When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
1 answer:
Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂
Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
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Answer:
θ = 19.66°
Explanation:
To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:
h: distance to the ground
g: gravitational acceleration = 9.8m/s^2
m: mass of the first person = 55 kg
In the image attache below you can notice that the height h is:
Then, the potential energy is:
When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:
From the previous equation you can get h':
Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:
hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°
