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Inelastic Collision

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A 10-g bullet moving at 1000 m/s strikes and passes through a 2.0 kg block initially at rest as shown in the diagram. The bullet emerges from the block with a speed of 400 m/s.

    [PLAIN]http://img34.imageshack.us/img34/554/82265240.gif [Broken]

    To what maximum height will the block rise above its initial position?

    3. The attempt at a solution

    It's not a perfectly inelestic collision since the bullet doesn't embed in the block. Therefore this is an inelastic collision. During the collision some momentum is conserved so,

    Momentum just before = momentum just after

    And I think:

    Kenetic energy just after the collision = final potential energy

    [tex]\frac{1}{2} 0.01 (1000)^2 = 2 (9.81) h[/tex]


    But this is FAR larger than the actual answer. What's wrong?? :confused:
    This is not a homework problem, just revision for my exam. So any help with this is really appreciated.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 14, 2010 #2


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    Bullet is traveling at 1000 m/s, passes through the block, and then comes out at 400 m/s.

    So what will the change in kinetic energy be equal to?
  4. May 14, 2010 #3


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    In your calculation you have not used the final velocity of the bullet.
    Here change in KE = change in PE.
    Last edited: May 14, 2010
  5. May 14, 2010 #4
    In finding the change in Kenetic energy

    [tex]\Delta E_K = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{i}^2[/tex]

    So for initial mass should I use "mblock + massbullet", and for the final mass just use "mbullet", right? And do the same thing for change in PE?
  6. May 14, 2010 #5


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    Change in KE of bullet = Change in PE of the block.
  7. May 15, 2010 #6

    Andrew Mason

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    Hang on. Bullet kinetic energy is not conserved. This is an inelastic collision. This is a conservation of momentum (collision) problem followed by an energy conservation (post-collision) problem.

  8. May 15, 2010 #7
    :confused: I'm confused... then what am I supposed to be doing?
  9. May 15, 2010 #8
    Out of curiousity what's the nswer supposed to be?
  10. May 15, 2010 #9

    Andrew Mason

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    Apply conservation of momentum to the impact of the bullet on the block. Can you figure out what the block's speed is after the bullet passes through? How is the gain of momentum of the block related to the loss of momentum of the bullet? Can you work out the resulting speed of the block (immediately after the bullet passes through)? How much kinetic energy does the block have with such speed? What happens to that kinetic energy as the block rises? Can you work out from that how high the block rises?

  11. May 15, 2010 #10
    The velocity of the block immediately after the collision is

    [tex]v=\frac{0.01 \times 1000}{2+0.01}=4.97[/tex]

    Then do I use the equation (with initial gravitational potential =0)?

    [tex]E_{Ki} + U_i = E_{Kf} + U_f[/tex]

    The correct answer is supposed to be 46 cm.
  12. May 15, 2010 #11

    Andrew Mason

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    No. This is where you are going wrong. How much momentum does the bullet have before the collision? How much momentum does the block + bullet have afterward? So how much speed does the block have? Write out the equation. You know two of the three momenta so you can solve for the unknown block momentum.

    Yes. Final kinetic energy is __?


  13. May 16, 2010 #12

    If you apply the Principal of Conservation of Momentum, the momentum of the system before will equal the momentum after.
    The momentum of the system before is just the momentum of the bullet, (0.01)(1000), while the momentum after is (0.01)(400) + 2V.
    Solve this to get the velocity of the block, and apply the Principal of Conservation of Energy.
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