Momentum: Notes and Such

Physics Index

Where are we going with this? The information on this page connects to standards such as:

• Using experimental evidence and investigations, determine that Newton’s second law of motion defines force as a change in momentum, F = Δp/Δt.

• Develop and apply the impulse-momentum theorem along with scientific and engineering ideas to design, evaluate, and refine a device that minimizes the force on an object during a collision (e.g., helmet, seatbelt, parachute).

•  Assess the validity of the law of conservation of linear momentum (p=mv) by planning and constructing a controlled scientific investigation involving two objects moving in one-dimension.

Momentum: Notes and Such

(Wow! Now, that's a fancy title!)

What is momentum? Well… It's not a force… And it's not energy… Basically, it's a thing that helps us understand motion. And collisions.

(You gotta do better than that!)

A more formal definition would say that momentum is the quantity of motion, the product of an object's mass and its velocity, and it can be quantized as

p = mv

  

where is p is momentum, m is mass, and v is velocity.

This means that the units for momentum are

kg•m/s

Momentum is… boring… unless something changes. Just watching a football player run at a constant velocity isn't very interesting.

Where momentum becomes interesting is when something happens. Momentum concepts can then help us describe it.

The first application is when momentum changes… So, you got a thing with some momentum. Then something happens and there is less mass. Or more. Or the velocity changes.

This is sort of an abstract concept that helps us illustrate that (this is important!)…

within any closed system, momentum is conserved. Unless something different is introduced to the system, momentum is a constant.

Hence, within a closed system, any changes are limited to within that system, so the momentum can be thought of as having a first and second state (and 3rd, 4th, etc… if you so desired). So, we can formulize that to say that

p1 = p2

And since p = mv, then it follows that

m1v1 = m2v2

This will lead to questions where WITHIN THE CLOSED SYSTEM the mass or velocity will change (by means that somehow do not violate the "closed system" concept).

A far more realistic application is one in which something affects the moving object and thus causes a change to the momentum. This gives rise questions that examine the m1v1 = m2v2 in momentum from a first condition to a second (and third, fourth, fifth… if so desired).

This can be represented as…

∆p = p2 - p1

∆p =  m2v2  -  m1v1       <--- plug in                (eq. 1

Frequently, this will be seen when the velocity of an object changes. Since it is the same object

m1 = m2

So the formula can be represented as:

∆p = m(v2 - v1)                         (eq. 2

The astute observer will recognize that 

v2 - v1 = ∆v

Leading to the fancified version of equation 2 shown below

∆p = m∆v                         (eq. 3

A less frequent possibility is where mass changes (or when both mass and velocity change). Equation 1 can be a starting point for those situations.

IMPORTANT NOTE

Momentum is the product of velocity and mass, and velocity has a direction, so therefore, momentum has a direction (it is a vector). You will have to attend to the positive and negative values as directions.

Further, (LIKE ALL VECTORS), momentum need not occur on a single axis. Vectors can be deconstructed into orthogonal components, then all of the components can be recombined into resultant vectors.

Momentum In Action—A First Look: Inelastic Collisions

Key to understanding what's going when two (or more) objects collide is to know that, within a closed system momentum is conserved. 

That means that, at any point within the closed system the sum of all the momentums of all the objects in the system stays the same. You can't end up with more momentum than you start with!

So if there are 2 objects, and one has a momentum of 5 kg•m/s and the other has a momentum of 3 kg•m/s the total momentum will always be 8 kg•m/s. If the two objects interact with each other (collide) the distribution of momentum can change, but the total will always be 8.

Collisions

There are two types of collisions. 

In an elastic collision, two things collide and bounce off each other, going off separately after the collision.

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Collisions (Both types)

In a system involving a collision, we look at the two (or more) objects as they interact. We look at them in isolation, generally. That is to say, we only consider how they are interacting with each other.

In this case, we have a closed system in which momentum is conserved. That means that, where case 1 is before the collision and case 2 is after, 

pinitial = pfinal            (Eq. 1

Since the total momentum of a system is the momentum of all the objects in it, for any case we can say that

ptot = p1 + p2 + …

Inelastic Collisions

In an inelastic collision, two things collide and stick to each other, going off as a single thing after the collision.

Recalling that momentum is conserved, it follows that the momentum of the two objects before the collision must be equal to the momentum of the two objects after the collision. This…

pinitial = pfinal 

So, if there are 2 objects, object 1 and object 2, then the initial momentum is

pinitial = p1 + p2 

And since momentum is found as

p = mv

we can substitute in for p1 and p2

pinitial = p1 + p2 

pinitial = m1v1 + m2v2  

Now, for inelastic collisions, m1 and m2 become stuck to each other. They go off as one thing, with only one mass and only one velocity.

Okay… so, we have 2 objects and a before (initial) and after (final) situation. We could pile up subscrips such as…

m1v1i + m2v2i …

OR… we can change the symbol for initial velocity and keep things less messy… Let's do that…

 So, what we get is…

let u be initial velocity and v be final velocity

 m1u1 + m2u2 = (m1+ m2)v            (Eq. 3

I like to think of it as…

m3 = m1 + m2 

This makes the the equation

 m1u1 + m2u2 = m3v            (Eq. 3 alt

___________________________

We will look at inelastic collisions after we examine the relationship between force, time, mass, and momentum.
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Force and Momentum

Another important consideration of momentum is the forces involved in creating momentum change.

This concept assumes that, for some period of time, a force of some magnitude and direction act on an object causing a change in velocity which causes a change in momentum. Let's have a look!

            p = mv            (Fundamental momentum equation)

∆p = m∆v         (Eq. 1

Now, what causes velocity to change? Acceleration! 

∆v = a∆t        (Eq. 2

 

Now, how is force related to acceleration?

F = ma

a = F/m         (Eq. 3

Plug a from Eq. 3 into Eq. 2 we get

∆v = (F/m)∆t        (Eq. 4

Now, plug Eq. 4 into Eq. 1 to get…

∆p = m • F/m • ∆t        

∆p = F∆t      <--- Cancel the m in above               (Eq. 5

One more version…  Substitute Eq. 1 into Eq. 5:

m∆v = F∆t            (Eq. 6

Different situations will call for the use of different equations, so look at what is given and pick accordingly.

Up to this point, we have examined momentum only from the perspective of a single object. However, the interaction of more than one object leads many interesting outcomes. 

This is usually considered collisions

Elastic Collisions

Since momentum for any object is 

p = mv

then

ptot = m1v1 + m2v2 + … (Eq. 2

If we limit our discussion to two objects, we can roll Equation 2 back into Equation 1 and get

pinitial = pfinal 

m1v1i + m2v2i = m1v1f + m2v2f

Woooo… That's a lot of subscripts!

We can make the formula easier on the eyes if we do this…

let u be initial velocity and v be final velocity

Since mass before and after does not change, then:

pinitial = pfinal 

m1u1 + m2u2 = m1v1 + m2v2            (Eq. 3

 

Equation 3 is easy to solve if you have seven of the eight values given. 

Often, though, you need to find one of the final velocities, but know the value of neither. Yikes!

Through a fairly tedious algebra process, you can arrive at the following formulas:

${\displaystyle }$v1=(m1-m2 ) / ( m1+m2) • u1    +    2m2 / (m1+m2) • u2              (Eq. 4

${\displaystyle }$v2=(m2-m1) / (m1+m2) • u2  +    (2m1 / (m1+m2) • u1                    (Eq 5

You will see this in different forms, if you do research. You might see the second arranged like this:

${\displaystyle }$v2= (2m1 / (m1+m2) • u1   +  (m2-m1) / (m1+m2) • u2                    (Eq 5 alt

It's math! You can rearrange things following the… you know… math rules!