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A 35.0- g bullet strikes a 5.0- kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and the bullet fly off together at 8.6 m/s. What was the speed of the bullet before it struck the lumber? Define the bullet and the wood as a system.

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Respuesta :

Answer:

1237.17 m/s

Explanation:

Applying the law of conservation of momentum,

  • m₁v₁ + m₂v₂ = (m₁ + m₂)v
  • 0.035v₁ + 0 = 5.035 x 8.6
  • 0.035v₁ = 43.301
  • v₁ = 43.301/0.035
  • v₁ = 1237.17 m/s

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