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Bullet of mass physics problem

  1. Nov 17, 2004 #1
    A bullet of mass 0.0021 kg is shot into a wooden block of mass 0.197 kg.

    They rise to a final height of 0.546 m as shown. What was the initial speed (in m/s) of the bullet before it hit the block?

    I figured out this one
    I got another one
    A 5.0 kg ball is attached to a vertical unstretched spring with k = 762 N/m and then released. What distance does the ball fall just as it momentarily comes to rest?
    I use F=-kx with F=mg therefore mg=-kx from there I solve for x but I got the wrong answer. What am I wrong. Thank you
     
    Last edited: Nov 17, 2004
  2. jcsd
  3. Nov 18, 2004 #2

    Andrew Mason

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    Think of it as a conversion of gravitational potential energy into spring potential energy and ball kinetic energy, as the ball falls.
    [tex]mgx = \frac{1}{2}(kx^2 + mv^2)[/tex]

    Since maximum displacement occurs when v=0,
    [tex]mgx = \frac{1}{2}kx^2[/tex]

    [tex]x = 2mg/k[/tex]

    AM
     
  4. Nov 18, 2004 #3
    thank you
    A 1.72 kg object is suspended from a spring with k = 19.1 N/m. The mass is pulled 0.305 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinentic energy of the object in J?
    1/2 kx^2=1/2 mv^2
    and I got it wrong too
     
  5. Nov 18, 2004 #4
    Try to use 1/2 kx^2=1/2 mv^2+mgl, l - heigt
    Because when the mass oscillates, it's potential energy changes
     
  6. Nov 18, 2004 #5

    Andrew Mason

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    The condition for maximum kinetic energy is dK/dt = 0 (when the rate of change of kinetic energy = 0). Since [itex]K = \frac{1}{2}mv^2 = \frac{1}{2}m(dx/dt)^2[/itex] the maximum kinetic energy occurs when [itex]d^2x/dt^2 = 0[/itex] (ie. when a = 0).

    The equation of motion is:
    [tex]F = mg-Kx = ma[/tex] where x = the displacement from equilibrium

    So when a=0
    [tex]kx=mg[/tex]
    [tex]x=mg/k[/tex]

    Note that it is independent of the maximum amplitude. To find the speed when x=mg/k, use an energy approach:
    [tex]U_g + KE + U_k = U_{ki}[/tex]
    [tex]mg(A-x) + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2[/tex]


    AM
     
    Last edited: Nov 18, 2004
  7. Nov 18, 2004 #6
    I'm afraid, that I didn't understand all correctly, but the question is "What is the maximum kinentic energy of the object in J?", isn't it?
    I think You are right speaking about condition for maximum kinetic energy, however it depends on amplitude x.

    1/2 kx^2=1/2 mv^2+mgx (conversion of energy)

    1/2 mv^2=x(kx/2-mg)
     
  8. Nov 18, 2004 #7

    Andrew Mason

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    see my edited reply above.

    AM
     
  9. Nov 18, 2004 #8
    But i have
    mgx+(1/2)mv^2=(1/2)k(A-x)^2,
    where A=(mg/k+x)
    What do You think?
     
  10. Nov 18, 2004 #9

    Andrew Mason

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    A= initial amplitude.
    Try:
    [tex]mg(A-x) + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 [/tex]

    substituting x=mg/k:

    [tex]\frac{1}{2}mv^2 = \frac{1}{2}kA^2 - mg(A - mg/k) - \frac{1}{2}m^2g^2/k[/tex]

    [tex]v^2 = KA^2/m - 2gA + mg^2/k[/tex]
    [tex]v = \sqrt{KA^2/m - 2gA + mg^2/k}[/tex]

    AM
     
    Last edited: Nov 19, 2004
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