I am lost, dont know how to solve part 2

[asmasky]

Homework Statement

part -1) A 30-06 caliber hunting rifle fires a bullet of mass 0.0113 kg with a velocity of 765 m/s to the right. The rifle has mass of 4.1 kg.
what is the recoil speed of the rifle as the bullet leaves the rifle? answer in units of m/s. answer is 2.1084 /s

part -2) If the rifle is stopped by the hunter's shoulder in a diatance of 2.52 cm, waht is the magnitude of the average force exerted on the shoulder by the rifle? answer in units of N.

Homework Equations

not sure, new user on this website

The Attempt at a Solution

not sure, new user on this website

 

[Integral]

You need to show us some effort.

What do you know about conservation of momentum?

 

[asmasky]

part1) V(bullet) = (m(bullet)v(bullet))/(M(rifle)) = 2.1084 m/s

Part2) J = (F) (deltaT)
J = P(final) - P(inital) = MV(final) - MV(initial) = M(V(final) - V(inital))

 

[mgb_phys]

For the second part you might be better using conservation of energy.
Energy = force * distance.
The hunters shoulder is essentailly a spring receiving the kinetic energy of the moving rifle. You know the mass and initial velocity of the rifle and the final velocity=0.

 

[asmasky]

do u mean,
k = (.5)(m)(v^2) or ENERGY = (.5)(m)(v^2)+(.5)(k)(x^2)

 

[mgb_phys]

You don't need to calculate the spring - I was just trying to describe what happened.
You know the velocity of the rifle when it first touches the shoulder, so you can find the KE.
Since work (or energy) is force * distance and you know the distance you can work out how much force it takes to stop this energy in this distance.