Question

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 with a mass of 0.04 kg and a speed of 400 m/s, What is the recoil speed of the shotgun & arm–shoulder combination? What energy is transmitted to the shotgun & arm-shoulder combination? What energy is transmitted to the bullet?

Answer 1

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

     $p_{f}$ = 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 =  $p_{f}$

       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