# Little Help - Every action equal and opposite reaction?

• JasonKnight
In summary, the conversation discusses the force and recoil of a bullet fired out of a gun, the distance it would travel before gravity brings it to the ground, and the average speed of shotgun bullets after being fired. The conversation also delves into the details of a potential murder involving a gunshot wound and the placement and recoil of the gun. The conversation concludes with a mathematical approach to calculating the velocity and force of a gun based on the weight and velocity of the bullet.
JasonKnight
Ok, I understand what this phrase means I have no problem with that, But what I am trying to figure out is the following.

The equations I am looking for would be somehting like this:

1. BUllet being fired out of a gun and the force that the gun would recoil back if there was little or no friction.

2. after the gun was fired, and if the gun was at a set angle and a set hieght how far would it fly before gravity made it hit the ground.

Also, what is the averate speed of shotgun bullets after they are fired? I can't seem to find this information.

JasonKnight said:
1. BUllet being fired out of a gun and the force that the gun would recoil back if there was little or no friction.

2. after the gun was fired, and if the gun was at a set angle and a set hieght how far would it fly before gravity made it hit the ground.

Also, what is the averate speed of shotgun bullets after they are fired? I can't seem to find this information.
Hi Jason;
I can't help with any of the math stuff, but I know a bit about guns. First, by shotgun 'bullets' do you mean regular shot loads or slugs? If slugs... standard, Breneke style or sabotted? In any event, all firearms velocities and energies vary according to how they're loaded. On average, figure a shotgun to output about 1,200fps.
I don't know whether or not you're supposed to ignore this, but a very large part of the recoil in a gun is due to reaction torque from the rifling digging into the bullet (not applicable to smooth-bores like shotguns). It's not as noticeable in heavier rifles, but my .44 mag Super Blackhawk revolver twists a good 50-60 degrees with a relaxed grip, and my friend's .50AE Desert Eagle auto like to screw his arm out of its socket.
How far the gun would fly depends upon how heavy it is and how it was held in the first place, along with what effect the torque has upon it.

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Its just conservation of momentum: whatever momentum the bullet has, the gun has: m1v1=m2v2

But bear in mind that the directions are opposite, otherwise you might calculate the speed towards the ground of the gun when you fire the bullet down instead of the speed upwards, making you're calculation useless. (been there done that (in an exam :P)

ok, IDK how to throw out the idea of what I am tryig to get at without explaining myself.

To put the information blunt, My unchle was killed by a 12 gauge shotgun. The records indicate that it was a suicide, but from looking at information and how everything was placed it looks more like a murder... Anyway, this is my delema

this is how the set up was.

My Unchle was on his back on the bed, (shot in head) won't go into any details. anyway... Gun is sitting on the ground leaning straight up (like a 80 degree angle) and just sitting there...

My problem is, THe shot gun looks like it didnt recoil... I have shot a 12gauge before and it kicks enough to thro you back on your rear if your not holding it right. but when there is nothing to hold the gun still, the gun should have bounced or at least through it's self backwards...

basically, if my theory correct to assume that how the gun was placed, having nothing holding onto it, except the floor holding it up and it leaning on the bed, that it looks "PLACED" I mean its not even leaning up against his legs. like if it would have fell over to one side...

Any ideas, theorys, anything, would be great.

JasonKnight said:
Any ideas, theorys, anything, would be great.
In all seriousness, your best approach would be to talk to the scene investigators and maybe the coroner to find out how and why they arrived at their conclusion. They have far more knowledge about both the field of expertise and the scene itself than any of us.
If you really, really want help from here, you'll have to give a lot more detail, including accurate scale drawings of where everything was, the exact size, weight and design of the gun, the exact loading data for the shells (if any remained unfired), your uncle's physiological measurements, the exact angle and placement of the wound, and a lot of other things that I can't even think of right now. To try and give you any kind of an answer without that information would just be pure guesswork. Sorry.

like I asked before, I wasnt asking for answers, just the exact equations I would need to figure it out. I can do everything else my self.

Any object in freefall follows the path:

$$x(t) = x_0 + v_0t + \frac{gt^2}{2}$$

For conservation of momentum, provided there are no forces acting on the system:

$$m_{bullet}v_{bullet} = m_{gun}v_{gun}$$

Solving for the velocity of the gun:

$$v_{gun} = \frac{m_{bullet}v_{bullet}}{m_{gun}}$$

A Glock 18 weighs 9.87oz = 0.28kg, and a bullet weighs 7.7g = 0.0077kg, and fires at 2000ft/sec = 609m/sec. Then:

$$v_{gun} = \frac{m_{bullet}v_{bullet}}{m_{gun}} = \frac{(0.0077kg)(609m/sec)}{0.28kg} = 16.7m/s$$

This is completely ignoring any frictional effects and assuming a perfect barrel.

The gun then has momentum $16.7m/s * 0.28kg = 4.676$ which must be dissipated by a force, your hand.

$$F t = mv = 4.676$$

Lets say it takes about 0.1s to completely absorb the shock of the gun, so then:

$$F t = m v$$

$$F = \frac{mv}{t} = \frac{4.676}{0.1} = 46.76N$$

Which is equivalent to holding a 4.7kg (10lb) object against gravity.

JasonKnight said:
like I asked before, I wasnt asking for answers, just the exact equations I would need to figure it out. I can do everything else my self.
Good in princple, but everything is abolutely useless if you don't know the original conditions.

Without reading the other 8 posts, my suggestion would be to go to the store and buy or "flip thru" some guns & ammo mags; the ones that have handguns, or more to your interests perhaps, shotguns.
These should answer these question, in particular, without reading too far. There are web-sites for gun lovers, like me, where you may get your questions answered quite readily.
Good luck.
L8R

## 1. What is the law of "every action equal and opposite reaction"?

The law of "every action equal and opposite reaction" is Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.

## 2. How does the law of "every action equal and opposite reaction" apply in everyday life?

This law can be seen in many everyday situations, such as when you push against a wall, the wall pushes back on you with an equal force. When you walk, your feet push against the ground and the ground pushes back, propelling you forward. It also explains the recoil of a gun when a bullet is fired.

## 3. Is the law of "every action equal and opposite reaction" always true?

Yes, this law is a fundamental principle of physics and has been proven to be true in countless experiments and observations. It applies to all types of forces, including gravitational, electromagnetic, and contact forces.

## 4. Can you give an example of the law of "every action equal and opposite reaction" in action?

One example is when you throw a ball. As you release the ball, your hand exerts a force on the ball, propelling it forward. At the same time, the ball exerts an equal and opposite force on your hand, causing your hand to move backwards slightly.

## 5. How does the law of "every action equal and opposite reaction" relate to the conservation of momentum?

The law of "every action equal and opposite reaction" is closely related to the law of conservation of momentum. This law states that the total momentum of a system remains constant, as long as there are no external forces acting on the system. This means that the momentum of the first object before the interaction is equal to the momentum of the second object after the interaction, taking into account the direction of the forces.

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