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A bullet flying horizontally

  1. Mar 31, 2017 #1
    1. The problem statement, all variables and given/known data
    A bullet flying horizontally hits a wooden block that is initially at rest on a frictionless, horizontal surface. The bullet gets stuck in the block, and the bullet-block system has a final speed of [itex] V_f[/itex]. Find the final speed of the bullet-block system in terms of the mass of the bullet [itex] m_b [/itex], the speed of the bullet before the collision, [itex] v_b[/itex], the mass of the block [itex] m_wb [/itex] and the amount of thermal energy generated during the collision [itex] E_t[/itex]

    2. Relevant equations
    [itex] K_{final} = (1/2)mv^2 [/itex]
    [itex] K_{inital} + P_{inital} + W_{total} = K_{final} + P_{final} + E_t [/itex]

    3. The attempt at a solution

    So it is asking for final speed so I use my first equation and rewrite it:

    [itex] v = \sqrt {(2K_f)/(m)} [/itex]

    in this case, m is going to equal the mass of the block + the bullet, so I write:

    [itex] v = \sqrt {(2K_f)/(m_{wb} + m_b ) } [/itex]

    Now all I have to do is find [itex] K_f [/itex]

    Using equation 2 I can rewrite to:

    [itex] K_f = K_i + W_t - E_t [/itex] I drop the potential energies because there is no potential energy in the system

    K_i = initial kinetic energy, right? So (1/2) (mass of bullet)(velocity of bullet)^2 = (1/2)(m_b)(v_b)^2

    I just left W_t as W_t and looked to E_t.

    Solving [itex] K_f = K_i + W_t - E_t [/itex] for E_t I get:

    [itex]E_t = K_i - K_f + W_t [/itex]
    [itex]E_t = -\Delta K + W_t [/itex] if [itex] K_f - K_ i = \Delta K [/itex] thats why I have [itex] -\Delta K [/itex].

    Now I plug in E_t and it cancels a few things...

    [itex] 2K_f = m_bv_b^2 + 2W_tot -2(-\Delta K + W) [/itex]

    [itex] 2k_f = m_bv_b^2 + \Delta K [/itex]

    for my final answer of

    [itex] v = \sqrt {(m_bv_b^2 + 2\Delta K) / (m_b + m_{wb})} [/itex]

    but my books answer is...:

    [itex] v = \sqrt {(m_bv_b^2 - 2\Delta E_t ) / (m_b + m_{wb})} [/itex]

    does anyone know where I went wrong??
     
  2. jcsd
  3. Mar 31, 2017 #2

    kuruman

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    Energy is not conserved, momentum is.
     
  4. Mar 31, 2017 #3
    Hmm I don't think I understand.. Do you know where in my logic I went wrong? I was using only the formulas I was given so I don't see where I went wrong :/
     
  5. Mar 31, 2017 #4

    kuruman

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    OK, if set aside momentum conservation for the time being, what does ##W_t## represent in your equations?
     
  6. Mar 31, 2017 #5
    Energy going from one system (K_e of bullet) to the next (Bullet block).


    Correct??
     
  7. Mar 31, 2017 #6

    kuruman

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    Consider writing expressions for the energy of your system of bullet + block before the bullet is embedded and after it is embedded. You have
    ##E_{before} = K_{bullet}## and ##E_{after} = K_{bullet + block} + E_t##
    If you start from this and do the algebra correctly, you should get the book's answer. However, I think this problem is over-determined because the answer does not conserve momentum. Have you studied momentum conservation?
     
  8. Mar 31, 2017 #7
    Not yet, we are not on momentum conservation. I will do the problem and report back!
     
  9. Mar 31, 2017 #8
    @kuruman is correct. Momentum conservation should be the way to go. This is a perfectly inelastic collision.
     
  10. Mar 31, 2017 #9

    gneill

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    The OP wants to match the book's solution which specifically involves the use of the energy lost to heat over the collision. Provided that you account for this energy you can write the energy balance equation.
     
  11. Mar 31, 2017 #10
    So I CAN use the equations I listen in the OP? If so do you know where I went wrong in my method?

    I have a hunch that:

    Solving [itex] K_f = K_i + W_t - E_t [/itex] for E_t I get:

    [itex]E_t = K_i - K_f + W_t [/itex]
    [itex]E_t = -\Delta K + W_t [/itex] if [itex] K_f - K_ i = \Delta K [/itex] thats why I have [itex] -\Delta K [/itex].

    was the wrong step, but I'm not too sure..
     
  12. Mar 31, 2017 #11

    gneill

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    I don't know why you're distinguishing two energies Wt and Et. There's only one value that is the difference between the initial and final kinetic energies, so use one variable for it.
     
  13. Mar 31, 2017 #12

    kuruman

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    When you conserve energy you have to stay with one system. Here your system is bullet + block. There is no ##W_t##
    Look at post #6. Set Ebefore = Eafter and proceed.

    BTW, I stand corrected. Momentum conservation is not an issue here as gneill noted.
     
  14. Mar 31, 2017 #13
    Wow.. I think I understand now. So W_t and E_t are the same thing? Since the definition of work is energy transferred and E_t is the only energy being transferred in the system?
     
  15. Mar 31, 2017 #14

    gneill

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    Well, it's up to you to define whatever quantities you introduce. Originally you wrote your equation with these two varables but didn't spell out explicitly what they stood for.

    You could, for example, account for the heat energy in the block and the bullet and the air and the horizontal surface all as separate quantities. They'd still all have to sum to one value that is the difference between the initial and final kinetic energies.
     
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