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Momentum
Momentum is defined as the mass multiplied but the velocity of an object. It is a vector quantity
used to describe the relationship inherent in the motion of an object or system and its mass.
Momentum uses Newton seconds as its units which look like “Ns”.
The formula for momentum is simply:
Where „m‟ is the mass of the system and „v‟ is the velocity of the system
Let us find the momentum of a car by assuming the car is moving at 30 m/s and weighs 900 kg.
We can use the simple formula above to solve for the momentum.
A car moving at 30 m/s and weighing 900 kg has a momentum of 27 000 Ns. This number
might not mean anything yet, but the higher the number the higher the energy of the system. A
car with that much momentum can plow through a building or demolish another car. A system
at rest has no momentum and thus cannot move anything with its own power.
Conservation of Momentum
In nature there is a law that states that the total momentum in a system will be constant as long
as there are no external forces acting on the system. This means that the initial momentum will
be equal to the final momentum which can be mathematically written as:
In the formula above „pI‟ and „pF‟ are the initial momentum and the final momentum
respectively. If we use billiards balls as an example we can investigate this a little bit easier.
Let us imagine two billiard balls weighing 1 kg each, with one moving at 5 m/s towards the
other stationary ball as shown below.
1
5
m/s
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2
If we assume that the first ball transfers all of its
momentum to the second ball we can then evaluate
for the velocity of the second ball.
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We can do this by using the formula for momentum and the fact that all momentum is
conserved. To find the initial momentum of the system we must find the momentum on ball 1
before it acts on ball 2 and the momentum of ball 2 before ball 1 acts on it. We can do this
mathematically by writing:
The subscript „I‟ simply means these are initial values as the collision between the two balls has
not taken place yet. It can be observed that the initial momentum of the system is simply 5 Ns
and it all comes from the first ball as it is the only one that is moving. The second ball has a
velocity of 0 m/s meaning that it has no momentum.
Using the law of conservation of momentum we now know that the initial momentum is equal
to the final momentum. This means in our case that the final momentum will also be 5 Ns. We
can use this value to find out the velocity of the second ball after the collision has occurred and
all of the energy was transferred to the second ball.
Therefore the second ball now has all of the momentum and is now travelling at 5 m/s. The
diagram of the system has changed from the one above to:
1
2
5 m/s
Before we move on, lets investigate what would happen if the first ball was travelling with 1 m/s
after the collision with the second ball. We can start from the original scenario and use the same
initial momentum of 5 Ns. Using the formula for the final momentum in this case we can put in
all of our known values and solve for the unknown speed of the second ball.
This is slightly different then above because now both balls have a velocity and momentum after
the collision has taken place.
1
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1 m/s
2
4 m/s
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Work
Work is a measure of energy. The work done on an object is found to be the force multiplied by
the distance travelled. The formula for work is given as:
The „F‟ is the measure of force in the system in question, „d‟ is the displacement vector for the
system and „Ɵ‟ refers to the angle in between the force and displacement vectors. The work is
measured in joules „J‟ and is a scalar unit. Here are a few examples of work being done on an
object.
Example 1: Force and displacement vectors are parallel
F=
5N
Ɵ=
0o
d=5m
In this example we do not have to worry about the
direction of both vectors because they are parallel. We
can use the formula above in order to find the work that
was done on this object.
Example 2: Force and displacement vectors are none parallel
F=5N
Ɵ=
45o
d=5m
This example is different because the force and
displacement vectors are now at different angles. We
can again use the formula and we will investigate how
the work changes due to a difference in angle
From these two examples we can see that as the angle between both vectors increases the work
done on an object decreases. This means that there must be a larger force in order to do the
same amount of work if you are increasing the angle. Once the difference in angle reaches 90o
the work done will equal zero all of the time.
With the idea of work we can think of energy. Energy is how your work is measured. For
example if we have a rigid object such as our box and we change its speed, we can find the work
done on the box.
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Energy
When talking about the motion of an object the energy of a system should also be discussed.
There are three types of energy that must be discussed: kinetic energy, elastic potential energy and
gravitational potential energy. All energy is measured in joules „J‟. Unlike momentum, energy is a
scalar quantity, so direction does not matter.
Kinetic Energy
The first type of energy can be used to evaluate the energy in a system that has a velocity or in
other words a system that is in motion. The formula for kinetic energy is:
Where „m‟ is the mass of the system and „v‟ is the velocity of the system. If you look closely at
the formula for kinetic energy you can see the formula for momentum. This is interesting
because both formulas describe the same type of system so you would expect them to be similar.
Let us find the kinetic energy of the car that weighs 900 kg and is moving at 30 m/s. We can
use the formula that we have just discussed above in order to solve for this value.
This is a different number then what was achieved for the momentum of the same car. This is
to be expected because we are now describing something different. This number gives the exact
energy in the system allowing us to show the work that can be done. The true power of this
value will be shown when the conservation of energy is discussed.
Going back to the idea of work we can find the work done on an object using this formula:
As we can see, the change in kinetic energy on a rigid body describes the amount of work energy
that was used to move that object.
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Elastic Potential Energy
The second type of energy has a similar formula to that of kinetic energy but differs in that it
describes a system that has some sort of spring or elastic in it. The formula looks like:
Where „k‟ is the spring constant and „x‟ is the distance away from the equilibrium position. The
figure below shows a spring in its equilibrium position as well as in a stretched form. The units
of elastic potential energy are still the joules „J‟, the units for the spring constant is Newton‟s per
meter „N/m‟ and the distance is simply in meters.
The distance „x‟ that the string is stretched is
put into the above formula to find the elastic
potential energy. Unlike kinetic energy the body
or system does not need to be moving in order
for there to be elastic potential energy. The
word potential means that something could
happen, therefore having the spring either
compressed or stretched means movement
could happen and the energy involved in the
system could be converted from elastic
potential energy to kinetic energy.
Let us find the potential energy of a spring that is compressed 0.5 m and has a spring constant of
100 N/m. We can use the formula above for this example.
Again this is just a number right now until we discuss the conservation of energy.
Gravitational Potential Energy
The third formula for energy is different from the above formulas, but it is similar in concept to
the elastic potential energy. The gravitational potential energy describes a system or body that is
suspended in the air and the potential energy it has as it falls towards the earth. The formula for
this looks like:
Where „m‟ is the mass of the system or body, „g‟ is the acceleration due to gravity and „h‟ is the
height the object will descend. Again, the units of this energy is the joule „J‟, the mass of the
system is in kilograms „kg‟ and the acceleration due to gravity is in meters per second squared
„m/s2‟, and the height is in meters „m‟.
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Let us find the gravitational potential energy of a rollercoaster that weighs 100 kg on top of a hill
10 m tall. We can use the formula discussed above in order to solve for the energy.
For the final time this is just a number right now until we discuss the conservation of energy.
Conservation of Energy
In nature, there is a law that says new energy cannot be created and energy cannot be lost, only
converted into a different form. Therefore, a rollercoaster at the top of a hill and commencing
its roll downwards will lose some of its gravitational potential energy, but start to gain kinetic
energy. So it can be seen that the initial energy of a system has to be the same as the final energy
of a system which can be seen mathematically as:
Where „EI‟ is the initial energy of the system, so in our case, with the rollercoaster at the top of
the hill, „EI‟ will be whatever gravitational potential energy it has, and „EF‟ is the final energy of
the system once it reaches the bottom of the hill, or in our case the kinetic energy.
In fact, let us look at an example where we have a rollercoaster that weighs 100 kg at the top of
10 m high hill, and we would like to know its speed at the bottom of the hill. We can use the
law of conservation of energy in order to solve for the speed. We can assume that there will be
no velocity at the top of the hill and at the bottom of the hill there will be a height of zero. For
our initial energy, we can add together both the kinetic energy and gravitational potential energy
of the system at the top of the hill.
Now we can use this initial energy of 9800 J, and substitute it as our new final energy saying now
that the system has some velocity and now no height. The new equation will look like:
So we know that the rollercoaster that went down a 10 m high hill and weighs 100 kg will be
going 14 m/s when it reaches a height of zero.
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Let us look at a new example with a toy car weighing 0.5 kg and moving at 5 m/s going straight
into a spring with a spring constant of 25 N/m. We can now find how much the car will
compress the spring by using conservation of energy.
Finding the initial energy will be the same idea using both the elastic potential energy and adding
the kinetic energy of the system.
When the car is travelling all of the energy is found in the kinetic motion of the car and the
spring is not contributing at all because it is in its rest position. Now we can use our initial
energy of 6.25 J as our final energy and solve for the compression on the spring when the car
has stopped moving.
Therefore, the car compressed the spring a total of 0.707 m or approximately 71 cm.
Now we can see how the laws of conservation of energy and momentum can be used to solve
for lots of unknowns in various systems and scenarios. Try and look around you to see where
you can use the conservation laws in order to understand the actions and reactions of the world.
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