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Homework Help: Ballistic Spring System

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    You have been asked to design a "ballistic spring system" to measure the speed of bullets. 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. The spring's maximum compression d is measured.

    Find an expression for the bullet's speed vb. Express your answer in terms of the variables m, M, k, d, and constant g.

    2. Relevant equations


    3. The attempt at a solution

    I just have absolutely no idea how to approach this problem. I know I am supposed to have an attempt at a solution but I've never felt so lost on a physics problem before. Please help!
  2. jcsd
  3. Nov 30, 2009 #2
    I hate to bounce, but I am really in need of some help here. It seems like whenever I get going on starting to solve it I just get overwhelmed with variables and my work gets jumbled up into a mess. This problem is due tomorrow morning and I'll be up all night working on it. Since I know I'm not getting help because I haven't posted an attempt at a solution here goes:

    Because in the collision momentum is conserved, we can use the conservation of momentum equation (the only reason I am doing this is because someone else somewhere told me it might be a good idea....I have no clue why I am doing this):
    mv_B + Mv_A = (m + M)v_AB

    Now we can solve for v_B:
    v_B = ((m + M)v_AB - Mv_A) / m

    Now I have v_B on one side and some random gibberish on the other side, which doesn't take in to account the spring, gravity, compression, etc. and A) I have no idea where to go from here and B) even if I did I am probably 100% wrong so far.

    Please guys, I'm really trying here even if it doesn't look like it. If you can help me I would be infinitely grateful!
  4. Nov 30, 2009 #3

    Doc Al

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    Staff: Mentor

    Think of the problem in two stages:
    (1) The collision of the bullet with the block. This is an inelastic collision, momentum is conserved:
    (2) The subsequent rise of the block with spring compression. What's conserved here?

    (If you knew the speed of the 'bullet+block' immediately after the collision, how could you find the maximum compression?)
  5. Nov 30, 2009 #4
    Thank you, Doc Al, for your response.

    OK, I understand why momentum is conserved now. Is my method valid? I am terrible at working with only variables.

    Energy is conserved here. From spring potential to kinetic and back to spring potential (I believe). I do not understand how to merge each 'stage' of the problem in equation form.

    I guess this is where I'm having the biggest problem. Although what you said in parenthesis greatly simplifies things, I still have no idea how I'd find the max compression if I knew the speed of the bullet+block (which I think I do now know). I am convinced we were never taught this in class, or I am missing some key piece of information here.

    For the velocity of the bullet+block (v_AB), I solved the equation in my previous post and got:
    v_AB = (mv_B + Mv_A) / (m + M)

    And I then plugged this into the kinetic energy equation to kind the kinetic energy of the bullet+block right at the impact:
    K = 0.5(m + M) * ((mv_B + Mv_A) / (m + M))^2

    Now I'm not sure where to go. I'm beyond being able to picture what's going on in my head, which doesn't help.
  6. Nov 30, 2009 #5

    Doc Al

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    Your momentum equation can be simplified. What's the initial speed of the block?

    Immediately after the collision, the 'bullet+block' has some speed and thus kinetic energy. Once it rises to the maximum compression, the only energy is potential since the speed is zero at that point. Note that there are two kinds of potential energy here: elastic and gravitational.

    Simplify. What's v_A?

    First get an expression relating the maximum compression to v_AB using conservation of energy.
  7. Nov 30, 2009 #6
    OK, now I think I understand everything you are saying except the last part:

    Maximum compression (d) can be expressed by using the spring potential energy (Us) formula:
    Us = 0.5kd2

    And I know that the Us at d is equal to the kinetic energy (K) right after the collision, so:
    K = Us
    0.5(m + M)((mvB/(m + M))2 = 0.5kd2

    Then do I just simplify? Or are there steps that I am forgetting? Thanks again for your help Doc, I wouldn't have been able to understand this without it.

    I just realized that I did not account for the fact that the spring was already compressed before the bullet hit the block. Hmm... another complexity. Where does this come into play?
  8. Nov 30, 2009 #7

    Doc Al

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    Staff: Mentor

    One way to set up conservation of energy for stage 2 is like this:
    KE1 + Spring PE1 + GPE1 = KE2 + Spring PE2 + GPE2

    (Some of those terms will be zero.)

    Measure GPE from the starting point; Measure Spring PE from the unstretched position of the spring. (When the block is hung from the spring, the spring stretches. By how much?)
  9. Nov 30, 2009 #8
    I know GPE1 will be zero (because origin is where block was initially) and KE2 will be zero (because that is when the velocity is zero).

    I already calculated KE1 in my previous post.
    KE1 = 0.5(m + M)(mv_B/(m + M))^2

    For Spring PE1, I calculated it using Hooke's Law (not sure if this is correct):
    -(m + M)g / k = x
    Spring PE = 0.5kx^2
    Spring PE1 = -0.5k((m + M)g / k)^2
    Should Spring PE1 be negative or positive?

    After this I do not know how to find either the Spring PE2 or the GPE2. I know the GPE2 should just be (m + M)gh but I don't know how to find the height.

    Thanks for your patience with me... I'm slowly but surely getting it.
  10. Nov 30, 2009 #9

    Doc Al

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    Staff: Mentor


    Good. (You can simplify that a bit.)

    The initial position is where the block alone stretches the spring, so use Mg, not (m + M)g. Spring PE, measured from the unstretched position, is always positive.

    (And since there's no acceleration, it's ΣF = Ma = 0. Thus kx = Mg.)

    That height is given as 'd'.

    You're getting there. :wink:
  11. Nov 30, 2009 #10
    Spring PE2 should just be the same as Spring PE1 except with the addition of the mass of the bullet, yes? This doesn't quite make sense to me, because the spring probably won't be equally compressed at the top and at the bottom, am I right? What is the correct way to calculate Spring PE2?

    Once I get Spring PE2 I will put all the terms together and then solve for v_B, correct? Hopefully I don't get lost in the algebra.
  12. Nov 30, 2009 #11

    Doc Al

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    The mass of the bullet has nothing directly to do with calculating the spring PE. And there's no reason whatsoever to think that the final Spring PE is equal to the initial Spring PE. To get an expression for Spring PE2, you need to know how much the spring is compressed from its unstretched length. You know the total amount the block moves is 'd' and you know how much of that is from the weight of the block stretching the spring down at the initial point--so subtract those two.
    Right. (Don't forget about gravity.)
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