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Impulse and Momentum
Chapter 7
Expectations
After chapter 7, students will:
 be able to define and calculate the impulse of a
force.
 be able to define and calculate the momentum of
a moving object.
 understand and apply the impulse-momentum
theorem.
 understand the conditions in which the
momentum of a system is conserved.
Expectations
After chapter 7, students will:
 distinguish between elastic and inelastic
collisions.
 know what quantities are conserved in what sorts
of collisions.
 be able to calculate the location and velocity of
the center of mass of a system of objects.
Impulse and Momentum
In the last chapter, we defined a force-related
quantity (work) and a motion-related quantity
(kinetic energy), and showed how they were
linked:

1
2
2
W  Fs cos   KE  m v f  v0
2
This was the work-energy theorem.

Impulse and Momentum
Now, we define a new force-related quantity: the
 
impulse of a force.
I  Ft
The impulse is the product of the net force and the
time over which it is exerted.
It is a vector quantity, having the dimensions of
force · time. SI units: N·s
Impulse and Momentum
The net force F, acting on an object of mass m,
produces an acceleration:
F  ma
Substituting that into the definition of impulse:
Solve for at:
I  Ft  mat
I
at 
m
Impulse and Momentum
I
at 
m
Substitute into the kinematic equation:
I
v f  v0  at  v0 
m
Multiply through by m:
mv f  mv0  I 
I  mv f  mv0
Impulse and Momentum
I  mv f  mv0
Again, we have a relationship between a forcerelated quantity, I, and a motion-related quantity,
mv.
We define this quantity as linear momentum:


p  mv
SI units: kg·m/s
Note that linear momentum is a vector quantity.
Impulse and Momentum


 


I  mv f  mv0  p f  p0  p
The above relationship is called the impulsemomentum theorem: the impulse of a net force
acting on an object is equal to the resulting
change in its linear momentum.
 


I  p f  p0  p
Impulse and Momentum
W  Fs cos  KE
 

I  Ft  p
Bottom line: a net force changes an object’s state of
motion. The work-energy theorem and the
impulse-momentum theorem are two different
descriptions of the same change.
Which description is more useful to you depends on
the situation: what things you know, and what
things you want to find out.
Systems of Objects
A system of objects is any collection of objects.
You define the system. You can say what objects
are part of it, and what objects are outside of it.
Isolated Systems
A system of objects is called isolated if the net
external force acting on the system is zero.
An external force is a force exerted on the system
(any object in it) by an object that is not a part of
the system.
A force could be external or internal, depending on
how the system is defined.
Conservation of Momentum
The conservation of linear momentum principle
tells us:
If a system is isolated, its total linear momentum
does not change.
Why not?
Conservation of Momentum
The only force experienced by any part of an
isolated system is an internal force – a force
exerted on one part of the system by another.
Newton’s third law tells us:


FAB   FBA
If objects A and B are both part of the system, then
the net force on the system as a whole is





Ftot  FAB  FBA  FAB   FAB  0


Conservation of Momentum
If the net force is zero, its impulse is zero:


I tot  Ftott  0
... and if the impulse is zero, the change in (total)
momentum is zero:


ptot  I tot  0
... and the total linear momentum is conserved.
Conservation of Momentum
We said the change in total (i.e., system) momentum
was zero. Notice that the linear momentum of
any particular object within the system will, in
general, change due to internal forces. But the
sum of the individual momenta of the objects in
the system, added as the vectors that they are,
does not change – as long as the system is
isolated.
Collisions
If we define a system of two or more objects, it is
usually because those objects are busy colliding
with each other.
In every collision among the objects in an isolated
system, the total linear momentum of the system
is conserved. We established that earlier, with the
linear momentum conservation principle.
Elastic Collisions
We now want to consider a special class of
collisions: elastic collisions. In an elastic
collision between objects in an isolated system,
the total kinetic energy of the system is also
conserved.
Why should that be?
Elastic Collisions
The work-energy theorem tells us that the total or
net work due to conservative forces does not
change the total energy of the system.
The gravitational potential energy of the system
does not change unless the system experiences a
net (external) gravitational force ... in which case,
it is no longer an isolated system.
Elastic Collisions
This means that, if only conservative forces do work
within the system, its kinetic energy must be
constant ... since we already know its potential
energy is constant.
What conservative forces do work within the
system?
Spring forces (as long as deformations are elastic).
Elastic Collisions
If objects in the system are elastically deformed by
the forces they exert on each other, the work done
by those forces is stored as elastic potential
energy. When the objects’ pre-collision shapes
are restored, that elastic potential energy is reconverted into kinetic energy.
So, the total kinetic energy of the system after the
elastic collision is the same as before the
collision.
Collision Summary
If the system is isolated (no net external force acts
on the system), momentum is conserved:


p f  p0
If only conservative forces do work in the collision,
the collision is elastic, and kinetic energy is
conserved:
KE f  KE0
Collision Summary
Collisions are not, in general, either perfectly elastic
or perfectly inelastic. Those two conditions are
idealizations that can give us insights into real
collisions.
Center of Mass
A system of several objects has its total mass
distributed among those objects.
When we consider the conservation of a system’s
total momentum, it will be useful to us to define a
single location where the system’s total mass is
effectively located: the center of mass.
Center of Mass
Consider a system of two objects, both located on
the X axis.
xCM
m1 x1  m2 x2

m1  m2
If m1 = m2 = m, xCM
mx1  mx2 mx1  x2  x1  x2



mm
2m
2
Center of Mass
Now, we let the objects in our system move
(independently):
xCM
m1x1  m2 x2

m1  m2
Center of Mass
The motion takes place in a time t:
x1
x2
m1
 m2
xCM
t
t

t
m1  m2
or, vCM
m1v1  m2v2
: velocity of the center of mass

m1  m2
Center of Mass
Notice that the center-of-mass velocity is the total
momentum of the system, divided by its total mass.
If the system is isolated,
vCM f  vCM 0
its total momentum is
constant … and so is its
… for isolated systems!
total mass. So, the
center-of-mass velocity of an isolated system is
constant, regardless of intra-system collisions.