Introduction to Momentum: Interactions between particles.
What is momentum?
Is it the same as the vernacular use, like “the whole momentum of the game has
shifted”?
We’ll see.
The word momentum comes from the Latin word meaning movement (around 1600).
In physics (science) the concept of momentum is defined specifically from how an
outside agent (an outside force) can change the uniform motion of an object.
We learned in studying Newton’s laws of motion, that the size of an objects change in
uniform motion (v) is directly related to the inertia (or mass) of the object.
Symbolically Newton’s 2nd law of motion shows that F = ma or by substituting v/t in
for acceleration, we get F = m v/t.
The momentum of an object is the product of the object’s mass (in kilograms) and its
velocity (in meters/second). Or symbolically…
P = mv
(Note that the bold-faced P indicates that momentum is a vector quantity.)
Example 1: What would be the momentum (P) of a 1200 kilogram car traveling at a
constant velocity of 20 meters/second due east?
Solution: P = mv = (1200 kg)(20 m/s due east) = 24,000 kg m/s due east
Example 2: Which has a larger momentum, a Ping-Pong ball with a mass of 0.002 kg
traveling at 90 m/s to the right, or a golf ball with a mass of 0.045 kg traveling at 4 m/s?
Solution:
If an object encounters another object, then the momenta (plural) of the two objects
changes. In other words, each object experiences a change in momentum. Symbolically
this can be written as P.
Assuming that the mass of each object isn’t altered, the change in momentum (P) of
the object must be due to its change in velocity (v).
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Therefore this means that the change in momentum (P) is equal to the product of the
mass of the object and its change in velocity, or simply P = mv. This is also known as
the impulse on the object.
Impulse = Change in Momentum
P = mv
Now notice if we divide both sides of this equation by t, we get an equation that looks
like…
P/t = mv/t (What is this side equal to?)
That’s correct! It is equal to a force!
Force P/t = mv/t = ma = Force
So F = P/t (the rate of momentum change) = mv/t = ma (mass x
acceleration)
The final note on impulse or change in momentum (P) is that it may be
determined two ways as shown below. Can you tell how the other way was
derived using the previously outlined equation above?
P = mv = Ft
Examples 3 & 4: A 4.0 kg bowling ball traveling initially at 5 m/s straight down the
lane collides with the head pin. Right after the encounter with the head pin the
bowling ball’s velocity changes to 4.5 m/s in the same direction (straight down the
lane).
What is the change in momentum of the bowling ball?




Note that the initial velocity, vi (or if you prefer vo) is 5 m/s, and the final
velocity, vf is 4.5 m/s.
Next you should realize that v = vf – vi.
And then the mass of the ball doesn’t change, so m = 4.0 kg.
Now it is simply a matter of plugging in the given amounts into the equation like
so…
P = mv = m(vf – vi)=(4.0 kg)(4.5 m/s – 5.0 m/s) = (4.0 kg)(-0.5 m/s) = -2.0 kg m/s
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If the collision between the bowling ball and the head pin lasted only 0.005 seconds (5
milliseconds), then what force was exerted on the bowling ball?

Now recall that there is another way to look at the change in momentum or
impulse…
P = mv = Ft
And t is given to be 0.005 seconds, therefore if we solve this equation for force (divide
by t) and plug in the time this is what we get…
mv/t = F = (-2.0 kg m/s)/(0.005 s) = -400 kg m/s2 = - 400 N
This answer means that the bowling ball experiences a force of 400 N in a direction
opposite (the negative sign) of its current motion. As a result, its momentum changes
by an amount of -2.0 kg m/s (notice the negative sign here too – what does this negative
sign mean?).
The Law of Conservation of Linear Momentum
This is another of the conservations laws we find in nature. As a reminder, for
something to be conserved, it means that the amount of the quantity doesn’t change.
Therefore in this case, the law of conservation of linear momentum states that the total
amount of momentum in a system does not change. There may be many interactions
between objects/particles in a system where many particles interact/collide and thereby
experience changes in momentum. However the sum of all the momenta of the
objects/particles in the system remains constant. This can be stated symbolically as:
P1 + P2 + P3 +… (before any interactions) = P1 + P2 + P3 … (after any number of
interactions), or more succinctly,
Pbefore =  Pafter
This like the law of conservation of mechanical energy, is a very powerful concept and
tool as we’ll see in the following example.
Example 5: Let’s visit the previous example with the bowling ball colliding with the head
pin. Every quantity in the previous example will be the same, except now we’ll add the
fact that the mass of the head pin is 1.5 kg.
We’ll use the law of conservation of linear momentum to answer the question: what is
the velocity of the head pin immediately after the collision with the bowling ball?
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Solution: The general approach is to –
 first identify the momentum of each object prior to the interaction,
 next find the sum of these momenta. Then identify and find the momenta
possible after the interaction,
 now note whether or not the interaction was elastic (the objects/particles
involved in the interaction did not stick together or combine) or inelastic (the
objects/particles involved did stick together into one mass), and finally
 use the law of conservation of momentum to solve for the specified unknown.
In this case the momenta of the objects before the interaction are:
the momentum of the bowling ball is –
p = mv = (4.0 kg)(5 m/s) = 20 kg m/s straight down the lane,
the momentum of the head pin is –
p = mv = (1.5 kg)( 0 m/s) = 0 kg m/s
Before the collision the total momenta is:
 p = 20 kg m/s straight down the lane + 0 kg m/s = 20 kg m/s straight down the lane.
Now, note that the bowling ball and the pin bounce off one another – they do not stick
together. This means we don’t have a combined mass, so we just find the momentum
or expression for the momentum of each object after the collision and set that sum
equal to the previous sum.
 Pafter = p bowling ball + p pin = (4.0 kg)(4.5 m/s down the lane) + (1.5 kg)(v bowling ball)
 Pafter = 18 kg m/s down the lane + (1.5 kg) (v bowling ball)
Pbefore=  Pafter
20 kg m/s down the lane = 18 kg m/s down the lane + (1.5 kg) (v bowling ball)
Let’s take out the units to see just the algebra…
20 = 18 + 1.5v
Solving for v…
20 – 18 = 1.5v
2 = 1.5v
2/1.5 = 1.33 m/s = v
How would this change if the collision was inelastic?
You will answer this question and others by completing the online assignment using The
Physics Classroom.
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Practice Assignment
Due Monday (A-day) February 3rd
Directions: After reading, taking notes, and asking Mrs. Painter any questions on the
previous introductory information and examples, complete the following tasks:
1. Go to the Physic Classroom tutorial (see URL/link below) on momentum and
read through the examples shown there.
2. Do the practice sheet on impulse and change in momentum found at the second
Physics Classroom link below.
3. Do the practice sheet on conservation of momentum found at the third Physics
Classroom link below.
4. Submit your completed practice sheets either electronically (by scanning and
emailing to painter@skitsap.wednet.edu ) or in-person by the due date above.
The Physics Classroom – Tutorial on momentum & conservation of linear momentum
http://www.physicsclassroom.com/Class/momentum/
Impulse or Change in Momentum Practice Sheet
http://www.physicsclassroom.com/curriculum/momentum/mom3.pdf
Conservation of Momentum Practice Sheet
http://www.physicsclassroom.com/curriculum/momentum/mom6.pdf
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