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

**Momentum: Notes and Such**

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

*(You gotta do better than 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 ispis momentum,mis mass, andvis velocity.

kg•m/s

*unless something changes*. Just watching a football player run at a constant velocity isn't very interesting.

*this is important!*)…

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

p1 = p2

*p = mv*, then it follows that

m1v1 = m2v2

**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 = m2v*

*2*in momentum from a first condition to a second (and third, fourth, fifth… if so desired).

∆p = p2 - p1∆p =m2v2 -m1v1<--- plug in(eq. 1

m1 = m2

∆p = m(v2 - v1)(eq. 2

v2 - v1 = ∆v

∆p = m∆v(eq. 3

*IMPORTANT NOTE*### Momentum In Action—A First Look: Inelastic Collisions

**within a closed system momentum is conserved.**

### Collisions

**elastic collision,**

*two things collide and bounce off each other, going off separately after the collision.***inelastic collision,**

*two things collide and stick to each other, going off as a single thing after the collision.*### Collisions (Both types)

pinitial = pfinal (Eq. 1

ptot = p1 + p2 + …

### Inelastic Collisions

**inelastic collision,**

*two things collide and stick to each other, going off as a single thing after the collision.*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 + m2This 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.*___________________________

### Force and Momentum

**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

∆v = a∆t (Eq. 2

F = maa = F/m (Eq. 3

∆v = (F/m)∆t (Eq. 4

∆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

### Elastic Collisions

p = mv

ptot = m1v1 + m2v2 + … (Eq. 2

pinitial = pfinalm1v1i + m2v2i =m1v1f + m2v2f

*Woooo… That's a lot of subscripts!*

let u be initial velocity and v be final velocity

pinitial = pfinalm1u1 + m2u2 =m1v1 + m2v2(Eq. 3

*Yikes!*

( 2 m 1 / ( m 1 + m 2 ) • u 1 + ( m 2 - m 1) / ( m 1 + m 2 ) • u 2 (Eq 5 alt