What is the optimal weight for a golf ball? I am 32 now, and have forgotten the basic physics formulas we used in school. I am sitting with a question that is bugging me.
If I had to hit two golf balls of different weight with the same force, which one travels further?
I imagine that a $200\;\rm g$ golf ball travels farther than a $1\;\rm g$ ball, and a $200\;\rm g$ ball also travels farther than a $2\;\rm kg$ ball, when all balls are acted on with the same force.
How can I use to work out the optimal mass to get the most distance?
 A: Mark's answer is fantastic and steps through much of the physical reasoning involved wonderfully.  But annoyingly, in this particular case the region of interest is pretty much right where both air resistance starts to become nontrivial and important and the details of the velocity of the ball off of the club also has non trivial behavior, so I decided to numerically simulate some golf drives.
To model the speed of the ball off of the club, I will treat it as a inelastic collision
$$ v_{\text{ball}} = \frac{ ( 1 + r ) v_{\text{club}} }{ 1  + \frac{ m_{\text{ball}} }{ m_{\text{club}} } } $$
Where $r$ is the coefficient of restitution of the ball (which is apparently regulated and set to be at maximum 0.83), $v_{\text{club}}$ is the velocity of the club when it hits the ball, for which I've taken 45 m/s or 100 mph, and $m_{\text{club}}$ is the mass of the club head, for which I've taken 227 grams.  I got these numbers from googling around.
After that I model the dynamics as just projectile motion with air resistance, where I take for my coefficient of drag 0.25, a golf ball diameter of 1.680 inches (42.67 mm).  I obtain for the length of the drive as a function of golf ball mass:

Which has as its optimum something around 80 grams, not the 300 grams a naive scaling would suggest. Though the true answer seems to be fairly dependent on the precise input values, and so you ought to do your own runs for your own swing speeds to figure out your own ball mass.
But the broader point is that taking into account air resistance has a large and nontrivial effect, as we can demonstrate by showing the drive lengths with and without air resistance against one another.

Where here we see that the region of interest and the reason there is a cross over at all is because of the effect of the details of modelling air resistance.
Additionally, as the blue line demonstrates, in the region of interest we are also in the region that that the details of our velocity off of the club has nontrivial cross over behavior.  
A: The most important thing to take away is that you don't want to approach physics in terms of a long list of equations, where the goal is to choose the right one for the question you have at hand. (That's why I edited the title to your question, so it no longer simply asks for an equation.) Physics is a method, and the equations are a tool for answering the question. They aren't all that physics is about. I'll try to illustrate this below.
When you hit a golf ball, there are a few different parts involved. First, there's the impact of the club with the ball. Next, there's the path of the ball through the air, which is affected by both gravity and aerodynamics.
Let's do the easiest part first - gravity. We want to know what happens if you take a bunch of golf balls of different sizes and shoot them on the same trajectory. It's a known fact that everything falls at the same speed, ignoring air drag. Heavy things and light things take the same amount of time to fall, so if you shoot them out at the same angle and speed (i.e. the same "velocity vector"), they will take the same amount of time to come back to the ground. (This time depends only on their initial vertical speed because gravity is a force that pulls only vertically.) They also have the same horizontal speed because we set it up that way, so all the balls follow the same path. Thus, the mass of your golf ball doesn't really matter for gravity.
Next we'll look at air drag. Suppose we take a bunch of balls of different sizes and shoot them all into the air at the same speed and angle, as before. They'll start slowing down because they're hitting the air and the air they slam into slows them down. When is this important, though? Maybe the air drag only slows them down by 1% before they land again, or maybe it slows them down by 50%. We should only worry about air drag if it's significant; if it's only 1% we can ignore it.
Air drag works because golf balls have momentum, and as they travel they give some of that momentum to the air. Momentum is mass*velocity, and we can assume that when a golf ball hits some air, the velocity it gives the air is pretty similar to its own velocity. That means that in order to give away a significant amount of its momentum, the golf ball needs to interact with about the same amount of air as its own mass. If the ball weighs 100g, it needs to interact with around 100g of air before it slows down too much.
Golf balls are about the same density as water (I'm not sure if they sink or float, but they don't do either very aggressively), which is about 1000 times as dense as air. Therefore, a golf ball needs to be hit around 1000 times its own distance before air stops it. The size of a golf ball is a few centimeters, so air drag is important for hits of tens or hundreds of meters. 100 meters is a good, long golf stroke. (edit: actually it's not so long, my mistake) We see that air drag is very important for such a strike; in fact it is a limiting factor. Bigger golf balls will be able to go further, since they all go about 1000 times their own length. We could say that in the air-drag limited regime, golf balls go about $l \propto m^{1/3}$. $l$ is the length, how far they go. $m^{1/3}$ is the cube root of the mass, which tells us the size of the golf ball.
Finally, we need to think about striking the golf ball. What happens is that a golf club comes in and smacks the stationary ball. If the ball is very light, it just bounces right off the golf club at something a little less than twice the golf club's speed. (Roughly this is because it runs away from the club at the same speed the club is going, but I don't want to get into too much detail; the factor of 2 isn't extremely important.) All very small balls behave roughly the same way because they aren't heavy enough to affect the club much. This lets us come to our first result: if all our golf balls are very small, the largest ones will go the furthest because they will all be hit the same speed by the club, but the larger ones go further before air drag stops them.
When the mass of the golf ball becomes comparable to the mass of the club, it doesn't go flying off so quickly any more. If the mass of the club and ball were equal, we might expect the club to come to a dead stop and the ball to go flying off at the same speed - all the motion would be transferred to the ball. (This isn't exactly true since it's never a perfect collision, but good enough for now.) Such a ball would shoot away half as fast as a small ball.
When you shoot out half as fast, how much does that hurt your flight? Not much if air drag was just going to stop you anyway, but a lot more otherwise. If you shoot out half as fast, you'll only be in the air half as long, and going at half speed for half the time go one quarter as far.
For golf balls heavier than the club, the club transfers most of its momentum to the ball. This means its velocity is inversely-proportional to its mass, and the distance it goes is $l \propto m^{-2}$
Thus, we're trying to find a balance. Make the golf ball too big and it won't pop off the club as fast. Make it too small and air drag will stop it. We could try to work out a bunch of equations, but it's usually more useful just to start off with an estimate. Let's try to estimate a distance where air drag and slower-initial speed are just becoming important. Balancing these effects should give a rough estimate of the optimal distance.
The ball should go about 1000 times its own diameter
$$l \approx 1000 d$$
This distance is also the square of the initial velocity, as we already worked out. That's not actually a distance though. We need to divide by gravity to get the units right.
$$l \approx \frac{v^2}{g}$$
Where $v \approx v_{club} \frac{m_{club}}{m_{ball}}$, since we might as well make the ball heavier until being heavier is slowing down its initial speed, and this is the approximate relation we found for that regime.
Converting from $d$ to the mass $m_{ball}$ by $d^3 \rho = m$ ($\rho$ is the density), we get
$$l \approx \frac{v_{club}^2 (m_{club}/m_{ball})^2}{g} \approx 1000 \left(\frac{m}{\rho}\right)^{1/3}$$
which gives
$$ m_{ball} \approx \left(\frac{m_{club}^2 \rho v_{club}^2}{1000 g}\right)^{3/7}$$
We've already approximated the density as roughly that of water, $\rho \approx 1 g/cm^3$, and $g = 1000 cm/s^2$ The velocity of the club shouldn't be wildly different from the faster speeds in ball sports, around 100 mph or 50 m/s or 5000 cm/s. Club heads are around 200g. Plugging these in gives
$$m_{ball} \approx 370g$$
about twice as much as the golf club itself. This would make golf rather awkward, since the club would bounce off the ball. We get a distance of around 80 meters out of this, suggesting that although a golf ball heavier than the 45g actual ball wouldn't really fly all that much further. The effects of air drag and slower takeoff speed fight each other, so that a decent-sized range of golf balls do about the same thing. The 45g ball plays much better without sacrificing too much range, but the weight of golf balls is regulated to prevent getting a distance bonus from heavier balls. In fact, golfers regularly hit balls further than our estimated range, which is largely because it was just an approximation, but could also be partially because the aerodynamics of a golf ball are more complicated than just simple air drag, and the lift generated as they fly and spin can boost their range a bit.
This answer wasn't intended to be the final word; I guess it's pretty likely that someone in the comments will find a significant mistake. My main point for you is that physics isn't about just asking "what's the equation for this? What's the equation for that?" over and over every time a new situation pops up. Instead, it's about understanding the various interactions in a system and finding ways to model their various importances and influences, which can be done either with a great deal of formulas and computation or with very little.
