Conservation of Momentum
Let's set up this situation first: two people, on skates, are in the center of an ice rink
facing each other. One is 140 lbs the other weighs 280 lbs. What happens to them if the heavy
guy pushes off the other, smaller guy?
Well, the lighter person will move away from the heavier person with the greatest velocity.
Correspondingly, the heavy person will move away from the lighter person but much slower. In
other words, they will move in opposite directions with different velocities.
If we analyze this "system" we see that the force that the 280 lb guy pushes with is the same
force he himself feels (in the opposite direction)...what law is that?
Newton's 3rd law: for every action there is an equal but opposite reaction.
So, we can now write:
or just,
Force on guy 2 = - Force on guy 1
F2 = - F1
That's interesting, but we can't get much further insight unless we consider the TIME over which
the force was administered. By pure reasoning we can say that if the heavy guy pushes on the
little guy for 2 seconds, he himself will experience the force for...2 seconds also. So the time of
the forces is the same for both. This allows us multiply the above force relation by "t", or time,
because it's just a constant:
F2(t) = - F1(t)
Do you recognize these terms?
I know you will immediately recognize these as IMPULSE (force x time)...(well, I'm always
hopeful). This means we can gain some insight by substituting from the Impulse-Momentum
Equation (which you all know):
F2(t) = m2v2f - m2v2i
and -F1(t) = -[m1v1f - m1v1i],
then
m2v2f - m2v2i = -[m1v1f - m1v1i]
m2v2f - m2v2i = -m1v1f + m1v1i
We'll put all the INITIAL terms on the left and all the FINAL terms on the right:
m1v1i + m2v2i = m1v1f + m2v2f
This is the famous Conservation of Momentum Law. In words, it says:
m1v1i + m2v2i
The totaled momentum
of all the objects in the
“system” (for example,
the “system” could be
our two skaters)
=
m1v1f + m2v2f
=
The momentum at some
other time
Numerical Example:
QUESTION: A 450 kg cannon fires an 8 kg shell at 110 m/s. What is the recoil velocity of the
cannon?
ANSWER: If we imagine the cannon on a frictionless track, when the charge fires, the cannon
will recoil with the velocity we calculate.
m1v1i + m2v2i = m1v1f + m2v2f
The initial velocity of the cannon is ZERO. So is the initial velocity of the shell, so
0
+ 0
= m1v1f + m2v2f
The final velocity of the shell is 110 m/s, so....
0
+ 0
= (8 kg)(110 m/s) + (450 kg)(v2f)
0
+ 0
= 880 kgm/sec + 450v2f
- 880 kgm/sec
-880 kgm/sec / 450
= 450v2f
= v2f
v2f = - 1.96 m/s
What are the limitations of the conservation of momentum law?
The law breaks down in the presence of “external” forces. Three definitions are in order:
1. Closed System - For our purposes this will constitute two objects that are isolated from outside
influences, enumerated below as ‘external’ forces.
2. External forces - forces caused by influences external to the "system"; examples are friction,
air resistance, gravity. These forces introduce unbalanced losses in energy that invalidate the
conservation law.
3. Internal forces - forces that only act BETWEEN the objects in the system; such as the 'push'
given by one skater to the other, the explosive force between the cannon and a cannon ball.
These forces do not alter the total momentum of the two objects in the 'system'. Friction CAN be
present between the objects, but only between the objects.
Why? Internal forces are forces between the two objects that collide; they act on both
objects equally and in opposite directions and cancel out. What one object loses the other
gains. So momentum gets conserved. But air resistance for example, as an external force, acts
between an object and its surroundings. So that object loses its momentum to the air (as
heat)...and not to another other object. The heat loss isn’t recoverable.