Suppose a 3.5-kg shotgun is held tightly by an arm and shoulder with a combined mass of 12.5 kg. When the gun fires a bullet wit
1 answer:
Answer:
v₂ = - 1.29 m / s
, K₁ / Em = 0.996
, K₂ / Em = 0.004
Explanation:
This is an exercise for the moment, where the system is formed by the shotgun, the man arm and the bullet, in this case the forces are internal therefore the moment is conserved
Initial. Before shooting
p₀ = 0
Final. After shooting
= m v₁ + M v₂
Where index 1 is for the bullet and index 2 for the shotgun, arm and man set
The mass of the bullet is m = 0.04 kg
The mass of the set M = 3.0 + 12.5 = 15.5 kg
p₀ =
0 = m v₁ + M v₂
v₂ = - m / M v₁
v₂ = - 0.04 / 15.5 400
v₂ = - 1.29 m / s
The mechanical energy is equal to the kinetic energy, let's calculate the energy for each of the two elements
K₁ = ½ m v₁²
K₁ = 1/2 0.04 400²
K₁ = 3200 J
K₂ = ½ M v₂²
K₂ = ½ 15.5 1.29²
K₂ = 12.90 J
The total energy of the system is the sum of the energy of each component
Em = K₁ + K₂
Em = 3200 + 12.90
Em = 3212.9 J
The fraction of transmitted energy is
To the bullet
K₁ / Em = 3200 / 3212.9
K₁ / Em = 0.996
To the system
K₂ / Em = 12.9 / 3212.9
K₂ / Em = 0.004
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