- #1
gld_tlh
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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.
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 energy 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:
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.
- 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 ...
- 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 energy 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:
- 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.).
- This energy that's left over to break things - is it equally applied to each object involved in the collision?
- 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?
- 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.