Explaining Gun Recoil Using Newton's Second Law

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Homework Help Overview

The discussion revolves around explaining the recoil of a gun using Newton's laws of motion, specifically focusing on the relationship between the firing of a bullet and the resulting backward motion of the gun.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of Newton's second and third laws, questioning which law best explains the recoil phenomenon. There are attempts to articulate the mechanics of recoil and its relation to action and reaction forces.

Discussion Status

Some participants have identified Newton's third law as relevant to the explanation of gun recoil, while others are still considering how to articulate this relationship effectively. Suggestions have been made to refine the explanation regarding the timing of the bullet's acceleration and the forces involved.

Contextual Notes

There is an ongoing exploration of the factors contributing to recoil, including the timing of bullet acceleration and the forces acting on both the bullet and the gun. Participants are encouraged to clarify their reasoning without reaching a final consensus.

tim_mannire
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Homework Statement



A gun that is fired "recoils". Explain using one of Newton's laws.

Homework Equations



??

The Attempt at a Solution



Newton's second law?? F=M/A
 
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tim_mannire said:

Homework Statement



A gun that is fired "recoils". Explain using one of Newton's laws.

Homework Equations



??

The Attempt at a Solution



Newton's second law?? F=M/A
Nope, not the second law. Think about what happens, why does the gun recoil?
 
Hootenanny said:
Nope, not the second law. Think about what happens, why does the gun recoil?

every action has an equal and opposite reaction. there for, it is related to Newton's third law. I'm not sure how to explain this scenario using Newton's third law.
 
tim_mannire said:
every action has an equal and opposite reaction. there for, it is related to Newton's third law.
Correct.
tim_mannire said:
I'm not sure how to explain this scenario using Newton's third law.
What happens when you pull the trigger?
 
Hootenanny said:
Correct.

What happens when you pull the trigger?

The gun has an equal and opposite reaction, when the bullet is fired it gains speed and momentum instantly, causing the gun to lunge backwards towards the shooter.

are there any more contributing factors?
 
tim_mannire said:
The gun has an equal and opposite reaction, when the bullet is fired it gains speed and momentum instantly, causing the gun to lunge backwards towards the shooter.

are there any more contributing factors?
Nope sounds good to me. However, I would suggest that "in a very short period of time" would be better than "instantly". I would also mention that this change in momentum requires a force, the reaction of which is the recoil of the gun.
 
Hootenanny said:
Nope sounds good to me. However, I would suggest that "in a very short period of time" would be better than "instantly". I would also mention that this change in momentum requires a force, the reaction of which is the recoil of the gun.

Ok, thanks very much for your help.
 
tim_mannire said:
Ok, thanks very much for your help.
A pleasure :smile:
 
tim_mannire said:
The gun has an equal and opposite reaction, when the bullet is fired it gains speed and momentum instantly, causing the gun to lunge backwards towards the shooter.

are there any more contributing factors?

You might want to show how the recoil force is calculated: The bullet accelerates down the barrel such that its maximum speed (at the muzzle) multiplied by its mass is equal to the average force x time it was accelerating. This is also the average force on the gun (and on the person holding the gun) during that time:

[tex]m_{bullet}v_{muzzle} = \bar F t[/tex]

Since the bullet accelerates from 0 to muzzle speed in time t, its average speed during acceleration is half of its muzzle speed:

[tex]\bar v = \frac{d_{barrel}}{t} = \frac{1}{2}v_{muzzle}[/tex]

From that you should be able to work out the time as a function of muzzle speed and barrel length and use that to find the expression for the average force [itex]\bar F[/itex]

AM
 

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