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Problem with bullet collision exercise

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello! I'm trying to solve this exercise:
    A bullet with initial velocity v and mass m pass through a block of mass M in a pendulum of length L and leave the block with velocity v/2. Which value of v will make the pendulum describe a circumference?
    As you can see, English is not my native language.

    2. Relevant equations



    3. The attempt at a solution
    I think this is a inelastic collision so this is what I did.
    [itex]K_{1i}+K_{2i}=K_{1f}+K_{2f}[/itex] but the pendulum has velocity 0 when the bullet hit it so [itex]K_{1i}+0=K_{1f}+K_{2f}[/itex].

    Using the information above. [itex]\frac{mv^2}{2}=\frac{mv^2}{8}+\frac{Mv^2}{2}[/itex]. I did this because the kinetic energy is conserved so I think I can use this to get to the mechanical energy which will give me [itex]Mg2L=K[/itex]. I think this is the way but honestly I don't know how to get the K to make it equal to the potential energy.
     
  2. jcsd
  3. Jun 7, 2012 #2
    What is meant by 'describe a circumference'?
     
  4. Jun 7, 2012 #3
    The bullet will hit the block and make it's center of mass describe a circumference. But it will only happen if the velocity v has a certain value or bigger than this value. I want to know what value is this. I can post a picture or post the .pdf link so you can see the situation.
     
    Last edited: Jun 7, 2012
  5. Jun 7, 2012 #4
    Again you mention 'describe a circumference', but I don't get what that refers to.

    Move in a complete circle?
     
  6. Jun 7, 2012 #5
    Exactly, it moves in a complete circle.It's path is a complete circle.
     
  7. Jun 7, 2012 #6

    gneill

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    Staff: Mentor

    The bullet passes through the block, so what can you say about the type of collision involved? What is conserved in that sort of collision?
     
  8. Jun 7, 2012 #7
    I think it's a inelastic collision. If I'm correct them the kinetic energy and linear momentum are both conserved. I think I can solve this exercise using the conservation of energy of the system after the collision but I'm not sure how to do this.

    PS: Is this situation possible? I mean, I can't see it happening in my mind.
     
  9. Jun 7, 2012 #8
    Assuming that the collision is inelastic is a gamble, but momentum is always conserved and is a more solid base for this problem.
     
  10. Jun 7, 2012 #9

    gneill

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    It's a form of inelastic collision where one of the bodies does not remain "attached". The only thing you can count on in an inelastic collision is that momentum will be conserved. Momentum is ALWAYS conserved.

    Why don't you consider the momentum of the bullet pre-collision and post-collision? Where do you think the difference will end up?
     
  11. Jun 7, 2012 #10
    In my mind the collision was inelastic because there was no external forces. But there are things like heat and sound so it's not a inelastic collision.

    I tried two things now. First way: by conservation of momentum: [itex]P_{1i}+P_{2i}=P_{1f}+P_{2f}[/itex] but we know that [itex]P_{2i}=0[/itex]. Then [itex]P_{1i}=P_{1f}+P_{2f}[/itex].
    Using the data the exercise gave to us [itex]mv=m\frac{v}{2}+Mv_{2f}[/itex].

    Second way: by conversation of total mechanical energy:
    [itex]\frac{1}{2}mv^2=Mg2L[/itex] solving for v I got [itex]v=2 \sqrt{\frac{MgL}{m}}[/itex]. Wrong again. What's wrong with this equation? Because in this situation the mechanical energy will be conserved right? So the kinetic energy of the bullet is equal to the potential energy of the block.
     
  12. Jun 7, 2012 #11
    Your second way is incorrect because some initial kinetic energy is lost to heat and sound.
     
  13. Jun 7, 2012 #12
    Ok, I Agree and understand why but if this is incorrect I can't get the right answer. The conservation of momentum won't give me a relation between the velocity v and L.
    By the first way I got this answer [itex]v=\frac{2Mv_{2f}}{m}[/itex] and according to the textbook that's wrong.
     
  14. Jun 7, 2012 #13

    gneill

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    Staff: Mentor

    First find an expression for the initial velocity of the block as a function of the initial velocity of the bullet, v. Then consider conservation of energy for the motion of the block as it moves around its trajectory.
     
  15. Jun 7, 2012 #14
    Using conversation of momentum: [itex]P_{1i}+0=P_{1f}+P_{2f}[/itex] and from this equation we get [itex]v_{2}(v)=\frac{vm}{2M}[/itex] and [itex]v_{2}[/itex] is the velocity of the block as a function of the initial velocity of the bullet.
    Now considering the conversation of energy for the motion of the block:
    [itex]K_i + U_i = K_f + U_f[/itex] using the data we have
    [itex]\frac{m(v_{2})^2}{2}=Mg2L[/itex]
    [itex](\frac{vm}{2M})^2=\frac{Mg4L}{m}[/itex] solving for v
    [itex]v=4\frac{M}{m}\sqrt{\frac{MgL}{m}}[/itex]
    Is this correct??
     
    Last edited: Jun 7, 2012
  16. Jun 7, 2012 #15

    gneill

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    I think you have to consider what velocity the block must have at the apex of its trajectory in order for it to describe a circular motion...
    Hint: The velocity (kinetic energy) will not be zero.
     
  17. Jun 7, 2012 #16
    The only idea I had until now was that at the apex of its trajectory, the block will have velocity [itex]v=\sqrt{gL}[/itex] so its kinetic energy is [itex]K=M\sqrt{gL}[/itex] but when I plugged this on the equation [itex]\frac{m(v_{2})^2}{2}=Mg2L+M\sqrt{gL}[/itex] ... well guess what... wrong again. *Sigh*
     
    Last edited: Jun 7, 2012
  18. Jun 7, 2012 #17

    gneill

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    The KE at the apex is the initial KE MINUS the PE due to its gain in altitude. Also, KE is ##\frac{1}{2}M V^2##, not ##MV##.
     
  19. Jun 7, 2012 #18
    Oh, my bad. So you are saying that ##K_f## is ##0-2MgL##? Because the initial kinetic energy of the block is zero since its velocity was zero before the bullet hit it. And PE at the apex is ##2MgL##.
     
  20. Jun 7, 2012 #19

    gneill

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    No, the "initial" KE of the block in this case is its KE immediately after the "collision" with the bullet.

    There are two distinct phases of the problem: The first is the analysis of the bullet-block collision wherein momentum is transferred to the block. The second is the resulting motion of the block given the "kick" it received from the bullet. The "initial" KE for this second phase is the KE the block has according to the momentum that resulted from the collision. It's this "initial" KE that you need to work with for the continuing motion of the block.
     
  21. Jun 7, 2012 #20
    Ok! Now I'm trying this:
    [itex]K_i + U_i=K_f + U_f[/itex] and using what you told me (KE at the apex is KE initial minus PE)
    [itex]\frac{m(v_2)^2}{2}=(K_i - U_f)+U_f[/itex] doing the algebra
    [itex]\frac{v^2 m^3}{4M}=\frac{v^2m^3-8M^2L}{4M}-Mg2L[/itex]
    I understand the situation now but still I can't solve this exercise....
     
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