Momentum is the driving force that a moving object has. It is the force with which the
object moves against resistance. The multiplication of an object's inertia and velocity equal
its momentum; and, since linear inertia equals mass, we come up with this equation.
p = m * v
You're probably wondering why I didn't just say that momentum equals mass times velocity,
instead of involving inertia. The reason for this will be evident in the last lesson of this
section, when I talk about What is the momentum of a 5kg object traveling at 15m/s?
p = 5kg * 15m/s
p = 75Ns
The "Ns" after the 75 is not some new unit, its just stands for That's where this neat thing called conservation of momentum comes in. What does it mean
that momentum is conserved? It means that if you have a certain amount of momentum before a
collision, then afterwards you will have the exact same amount.
For the two sides to even out though, you need to make sure that you use a closed set of
objects. That means that if you use a bat and a baseball at the beginning of the equation, you
have to end the equation with a bat and a ball, you can't end it with a bat and a glove. You
have to do another equation for a ball and a glove. That is, unless you start with a ball, a bat,
and a glove. It all depends on what you want your set of objects to be. But, as long as you
use the same ones before as after, your momentum will always be conserved.
We'll get to a sample problem in a minute, but first I want to talk about two different types
of collisions. They are elastic collisions and inelastic collisions. In elastic collisions, the
objects make impact and then join to become a larger object. For instance, if two balls of clay
are thrown at each other and when they hit, they become one big ball of clay. Inelastic
collisions occur when the objects collide and bounce off each other. An example of this would be
two pool balls hitting each other and then ricocheting off into the pockets. (Hopefully!)
There is also a situation where two objects are together, and then they break off and go
their separate ways. For example, a person is riding on a sled and then jumps off the back of it
separating the two.
Now let's use all this newfound knowledge in a sample problem.
A 3kg gun fires a 5g bullet at 300m/s, assuming there is no friction to slow it down, what
is the guns recoil velocity?
Okay, in this situation, the two objects start together, so let's write down that part of the
equation, remembering that you have to convert grams to kg.
3kg(0m/s) + .005kg(0m/s)
Now after the gun is fired, we know that the bullet goes off at 300m/s, and the gun recoils at
some unspecified velocity. Let's write that down.
3kg(v) + .005kg(300m/s)
According to conservation of momentum, these two are supposed to be equal, so let's set them
equal to each other and solve for "v".
3kg(0m/s) + .005kg(0m/s) = 3kg(v) + .005kg(300m/s)
0Ns = 3kg(v) + 1.5Ns
-1.5Ns = 3kg(v)
-0.5m/s = v
The gun recoils at -0.5m/s. See how useful momentum is?
Next we get into Impulse, which is the change in momentum.
