If the rifle is stopped by the hunter's shoulder in a distance of 3.23 cm, what is the magnitude of the average force exerted on the shoulder by the rifle? answer in units of n

By conservation of momentum, the initial recoil speed of the rifle is:

u = 904kg * 0.0106 kg / 3.53 kg = 2.7146 m/s

The final recoil speed v is zero in a distance of d = 0.0346 m

So the rifle's acceleration is a = (0 - u^2) / 2*d = - 2.7146^2 / 2*.0323 m/s^2

a = - 114.07m/s^2