Is more force imparted during an perfectly elastic collision or perfectly inelastic collision? Would more force, less force, or the same amount of force be imparted to another object in an elastic collision compared to an inelastic one? 
The set up is if I was playing pool and could have a perfectly elastic collision when I hit the cue ball into another ball and then changed the cue ball out for something with the same mass, but would only make perfectly inelastic collisions, which would I want to use to have the most force on the stationary ball?
 A: There is no direct relationship between the forces involved and degree of elasticity.
As an example, let's roll two clay balls at each other.  The collision is completely inelastic as they stick to each other.  If we surround the clay balls with a hard plastic shell, the collision becomes almost completely elastic.  The time of interaction is tiny, so the forces felt are very high.
A different possibility is to place a light spring between the two balls.  Again the interaction is very elastic, but this time the forces involved are much lower than the clay ball collision.
What you can say is that the impulse (change in momentum) is greater in the elastic collision, and (assuming mass and speed are held constant) determines the speed of the object ball after the impact.  High force or low force, the object ball will be moving more quickly after the elastic collision.
A: The force on the stationary ball doesn't change, regardless of whether its a elastic or inelastic collision.
The only difference is that in an elastic collision (perfect), the kinetic energy of the system remains the same. Meaning the velocity of the two balls (as a total system) doesn't change (I also assume the mass of the balls is constant). 
In the inelastic collision, some the kinetic energy is turned into sound, very slight deformation of the ball, etc  
So in the second case the total kinetic energy of the system after the collision will be less. BUT the initial force imparted to the ball stays the same, regardless of the collision type.
Hope that helps.  
A: Forces can easily be very high or very low in both types of collisions.
But the question is awkward. “Force” is not really “imparted” during collision in the sense of something being ”passed” from first object to the second however momentum or kinetic energy can be considered as being imparted. The impacting ball also receives an equal and opposite force from the target ball slowing it down. Also there isn’t just one numerical value for force but rather the interaction force varies with time in a messy way depending on many factors including the shape of the body and many other variables and so the average force can take almost ANY value for both extremes of collision either perfectly elastic or perfectly inelastic. The average force multiplied by the time of impact duration is however an easy thing to analyse as it equals the momentum change. And momentum change is easy to describe. However the average force and worse the maximum force (which could cause breakage of an object) could take almost any value depending on many factors.
INELASTIC. Consider a sticky wet lump of chewing gum impacting on your pool ball. The impact force as function of time will vary greatly depending for example on precise shape of gum. If the gum is say elongated in direction of motion the force will be less but last longer. It is a very messy situation in many regards.
ELASTIC. Even perfect elasticity is messy as regards “force”. For starters there are two different ideas behind the word elastic and you can have one or the other or both. There is high elasticity in the sense that all or most the incident kinetic gets stored as elastic energy and returned back as kinetic energy after collision is over. Also even for such perfectly elastic materials then there is the concept of how much a material bends for a given force which is the opposite of stiffness. For example
TWO BILLIARD BALLS conserve kinetic energy (high elasticity type 1) but are very hard with high stiffness and low elasticity type 2. There will be extremely high forces here and as such a danger of shattering.
TWO RUBBER BALLS. Again all kinetic energy can be conserved (elastic type1) but now also the forces are lower but last longer as much more squashing before springing back out (high elasticity type2).
Unfortunately there are two kinds of elastic in physics. Both rubber and billiard ball material are elastic type1 but rubber is obviously more elastic type2.
Actually rubber may be less elastic type1 as it may lose a bit more energy as heat but one can imagine and there exist some really super bouncy balls which have high elasticity type1 and are still no where near the incompressibility of billiard ball.
