The law of conservation of momentum is the second of the
major conservation laws that we'll use this year. The law in its most
general form is written in a way that looks similar to the law of
conservation of energy.
This looks suspiciously like the impulse-momentum theorem. However, there's an important difference. The impulse-momentum theorem applies to a single object. The object could, for example, be a ball colliding with the floor. In that case, Fnet Δt = Δp, where Fnet is the net force on the ball, and Δp is the change of momentum of the ball. The equation above applies to a system of objects. The system could, for example, be two colliding objects. In that case, Fnet,ext represents the net, external force acting on the system, and Δpnet represents the total change of momentum of all the objects in the system. We list important points below.
Conservation of momentum is a vector law. Therefore, one must apply
the law independently along each coordinate axis.
Fnet,ext is the vector sum of all the external forces acting on the system. Δt is the amount of time that the interaction lasts.
Δpnet is the vector sum of the momenta of all the objects in the system.
The system is usually selected to be one for which Fnet,ext = 0.
Such a system is termed isolated.
In all of the problems that you will do with conservation of
momentum in this course, you will be able to define the system in such a way
that Fnet,ext = 0.
In that case, the law of conservation of momentum can be written as follows.
We will simplify the notation by letting upper case P represent the net momentum of the system. Therefore, we have for an isolated
system the following.
In two dimensions, the vector equation becomes two component
Pix = Pfx
Piy = Pfy
For 1-dimensional problems, we omit the x or y subscript.
One selects the system so
that the net, external force on the system is 0.
Momentum is treated as a vector. Therefore, direction
Now for some examples.
Two ice skaters push off against one another starting from a stationary
position. The 45-kg skater acquires a velocity of 0.375 m/s to the right.
What velocity does the 60-kg skater acquire?
v1i = v2i = 0
v2f = +0.375 m/s
m1 = 60 kg
m2 = 45 kg
System: ice skaters
External forces: friction (assumed to be
negligible), normal and weight add to 0
Initial state: skaters motionless ready to push apart
Final state: skaters moving in opposite directions
||The skaters exert
internal forces on each other, but the external forces of normal and
weight balance. We're assuming no friction. Actually, if we
look at the situation immediately after they push off and haven't
separated far, then any influence due to friction will be minimal.
Therefore, we've selected the system so that Fnet,ext = 0.
In the diagram, note that:
- +x is to the right.
- The skaters are denoted 1 and 2 to
- The initial and final velocities
have double subscripts. One subscript is for the object and the
other is for the state. (The textbook inserts a comma between
||We use upper-case P to
denote the total momentum of the system and lower-case p to denote the
momentum of a single object. The total momentum is the sum of the
individual momenta of the objects in the system.
|The total initial
momentum of the system is 0, because the skaters are motionless.
|Substitute the definition
of momentum .
|Solve for the unknown.
|Substitute values and
units and reduce. The sign of the result is negative, indicating
motion to the left as expected. The magnitude of the velocity of
skater 1 is less than skater 2. This is also expected, because
skater 1 has the greater mass.
Example 2: A 0.60-kg
glider traveling at 8.0 m/s on a level air track undergoes a head-on
collision with a 0.20-kg mass traveling toward it at 4.0 m/s. The two
gliders stick in the collision. What is the velocity of the combined
gliders after the collision?
v1i = +8.0 m/s
v2i = -4.0 m/s
m1 = 0.60 kg
m2 = 0.20 kg
Goal: Find vf
External forces: friction with track (assumed to be negligible),
normal and weight add to 0
Initial state: gliders moving toward each other before
Final state: gliders moving together after collision
||We have to be careful to
assign the correct signs to the initial velocities.
There is only one final velocity, so a
single subscript is all that is needed on that velocity.
The direction of vf isn't
given, but we're guessing it's to the right. The sign of the final
result will tell us whether that choice is correct.
||We proceed as in the last
example setting the total initial and final momenta equal to each other
and then writing each as the sum of the momenta of the objects in the
|The mass of the combined
gliders is m1 + m2. They have the same final
|Solve for the unknown.
|Substitute values and
units and reduce. The sign of the result is positive, indicating
motion to the right as we had guessed.
Example 3. A glider with mass m and speed v moving along a horizontal, frictionless air track collides with a stationary glider
with a mass m/3. After the collision the first glider has a speed v/2.
What is the velocity of the second glider?
v1i = v
v2i = 0
v1f = v/2
m1 = m
m2 = m/3
Goal: Find v2f
External forces: friction with track (assumed to
be negligible), normal and weight add to 0
Initial state: gliders before collision
Final state: gliders after collision
||In order to solve this problem, we
need to be given both masses and 3 velocities. That leaves the fourth
velocity as the only unknown. Later, we'll see how to solve
problems like this when both of the final velocities are unknown.
||After the collision, both
gliders are moving to the right. Glider 1 is slower than glider 2.
Therefore, the gliders separate after the collision.
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