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Physics
Mr. Tumolillo
Momentum and Center of Mass
Center of Mass Definition: the center of mass of an object is the single point where the entire
mass of the object would be located to produce the same linear motion. If you were able to place
the object on a point precisely below the center of mass, the object would balance. We are
assuming a uniform distribution of mass within the object.
Example: for a cube of length l, the center of mass is at l/2 in each dimension. See diagram.
l

 =center of mass of cube at
point l/2 from left (right) side,
l/2 from bottom (top) and l/2
from front (back).
l
l
For a sphere, the center of mass is at the center of the sphere. When we looked at the
Gravitational Force between an object on the surface of the Earth and the Earth itself, we used as
the distance the radius of the earth, rE. This definition of the Gravitational Force assumes the
center of mass of the each object, Earth and you for example, are a distance rE apart. Instead of
the picture of a person standing on the surface of the earth, we replace it with the one shown on
the right side in the diagram below.

Using the center of mass of an object is something we implicitly do in most motion problems.
We will not do much in the way of calculating the center of mass of objects that are not regular
in shape. Rectangular solids, cylinders and spheres are easy; objects that are collections of these
regular shapes are likewise easy to analyze. The center of mass of two cubes, for example, is
found by finding the center of mass of each cube then finding the center of mass of those two
points.
For example, suppose we have two cubes, one with side l1 = 6m and the second with side l2 =
4m. They are sitting adjacent to each other as shown below with the front faces flush. Where is
the center of mass of the two cubes?
The center of mass of the cube on the left as at the point: x1 = 3m from the left face, y1 = 3m
from the bottom face, and z1 = 3m from the back face. Assume we have a coordinate system
centered at the back left hand corner of the left cube as shown in the diagram on the right below.
y

6
4

6
4
6
z
x
4
The center of mass of the second cube is 2m from the left face, 2m from the back face and 2m
from the bottom face. That puts the center of mass of the right hand cube at the following
location with respect to the origin of our coordinate system. Remember the smaller cube is
already 2m away from the back face of the first cube, i.e., its distance in the z-direction is 2m.
x2 = 6 + 2 = 8m from left face of first cube
y2 = 2m above bottom face
z2 = 2 + 2 = 4m from back face of the larger cube.
So, thus far we have the center of mass of the two objects located at:
Left Cube
x1 = 3m
y1 = 3m
z1 = 3m
RightCube
x2 = 8m
y2 = 2m
z2 = 4m
We do not simply add the coordinates and divide by 2. Suppose the first cube is made of iron and
the second is made of feathers. The center of mass of the combined two cubes would be very
close to the center of mass of the iron cube. So we need to “weight” the centers of mass as
follows. Multiply each coordinate by the respective mass of the object. Then sum the
corresponding weighted x, y and z coordinates. Then divide those values by the total mass for
each of the coordinates, x, y and z.
Let m1 = 100kg and m2 = 200kg, i.e., the second cube is much denser than the first cube. The
total mass is then 300kg.
Left Cube
m1x1 = 300kg-m
m1y1 = 300kg-m
m1z1 = 300kg-m
Total = 900kg-m
Right Cube
m2x2 = 1600kg-m
m2y2 = 400kg-m
m2z2 = 800kg-m
Total = 2800kg-m
Calculations for CM
xCM = (300+1600)/300 = 1900/300 = 6.33m
yCM = (300+400)/300 = 700/300 = 2.33m
zCM = (300+800)/300 = 1100/300 = 3.67m
From this analysis the center of mass of the two cubes is 6.33m from the left face of the first
cube, i.e., it is just inside the left face of the second cube. Likewise, the center of mass is 2.33m
above the bottom face, a little above the location of the second cube’s own center of mass but
below the location of the first cube’s center of mass. Finally, the center of mass of the system is
3.67 meters from the back face of the first cube. Remember, the second cube is displaced
forward a distance of 2m from the back face of the first cube.
CM of the two
cubes combined
y
xCM = 6.33m

6
4
 
6
4
6
z

yCM = 3.67m
yCM = 2.33m
x
4
Center of Mass and Momentum
We will assume we can find the center of mass of any object and investigate the motion of the
center of mass of a system during a collision. Objects like cars are very difficult for calculating
the center of mass. The center of mass of a system is found just as we did above for finding the
center of mass of the two cubes. So, we will simply assume we know where the center of mass of
the system is. I wish to make a point about the behavior of systems during collisions.
The point you should remember and appreciate is the following:
during a collision, the momentum of the system’s center of mass remains unchanged. While the
two objects may bounce off each other, the center of mass of the combined system continues its
motion undisturbed until additional outside forces are brought to bear.
An example will help.
Suppose we have two masses: m1 = 10kg and m2 = 15kg. Mass 1 is moving to the right at a
speed v1 = 10m/s and mass 2 is at rest, i.e., v2 = 0m/s. Initially we have a momentum only to the
right of m1v1 = 100kgm/sec.
We have, in equation form:
m1v1 + m2v2 = pbefore
And we can write pbefore as follows:
pbefore = (m1+m2)v* where v* is the resultant weighted velocity of the two masses, weighted
according to their respective masses
m1v1 + m2v2 = (m1+m2)v*
v* = [m1v1 + m2v2]/[m1+m2] = [m1/(m1 + m2)]v1 + [m2/(m1 + m2)]v2
The “weights” applied to each of the initial velocities are m1/(m1 + m2) and m2/(m1 + m2)
Using numbers:
v* = [10*10 + 0]/[25] = 4m/sec to the right.
In the above example, the “weight” applied to the initial velocity of mass 1 is m1/(m1 + m2) =
10/25 = 0.4. The weight applied to mass 2’s initial velocity is m2/(m1 + m2) = 15/25 = 0.6 and
since v2 = 0m/s, it drops out.
After the collision we know that momentum before = momentum after, so we have:
m1v1+m2v2 = [m1+m2]v* = m1v1 + m2v2
Since momentum is conserved, v* remains unchanged and it equals the weighted average of the
two speeds after the collision as well as before, with the final speeds each getting the same
appropriate “weights” of m1/(m1 + m2) and m2/(m1 + m2):
v* = [m1v1 + m2v2]/[m1+m2] = 4m/sec to the right
Suppose after the collision we note that v1 = -2m/sec then:
v* = 4 = [(10)*(-2) + 15*v2]/[25] and solve for v2:
v2 = [25*4 + 20]/15 = 120/15 = 8m/sec to the right
So, before the collision we have m1 moving at 10m/sec to the right and m2 at rest. After the
collision we have m1 moving to the left at -2m/sec and m2 moving to the right at 8m/sec. In both
cases the total momentum requires the combined masses to be moving at 4m/sec to the right. In
other words, the center of mass of the two objects remains moving at 4m/sec to the right no
matter what the two objects do.
In pictures:
m1 = 10kg
v1 = 10m/sec
m2 = 15kg
v2 = 0m/sec
vCM
Center of mass = location where total
mass of 25kg is located
CM is moving right at 4m.sec
Collision and CM is “inside” object 2
since object 2 is more massive.
m2 = 15kg
v'2 = 8m/sec
m1 = 10kg
v'1 = -2m/sec
Center of mass = location where total
mass of 25kgis located
CM is still moving right at 4m.sec
Reformulation of the Law of Conservation of Momentum
The momentum of the combined masses, located at the center of mass of the system, remains
unchanged. The center of mass of the system moves along at a constant speed immediately
before, during and after the collision.
Using Charts and Center of Mass
We have been using a chart (or matrix) to solve complex collision problems. One is replicated
below for the problem outline above – m1 = 10kg, v1 = 10m/s, m2 = 15kg, v2 = 0m/s and v'1 = 2m/s. The full chart shows that v'2 = 8m/s.
In the template the values at the bottom of each of: v, v', v, p, p', p, F, a and Impulse
columns all apply to the Center of Mass of the system. To find the velocity of the Center of Mass
all you do is divide the total momentum by the total mass. In the chart below that yields a
velocity of 4m/s. And it is positive in this problem, i.e., moving east. You must get the same
number for the velocity of the Center of Mass after the collision since the total mass does not
change nor does the total momentum.
Notice that the total Force acting on the Center of Mass is 0 as is p. Hence, by Newton’s 1st and
2nd Laws, the acceleration is 0m/s2 for the Center of Mass. One can read the bottom row of the
charts as the situation for the Center of Mass – it moves at constant velocity hence F, a, v, p
and Impulse are all 0.
The last column is the distance. It says mass 1 moved d1 = (10)*(0.5) – (0.5)(24)(0.5)2 = 5 – 3 =
2m during the half second it took to complete the collision. Mass 2 moved d2 = (0)*(0.5) +
(0.5)(16)(0.5)2 = 2m. The Center of Mass likewise moved 2m since the time interval was 0.5sec
and its speed was 4m/s so dCM = vCMt = (4)(0.5) = 2m/s. Not all problems are quite so neat!
Collision Template -- Center of Mass Added in
Object
Mass
(kg)
v
(m/s)
v' (m/s)
Formulas
p
(kg-m/s)
p'
(kg-m/s)
v
p
(kg-m/s)
Force
(N)
acc.
(m/s2)
p = mv
p' = mv'
v'-v
p' - p
F = p/t
a = F/m
time
(sec)
Impulse
(kg-m/s)
Imp = p =
Fxt
Distance
(m)
d = vt +
½ at2
1
10
10
-2
100
-20
-12
-120
-240
-24
0.5
-120
2
2
15
0
8
0
120
8
120
240
16
0.5
120
2
Total
25
4
4
100
100
0
0
0
0
0.5
0
2
m1 =
m2 =
v1 =
v2 =
v'1=
t=
10kg
15kg
10m/s east
0m/s
2m/s west
0.5sec
v1 = 10m/s
After the collision, the two objects move apart.
v2 = 0m/s
vCM = 4m/s
v'1 = -2m/s
v'2 = 8m/s
Reference Frames
If you imagine for a moment that you are at the center of mass of the two objects in the above
example and you are moving along at 4m/sec. From your point of view, you are at rest and the
two objects are moving toward you. Object 1 is moving toward you at v1 = v1 – vCM = 6m/sec
and v2 = 0-vCM = -4m/sec. From where you sit, object 1 is moving toward you at 6m/sec and
object 2 is moving toward you at 4m/sec. The closing speed is v1 – v2 = 6-(-4) = 10m/sec.
Afterwards, object 1 is moving away from you at v1 = -2-vCM = -2-4 = -6m/sec and object 2 is
moving away in the opposite direction at a speed of v2 = 8-vCM = 8-4 = 4m/sec. The speed of
separation is 10m/sec = v1 – v2 = -6-4 = -10m/sec (the minus sign indicates separation and a
plus sign indicates closing).
Coefficient of Restitution
In physics we define a term called the Coefficient of Restitution. It is symbolized by the Greek
letter  (“epsilon”) and it is the following ratio:
 = (v2 – v1)/(v1 – v2) = velocity of separation/velocity of approach
o
o
If  = 1, we also have conservation of kinetic energy.
If  < 1, then there is a loss of kinetic energy.
In the above example we have:
 = (v2 – v1)/(v1 – v2) = 10/10 = 1
As this problem is set up, Kinetic Energy is conserved ( = 1).
Kinetic Energy is defined as ½mv2 and in this problem we have:
KEbefore = (0.5)*(10)*(10)2 = 500 joules (units of KE = joules)
KEafter = (0.5)*(10)*(-2)2 + (0.5)*(15)*(8)2 = 20 + 480 = 500joules
KEbefore = KEafter
Case where KE is Not Conserved
Suppose we have the same initial conditions as in the above problem except this time the
collision time is only 0.1sec. This time the two masses stick together. As you can calculate, the
initial KE is as before, 500j. After the collision each object has a speed of 4m/s (as does the
Center of Mass) and the final KE is 200j – a loss of 300j which goes off as heat. One thing that
happens, and we do not have all the methods that we need to do it full justice, is the objects
crumple when heat is absorbed, sometimes a lot. We have to consider the objects as “point”
objects with no real dimensions given. However, we do know that acceleration on cars, for
examples, when they collide, is highest at the front end and the least at the rear and. In the chart
below for this case, we see object 1 moves 0.7 meters and object 2 only 0.2 meters, suggesting
that when collisions occur, it gets messy and we would need a far more robust analysis of
collisions. The center of mass, as before, still moves at 4 m/s and the CofM moves 0.4 m.
Center of Mass Case with KE Not Conserved
Object
Mass
(kg)
v
(m/s)
v' (m/s)
Formulas
p
(kg-m/s)
p'
(kg-m/s)
v
p
(kg-m/s)
Force
(N)
acc.
(m/s2)
p = mv
p' = mv'
v'-v
p' - p
F = p/t
a = F/m
time
(sec)
Impulse
(kg-m/s)
Imp = p =
Fxt
Distance
(m)
d = vt +
½ at2
1
10
10
4
100
40
-6
-60
-600
-60
0.1
-60
0.7
2
15
0
4
0
60
4
60
600
40
0.1
60
0.2
Total
25
4
4
100
100
0
0
0
0
0.1
0
0.4
Summary
In summary, the Law of Conservation of Momentum requires the total momentum before a
collision to equal the total momentum after a collision. We can use the center of mass argument
to re-define this as saying that the momentum of the center of mass of the system is constant, i.e.,
the point where the combined masses would be for the center of mass moves at a constant
velocity before, during and after the collision. From the vantage point of the center of mass, the
objects move toward the center of mass before the collision and away from the center of mass
after the collision. If the objects stick together, well then three’s a crowd.
Mr. Tumolillo
Physics
Homework Problems
NAME: _________________________
1. A cylinder of height = 5m has a radius = 3m. Where is the center of mass located? Assume the
cylinder has a uniform density.
h = 5m
r = 3m
2. Two cylinders are stacked on top of each other. Cylinder 1, the one on the bottom, has a mass
of 100kg and a height = 5m and radius = 3m. The upper cylinder has a mass of 300kg, a radius of
2m and a height of 13m. Where is the center of mass located for the two cylinders combined?
Assume the cylinders have uniform densities. Assume the two cylinders are stacked so that they
are centered (i.e., not offset from the centers.)
h = 13m
h = 5m
r = 3m
3. A bowling ball of mass = 10kg is moving to the right with an initial speed of 8m/s. A second
bowling ball of identical mass is moving to the left at 4m/s. They collide head-on. The collision
takes 0.5sec. After the collision, the second bowling ball is moving to the right at a speed of
3m/s. Using the chart on the next page identify the following:
a) Find the final speed of the first bowling ball;
b) Find the speed of the center of mass of the two bowling balls before the collision;
c) Find the speed of the center of mass of the two bowling balls after the collision;
d) Find the Coefficient of Restitution and from this value determine if Kinetic Energy is
conserved.
e) Confirm your findings by calculating the KE of the system before the collision and then after
the collision. The formula for KE is ½mv2. KE is NOT a vector so there is no direction
associated with it. When you square the velocity, the directionality is lost.
f) Compare the change in KE to the Work = F*d approach.
f) Then find the KE of the Center of Mass before and after and the change in KE. Compare these
total values to the ones calculated for the system and explain why they are different.
Object
Mass
(kg)
v
(m/s)
v' (m/s)
Formulas
p
(kg-m/s)
p'
(kg-m/s)
v
p
(kg-m/s)
Force
(N)
acc.
(m/s2)
p = mv
p' = mv'
v'-v
p' - p
F = p/t
a = F/m
time
(sec)
Impulse
(kg-m/s)
Imp = p
= Fxt
Distance
(m)
d = vt +
½ at2
1
2
Total
Object
Formulas
KE
(j)
KE = ½
mv2
KE
(j)
KE' = ½
mv' 2
KE
(j)
Work
(j)
KE' - KE
F*d
1
2
Total
In the above chart you should calculate the KE initial and final and then find the change, which
is KE' – KE. Then, using the values you calculated for Force and d in the table, see if the
product, F*d, does indeed equal the change in KE, KE (don’t forget the sign for F and d –
either + or – as appropriate). The Work-Energy Theorem states that Work = F*d = KE.
Coefficient of Restitution: __________________
Is KE conserved based on the coefficient? _________________________
Compare the KE in the table to the KE of the Center of Mass. Are they the same? Does the KE
of the Center of Mass change during the collision?
KE CofM:
________________________
KE' of CofM: ________________________
KE of CofM: ________________________
Can you explain why these values are different than the ones in the table for KE, KE' and KE
of the system?