1. The problem statement, all variables and given/known data 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. 2. Relevant equations 1. Vaverage= O + Vi(1.6049) / 2 2.∆T = 2x/ Vi 3. F∆t=∆mv 4. F=MVi /∆t 3. 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?