If the rifle is stopped by the hunter’s shoulder in a distance of 3.16 cm, what is the magnitude of the average force exerted on the shoulder by the rifle?
answer in units of n.
mass of bullet= 0.0137 kg
velocity of bullet= 546 m/s to the right
mass of rifle= 3.82 kg
recoil speed of the rifle as the bullet leaves the rifle= 1.958167539 m/s

F = 231.77N

Explanation:

Given the following data

Distance of Hunter's shoulder (d) = 3.16cm = 0.0316m

mass of bullet (m1) = 0.0137 kg

velocity of bullet (v1) = 546 m/s

mass of rifle(m2)= 3.82 kg

Velocity of rifle (V2) = 1.958167539 m/s

Momentum = MV

Momentum is conserved

Since we are looking for the force exerted on the shoulder by the rifle

Work done = Force × distance (F×d)

The rifle possessed kinetic energy = 1/2mV²

Therefore, work done = kinetic energy

F×d = 1/2mv²

F = 0.5mv²/d

By substitution we have

F = 0.5×3.82×1.9582²/0.0316

F = 7.324/0.0316

F = 231.77N

with the invention of the automobile, the woman of the house could drive to the store and get groceries. women were also free to socialize more. during world war ii women were allowed to work in factories to the war effort and could drive to work or ride mass transit buses to their jobs.

thermal energy is the answer