Answer:
I calculated it several times and got the same answer, 40.38m/s, but the textbook says it's 28.6m/s.
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 = 231.77N
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