Re: A bullet CANNOT knock someone/thing over can it?
In the end, it is force that pushes/drops something and this can be calculated using either momentum or energy. Both are conserved during a collision. The kicker with energy is that it is TOTAL energy that is always conserved. Energy can change forms during the collision. Kinetic energy is not always conserved.
F=M*A
A=F/M
To determine how much the object gets "pushed", you need to determine it's acceleration. This is a function of its mass and how much force it experiences.
Fo=Fb (Force object experiences = Force bullet delivers)
Ao=Fb/Mo (Acceleration of object hit = Force bullet delivers / Mass of object getting hit)
Ao=Mb*Ab/Mo (Acceleration of object hit = Mass of bullet * Acceleration (deceleration) of bullet / Mass of object getting hit)
So, for a given impact force, the larger the object getting hit is, the less it will accelerate (all else being equal). The mass of the object is easy, the force the bullet delivers to it is not so easy. This is what needs to be calculated from the momentum equations. It is highly dependant on the bullet's terminal performance. How the bullet decelerate (negative acceleration) determines how much force the object experiences (i.e. energy gets transfered). This is the idea behind designing cars that crumple during an accident. The deformation of the body results is a smaller deceleration and thus, less force on the occupants of the vehicle.
In the case of your 50 BMG hitting the deer above, you would use the equation for elastic collision since the bullet does not stay in the deer. If you make some assumtions about the velocity of the bullet after the impact, you can calculate bullet deceleration based on the width of the deer.
A(avg) = (V2-V1)/dt
If the bullet gets stuck in the animal (varmint bullet), you need to use the equation for inelastic collision. You need to guess at how much it penetrates and how long it takes for its velocity and the objects velocity to become the same. Then you can determine its deceleration.
A(avg) = (Vf-Vb(i))/dt
The problem with doing this whole exercise using simple momentum equations is that it it requires you to make all kinds of assumptions and predictions about the terminal performance of the bullet. You're simplifying something that is in reality, very complex. A really detailed computer model that takes all things into account could probably predict it more reliably, but that isn't what the OP is looking for. If you want to take the quick and dirty practical approach, then do what I originally suggested: take known results on an animal/object of a given size, scale the weight and velocity of the second bullet based on the energy/bodyweight ratio. And most importantly; compare similar bullets.
*The bullet that passes through an object transfers less energy than one that stays in it (bullets of equal mass and velocity)
*The bullet that penetrates more, experiences a lower deceleration, and transfers less energy than the one that stops in the object sooner (bullets of equal mass and velocity)
A 750 gr. AMAX is a solid that doesn't expand; it doesn't transfer energy as efficiently as a varmint bullet. A frangible bullet is probably closer.
If you make a general statement that a bullet cannot knock someone/thing over, as the title says, that is incorrect. It is dependant on the size of the person/animal/object, the energy of the bullet and how well that energy gets transferred. It can also depend on where the object's center of mass is if the force is small relative to the object's mass. There isn't a black and white rule that can be applied for any bullet hitting any object.
If you site specific examples; size of person/animal/object along with weight, velocity and type of bullet, then you can make claims as to how much that person/animal/object will or will not be moved by the impact. The 308 does not make a moose budge, but it does send a gopher flying. Some size of animal in between them will simply get knocked down.
In the end, it is force that pushes/drops something and this can be calculated using either momentum or energy. Both are conserved during a collision. The kicker with energy is that it is TOTAL energy that is always conserved. Energy can change forms during the collision. Kinetic energy is not always conserved.
F=M*A
A=F/M
To determine how much the object gets "pushed", you need to determine it's acceleration. This is a function of its mass and how much force it experiences.
Fo=Fb (Force object experiences = Force bullet delivers)
Ao=Fb/Mo (Acceleration of object hit = Force bullet delivers / Mass of object getting hit)
Ao=Mb*Ab/Mo (Acceleration of object hit = Mass of bullet * Acceleration (deceleration) of bullet / Mass of object getting hit)
So, for a given impact force, the larger the object getting hit is, the less it will accelerate (all else being equal). The mass of the object is easy, the force the bullet delivers to it is not so easy. This is what needs to be calculated from the momentum equations. It is highly dependant on the bullet's terminal performance. How the bullet decelerate (negative acceleration) determines how much force the object experiences (i.e. energy gets transfered). This is the idea behind designing cars that crumple during an accident. The deformation of the body results is a smaller deceleration and thus, less force on the occupants of the vehicle.
In the case of your 50 BMG hitting the deer above, you would use the equation for elastic collision since the bullet does not stay in the deer. If you make some assumtions about the velocity of the bullet after the impact, you can calculate bullet deceleration based on the width of the deer.
A(avg) = (V2-V1)/dt
If the bullet gets stuck in the animal (varmint bullet), you need to use the equation for inelastic collision. You need to guess at how much it penetrates and how long it takes for its velocity and the objects velocity to become the same. Then you can determine its deceleration.
A(avg) = (Vf-Vb(i))/dt
The problem with doing this whole exercise using simple momentum equations is that it it requires you to make all kinds of assumptions and predictions about the terminal performance of the bullet. You're simplifying something that is in reality, very complex. A really detailed computer model that takes all things into account could probably predict it more reliably, but that isn't what the OP is looking for. If you want to take the quick and dirty practical approach, then do what I originally suggested: take known results on an animal/object of a given size, scale the weight and velocity of the second bullet based on the energy/bodyweight ratio. And most importantly; compare similar bullets.
*The bullet that passes through an object transfers less energy than one that stays in it (bullets of equal mass and velocity)
*The bullet that penetrates more, experiences a lower deceleration, and transfers less energy than the one that stops in the object sooner (bullets of equal mass and velocity)
A 750 gr. AMAX is a solid that doesn't expand; it doesn't transfer energy as efficiently as a varmint bullet. A frangible bullet is probably closer.
If you make a general statement that a bullet cannot knock someone/thing over, as the title says, that is incorrect. It is dependant on the size of the person/animal/object, the energy of the bullet and how well that energy gets transferred. It can also depend on where the object's center of mass is if the force is small relative to the object's mass. There isn't a black and white rule that can be applied for any bullet hitting any object.
If you site specific examples; size of person/animal/object along with weight, velocity and type of bullet, then you can make claims as to how much that person/animal/object will or will not be moved by the impact. The 308 does not make a moose budge, but it does send a gopher flying. Some size of animal in between them will simply get knocked down.