# Impact force (without a rope)

1. Mar 26, 2005

### leden

I wonder how could I calculate impact force of an object (eg. a stone) in a free fall impacting on the perfectly inelastic (perfectly hard) floor.
I know:
- m -mass of the falling object
- h -height of the fall
- v -speed at the time of impact
- t -time it takes from highest point to the lowest
- I can also calculate kinetic energy of an object at the impact point
I DO NOT know dwell time (how long the object and the floor are in contact).
So I cannot calculate the impact force with F = \delta p * \delta t.

How can I calculate the impact force?

2. Mar 26, 2005

### PBRMEASAP

You have answered your own question with this statement. You cannot determine the force, or even the average force, of the impact without knowing how long they are in contact. All you can figure out is the impulse delta p. In your example of a stone falling on a perfectly hard (rigid) floor, the stone isn't going to deform elastically, so it must bounce or shatter. In this case, the time spent in contact is nearly zero. Since the impulse is certainly not zero, the average force is practically infinite. So for elastic "billiard ball" type collisions, it is not meaningful to talk about force of impact--momentum/energy transfer is the best you can do.

edit: I should clear up a likely source of confusion. Even though one normally thinks of "elastic" as being stretchy or rubbery, that is not what it means in physics jargon. An elastic collision means that there as just as much kinetic energy after the collision as there was before. This can (and often does) apply to objects that are perfectly hard. An example of an inelastic collision is a lump of clay smashing into another lump of clay and sticking to it. Momentum is conserved, but kinetic energy is not. Some of it was used up in changing the shape of the lumps.

Last edited: Mar 26, 2005