Explaining Gun Recoil Using Newton's Second Law

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SUMMARY

The discussion centers on explaining gun recoil through Newton's laws, specifically identifying it as a demonstration of Newton's third law of motion. When a bullet is fired, it gains speed and momentum, resulting in an equal and opposite reaction that causes the gun to recoil backward. The participants clarify that the recoil force can be calculated using the relationship between the bullet's mass, its muzzle speed, and the time of acceleration, expressed as mbulletvmuzzle = &bar;F t. This highlights the importance of understanding both the forces involved and the time factor in recoil dynamics.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly the third law.
  • Basic knowledge of momentum and its calculation.
  • Familiarity with force and acceleration concepts.
  • Ability to manipulate equations involving mass, velocity, and time.
NEXT STEPS
  • Study the derivation of momentum equations in physics.
  • Learn how to calculate recoil forces in firearms using mbulletvmuzzle = &bar;F t.
  • Explore practical applications of Newton's third law in various physical systems.
  • Investigate the effects of barrel length on muzzle speed and recoil dynamics.
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in the principles of motion and forces related to firearms.

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:

m_{bullet}v_{muzzle} = \bar F t

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

\bar v = \frac{d_{barrel}}{t} = \frac{1}{2}v_{muzzle}

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 \bar F

AM
 

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