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

[redhot209]

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?

 

[Kurdt]

Can you not work out the average acceleration of the rifle and then the magnitude of the force from that?

 

[thomate1]

Your approach is correct, but there are a few errors in your calculations.

1. The average velocity should be calculated as (Vi + Vf)/2, not (O + Vi)/2. So it should be (0 + 1.6049)/2 = 0.80245 m/s.
2. The time should be calculated as ∆t = 2x/Vf, not ∆t = 2x/Vi. So it should be 2(0.0136)/1.6049 = 0.0213 s.
3. The force should be calculated as F = ∆p/∆t, not F = MVi/∆t. So it should be (0.00689*778)/0.0213 = 248.7 N.

So the magnitude of the average force exerted on the shoulder by the rifle is approximately 248.7 N, which is a much more reasonable value for a bullet hitting a shoulder.