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Bullet muzzle velocity

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img403.imageshack.us/img403/1751/problem1r.jpg [Broken]

    2. Relevant equations

    [itex]v_f = \frac{m v_1 + M v_2}{m+M}[/itex]

    3. The attempt at a solution

    Let "A" be the configuration immediately before the collision and "B" the configuration immediately after the collision, and "C" the very final state.

    I have used the above equation and solved for the velocity when the bullet enters the system (immidiently after the collision):

    [itex]v_B = \frac{m v_{1A}}{m+M}[/itex]

    The expression for the total kinetic energy of the system right after the collision is

    [itex]K_B = 1/2 (m + M) v_B^2[/itex]

    Substituting vB we get

    [itex]K_B = \frac{m^2v_{1A}^2}{2(m+M)}[/itex]

    We then apply the conservation of mechanical energy principal to the system to get:

    [itex]K_B+U_B = K_C + U_C[/itex]

    [itex]\frac{m^2v_{1A}^2}{2(m+M)} + 0 = 0+ (m+M) gh[/itex]

    [itex]v_{1A} = \left( \frac{M+m}{m} \right) \sqrt{2gh}[/itex]

    This is very close to the answer, but how can I bring μk (coefficient of friction) into my equation? The potential energy is given by mgh, so how do I write it in terms of μk and d? :confused:
    Any help is appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 9, 2012 #2
    Just applying conservation of mechanical energy will not work since there's an extra force acting. So, apply the work-energy conservation theorem, you need to include the total work done by all forces, including that done by the frictional force(missing in your equation)


    Whats the definition of work by a force?
     
  4. May 9, 2012 #3
    Okay I tried the work-kinetic energy theorem but it's still not working:

    [itex]W_{net}= K_f -K_i[/itex]

    Where work is force times d and fkkmg is the magnitude of kinetic friction.

    [itex](F-f_k)d = (F- \mu_k mg) d = \frac{m^2 v_o^2}{2(m+M)}[/itex]

    If we solve for vo, how come we don't end up with the correct expression? (I tried that) :confused: Is there any other way to factor μk into the equation?
     
  5. May 9, 2012 #4
    You're almost there....Though, where did that 'F' come in from!?! :confused: (there is no F!)
     
    Last edited: May 9, 2012
  6. May 9, 2012 #5
    So how should I write the equation? "F" is the force and fk is the friction slowing it down (W=Fxd)...
     
  7. May 9, 2012 #6
    What force? The only force acting after the collision is the friction force. So you only need the frictional force part of the work.
     
  8. May 10, 2012 #7
    Okay but there is another problem

    [itex]\mu_k mgd = \frac{m^2v_o^2}{2(m+M)}[/itex]

    [itex]v_o^2 = 2 \mu_k g d \frac{m+M}{m}[/itex]

    [itex]v_o = \sqrt{2 \mu_k gd} \sqrt{\frac{m+M}{m}}[/itex]

    But the term "m+M/m" should not be under the square root, since the correct answer must be:

    [itex]v_o = \sqrt{2 \mu_k gd} \frac{m+M}{m}[/itex]

    So, what's wrong? :confused: How can I get rid of this square root?
     
  9. May 10, 2012 #8
    Here's your mistake. Is the frictional force only acting on the 'm'?
     
  10. May 10, 2012 #9
    Thank you so much, I had totally forgot the mass of the block M... I got the correct answer now. Thanks! :)
     
  11. May 10, 2012 #10
    Glad to help :smile:
     
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