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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

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Answer:

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

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