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Homework Help: Bullet hitting block attached to spring

  1. Nov 13, 2006 #1
    A 0.0125 kg bullet strikes a 0.300-kg block attached to a fixed horizontal spring whose spring constant is 2.25 x 103 N/m and sets it into vibration with an amplitude of 12.4 cm. What was the speed of the bullet if the two objects move together after impact?

    E=(1/2)kA^2=(1/2)(2250 N/m)(0.124 m)2=17.30 J

    m=0.0125 kg + 0.3 kg=0.3125 kg

    K=(1/2)mv^2
    17.30 J=(1/2)(0.3125 kg)v^2
    v2=0.009
    v=0.095 m/s

    Conservation of momentum:
    (.0125 kg)v=(0.3125 kg)(0.095 m/s)
    v=2.375 m/s

    Have I done this properly? The velocity doesn't seem right to me. Have I missed anything?
     
  2. jcsd
  3. Nov 13, 2006 #2

    Doc Al

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    Staff: Mentor

    Redo this calculation.
     
  4. Nov 13, 2006 #3
    K=(1/2)mv2
    17.30 J=(1/2)(0.3125 kg)v2
    v2=110.72
    v=10.52 m/s

    Conservation of momentum:
    (.0125 kg)v=(0.3125 kg)(10.52 m/s)
    v=152.91 m/s

    Yes, this looks better. Thanks for the help!
     
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