and momentum
Suppose a single force F acts on a particle. Then
we know F = dp/dt, or
dp  Fdt
which we can integrate to figure out the change in
momentum when the force acts over some time
interval.
The impulse-momentum theorem
tf
I  p   Fdt
ti
The impulse of the force F acting on a particle equals the
change in momentum of the particle. Since the force imparting
an impulse generally varies with time, we can define impulse in
terms of the time-averaged force.
tf
1
F
Fdt

t ti
Therefore,
This is the mean value theorem of calculus.
t = tf – ti,
I  Ft
Often times, we can approximate the force acting on an
object as constant, especially if the force acts for a short
time. Then, I = Ft. This is called the impulse
approximation, in which we assume that one of the
forces exerted on a particle acts for a short time but
is much greater than any other force present.
Example: The collision of a baseball and the bat. The
time of the collision is very short (about 0.01 s) and
the average force the bat exerts on the ball is very
large (1000s of Newtons). Therefore we are justified
in ignoring the gravitational force when analyzing the
collision using the impulse-momentum theorem.
Quick Quiz
Two objects are at rest on a frictionless surface. Object 1
has a greater mass than object 2. When a constant
force is applied to object 1, it accelerates through a
distance d. The force is removed from object 1 and
applied to object 2. At the moment when object 2 has
accelerated through the same distance d, which
statements are true?
a)
b)
c)
d)
e)
f)
p1 < p2
p1 = p2
p1 > p2
K1 < K2
K1 = K2
K1 > K2
Quick Quiz
Two objects are at rest on a frictionless surface. Object 1
has a greater mass than object 2. When a constant
force is applied to object 1, it accelerates for a time
interval t. The force is then removed and applied to
object 2. After object 2 has accelerated for the same
time interval t, which statements are true?
a)
b)
c)
d)
e)
f)
p1 < p2
p1 = p2
p1 > p2
K1 < K2
K1 = K2
K1 > K2
Practice problem
A ball of mass 0.150 kg is dropped from rest from
a height of 1.25 m. It rebounds from the floor to
reach a height of 0.960 m. What impulse was
given to the ball by the floor?
Collisions in one dimension
An elastic collision between two objects is one in which
the total kinetic energy as well as total momentum of the
system is the same before and after the collision.
An inelastic collision is one in which the total kinetic
energy of the system is not the same before and after
the collision, even though the total momentum is
conserved.
Perfectly inelastic: the objects stick together after the
collision
Inelastic: the objects do not stick together
The momentum of the system is conserved in all collisions, but
the kinetic energy is conserved only in elastic collisions.
The perfectly inelastic collison
Before Collision
m1
v1i
v2i
m2
After Collision
m1
m2
vf
m1v1  m2 v 2  (m1  m2 ) v f
The perfectly elastic collision
m1
v1i
v2i
v1f
m1
m2
Before Collision
m2
v2f
m1v i1  m2 v i 2  m1v f 1  m2 v f 2
1
1
1
1
2
2
2
m1v1i  m2v2i  m1v1 f  m2v22 f
2
2
2
2
After Collision
(conservation of momentum)
(conservation of energy)
Quick Quiz
In a perfectly inelastic one-dimensional collision
between two objects, what condition alone is
necessary so that all of the original kinetic
energy of the system is gone after the
collision?
a) The objects must have momenta with the
same magnitude but opposite direction
b) The objects must have the same mass
c) The objects must have the same velocity
d) The objects must have the same speed, with
velocity vectors in opposite directions
The ballistic pendulum can be used to measure the
speed of a fast-moving projectile, such as a bullet. A
bullet of mass m1 is fired into a large block of wood of
mass m2 suspended from some wires. The bullet gets
stuck in the block, and the entire system swings through
a height h. Find the speed of the bullet in terms of h and
the two masses.
Two dimensional collisions
p1i + p2i = p1f + p2f
pix = pfx
piy = pfy
piz = pfz
The key to solving problems involving two dimensional
collisions is to realize that the momentum in each
individual direction is conserved.
Practice Problem
An unstable atomic nucleus of mass 17.0 X 10-27 kg
initially at rest disintigrates into three particles. One
of the particles, of mass 5.00 X 10-27 kg, moves in the
y-direction with a speed of 6.00 X 106 m/s. Another
particle of mass 8.40 X 10-27 kg, moves in the xdirection with a speed of 4.00 X 106 m/s.
a)Find the velocity of the third particle
b)Find the total kinetic energy increase in the
process.
Trickier practice problem
After a completely inelastic collision, two
objects of the same mass and same initial
speed move away together at half their
initial speed. Find the angle between the
initial velocities of the objects.
Hint: Choose your x and y directions
carefully!
The center of mass
y
m1
m2
...
mi
...
mN
x
Suppose we have N number of particles of different masses lined up on the x-axis as
shown. We define the center of mass as
N
xcm
m1 x1  m2 x2  ...  mN xN


m1  m2  ...mN
m x
i i
i 1
M
where M is the total mass of the system of particles. This can easily be
extended to 2-D and 3-D by computing ycm and zcm.
Practice Problem
Find the center of mass of the three particles shown in the figure:
p1 = (3.0, 2.0), m1 = 6.0 kg
p2 = (4.0, -4.0), m2 = 2.0 kg
8
p3 = (-2.0, 7.0), m3 = 4.0 kg
6
4
2
0
-2
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
Example: Earth-moon barycenter
Find the center of mass of the Earth-moon system.
ME= 5.98 X 1024 kg, MM = 7.36 X 1022 kg, and their
centers are 3.84 X 105 km apart. This point is
called the barycenter and the motion of the Earth
and moon around it has important consequences
for studying phenomena such as tides.
Symmetry
Symmetry can make calculating the center
of mass a whole lot easier.
(-x3,-y3)
(-x2,-y2)
(-x1,-y1)
(x1,y1)
(x2,y2)
(x3,y3)
Consider 6 particles all of mass m
arranged in a circle. Symmetry
ensures that everything cancels out
if we choose the origin at the center
of the circle. The center of mass is
at the center of the circle.
Motion of the center of mass
In general,

1
rcm 
M

 miri
i

ˆ
Where, rcm  xcmˆi  ycmˆj  zcmk
is the displacement vector of the center of mass.
By differentiating we can determine the velocity and
acceleration of the center of mass.

1
v cm 
M

 mi vi
i

1
acm 
M

 miai
i
Notice that these are vector equations, which means we can
break them up into components.
Suppose Fi is the sum of all the forces acting on the ith particle in our
system. Then, by Newton’s second law, Fi = miai.

1
acm 
M


1  
i miai  M (F1  F2  ...  FN )
 


Macm  F1  F2  ...  FN
Remember, each F is the sum of all the
forces acting on that particular particle
(1, 2, ..., N).
Now, we do something clever. We will break up the sum of the forces on
each particle into the sum of the external forces on that particle, FiE, and the
forces exerted by the other particles in the system. For instance,
 E 


F1  F1  F12  F13  ...F1N
So we can write...
 

E 


Macm  F1  F2  ...  FN  (F1  F12  ...F1N )  ...
E 

 ( FN  FN 1  FN 2  ...  FN ( N 1) )
And since we’re dealing with a sum we can rearrange like so...
E E
E
 
 



Macm  (F1  F2  ...FN )  (F12  F21)  (F13  F31)  ...  (FN ( N 1)  F( N 1) N )
I believe my third law of motion can help
simplify things here.
E E
E
 External

Macm  F1  F2  ...FN   F
The center of mass of a system of particles acts as if it were a
particle of mass M moving under the influence of just the
external forces on the system. This is true in all cases. The
particles could be a blob of silly putty, a gas, or a rigid body. It
can be anything!
Example: Two colliding asteroids
Two asteroids are initially at rest relative to each other. Due to
gravitational attraction, they eventually collide. If external forces are
negligible, where do the asteroids collide?
d = 1.0 X 107 m
M1 = 8.0 X 1010 kg
M2 = 2.0 X 1010 kg
Example: The exploding rocket
A rocket is fired vertically upward. At the instant it reaches an altitude
of 1000 m and a speed of 300 m/s, it explodes into three fragments
having equal mass. One fragment continues to move upward with a
speed of 450 m/s. The second fragment has a speed of 240 m/s
and is moving east right after the explosion. What is the velocity of
the third fragment right after the explosion?
Example: a proton-proton collision
A proton collides elastically with another proton that is initially at rest.
The incoming proton has an initial speed of 3.50 X 105 m/s and
makes a glancing collision with the second proton*. After the
collision, one proton moves off at an angle of 370 to the original
direction of motion, and the second deflects at an angle of  to the
same axis. Find the final speeds of the two protons and the angle
.
v1f
Before
After
v1i = 3.50 X 105 m/s
370

v2f
* The two protons do not actually come into direct contact with one another. They
interact through a repulsive electric force.