Solving Rifle Recoil & Force: A 30-06 Hunting Rifle Case Study

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SUMMARY

The discussion focuses on calculating the recoil speed and average force exerted on a hunter's shoulder when firing a 30-06 caliber hunting rifle. The recoil speed of the rifle is established at 1.6049 m/s based on the bullet's mass of 0.00689 kg and velocity of 778 m/s. The average force exerted on the shoulder, calculated using the rifle's mass of 3.34 kg and the stopping distance of 1.36 cm, is determined to be approximately 3.16 N. The participant expresses concern that this force seems too low, prompting a discussion on the correct approach to calculating average acceleration and force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with basic physics equations related to momentum and force
  • Knowledge of unit conversions, particularly between meters and centimeters
  • Ability to perform calculations involving mass, velocity, and distance
NEXT STEPS
  • Review the concept of momentum conservation in collisions
  • Learn how to calculate average acceleration using displacement and time
  • Study the relationship between force, mass, and acceleration (Newton's second law)
  • Explore real-world applications of recoil forces in firearms and their effects on shooters
USEFUL FOR

This discussion is beneficial for physics students, firearm enthusiasts, and anyone interested in the mechanics of recoil and force dynamics in shooting sports.

redhot209
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Homework Statement


Already Solved(Part 1) :A 30-06 caliber hunting rifle fires a bullet of mass of 0.00689 kg with a velocity of 778 m/s to the right. The rifle has a mass of 3.34 kg. What is the recoil speed of the rifle as the bullet leaves the rifle. Answer in units of m/s. Answer is 1.6049

Part 2: If the rifle is stopped by the hunter's shoulder in a distance of 1.36 cm, what is the magnitude of the average force exerted on the shoulder by the rifle? Answer in units of N.

Homework Equations


1. Vaverage= O + Vi(1.6049) / 2
2.∆T = 2x/ Vi
3. F∆t=∆mv
4. F=MVi /∆t



The Attempt at a Solution


1. 0+1.6049/2 = 0.80245
2. 2(1.36) / 1.6049 = 1.694809645
4. (3.34)(1.6049) / 1.694809645 = 3.162813013.
This was my answer by I know that the force is too low for a bullet hitting a shoulder...What did I do wrong?
 
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Can you not work out the average acceleration of the rifle and then the magnitude of the force from that?
 

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