Finding force exerted in an Inelastic Collision I did a lab today in Physics in which we launched ball from a spring loaded cannon directly into a pendulum that captured the ball, held it, and swung upwards with it (representing a totally inelastic collision).  One question in particular has confused me:

If the collision between the projectile and pendulum had lasted 1 millisecond, what would the average force have been which the projectile exerted on the pendulum for the long-range case?  

My attempt at a solution is as follows: From all the searching I've been doing online, I've found the equation $F = {{p_f}-{p_i}\over {t}}$.  I know $p_f$, $p_i$, and I'm given t.  Is my understanding right?  Can I go right ahead and crunch these numbers, or do I have an incorrect equation?
 A: Yep, average force $\langle F \rangle=\frac{p_f-p_i}{t}$
There are two things going on here:
Impulse
For any collision, it is convenient to define a quantity called "impulse". Most collisions consist of large, varying forces acting in a short time. These are hard to calculate, so they can be encoded into the "impulse". The impulse is the change of momentum on a body during a collision. Due to the identity $\vec F=\frac{\rm d\vec p}{\rm dt}$, we get:
$$\vec J=\Delta \vec p=\int\rm{d}\vec p=\int\vec F\rm{d}t$$
Impulses are pretty useful in multi-body problems. Especially when string/friction/etc are present; since we can use them in place of forces and conserve momentum.
Time-averaging
For any quantity $X$ dependant on time, the time average of the quantity is:
$$\langle X\rangle=\frac{\int X\rm{d}t}{\int \rm{d}t}$$
with limits of integration as the time you want it to be averaged over.
In this case, the numerator becomes the impulse (change in momentum).
Note that for any quantity $X=\frac{\rm{d}y}{\rm{d}t}$, $\langle X\rangle=\frac{\Delta y}{\Delta t}$
This is useful for average speed (total distance traveled by total time), average acceleration, and, of course, average force.
