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Initial velocity of a bullet

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A bullet of mass 12.0 grams is traveling with a speed V0 and strikes a block of mass M = 1.00 kg that rests on a horizontal surface (mu = 0.40). After the bullet imbeds in the block, it slides 2.60 meters before coming to rest. The initial velocity of the bullet must have been ? m/s.


    2. Relevant equations
    pi=pf
    We were told that energy was not lost to anything except friction so
    Ei+Wfriction=Ef


    3. The attempt at a solution
    Ei+Wfriction=Ef
    (1/2)mvi2+μmgd=(1/2)mvf2
    eliminated potential energy because no height is given
    (1/)mvi2=μmgd
    (1/2)(0.012kg)(vi2)=(.40)(1.012)(9.8)(2.60)
    vi2=1719.050667
    vi=41.461 m/s

    Which isn't right. I think my error may lie in my assumption that the final KE is when the block has stopped and not right after impact, but I'm not certain.
     
  2. jcsd
  3. Jan 23, 2013 #2

    TSny

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    Your calculation does not take into account the friction between the bullet and the block as the bullet embeds itself in the block. This causes additional loss of kinetic energy in the system during the collision ("inelastic collision"). Your calculation would be correct if Ei is the kinetic energy remaining just after the collision.

    The kinetic energy just after the collision involves the velocity of the system just after the collision. See if you can relate this velocity to the initial velocity of the bullet using momentum concepts.
     
  4. Jan 23, 2013 #3
    Would that be m1v1=(m1+m2)v'? And then solving for v', and using that velocity in (1/2)mv^2 to find the KE immediately after the collision?
     
  5. Jan 23, 2013 #4

    tms

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    Since the final energy is zero, the above implies that the initial energy is also zero.
    If that were true, then there would have been no loss to friction.
     
  6. Jan 23, 2013 #5

    tms

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    Since the problem gives no information from which such loss can be calculated, it must be assumed that the collision was elastic.
     
  7. Jan 23, 2013 #6

    Matterwave

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    Since the bullet embeds itself in the block, the collision is perfectly INelastic - certainly not elastic.
     
  8. Jan 23, 2013 #7

    Matterwave

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    Yes, you have the right approach here.
     
  9. Jan 23, 2013 #8

    tms

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    [strike]An elastic collision is one in which kinetic energy is conserved. While in real life there would certainly be some energy loss in this situation, for the purposes of an introductory physics problem one can assume either no loss or a negligible loss. Otherwise, how would you propose to calculate the loss in this case?[/strike]
    Disregard.
     
    Last edited: Jan 23, 2013
  10. Jan 23, 2013 #9

    HallsofIvy

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    No, Matterwave is correct- the bullet imbedding itself in the block is a "perfectlyinelastic" collision. You cannot use "conservation of energy", you must use "conservation of momentum" to determine the speed of the "block+ bullet" system immediately after the bullet lodges in the block. You can then use the kinetic energy of that system as the energy that must be lost due to friction.
     
  11. Jan 23, 2013 #10

    gneill

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    It is an inelastic collision. Kinetic energy will not be conserved through the collision, therefore you don't bother with the KE on either side of the collision and apply conservation of momentum instead. Momentum is ALWAYS conserved.
     
  12. Jan 23, 2013 #11

    TSny

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    As already noted, you can use momentum conservation to relate the initial velocity of the bullet to the velocity of the bullet-block system immediately after the collision. Then you can find the loss of KE. In the case of a bullet embedding in a block, there will be a great loss of KE. For the numbers in this problem, you will discover that 98.8% of the kinetic energy of the bullet will be transformed to other forms of energy in the collision.
     
  13. Jan 23, 2013 #12
    I went with this approach, found KE in terms of the initial velocity, set that equal to the work by friction, solved for the initial velocity and got the right answer. Thank you all.
     
  14. Jan 23, 2013 #13

    tms

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    Disregard. I'm an idiot.
     
    Last edited: Jan 23, 2013
  15. Jan 23, 2013 #14

    tms

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    Disregard. I'm an idiot.
     
    Last edited: Jan 23, 2013
  16. Jan 23, 2013 #15

    gneill

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    No, the collision is inelastic. The assumption is that the bullet comes to rest with respect to the block in a very short time so that the block's motion with respect to the floor is insignificant during the actual collision. An essentially infinitesimal time after the collision the block and bullet have the same momentum as the bullet alone had before the collision. It's only after the block begins moving that friction couples the system to the floor (the previously effectively isolated system becomes an open one, or at least now includes the rest of the planet!).
     
  17. Jan 23, 2013 #16

    tms

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    I'm wrong. Someone must have toggled the ON/OFF switch in my brain.
     
  18. Jan 23, 2013 #17

    gneill

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    No worries. Happens to everyone. Back on the horse! :smile:
     
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