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Momentum
The momentum of a body is its mass times its
velocity:
p = mv
.
Suppose there is more than one body on our
track. We then have a many-body system.
1
2
3
4
In such systems, the concept of momentum
becomes particularly useful. Let’s see why.
In general, two kinds of forces can act on the
bodies in the system. There are internal forces between the bodies, and forces due to external
agents acting on individual bodies.
In what follows, we will make a specific assumption about the nature of the forces acting between
bodies. We will assume that they satisfy
Newton’s “third law”:
The force of body j on body i is equal in magnitude and opposite in direction to the force of
body i on body j.
(Here, i and j stand for numbers which label the
bodies.) This means that if the sub-system
containing only those two bodies were placed in
isolation (no external forces acting), then no net
force would act on it.
Although this may seem obvious, it does not hold
for all types of force. Newton’s third law is not a
law at all; it is just a description of a “nice” kind
of force.
We’ll symbolize the force on body i due to body
j by Fij. In pictures, the third law looks like this:
F 13 1
2
3 F 31
4
and in symbols, like this:
F ij = − F ji .
We’ll assume that a body exerts no force on
itself, so F ii =0. (This is contained in the above
equation.)
Let’s also let F i be the external force acting on
body i. We make no assumptions about its
nature. Then Newton’s second law applied to
body i reads
m i ai = F i + 3 Fij .
j
system is zero. Then we have a conservation
law:
The right-hand side is the sum of the external
force and all the internal forces acting on body i.
Now let’s consider the total momentum of the
system; it is the sum of the individual momenta:
p tot = 3 m iv i .
i
How does the total momentum change in time?
d
3 m v = 3 m a =3 F + 3 F
dt i i i i i i i i i,j ij .
The second term on the right-hand side is zero
because the forces, taken in pairs, cancel out by
Newton’s “third law”. Hence, we find
dptot
dt
= Fext
,
where the sum of the external forces acting on
the system is
F ext = 3 F i .
dp tot
dt
=0.
This is not as trivial as it might seem, because it
is true even when the internal forces are nonzero.
The only thing that can change the total momentum is a nonzero total external force on the
system.
Example: collision of two bodies that stick
together
As our first example, suppose we have two
bodies on the track. They move freely, collide
and stick together, and the resulting body moves
freely once again. The question is: what is the
velocity of the resulting body?
Let’s draw two diagrams showing the velocities
before and after the collision:
m1 v1
m2
M
v2
w
i
The internal forces have no effect on the rate of
change of the total momentum. Now let’s
suppose that the total external force acting on the
We’ll let w be the velocity of the final body, and
M=m1+m2 be its mass. For consistency, we draw
all velocities in the positive direction. If a
velocity happens to turn out to be negative, then
it points in the direction opposite to that shown in
the diagrams.
The statement that the bodies move freely means
that no external forces act. Hence, the total
momentum is conserved.
m 1 v 1 + m2 v 2 = M w .
Solving for the final velocity w, we find
w=
m1 v 1 + m2 v 2
.
M
You might be wondering: the law of conservation
of momentum is all we needed, in order to solve
for w. But what about conservation of energy?
The answer is that kinetic energy appears not to
be conserved, if we analyze the process at this
level. To show this, simply calculate the final
kinetic energy minus the initial kinetic energy.
You will find
m m
ä1
ë
1
1
M w 2 − åååå m 1 v 21 + m 2 v 22 ìììì = − 1 2 (v 1 − v 2 )2 ,
2
2M
2
ã2
í
which is negative. Kinetic energy is lost. Where
does it go? It must go into heat energy in the
composite final system.
The bodies making up the final system are
macroscopic - they are made up of many atoms
held together by interatomic forces. When the
bodies collide and stick together, the atoms are
jostled about and begin to move faster than
before. This kind of energy is called heat.
Experimentally, we would detect a rise in the
temperatures of the bodies.
If we looked on this finer scale, considering the
kinetic energies of all the atoms making up the
composite final system, we would find that the
total kinetic energy is indeed conserved.
Example: elastic collision of two free bodies
Now let’s suppose that the two bodies, instead of
sticking together, bounce off one another. We’ll
assume that the collision is an ideal, elastic
collision. This means that no energy is lost
during the collision. (Our previous example was
a completely inelastic collision.) We’ll also
assume that the masses don’t change during the
collision.
We first draw the appropriate diagrams:
m1 v1
m1
m2
w1
m2
v2
w2
Velocities after the collision are denoted by w’s.
Suppose the initial velocities are known, and we
want to solve for the final velocities. No external
forces act, so we may invoke conservation of
total momentum and of total energy.
Conservation of momentum:
before = after
m 1 v 1 + m2 v 2 = m 1 w 1 + m2 w 2
Conservation of energy:
before = after
m1 2
m
m
m
v 1 + 2 v 22 = 1 w 21 + 2 w 22
2
2
2
2
These are two equations in two unknowns, the
w’s, so we have enough information to find a
solution. The equations are quadratic, so there
will be two solutions. Some very tedious algebra
reveals them to be
w 1 = v 1 and w1 =
(m 1 − m 2 )v 1 + 2m 2 v 2
,
m1 + m2
with w2 given by the same equations with 1 and 2
interchanged everywhere.
What is the physical significance of these two
solutions? Well, the first one just says that body
1 sails straight through without changing
velocities at all. This is a valid solution, representing no collision. But it is not very interest-
ing.
The second solution is the one we are after. How
can we check it? There are two ways.
First, suppose the masses were the same. Then
it’s more or less obvious that the two bodies just
trade velocities. Setting m1=m 2 in the second
solution, we indeed find w1=v2.
Second, suppose that one of the bodies, say
number 2, had zero velocity and were very
massive. Then it’s as though the first body were
hitting a brick wall. It would just bounce back
with final velocity equal and opposite to its initial
velocity. Setting v2=0 in the second solution and
then taking the limit as m2 becomes infinite, we
indeed find w1 approaches –v1.
So, we are convinced that the solution makes
physical sense. But if you tried to reproduce the
solution for w 1 , you will know that it was
obtained only after a lot of tedious algebra. In
physics, that sometimes means that we have not
found the best way of analyzing the problem. In
fact, there does turn out to be a much better point
of view from which to look at systems containing
many particles. Let’s see what it is.
• which is acted on by a force equal to the sum of
the external forces acting on the bodies of the
system.
The center of mass frame
We return to our equation for the rate of change
of the total momentum of the system. Written in
terms of the positions xi, it reads
d2
3 m x = Fext .
dt2 i i i
v cm =
We can make this look like Newton’s law for a
single body whose total mass is the sum of all the
masses in the system, by multiplying and
dividing by the total mass M:
M
d 2 x cm
dt2
= F ext
1
where x cm =
3m x
M i i i
In general, the center of mass moves as time goes
on. The velocity of the center of mass is just the
time derivative of xcm, or
.
This new quantity x c m is called the center of
mass. It is the average of the positions of all the
bodies, weighted by their masses.
The above two equations say that the motion of
the center of mass of the system is the same as
the motion of a single body
• whose mass is equal to the total mass of the
system;
• which is located at the center of mass of the
system; and
1
3m v
M i i i
Now, let’s do a trick. We will view the whole
system from a frame of reference which is
moving along with the velocity of the center of
mass.
Let’s say that a body is moving at 10 m s-1 in the
original reference frame, and that the center of
mass is moving in the same direction at 6 m s-1.
Then the velocity of the body when viewed in the
center of mass frame is 4 m s- 1. To get the
velocity in the center of mass frame, you just
subtract vcm.
We will symbolize quantities in the center of
mass frame by putting primes on them. Then
v iN = v i − v cm .
The momentum of a body in the center of mass
frame is just its mass times its velocity in the
center of mass frame:
p iN = m iv iN = m i(v i − v cm ) .
Now for the punch line. We can show that the
sum of the momenta in the center of mass frame
is zero.
3 p iN = 3 m i(v i − v cm)
i
i
= 3 m iv i − M vcm
i
=0 .
The last line follows directly from the definition
of the velocity of the center of mass.
Application to previous example
How does this help us in our previous collision
example? Well, there are only two momenta
before the collision, so they are equal and
opposite in the center of mass frame:
p 2N = − p 1N .
There are only two momenta after the collision
(we’ll call them q1 and q2), which are also equal
and opposite to one another in the center of mass
frame:
q 2N = − q 1N .
We now apply conservation of energy in the
center of mass frame:
p 1N 2 p 2N 2 q 1N 2 q 2N 2
+
=
+
.
2m 1 2m 2 2m 1 2m 2
There are thus only two possibilities; either
q 1N = p 1N and
q 2N = p 2N ,
or
q 1N = − p 1N and
q 2N = − p 2N .
The first says that both bodies just sail through
the collision without changing their momenta;
this is not interesting. The other says that the
momenta of both bodies just reverse their
directions, when viewed in the center of mass
frame. Since the masses of the bodies don’t
change during the collision, we therefore
conclude that the velocities just reverse their
directions, when viewed in the center of mass
frame. This fact simplifies the analysis considerably, as we will now see.
Let’s use this fact to get our solution for w1 the
easy way. The first step is to find the velocity of
the center of mass. It is
v cm =
m1 v 1 + m 2 v 2
.
m 1 + m2
The second step is to find the velocity of body 1
in the center of mass frame:
v 1N = v 1 −
m1 v 1 + m 2 v 2
m2
=
(v − v ) .
m 1 + m2
m 1 + m2 1 2
The third step is to use what we just discovered the final velocity in the center of mass frame is
just the negative of this:
w 1N = −
m2
(v − v ) .
m 1 + m2 1 2
The last step is to transform back to the original
frame by adding vcm. We find
w 1 = w 1N + vcm
=
(m 1 − m 2 )v 1 + 2m 2 v 2
,
m 1 + m2
which is the result we found before.
Example: explosive fragmentation of a body
Here’s another example. Suppose a body of
mass M is flying along with constant velocity v.
The body has a charge contained within it, and it
explodes, releasing some energy ∆E. The body
breaks into two pieces with masses m 1 and m 2
(such that m1+m2=M), and the energy released
during the explosion appears as kinetic energy of
the fragments. What are the final velocities of
the two bodies?
Let’s let the final momenta of the two bodies be
q1 and q2. We go into the center of mass frame,
in which we know that
q 2N = − q 1N .
q N = − q1N
2
m 2 m1
q1N
Before the collision, the total energy in this frame
was zero (because the original body was at rest in
the center of mass frame!) After the collision,
the total energy in this frame is ∆E . Hence,
q 1N 2 q 2N 2
∆E =
+
.
2m 1 2m 2
Solving, we find
q 1N 2 =
2m 1 m 2 ∆E
.
M
This has two solutions, corresponding to body 1
going in the positive or negative direction. Both
are physically equivalent, so we’ll choose the
positive one. Then the final velocity of body 1 in
the center of mass frame is
w 1N =
q 1N
=
m1
2m 2 ∆E
m1M
.
The velocity of body 2 in this frame is obtained
by switching 1 and 2 in this formula, and
multiplying by –1.
Transforming back to the original frame, we find
the result
2m 2 ∆E
w 1 = v cm + w 1N = v +
.
m1M
It is instructive to calculate the final total energy
of the system in the original frame. We find
E final =
=
and the fact that the sum of the momenta in the
center of mass frame is zero:
3 m iv iN = 0 ,
i
we find
3
i
1
1
m iv 2i = 3 m i ãäv 2cm + 2v cm v iN + v iN 2 ëí
2
i 2
äå 3 m ëì v 2 + v 3 m v N + 3 1 m v N 2
å
i ì cm
cm
i i
i i
ãi
í
i
i 2
1
1
m 1 w 21 + m 2 w 22
2
2
=
1
2
1
M v2 + ∆E ,
2
=
1
1
M v2cm + 3 m iv iN 2 .
2
i 2
as it should be. The total energy is increased by
∆E, no matter which frame we calculate it in.
Energy and the center of mass frame
This last fact is an example of an important
property of the center of mass frame:
the total kinetic energy of a system is equal to the
kinetic energy of the center of mass plus the sum
of the kinetic energies of the bodies of the sytem,
as calculated in the center of mass frame.
The proof is just a short calculation. Using the
velocity transformation law
v i = v cm + v iN ,
The first term on the last line is the kinetic
energy of the center of mass. Another way to
write the last equation is
E = E cm + E N
.
An example with an internal force
Consider a pair of masses joined by a spring.
The whole thing can move freely along the track.
How does it move?
Well, we know that the center of mass will just
move with constant velocity, since no external
forces act. And we expect that the length of the
spring will oscillate about its equilibrium value.
What will be the frequency?
L
x2 − x1 − L
x1
The complete motion of the system is an oscillation of the relative coordinate superimposed upon
the uniform motion of the center of mass.
x2
The equations of motion are
m 1 x¨ 1 = + k (x 2 − x 1 − L)
m 2 x¨ 2 = − k (x 2 − x 1 − L)
Adding these, we find
m 1 x¨ 1 + m 2 x¨ 2 = 0 ,
which is just the statement that the center of mass
moves with constant velocity.
We also find
ä k
k ìëì
x¨ 2 − x¨ 1 = − åååå
+
ìì (x 2 − x 1 − L) ,
ã m1 m2 í
which has the solution
x 2 − x 1 = L + A cos (ωt − δ) ,
where the frequency is
ω=
lates about its equilibrium value L. The constants
A and δ are determined by initial conditions.
k
k
+
.
m 1 m2
Hence, the relative coordinate x 2 –x 1 just oscil-