# Silly collision question; hit by turkey vs. hit by bullet

1. Jul 30, 2010

### gld_tlh

Hi, everyone.

Please help us work through a hypothetical situation that we came up with at work. We're chemists, not physicists, and none of us have had to do physics equations since undergrad. So, we're kind of hopeless without help.

We were thinking about the following scenario. In the woods, there is a hypothetical hunter (if you need details, assume that he weighs 80 kg, is motionless, forgot to wear orange, and has very bad luck). One of two unfortunate scenarios befalls him.

1. A very large wild turkey, confused, flies into him at full speed. The turkey is a full-grown tom of the largest size, weighing 25 pounds (11.34 kg). It is flying as fast as it can go, 55 mph (24.59 m/s, which we're going to assume is a constant speed). Or ...
2. His hunting buddy accidentally shoots him at point-blank range with a high-powered rifle. The rifle is a .30-06, probably the most powerful rifle in common usage. It fires a 150 grain (0.00972 kg) bullet at 2910 ft/s (886.97 m/s, again, let's assume a constant speed).

Obviously, either of these situations is going to be very painful, and possibly lethal, for our hapless hunter. Common sense tells us that the rifle shot, however, will be the much worse option of the two for him. We're trying to figure out if the math confirms common sense's hunch. If so, we'd like to know roughly how much worse the gunshot is than the collision with the turkey.

We started with kinetic energy. The turkey has 3427.63 J of kinetic energy, about 90% of the 3823.36 J of kinetic energy possessed by the bullet. Since those numbers are so close, we've made the assumption that kinetic energy alone is a poor predictor of the damage done by a collision.

The only other step we've been able to take is that not all of the kinetic energy is available to do damage to the hunter. Some of it is used to impart kinetic energy to him, by knocking him backwards. Assuming a perfectly inelastic collision, momentum and total enegy are conserved, but kinetic energy is not. So, we assume that any decline in kinetic energy is energy available to damage the objects involved in the collision. Since the turkey has more mass, it has more momentum, and therefore more of the collision's kinetic energy remains as kinetic energy. In inelastic collisions, we are calculating that the hunter+turkey mass will have 425.54 J of kinetic energy after the collision, while the hunter+bullet mass will have 0.46 J. This leaves 3002.09 J of energy to damage the hunter in the collision with the turkey, vs. 3822.89 J of energy to damage him in the collision with the bullet. So, now we're down to the turkey being about 79% as bad for the hunter as the bullet is, which still seems too high.

So, here are our questions:

1. These collisions aren't going to be perfectly inelastic. The collision with the bullet will be very close to perfectly inelastic (assuming that the bullet remains lodged in the hunter), but the turkey is going to bounce off. Can anyone give us a ballpark guesstimate of the coefficients of restitution for these collisions? That's not the sort of thing you can Google (OK, you can Google it - we tried - but you won't get useful answers that way.).
2. This energy that's left over to break things - is it equally applied to each object involved in the collision?
3. We have a hunch that what we need to look at is power, not energy. After all, if I flick you with my finger 100 times, I've transferred as much energy to you as if I'd punched you, but it doesn't hurt you nearly as much. Power is energy transferred over time. How can we determine how long it takes these collisions to transfer their energy into the hunter's body?
4. Does the size of the colliding object have anything to do with this? If I punch something vs. stab it with a knife, the energy transferred is about the same, but the knife is more effective at breaking things. We have no idea how this comes into play with the math, though. We have equations for pressure, but they all involve force, which involves acceleration, and we have no idea how that applies here.
Thanks so much for your help!

2. Jul 30, 2010

### mgb_phys

The bullet is bad for you because when it goes into you it cuts all sort of squishy internal plumbing that you would prefer to have unperforated.

Perhaps it would be more useful to think of being shot while wearing a bullet proof vest. The bullet will transfer the same KE to you as before but the damage will be much more consistent with being hit by a turkey.
Getting shot with any decent calibre round while wearing a bullet proof vest is still not pleasant.

3. Jul 30, 2010

### gld_tlh

Well, yes, a bullet puts a hole in you. That I knew. But, mathematically speaking, why?

Wound theory is controversial, since the science is still poorly understood, despite all of the research put into it by the world's militaries. But the prevailing theory is that most of the injury done by a high-speed bullet is caused by something caused "hydrostatic shock." This occurs when the bullet travels through tissue so fast that it treats the tissue like a liquid, resulting in waves traveling through the body. The disruption done by these waves is what injures or kills. All of this is a fancy way of saying that the bullet transfers its energy to its target as part of the collision. Which is back where we started.

A protective garment made of ballistic cloth works like a shock absorber. The web of fibers catch the bullet, slowing it down by absorbing its kinetic energy and dispersing this energy through the weave of the garment. That is, the bullet is left with less kinetic energy to transfer to the target via hydrostatic shock. The only type of body armor that is capable of reliably stopping a .30-06 projectile is NIJ Class IV, which contains a trauma plate. This is a hard plate that is physically interposed between the bullet and the wearer. Despite what you see in movies, it is a one-use device; it will shatter when shot. Much of the energy transferred in the collision is absorbed by this plate, hopefully leaving the bullet with so little kinetic energy that the ballistic fibers behind it will be able to slow down the bullet.

We may not know physics, but we do know guns. That's how we started on this conversation amongst ourselves.

So, again, we're back where we started. Why, in mathematical terms, does the bullet, which has about the same amount of kinetic energy as the turkey, transfer more of it to the target? How much more is transferred and how quickly?

4. Jul 30, 2010

### mgb_phys

Momentum, pressure and ultimate tensile stress!

You extract the momentum from the incoming round (or chicken)
Depending on how quickly you 'stop' the incoming body the greater the force - that's why padding helps - greater stopping time - less acceleration - less force.

Force divided by area is pressure - smaller area is larger pressure.
Pressure has the same units as stress - at some stress the target material yeilds.

5. Jul 30, 2010

### gld_tlh

OK,

So, in the examples we've got, is there a way to determine or guesstimate how quickly the projectile is going to come to a stop?

6. Aug 1, 2010

### PaulS1950

unless your hunter were wearing the very best in ballistic clothing the bullet will go through him. A 3006 round will go through a number of people before it stops. The penetration and resulting damage is what kills - not the energy. That is why an arrow kills almost as well as a bullet - it too cuts and tears vital tissues.
The very best way to decide which is worse is to calculate the momentum and divide by the cross-sectional area of the projectile.
The turkey collision would most probably be fatal due to trauma but without penetration. The extensive bruising of internal organs would likely cause enough damage that they would quickly shut down resulting in hemoraging and some nerve damage.