# Coservation of momentum in inelastic collisions

1. Jun 16, 2009

### kropac1

If the gun club members shoot at steel targets and the bullets are traveling between 800 and 1300 fps and weigh between 112 and 220 grains how far will the fragments travel if they are less than 10 grains in size. The impact on the steel is at approximately 90 deg.

So far testing has shown very little back splatter and most fragment dispersion has been radially from the point of impact. We have also tested with full metal jacketed bullets with no noticable difference in fragment size or dispersion pattern. We are concerned about the possibility of any fragments leaving the club property. Any help is most appreciated.

kropac1

2. Jun 16, 2009

### Andy Resnick

Interesting problem- there's a lot of real-world effects that can't easily be included (terminal velocity of the fragments, size, scatter direction & energy distribution of the fragments, etc), so you need to see how actual, real-world safety calculations are done for firing ranges. But, here's how to solve an idealized situation. The big unknown for me is the fraction of energy transferred to the target and lost during disintegration.

Say the bullet of mass 'm' is traveling with velocity 'v'. For the sake of argument, say 1/2 of the energy is transferred to the target- the remainder is all kept within the bullet fragments, and uniformly distributed amongst the fragments.

The fragment with mass m_f will have a proportional amount (m_f/m) of the total kinetic energy, and so the fragment will be ejected at sqrt(0.5) of the velocity as the incident bullet- call it v_f. (kinetic energy is 0.5mv^2) The maximum distance the object can travel is the fragment ejected at 45 degrees to ground level, and will travel a distance of (v_f)^2/g, where 'g' is gravitational acceleration = 10 m/s^2

Ballpark calculation, saying 1000 fps and 200 grain bullet, 10 grain fragment, I have a fragment ejection at 215 m/s for a maximum travel distance of 4600 meters. This seems long, but again, I only transferred 1/2 the energy to the target (and neglected the energy cost of shattering the bullet).

3. Jun 18, 2009

### kropac1

Andy,
I have to believe there is a lot of the energy used in the deformation and break up of the bullet. Dept. of energy range design recommends a safe zone greater than 15 degrees forward of steel plate point of impact and 50 feet radial distance. We did some testing Monday night by placing a 3/8" chip board enclosure totally surrounding the steel target except for the access port. 50 rounds of factory loaded S&W .40 180 gr bullets traveling at 960 fps were shot at the plate from 21 feet. The chip board verified that splatter conforms to the above pattern. Chip board was for most part failed to contain any fragments.

My thought is to build the enclosure with heavier material such as 3/4" or 1-1/2" plywood and contain all the splatter. Momentum analysis of fragments with the largest mass and depth of penetration into the plywood could be approximated by using a chisel and a drop hammer to approximate penetration and obtain a momentum vs penetration scale. Working back through mass to a velocity for the fragment may give us an idea of the potential maximum travel distance. Care to opine?

4. Jun 22, 2009

### Andy Resnick

My only comment is that the correct solution to your problem is most likely empirical; theoretical considerations are not likely to be sufficiently accurate. Also, safety margins usually have a factor of 3x or so to account for any statistically unlikely events.