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

Answer:F = 231.77NExplanation:Given the following dataDistance of Hunter's shoulder (d) = 3.16cm = 0.0316mmass of bullet (m1) = 0.0137 kgvelocity of bullet (v1) = 546 m/smass of rifle(m2)= 3.82 kgVelocity of rifle (V2) = 1.958167539 m/sMomentum = MVMomentum is conservedSince we are looking for the force exerted on the shoulder by the rifleWork done = Force × distance (F×d) The rifle possessed kinetic energy = 1/2mV²Therefore, work done = kinetic energyF×d = 1/2mv²F = 0.5mv²/dBy substitution we haveF = 0.5×3.82×1.9582²/0.0316F = 7.324/0.0316F = 231.77N

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