Answers

1. Answer:

   W = ½ m v²

   Explanation:

   In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

   We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

   initial instant. before separation

           p₀ = m v

   final attempt. after separation

           = m /2  0 + m /2 v_{f}

          p₀ = p_{f}

          m v = m /2

          v_{f}= 2 v

   this is the speed of the second part of the ship

   now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

        

   initial energy

            K₀ = ½ m v²

   final energy

           = ½ m/2  0 + ½ m/2 v_{f}²

           K_{f} = ¼ m (2v)²

           K_{f} = m v²

            

   the expression for work is

            W = ΔK = K_{f} – K₀

            W = m v² – ½ m v²

            W = ½ m v²

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