A 3 g bullet has a muzzle velocity of 290 m/s when fired by a rifle with a weight of 4 N. 1) Determine the recoil speed (in m/s) of the rifle.
2) it a marksman with a weight of 600 N holds the rifle firmly against his shoulder, determine the recoil speed (in m/s) of the shooter and rifle
1) To determine the recoil speed of the rifle, we can use the principle of conservation of momentum. The initial momentum of the system (bullet + rifle) is equal to the final momentum of the system.
Initial momentum = 0 (since the rifle and bullet are initially at rest)
Final momentum = (mass of bullet x velocity of bullet) + (mass of rifle x velocity of rifle)
Let's denote the mass of the bullet as m1, the mass of the rifle as m2, the velocity of the bullet as v1, and the velocity of the rifle as v2.
m1 = 3g = 0.003 kg
v1 = 290 m/s
m2 = 4 N / 9.81 m/s^2 ≈ 0.41 kg
Applying conservation of momentum:
0 = (0.003 kg x 290 m/s) + (0.41 kg x v2)
0 = 0.87 + 0.41v2
0.41v2 = -0.87
v2 = -2.12 m/s
Therefore, the recoil speed of the rifle is approximately 2.12 m/s in the opposite direction of the bullet.
2) When the marksman with a weight of 600 N holds the rifle firmly against his shoulder, the total mass of the system changes.
The total mass of the marksman, rifle, and bullet = m1 + m2 + m3
m3 = 600 N / 9.81 m/s^2 ≈ 61.22 kg
The total momentum of the system is still conserved.
0 = (0.003 kg x 290 m/s) + (0.41 kg x v4) + (61.22 kg x v4)
0 = 0.87 + 0.41v4 + 61.22v4
61.63v4 = -0.87
v4 = -0.014 m/s
Therefore, when the marksman holds the rifle firmly against his shoulder, the recoil speed of the shooter and rifle combined is approximately 0.014 m/s in the opposite direction of the bullet.