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Bullet going through a box

  1. Jul 15, 2017 #1
    Can someone check if I'm doing this problem correctly? Please tell me if I did anything wrong.

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

    A 12 g bullet passes horizontally through a 15 cm tall, 20 cm long .5 kg block which is initially at rest on a 1 meter tall table’s edge. The bullet is initially moving 230 m s , and it passes through the block. After the bullet passes through it, the block hits the ground 134 cm from the edge of the table.

    2. Relevant equations
    Δy = -4.9t2
    Kf = Ki + W


    a) How fast is the block moving after the bullet emerges from it?
    3. The attempt at a solution
    I solved this using projectile motion and got 2.97m/s:

    Δy = -4.9t2
    t2 = -1/-4.9
    t= 0.45s
    vx = 1.34/0.45 = 2.97m/s

    b) How fast is the bullet moving when it emerges from the block?
    3. The attempt at a solution
    p of bullet initial = p of bullet final + p of block final

    p of bullet initial - p of block final = p of bullet final

    speed of bullet final = (p of bullet initial - p of block final)/(mass of bullet)

    v = (0.012*230 - 2.97*0.5)/0.012 = 106m/s

    c) Assuming that the block does not move significantly while the bullet is inside it, determine the average force exerted on the block by the bullet while the bullet passes through the block. Use energy!
    3. The attempt at a solution
    Here is where I think I'm wrong:

    Kblock, final - Kblock, initial = Fd

    d = 0.2m because it goes through the length of the box.

    so (Kblock, final - Kblock, initial)/d = F

    F = (0.5*0.5*2.972)/0.2 = 11N

    My intuition tells me that 11N is SUPER low for a bullet flying through something. Maybe you would use 106m/s but I don't get why you would do that.
     
  2. jcsd
  3. Jul 15, 2017 #2

    haruspex

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    The 0.2m is the total distance for which the bullet experienced a reaction from the block, so Fd there is the work done by the bullet. Equating this to the gain in KE effectively assumes work is conserved - highly unlikely.
    Can you think of another way?

    By the way, obtaining the work done on the block by an Fd expression is a bit tricky. As far as the block is concerned, the force advanced only over the distance the block moved while the bullet passed through it. Since you are told to pretend that is zero, you will not get a useful equation that way.
     
    Last edited: Jul 15, 2017
  4. Jul 15, 2017 #3
    I've never heard of work not being conserved :/. I'm just doing introductory physics right now and I don't think we go into work not being conserved and stuff. Could you give me a brief explanation?

    I said that distance was 0.2m right? Since it passes through the length of the block? So, does the distance the block go while the bullet is in it count too? I don't get exactly what you're trying to say. If you assumed Kf to be 0.5*0.5kg*106m/s, would you get the right answer? That's what my teacher is saying to do: use 106m/s as the final velocity of the block.

    Thanks for the response! Are my other answers correct?
     
  5. Jul 15, 2017 #4

    haruspex

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    You need to distinguish between energy - which includes internal energy (crudely, heat), sound energy, etc. - and mechanical work, which is just kinetic energy and potential energy (gravitational PE, elastic PE, electrostatic PE....)
    When an object slides along a floor, losing speed to friction, the KE is being converted to internal energy etc., mechanical work is not conserved. Friction is called a non-conservative force.
    As the bullet passes through the block, some work is lost to friction but most in crushing the wood.
    When a force pushes on an object, and succeeds in moving it, we say the work it does is the force times the distance it advances. But it can be tricky working out exactly what distance to use. In the present case, it helps to think of the block in two parts:
    - a point mass (so incompressible)
    - a massless compressible region between the bullet and the point mass
    Since the compressible region is massless, it exerts a force F at each end.
    To make matters clearer, we need to drop the suggestion of ignoring the distance the block moves while the bullet is inside it. Say it moves some small distance s. I will use d for the length of the block.
    In passing through the block, the bullet moves d+s. The work done on it is -F(d+s), so F(d+s) will equal its loss of KE.
    Meanwhile, the block's point mass only moves distance s, so the work done on it is only Fs, and that will equal its gain in KE.
    If we subtract the one from the other we get Fd, and that will equal the net loss of KE in the system. Dividing by d yields F.
    That is clearly nonsense.
     
    Last edited: Jul 15, 2017
  6. Jul 15, 2017 #5

    scottdave

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    If you want to consider the final kinetic energy og the bullet, then you would need to compare that with the beginning KE of the bullet. Im not sure if this is the best way to figure average force. There was an Insights article which discussed this.
     
  7. Jul 15, 2017 #6

    haruspex

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    You are right that there is not enough information to determine average force unless we know it is constant. But if we know it is constant, why ask for average?
    The insights article (which I wrote) is https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ (see item 3).
    But I was planning to bring that up later after resolving the wildly incorrect answer in post #1.
     
  8. Jul 16, 2017 #7
    I honestly don't think the problem is that in-depth. I understand what you mean, but I think the compressibility of the box is most likely ignored. I don't also know the distance s unless we assume that while the bullet is inside the box, both are moving as a system at 106m/s, in which case the professor was correct?

    He told me the answer was 1256N, although the steps he showed were very vague. Here are the steps:

    W = Kf - Ki

    Vf = Vb

    Can you verify that that answer is correct or not? I tried plugging in 106m/s as final and I got *near* that answer.


    I'm pretty sure I'm worried only about the block, but I'm not sure. Could you tell me what you mean?

    Would you say 1256N is more correct? I'm still super new to physics, I'm doing basic high school AP physics. I think what he means by average force is just the work divided by distance it travels. I'm still unaware of any way to find that s value you mentioned before.

    Either way, thanks for helping. Your explanations are slightly beyond the scope of my knowledge, but I'm curious. If anyone has any advice of what to ask my teacher, please tell me. I know my answer is wrong... and although, my distance is definitely wrong, can someone tell me if I did the change in kinetic energy part correctly at least? Is it not 2.97m/s?

    Again, I'm just curious. It's not a homework problem and I already got this wrong on a test so I just want to know how to do something like this in my actual physics class next year.

    Thanks guys
     
  9. Jul 16, 2017 #8
    Also thanks dude. I legit saw that article the day before the forces test (one before the work one) and it helped me get a tension problem right (tip #2). I stopped reading there because I didn't know about work yet. I'll definitely give that list a read, but I don't really think high school physics is that advanced :P (tip 3). After all, the teacher used the same formula I did. I'm still pretty sure his answer is more accurate, but the way I got my answer (the bullet does some work on the object; after it leaves, the box gains KE due to work done by the bullet; the change in that is the work done). Is that way of thinking (ignore the average force aspect) about work correct? Is the Kf - Ki = 0.5*0.5kg*(2.97m/s)^2 = 2.2 J correct? It looks so wrong but I've had many problems that looked weird, but turned out to be right.
     
  10. Jul 16, 2017 #9

    haruspex

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    Sorry, but it is. See (*) below.
    That was just a model. The point is that while the bullet passes through the block the block moves a much shorter distance, which I called s.
    You don't need to. The method I outlined avoids calculating it. It turns out to be a few mm. See (**) below.
    Yes, that works approximately, but only if you pick the bullet as the object to work with.
    But that is not very accurate.

    Here's the full solution:
    Let the distance the block advances while the bullet passes through it be s. So while the force F is exerted (one way on the bullet, the other on the block) the bullet moves distance d+s and the block mass centre moves distance s.
    The work done by the bullet is -F(d+s), so
    KEbullet, final=KEbullet, initial-F(d+s)
    The work done on the block is Fs, so
    KEblock, final=KEblock, initial+Fs
    Adding these together:
    KEsystem, final=KEsystem, initial-Fd
    So Fd = KEsystem, initial-KEsystem, final = 0.5*0.012*2302-0.5*0.012*106.42-0.5*0.5*2.9662 = 247.3J
    Hence F = 247.3J/0.2m = 1236N

    Your teacher's solution is a bit inaccurate partly because it ignores the energy acquired by the block:
    (0.5*0.012*2302-0.5*0.012*106.42)/0.2 = 1247N
    and perhaps because of rounding errors.

    Also, as scottdave reminds us, dividing work by distance is not generally a valid way to calculate an average force. It is only sure to work if the force is constant, and goes quite wrong for e.g. simple harmonic motion. See the article I linked for details.

    (*) If you fail to understand the subtleties I described in post #4 then you might make the mistake of equating the energy gained by the block to Fd, as you did.

    (**) Having found F, we can find s:
    Fs = KE gained by block = 0.5*0.5*2.9662 = 2.20J
    s = 2.20J/1236N = 1.78mm
     
  11. Jul 16, 2017 #10
    I get what you're doing... but I would never have been able to think of it. I still don't get your thought process though... I thought: on the box, so it must be delta KE of the box, which is.. wrong. So, how did you arrive at the whole system aspect of things? I'm slightly less interested in the math (once I understand the concept of a question, I'm usually fine) and more so in how you arrived at that way of doing it. Why did you want to look at the bullet? or the system?

    Basically, where in my thinking did I screw up?

    Here you show a slightly less accurate version (yet still much better than mine): 0.5*0.012*2302-0.5*0.012*106.42/0.2 = 1247N

    I was very close to doing this on the test, but I remembered reading this: "The change in the kinetic energy of an object is equal to the net work done on the object." So I thought that the bullet was irrelevant, just the change in KE of the block must represent the work done on it.

    I'm probably coming off very adamant, but I really don't get where in my thinking I went wrong. That said, your mathematical explanation makes sense. Thanks for that :)
     
  12. Jul 16, 2017 #11

    haruspex

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    Yes, that is quite right. The trouble is that the work done on it is Fs, not Fd. That is, you must use the distance the block moved due to the force while it was being applied, not the distance the bullet moved.
    In fact, you do not know either distance to begin with. One is s and the other is d+s, but you know d, not s. The question statement tells you to ignore s, but that is a bit flaky since it can lead to contradictions. More usefully, it could say to assume s is small compared with d. That allows you to approximate that the work done by the bullet is -Fd. That is your teacher's method.
    My method says not to treat s as zero. By considering the block and bullet separately we can get two equations and eliminate s between them. This is standard "solving simultaneous equations".

    To sum up:
    - you cannot do it by only considering the KE change of the block because you do not know s
    - you can get an approximate answer only considering the KE change of the bullet by assuming s is very small
    - you can get an accurate answer by considering both KE changes
     
  13. Jul 16, 2017 #12
    Alright I think I got it. It's weird: I've been doing physics for a month or so now and math for a long time. I've done stuff like simultaneous equations in math and they're no big deal, but once it's a word problem in physics, it trips me up. I despise force/torque and force problems where you get a bunch of unknowns and you have to systematically figure each one of them out. But, once you figure it out, it seems super obvious and logical.

    With practice I guess.

    Thanks for the help!

    another quick note: usually, on the physics tests that I do (of which this was a question), this was by far the hardest one, but usually the harder ones are at the end. So, did this one take you even a little time to figure out? I assume not, but I just want to know if I'm alone in feeling that this one was fairly hard. I'm probably just dumb though haha
     
  14. Jul 17, 2017 #13

    scottdave

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    Here is one way to think about the amount of work done on the block. Maybe it will help you decide which object should be considered. Suppose you push on the block with some amount of force, for a small amount of time, but the block does not move. How much work is done? Even if you rub along the length of the block with an average force, if the block doesn't move, was work done?
     
  15. Jul 17, 2017 #14
    Nope, no work was done. d was equal to zero. Work is only done if the force causes displacement. Correct?
     
  16. Jul 17, 2017 #15

    scottdave

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    Correct, but the bullet is moving, and a force is acting on it while it is moving (slowing it down), so there is work done on the bullet. @haruspex showed how you can account for the change in energy of the block, even not knowing how much it moved.
    But I think what your test question was asking is "assume the block movement is so small that hardly any work done", then you assume that almost all of the work is on the bullet (which you know that distance).
     
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