My question is this: How do I calculated pounds per square inch at impact of a drop hammer on a coining die?

The weight is 50 pounds, the drop is 5 feet. It will impact a coining die that has an area of 1.25 square inches.

Keep it simple. I am 39 years away form high school physics, and didn't need it in college!

You have almost everything you need to find the answer, but the replies I have seen are correct. When you drop the 50 lb hammer onto something from a height of 5 feet it gains kinetic energy while falling. When it impacts the thing it is to strike, the thing takes on the weight of the mass and also must take on the additional force from the mass having energy that must be rid of through some means.

The actual equation to use that would tell you the peak force of impact when it strikes is the relationship between an impulsive force and a change in momentum. You've probably seen F = ma, well re-write it as F = m(deltaV/deltaT) and multiply both sides by deltaT getting: FdeltaT = mdeltaV. This states that the change in momentum of some mass yields the force applied times the time it took to change the momentum from what it was just before the strike to what it is just after. Just before is easy to find, as you can find the velocity the mass has from being dropped from 5 feet easily, and the velocity of the mass after the collision is zero. So you know deltaV it is the velocity just before it strikes the coin minus zero velocity. The mass is also known because you have the weight. That means you need only the time it takes for the hammer to come to a rest from what its maximum velocity was just before the strike to zero velocity.

Given that you are working with hard materials the time it takes for the hammer to come to rest from some speed will be small or very quick. Therefore the force it applies upon the coin is:

F = m (Vf - Vi)/(tf - ti) = m (0 - Vi)/(tf - 0) = -mVi/tf where tf is the final time or the time it takes just at the moment the hammer is striking the mass to come to a rest upon doing so; and Vi is the velocity of the hammer the moment just before it strikes the coin. It always looks as if the hammer simply stops when it hits the coin, but that isn't really true. The coin would have to be beyond the hardest substance known to man, it would have to be a completely rigid body that doesn't flex in anyway. The implications of such an event would be that the force it strikes the coin with would be infinitely large in value for a period of time that is zero seconds long. That doesn't make too much sense; we use the idea a lot in engineering as what we call the impulse function. It is a function that inputs an infinite amount of something into something else, but does so in no time. There is actually a value that can be obtained from this, because the infinitely sized peak is spread over no distance, the area under the curve of that peak is usually an actual value.

In your case, the coin will give a little and therefore the time it takes for the hammer to stop will be non zero. You need that time if you want to know what the force is. Since the time it takes for the hammer to stop will be very small the force it imparts upon striking the coin should be great. With every strike more thermal energy is transferred into the coin and hammer, heating them both up some, but the initial force it strikes with is the same every time if dropped from the same distance and is stopped in the same amount of time.

To mathematically find the time would be a nightmare. You would probably be better off putting a load cell where the coin goes and then drop the hammer upon it. The load cell will read the force it sees upon it and you can set it up such that the maximum force the cell sees is displayed.

I love complex mathematics, but I wouldn't touch trying to find the time it takes that mass upon striking the coin to come to a rest through a mathematical analysis, it would drive you crazy and you'd probably get the wrong value anyway; or even if you got something somewhat close, being it is such a small value, a small difference will relate back to large differences in the calculated force upon the coin.

If not a load cell, a simple scale would work. You can set the scale there in place of the coin, make sure the scale is one that when it feels the maximum force applied upon it; has a way of keeping that value for you to use. If it is a digital scale there is probably some button on there that can be pushed so that it registers and holds the maximum force seen upon it. If it is an analog scale, it can have something like another hand that moves with the first and stays at the maximum displacement the needle moves to.

Just a couple of ideas that may save you some time and effort.

Craig