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The man fire a 50-g arrow that moves at an unknown speed. It hits and embeds in a 350-g block that slides on an air track. At the end, the block runs into and compresses a 4000-N/m spring 0.10 m. A) How fast was the arrow traveling? Let the system be comprised of the arrow and the block. Assume that all the kinetic energy of the arrow-block system has been converted to the elastic potential energy of the spring.

Respuesta :

Answer:

v₀ = 252.8 m / s

Explanation:

In this exercise we must use the concepts of energy and momentum. Let's start with the energy part. Let's find the mechanical energy at two points

Final. Spring compression

     [tex]Em_{f}[/tex] = [tex]K_{e}[/tex] = ½ k x²

Initial after the arrow is embedded in the block

     Em₀ = K = ½ (m + M) v²

As there is no friction

      Emo =  [tex]Em_{f}[/tex]

      ½ (m + M) v² = ½ k x²

      v² = x² k / (m + M)

Let's calculate

      v = √ [0.10 4000 / (0.050 + 0.350)]

      v = 31.6 m / s

This is the speed of the arrow plus block set

Let's use moment conservation now

Initial. Before the crash

      p₀ = m v₀ + 0

Final. After the arrow is embedded in the block, but before starting to move

       [tex]p_{f}[/tex]  = (m + M) v

The system consists of the arrow and the block, so the forces in the crash are internal, and the moment is preserved

      p₀ =  [tex]p_{f}[/tex]

      m v₀ = (m + M) v

      v₀ = v (m + M) / m

      v₀ = 31.6 (0.050 + 0.350) /0.050)

      v₀ = 31.6 8

      v₀ = 252.8 m / s