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Bullets and springs

  1. Dec 15, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    A spring whose spring constant is ##k## is suspended from the ceiling. A block of mass ##M## hangs from the spring. A bullet of mass ##m## is fired vertically upward into the bottom of the block and stops inside of the block. The spring's max compression ##d## is measured.

    Find an expression for the bullets speed in terms of ##k, M, m, d##.

    2. Relevant equations

    Conservation of energy in an isolated system and conservation of momentum.

    3. The attempt at a solution

    This question is pretty straightforward, but I was curious about something. When I found my final answer it didn't depend on gravitational energy at all. Why isn't there any gravitational energy to consider in this scenario?

    Was the question just poorly made in the sense they didn't give a height?
     
    Last edited: Dec 15, 2013
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  3. Dec 15, 2013 #2

    haruspex

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    Do you mean that g does not feature in your answer? It does in my answer. Please post your answer, and preferably your working too.
     
  4. Dec 15, 2013 #3

    Zondrina

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  5. Dec 15, 2013 #4

    haruspex

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  6. Dec 15, 2013 #5

    Zondrina

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    That was my intention, but I'm having a bit of trouble with the equations.

    I'm figuring I need to use the conservation law to isolate and then the momentum law to solve for what I isolated for, but I can't seem to piece this one together.
     
    Last edited: Dec 15, 2013
  7. Dec 15, 2013 #6

    haruspex

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    Just do the impact first. What is the speed of block plus bullet immediately after impact? (You can assume that the bullet finishes burrowing its way into the block very quickly.)
     
  8. Dec 15, 2013 #7

    Zondrina

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    So all I had to do was re-arrange the equation I had:

    $$v_f = \frac{m{v_1}_i}{m+M}$$
     
  9. Dec 15, 2013 #8

    haruspex

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    Right. From that point you can use work conservation.
     
  10. Dec 16, 2013 #9

    Zondrina

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    The work conservation is a bit tricky.

    Initially there is only elastic potential and kinetic energy I think with zero gravitational potential (given where I've defined my "zero"). In the final phase there is gravitational and elastic, but zero kinetic.

    The initial kinetic energy is based off that value we calculated for ##v_f## I believe. So by writing everything out and plugging in the value for ##v_f## into ##v_i^2## I would get:

    http://gyazo.com/3c79b9b7d16ff744199e17672062fbb7
     
  11. Dec 16, 2013 #10

    haruspex

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    You have yf and Δyf. How are you defining these?
    What mass should you be using in the work conservation equation?
    You can calculate Δyi from initial conditions.
     
  12. Dec 16, 2013 #11

    Zondrina

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    ##y_i## is the initial position of the mass M (which it happens to be at position 0 with the way I've defined things). This is why the initial gravitational energy is zero.

    ##y_f## is it's final position, but it will also be the final position of the bullet in this case since the bullet gets imbedded.

    ##Δy_i = (y_i - y_e)## and ##Δy_f = (y_f - y_e)##.

    The mass I should be using for the final gravitational energy is ##m + M##.

    I changed the notation up a bit to make things a bit more clear and I have this now:

    http://gyazo.com/c10d40449ef3ac1c111463887064f35c

    Also I'm not sure if this will help with incorporating ##d##, but:

    http://gyazo.com/e35fbc71c122fdbfc33b21fa5beff188

    EDIT: Wouldn't ##y_f^2## just be ##d^2##?
     
    Last edited: Dec 16, 2013
  13. Dec 16, 2013 #12

    haruspex

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    All the masses in the energy equation should be M+m. m should only appear by itself in the expression for the initial (after impact) speed.
    It makes it a lot easier to comment on your equations if you put them inline instead of attaching images.
    Your choice of zero level for the height as the initial position of the block M has led you into a trap. It would make things clearer if you chose the relaxed position of the spring. Also, it does say the spring becomes compressed, so the bullet pushes it up beyond the relaxed position.
    Not sure how exactly you're defining Δyi and Δyf. Assuming those are spring extensions, Δyf is negative. With your choice of base level as it is, the total ascent of the block+bullet is d+Δyi, and Δyf = -d.
     
  14. Dec 16, 2013 #13

    Zondrina

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    This :

    It's a trap!

    Knowing this, I'll tackle the problem from scratch, with a different perspective. I'm always assuming up and right are positive. Here's the new diagram just for show:

    http://gyazo.com/a8ec1e2f046016c5502495e037fc424c

    First consider when the bullet strikes and embeds itself into the block. The momentum is conserved in a completely inelastic collision and the final speed of the block + bullet can be found:

    ##p_{T_i} = p_{T_f}##
    ##m_1v_{1_i} + m_2v_{2_i}=(m_1 + m_2)v_f##
    ##mv_{B_i} = (m+M)v_f##
    ##v_f = \frac{mv_{B_i }}{m+M}##

    Now, initially there is only kinetic energy transferred into the block and in the final phase there is elastic + gravitational.

    ##E_i = E_f##
    ##K_i = U_{s_f} + U_{g_f}##
    ##\frac{1}{2} mv_i^2 = \frac{1}{2}k(Δy_f)^2 + mgy_f##
    ##mv_i^2 = k(Δy_f)^2 + 2mgy_f##

    Now I'm not certain about this, but subbing what we know into the equation we get:

    ##(m+M)v_{B_i}^2 = kd^2 + 2(m+M)g(y_e - d)##

    Does this seem more reasonable than before?
     
    Last edited: Dec 16, 2013
  15. Dec 16, 2013 #14

    haruspex

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    I'll say it again - all those m's should be (M+m). Or maybe you're using m generically here?
    If you want to make up as positive then I would take the zero height to be the relaxed position of the spring. You have:
    yi = (minus initial extension of spring) = Mg/k (taking g to be negative)
    yf = (minus final extension of spring) (positive), i.e. at max height.
    As the bullet + block rise to maximum height:
    - what is the change in gravitational PE?
    - what is the change in spring PE?
    - what is the change in KE?
     
  16. Dec 16, 2013 #15

    Zondrina

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    I was using m generically up until I subbed things into the equation. You're telling me that equation should read this instead:

    ##(m+M)v_{B_i}^2 + 2(m+M)gy_i = kd^2 + 2(m+M)gy_f##

    Makes sense. There initially has to be gravitational energy, otherwise how else would the bullet be fired vertically upward?

    Re-arranging the above equation I get:

    ##v_{B_i} = \sqrt{\frac{kd^2 + 2g(m+M)(y_f - y_i)}{m+M}}##

    The only thing left is ##(y_f - y_i)##, which represents (the final position of the mass) - (the initial position).

    http://gyazo.com/a8ec1e2f046016c5502495e037fc424c

    When I say ##y_e## I mean the equilibrium position of the spring. Given what I've drawn, I believe ##y_i = y_e## and ##y_f = y_e + d##. That would mean ##(y_f - y_i) = d##? Then the equation would be:

    ##v_{B_i} = \sqrt{\frac{kd^2 + 2gd(m+M)}{m+M}}##
     
    Last edited: Dec 16, 2013
  17. Dec 16, 2013 #16

    haruspex

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    No, ##y_i ≠ y_e##. Before the impact, the spring will be under tension Mg.
     
  18. Dec 16, 2013 #17

    Zondrina

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    You're telling me ##F_G = F_T## at the point of equilibrium, which is justified, but I don't see how it helps me determine the initial and final positions of the mass.

    Initially wouldn't the bullet be colliding with the mass when it's at the equilibrium rest position?

    EDIT:

    In case you think that I didn't see you post this:

    I did, but I don't see................. entirely why?

    Wait wait wait hold the phone.

    ##F_G = F_T##
    ##Mg = -k \Delta y##
    ##- \frac{Mg}{k} = y_f - y_i##

    That sir is a dirty trick I am never going to forget. Hooke's law.

    The final equation would be:

    ##v_{B_i} = \sqrt{\frac{kd^2 - \frac{2Mg^2(m+M)}{k}}{m+M}}##

    Lookin good?
     
    Last edited: Dec 16, 2013
  19. Dec 16, 2013 #18

    haruspex

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    That's at the original point of equilibrium, when there's only the block mass M suspended. It is below the relaxation point of the spring. The given d is the maximum compression of the spring, i.e. a distance above the relaxed spring point. The total vertical movement of the block is the sum of these: the distance it has to go from the initial stretched position of the spring, through the relaxed position, to the 'final' compressed position.
    In the first part of that, the spring is losing potential energy as it goes from stretched to relaxed. In the second part, as it becomes compressed, it is regaining PE. You need to allow for the net change in spring PE as well as the net change in gravitational PE.
     
  20. Dec 17, 2013 #19

    Zondrina

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    I re-did the diagram. Is the positioning for the ##y_e## where it's supposed to be now?

    http://gyazo.com/4baca1b044d8b31a8fc11003b4b6dd74

    I don't even know how to draw a proper diagram.

    My gut is PULLING on me that I need to define when the spring was completely relaxed BEFORE the mass was attached onto it. Otherwise how could I possibly define ##\Delta y_i##? Call this relaxed position ##y_0##. I would define ##\Delta y_i = y_e - y_0## and THAT would be my initial displacement from the spring's relaxed position.

    I'm back to square one again, I REALLY think this is the case:

    ##U_{s_f} + U_{g_f} = K_i + U_{s_i} + U_{g_i}##
    ##k(Δy_f^2 - Δy_i^2) + 2mg(y_f - y_i) = mv_i^2##

    I'm admittedly lost a this point though, conceptually. I don't even understand whether or not I have the correct energy equation because I don't even know what the problem wants me to consider.

    It was established that:

    ##v_i = v_f = \frac{mv_{B_i}}{m+M}## - I understand this

    ##y_f - y_i = - \frac{Mg}{k}## - I understand this too
     
    Last edited: Dec 17, 2013
  21. Dec 17, 2013 #20

    Zondrina

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    Tossing everything I've said and done prior to this, I have a serious question now, because I desperately want to understand this problem conceptually at this point. I'm always assuming up and right are positive.

    Why on earth should I not define the relaxed position of the spring BEFORE the mass was attached to it? I couldn't possibly define any form of initial displacement ##( \Delta y_i )##, which is in the down direction, without knowing the initial relaxed position of the spring. Suppose I called this relaxed position ##y_0##.

    Then lets say I attach the mass ##M## and it drags the spring down until the restoring force equals the force of gravity so that the net force is zero. The restoring force will now keep the mass and spring at a new equilibrium position, lets call it ##y_e##. I can NOW define ##\Delta y_i = y_e - y_0## and if this is incorrect please explain why. In case there is any confusion, ##\Delta y_i## is the distance the spring is initially EXTENDED downwards in this case, which will be used to determine the initial elastic potential. At this time, there is gravitational potential energy and elastic potential energy stored in the system.

    Then in comes the bullet, which we have to conserve the momentum for. The final speed of the block + bullet should be used to determine the initial kinetic energy of the system. So in the initial instant of the system, there should arguably be kinetic, gravitational and elastic energy.

    As the bullet + mass travel up after the impact, kinetic energy and elastic energy are being lost (the spring is returning to the relaxed position ##y_0##), while the gravitational energy is increasing. By the time the mass + bullet have reached the maximum height, all of the kinetic energy is gone and in the form of elastic + gravitational. We know that the spring compressed by a factor of ##d##, so that it's "new equilibrium position for an instant" is ##y_{e_2} = y_e + d##. Would that not mean the final extension of the spring ##\Delta y_f = y_e + d - y_0##? If not, why not?
     
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