# how does the force on the rifle compare with the force on the bullet and why?

Hi everyone! On negarinfo and in this post we are going to talk about **“how does the force on the rifle compare with the force on the bullet and why?”**

Thank you for choosing and reading our website. stay with us to the end and find the answer.

## How does the force on the rifle compare with the force of the bullet, and why?

Newton’s First Law:

A body at rest stays at rest unless acted upon by an outside force.

Newton’s Second Law:

F=ma (force equals mass times acceleration.

Newton’s Third Law:

For every action, there is an equal and opposite reaction.

That means if you want to push a 147 grain (.336 ounce) bullet at 2,800 feet per second then you’re going to exert that same force on your rifle. The math on how much force you FEEL gets a little advanced, and depends on the weight of the rifle and the area of the buttock in contact with your shoulder.

The reason most rifles (outside of something like the .577 T-Rex) don’t knock the shooter down is because the forces exerted only last for a few microseconds. For instance, at 2,800 feet per second with a two foot long barrel, the bullet takes about .0007 seconds for the bullet to leave the barrel. If your rifle weighs 11 pounds then you’ll be experiencing roughly 30,800 pounds of force for the .0007 seconds the bullet is in the barrel, while the bullet is experiencing roughly 58.8 pounds of force (if I did my math right). Since the rifle is so heavy, it takes longer to accelerate. This means it’s quite likely that the bullet is out of the barrel before you feel any force at all.

*The Laws of Motion at the top are simplified to relate directly to the question.

MORE POSTS TO READ:

- to apply the fitt principles to eating habits we use
- the core of a computer’s operating system is called the
- difference between sociopath and psychopath
- the return of swallowed food to the mouth is known as
- on the windows desktop what’s the globe or bottom left menu called?

## how much force does a bullet have on impact?

By force, I assume you are talking about the kinetic energy a bullet imparts to its target upon impact. This is expressed in foot-pounds.

That force is infinitely variable because it depends on a number of different things in order to calculate said force. In other words, it depends on how heavy the bullet is, and how fast it’s going when it makes its acquaintance with the target.

This can be calculated by multiplying the weight of the bullet (in grains) by the velocity squared and then divided by 450240 (which has something to do with the gravitational acceleration constant which is essentially the same all over the world and gives me a headache trying to figure out how to explain it in English that is easily understood). Just trust me, it works.

Suppose you have a 9mm pistol and are firing a 115 grain bullet. That establishes the mass.

Now multiply that number by the velocity squared. Let’s pretend the velocity is 1,200 ft/s., so that number would be 1,440,000.

Divide that number by 450240 and you’d get just a smidge under 368 ft.-lbs. of energy.

**Mass x Velocity x Velocity / 450240.**

(115 x 1,200 x 1,200 = 165,600,000 / 450240 = 367.803838)

## another way of calculating the impact of bullet

I can think of two ways to find the ‘average’ force of a bullet ‘on impact’. The first is to calculate the force averaged over the time required to stop the bullet. This time interval can be found by estimating (or measuring) the initial velocity of the round as it first strikes the target. The average stopping speed is then 1/2 of this value and the stopping Time is Distance / average speed.

Now we need the momentum of the round which is mass times speed = momentum.

Now we can calculate the average force as = momentum/ Time.

If you are careful to use proper units everywhere you will get the correct answer. (Easy, proper etc. units would be kg for mass, meters for distance, seconds for time, giving force in newtons.)

The second method of Averaging over Distance is a bit more straightforward but less commonly done.

Calculate the kinetic energy of the round = 1/2 m v^2 and divide this by the stopping distance. This will give the average force in newtons averaged over Distance.

**why does a bullet move from a gun according to the third Newton’s law?**

#### ANSWER 1 :

**Newton’s third law is a law of the forces acting on an object**, which are the product of mass and acceleration.

So let’s take a closer look at your bullet example.

Initially the bullet and gun are in contact with each other, so we consider them as a **single closed system.** When the gun is fired, a combustion reaction heats up the air causing it to expand rapidly and shooting the bullet.

We can recast Newton’s third law from forces to change in **momentum. Momentum is defined as mass times velocity.**

Before the gun is fired, everything is at rest. The bullet experiences a change in momentum that must be matched by a change in momentum of the gun. This is Newton’s third law in action. **The change in momentum of the gun is called recoil.** Because the gun is much more massive compared to the bullet, it does not recoil as noticeably. You can see this whenever someone shoots a gun; they need to ground themselves to hedge against the recoil.

You can apply the same principle to how a rocket moves in space. The fuel that is combusting at the tail-end of the rocket is what is pushing on the rocket and propelling it forward (the same is true when the rocket is being launched, unlike the misconception that the rocket pushes off the ground).

#### ANSWER 2 :

You have almost stated Newton’s Third Law correctly! What you forgot is that **the action is applied from object A onto object B and the reaction arises from object B to A!** So, what we refer to as **“action-reaction”** is a pair of forces that is applied to two objects, not one, and this is why they cannot cancel.

Let’s think of two examples to convince you. Let’s assume you play pool (billiards) and all the balls have the same mass. And let’s say I hit a striped ball on a solid one. Initially, the striped ball is moving, and the solid is stationary. But after the collision, the striped will be stationary and the solid will move. *How did the solid ball move?* **It moved because the striped ball applied a force on it, in Medieval terms called “an action”.** *How did the striped ball stop?* **It stopped moving because the solid ball applied a force on it, in Medieval terms, “a reaction”.**

“Action” and “reaction” are nothing more than two forces. But Newton gave them this name because they are **special forces, they have same magnitude and opposite directions (and are applied at different centers of mass)**. And this is the reason why if the two billiard balls have the same mass, the striped ball will stop immediately and the solid will move with the velocity that the striped had, assuming the collision happened at the center line. If not, new physics of rotation had to be discovered!

Now, the second example, is more complex in reality but may help you to think of “action-reaction” pairs is easy. Imagine you are taking your dog (or other favorite animal) for a walk, so there is a leash in between you. The leash can only sustain tension, not compression. You pull the dog towards you because you want to keep moving (you perform the action). But the dog resists (reacts) and pulls towards its favorite tree / street corner.

The dog experiences your force (action) via the leash, and you experience its force (reaction). But these forces can only be the same in magnitude! But you are definitely getting pulled by the leash (and the dog) and only the friction of the ground and your muscle power helps you to stay put and not fall ! So, definitely, you see that you who applies the action, experiences a force that is not being cancelled.