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Muzzle velocity of a pistol

  1. Oct 30, 2005 #1
    In an arrangement for measuring the muzzle velocity of a rifle or pistol, the bullet is fired up at a wooden mass, into which it embeds. The wood is blasted straight up into the air to a measured height h. Assuming negligible losses to friction, write an expression for the velocity in terms of the known masses and height. Use mb for the mass of the bullet, mw for the mass of the wood, h for height and g for gravity.

    I got this question set in my college course, and we haven't even begun to cover momentum etc. yet, so I don't really have any ideas of how to go about answering this...any help please? Thanks a lot.
     
  2. jcsd
  3. Oct 30, 2005 #2
    i would have thought that was an energy conservation question,
    think KE and PE
     
  4. Oct 30, 2005 #3
    So m x g x h? And/or 1/2mvsquared?
     
  5. Oct 30, 2005 #4
    Ah ok I cracked it, I got the formula:

    sqrt(2gh)+(mw*sqrt(2gh))/(mb)

    which appears to be correct. However, part b) of the question is this:

    If a 100 grain (6.48 g) 25-06 Remington rifle bullet is fired into a 4.72 kg block that then rises 4.4 cm into the air, what was the muzzle velocity of that bullet?

    I tried feeding all the values into the above formula:

    g = 9.81
    h = 0.044
    mw = 4.72
    mb = 0.00648

    And I got a velocity answer of 820m/s which my teacher tells me is wrong....anyone know what I'm doing wrong? Thanks
     
  6. Oct 30, 2005 #5
    Please, does anyonw know why 820m/s is wrong? This is my working:

    sqrt(2gh)+(mw*sqrt(2gh))/(mb)

    sqrt(2 x 9.81 x 0.044) + (4.72 x sqrt[2 x 9.81 x 0.044])
    -----------------------------------------------------
    0.00648

    0.929 + 4.285
    --------------
    0.00648

    5.314
    ------
    0.00648

    = 820 m/s

    But this answer isn't right! How is it not right?! Am I using all the correct units? Thanks
     
  7. Oct 30, 2005 #6
    im not sure where u got that formula

    but the formula i get is 0.5mv^2(of bullet) + 0.5mv^2(of wood) = mgh(of bullet) + mgh(of wood)

    the kenetic energy of the wood initially is zero so u can simplify equation to..

    0.5mv^2(of bullet) = mgh(of bullet) + mgh(of wood)
    further simplify to..
    0.5mv^2 = (Mb + Mw)gh

    and you have to remember that if the bullet is embeded in the wood then for mgh you must use mass of both bullet and wood.

    can u explain how you got your equation?
     
    Last edited: Oct 30, 2005
  8. Oct 30, 2005 #7
    0.5mv^2 = (Mb + Mw)gh

    So 0.5mv^2 = 2.04

    What does the '^' represent in your equation?
     
  9. Oct 30, 2005 #8

    Päällikkö

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    Homework Helper

    Energy cannot be assumed to be conserved in collisions (unless otherwise stated, as in elastic). Therefore the problem is not purely about KE and PE.

    To the original poster: Have you ever used the conservation of momentum principle?
     
  10. Oct 30, 2005 #9
    I got my equation with some help from my teacher, although I didn't fully understand his explanation of why...but he said that was defintely the correct forumla, which doesn't explain why I'm getting the wrong answer. Maybe your formula will help...
     
  11. Oct 30, 2005 #10
    '^' represents 'to the power of..'
    wel if ur teacher says it's definatly the correct equation, who am i to argue, im only a 6th form student.
    conservation of momentum would work actually i think mu+mu = mv+mv
    hang on you would need finial velocity would that be '0'? that complicates things
    b right back
     
    Last edited: Oct 30, 2005
  12. Oct 30, 2005 #11
    Yes, is that what I should be using for this?
     
  13. Oct 30, 2005 #12
    By using your original formula I got an answer of 25, which I'm fairly sure is wrong, and as I only have 1 submission left, I don't want to use it unless I'm absolutely sure! The Conservation of momentum sounds good, but isn't it mu1 + mu2 = vu1 + vu2? Because surely mu + mu = vu + vu is equal ro mu = vu which means m = v?
     
  14. Oct 30, 2005 #13

    Päällikkö

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    That's what I get using the conservation of energy principle, which in this case ought to be wrong.

    No.
    I've got no idea what the v's and u's are in your equation (normally used for velocities in conservation of momentum). Momentum: [itex]\vec{p}= m \vec{v}[/itex]
     
  15. Oct 30, 2005 #14
    Ah I'm even more confused than ever now, so which formula should I be using to solve this question?:

    If a 100 grain (6.48 g) 25-06 Remington rifle bullet is fired into a 4.72 kg block that then rises 4.4 cm into the air, what was the muzzle velocity of that bullet?

    Thanks
     
  16. Oct 30, 2005 #15

    Päällikkö

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    Normally, you are supposed to use conservation of momentum (and after collision, conservation of energy). See the case "totally inelastic collision" (refer to your physics book or google). Energy is lost during collisions of that type.

    I'm confused too: Your teacher has helped you with what seems to be an equation of conservation of energy. This is not the way to approach a collision problem, but then again, if you've not been teached the conservation of momentum, how could you use it?
     
  17. Oct 30, 2005 #16
    from the formula you given ....

    ( sqroot(2gh) + Mw(sqroot(2gh) )/Mb = v

    i rearrage..

    vMb = sqroot (2gh) + Mw(sqroot(2gh)) ...to..

    vMb = (sqroot(2gh)) *( 1 + Mw)

    we know from simplifiying mgh = 0.5mv^2 that v = sqroot(2gh)
    so..

    vMb = V ( 1 + Mw)

    vMb = V + vMw
    and that is as much as i can get close to conservation of momentum

    and i think its wrong too. i dunno it might help u figure something out
     
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