### MOMENTUM & CoM

Centre of Mass
NCEA Level
3
Physics
This is a useful tool for studying collisions, explosions
and other forms of motion.
 Centre of mass = point where the total mass acts.
 Another name for it is centre of gravity.
 This is of paramount importance in architecture:

CoM
CoM
Centre of mass acts
through base building
stable.
E.g. the leaning tower of Pisa
Centre of mass is not supported
by the base, result is building
falls
CoM
The foundations subsided
which caused the tower to
lean. However, because the
CoM is over the base the
tower will not topple over.
As people walk up to the
top of the tower the CoM
shifts over time. As a result
the CoM moves further
along until eventually the
base will not support it
and it will fall.
When analysing the motion of more than one
object, it is often useful to be able to consider
the group of objects as a single object called a
system.
Calculating the position of the CoM:
The CoM can be calculated as a ratio
along the distance between two masses:
MgX = mgx
MX = mx
4kg
1kg
M/m = x/X
MX = m(d-X)
•CoM is four times closer to the 4kg mass than the 1kg
mass.
•If the distance between the two masses is 12m then the
CoM is 2.4m from the 4kg mass.
MgX = mgx
MX = mx
M/m = x/X
MX = m(d-X)
4x = 1 ( 12 – x)
4x = 12 – 1x
5x = 12
x = 2.4 m from 4kg
1kg
4kg
12m
Example 1:
The Earth and Moon are like a two particle system, if we
imagine all their mass as concentrated at their CoM.
Calculate the distance of the CoM of the Earth-Moon
system from the centre of the Earth.
Mass of Earth = 600 x 1022 kg
1022 kg
Mass of moon = 7.3 x
Distance between Earth & moon = 3.8 x 105 km
SOLUTION:
Distance between Earth & Moon = 3.8 x 108 m
Earth:
600x1022kg
X
Moon:
CoM
7.3x1022kg
d=X+x
MX = m(d – X)
X is the distance from the centre of the earth
X = md/(M + m)
Rearranging
X = 7.3x1022 x 3.8x108 / (600x1022 + 7.3x1022)
X = 4.6 x 106 m
As the radius of the earth is 6.4 x
106m, the centre of mass of the earthMoon system is beneath the earth’s
surface. While it is usually said that
the Moon orbits the earth, it would be
more correct to say that the Moon
orbits the earth each orbit around
their common CoM.
CoM & Conservation of Momentum
Any form of calculation involving a collision, as you
learnt last year, can be solved using momentum.
 = mv
The law of conservation of momentum can also be
used for isolated systems. The behaviour of the
system can be analysed by looking at what happens
to its CoM.
system = msystem x vCoM
system = (M + m) x vCoM
If the system is isolated then the momentum will not
change. The masses of the individual particles also
remain the same thus the vCoM must therefore
remain constant.
The momentum of a system is also the total momentum
of the particles that make up the system.
system = Σparticles of system
system = (M + m) x vCoM
(M + m) x vCoM = MvM + mvm
As two objects approaching each other collide they can
change both their speed and direction, but their CoM
moves with a constant velocity. The velocity of the
CoM is unchanged by the collision.
Example 2:
A 2000kg truck moving at 8.0ms-1 hits a stationary car of mass
1200kg. After the accident, the two vehicles are locked together.
Calculate the velocity of the truck and car after the collision.
BEFORE
2000kg
AFTER
vms-1
8ms-1
1200kg
2000kg 1200kg
SOLUTION:
The total momentum before the collision = (2000 x 8.0) + (1200 x 0)
= 16000 kgms-1
The velocity of the CoM before the collision is given by:
vCoM = total momentum / total mass
vCoM = 16000 / (2000 + 1200) = 5.0 ms-1
This will also be the velocity of the CoM after the collision. Since the
vehicles are locked together their velocity after the collision will also be
5.0 ms-1.
COMPLETE EXERCISESRUTTER
PAGE 34 - 38
Momentum
NCEA Level
3
Physics
Why can’t it stop easily ??
1. It is MASSIVE
2.
It is FAST
IT has a lot of MOMENTUM
 = mv
 = f - i
Example 3: Conservation of  in 2-dimensions
The diagram shows a radioactive nucleus, which was
initially at rest, immediately after it has decayed. An emitted
electron moves off in a northerly direction, a neutrino
moves off in an easterly direction, and the decayed nucleus
moves off in a different direction.
Electron
Decayed
nucleus
Neutrino
The momentum of the electron is 1.2 x 10-22 kgms-1, and
that of the neutrino is 6.4 x 10-23 kgms-1.
Calculate the size of the momentum of the nucleus after
the decay.
SOLUTION:
Σ (before) = Σ(after)
Σ (before) = 0
Σ(after) = 1.2 x 10-22kgms-1  + 6.4 x 10-23kgms-1  + (nucleus)
(nucleus) = 0 – 1.2 x 10-22kgms-1  - 6.4 x 10-23kgms-1 
 (nucleus)
= 1.4 x 10-22kgms-1
6.4x10-23kgms-1
1.2x10-22kgms-1
(nucleus) = (1.2x10-22)2 + (6.4x10-23)2
Impulse: If we change the velocity of an object
then it has been accelerated and thus has been
acted upon by an unbalanced force. A change in
velocity thus results in a change in momentum.
The change in momentum is called IMPULSE.
F = ma
a = v/ t
Substituting both
OR
F = mv/ t
F t = mv
F t =  
Example 4: Impulse in 2-dimensions
A ball of mass 100g is moving north at 5.0ms-1. Raj hits
the ball with his stick so that it travels west at 12ms-1.
The diagram is a view from above and shows the
position of the stick at the instant the ball is hit.
a. Find the size and
direction of the change in
momentum of the ball.
b. If the stick is in contact
with the ball for 0.020s,
what is the magnitude
and direction of the
average force acting on
the ball.
12ms-1
5ms-1
SOLUTION:
a. Initial i = mv
= 0.10 x 5.0
1.2kgms-1
0.50kgms-1
= 0.50 kgms-1
Final f = mv
= 0.10 x 12
= 1.2 kgms-1
 = f - I
1
= 1.2 kgms-1 - 0.50 kgms-
Using
trigonometry:
tan  = 0.50 / 1.2
 = 23o
 =  1.22 + 0.502
= 1.3 kgms-1 23o south-west


b. The average force F on the ball can be calculated from:
Ft = 
F x 0.020 = 1.3
F = 1.3 / 0.020
F = 65N (in the same direction as )
DO
EXERCISE
10 FROM
Page 15
S&C
Just like with force components horizontal and vertical components for
momentum can be conserved.
This means that all the horizontal components of vectors before a collision
must equal all the horizontal components of vectors after the collision.
The same is true for all vertical components.
BEFORE
x1
AFTER
y3
x2
y1
A
y2
A+B
x3
B
HORIZONTALLY: x1+ x2 = x3
RESULTANT = x3 + y3
VERTICALLY: y1 + y2 = y3
Use trigonometry
All energy is also conserved as well as momentum. Thus
energy before is the same as energy after as long as there
has been no external input. The main energies we may
have to look at are:
Kinetic energy
Gravitational potential
energy
Work done
Power
Ek = ½mv2
Ep = mgh
W = Fd
P = W/t
Remember if all energy is conserved then the collision
is ELASTIC. If energy is lost then it is INELASTIC.