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Interesting Impulse Problem!

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass m = 17.6 g traveling horizontally at speed vo = 565 m/s strikes a block of mass M = 678 g sitting on a frictionless, horizontal table. This time, however, it comes out the other side of the block at speed v = 524 m/s.

    Calculate J, the magnitude of the impulse exerted by the bullet on the block.

    2. Relevant equations

    Hm...

    Impulse = Momentum = Ft = mv_f² - mv_i²

    3. The attempt at a solution

    I think that the impulse is same as the momentum. I think that is...

    Impulse = (M + m)v_f² - mv_i²

    But it seems like I'm on the wrong path.
     
  2. jcsd
  3. Oct 18, 2012 #2

    ehild

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    The momentum is mV. When a force F acts for time Δt on a point mass m the momentum will change and the change of momentum equals to the impulse of the force: FΔt=mVf-mVi

    ehild
     
  4. Oct 18, 2012 #3
    Does this means that only the mass of the bullet takes in account of this situation?
     
  5. Oct 18, 2012 #4

    ehild

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    It means that the momentum is proportional to the velocity instead of the square of the velocity as you wrote.

    ehild
     
  6. Oct 18, 2012 #5
    This is the revised work.
     
  7. Oct 18, 2012 #6
    You didn't answer my question. Does the mass of the bullet takes in account of the momentum or the mass of both objects?
     
  8. Oct 18, 2012 #7
    Lacking approach and explanation here. Not helpful. -__-
     
  9. Oct 18, 2012 #8

    ehild

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    Sorry.


    ehild
     
  10. Oct 18, 2012 #9
    Waiting for other response. Otherwise, it seems that I need to answer myself.
     
  11. Oct 18, 2012 #10
    Did you try using I = pfinal - pinitial?
     
  12. Oct 18, 2012 #11
    Yes, I did, but it seems that I made the wrong approach. I believe that is...

    (m + M)v_f² - (m + M)v_i²
    = - (m + M)v_i²
     
  13. Oct 18, 2012 #12
    It doesn't state that part of the mass of the block sticks to the bullet so you should just be able to use the bullet mass and the final and inital velocities for change in momentum
     
  14. Oct 18, 2012 #13
    Then, it's mv_f - mv_i, where m is the mass of the small bullet.
     
  15. Oct 18, 2012 #14
    Then, this is the negative impulse. We have m(v_f - v_i)
     
  16. Oct 18, 2012 #15
    This is what I have:

    I = mv_f - mv_i

    If I only use the mass of the bullet as the part of this equation, then we have...

    I = 0.0176 * (524 - 565) = -0.722

    Is that true?
     
  17. Oct 19, 2012 #16
    All you were supposed to calculate was I = Δp , where Δp is linear momentum and I is impulse.....

    Your work seems correct to me. Mass of the block was only to confuse you. This is because from the given parameters , you cannot solve for deceleration in bullet and time period..
     
  18. Oct 19, 2012 #17
    But it seems that the answer is not right. I entered it, but not right.
     
  19. Oct 20, 2012 #18
    Well consider right x axis to be positive and left to be negative. (Choice of sign convention lay in user's choice , but then he has to stick to that convention throughout the problem.) So by Newton's third law , force which block exerts on bullet will be equal and opposite to force exerted by bullet on block. Since Δt is constant , so impulse exerted by block on bullet will be negative of that by bullet on block. Thus I=Δp , simply.

    Why you say , this isn't correct ?
     
  20. Oct 21, 2012 #19
    Based on what I know, impulse can't be negative. It's force times time. Even though impulse is same as the momentum, impulse can't have negative magnitude. I spoke with my instructor about this, and he said that impulse can't be negative.
     
  21. Oct 21, 2012 #20
    Adding in to what I said before, impulse is Force times time. Though it's the same as momentum, the impulse can't be negative. That is what is different from the momentum.
     
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