Finding average force on shoulder from 30-06 rifle recoil

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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?