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Bullet strikes a wooden block ()

  1. Dec 3, 2013 #1
    I'm stuck in in school suspension and I have to finish this quiz (it's open note/technology, not cheating!) by the end of school today, and I can't seem to find a problem similar to this on the internet, so I'm coming here. If you want like, I dunno, a poem or a drawing or something, I'll give ya what I can if you help me quickly.

    1. The problem statement, all variables and given/known data
    A 0.017 kg bullet with a velocity of 380 m/s strikes a 2.5 kg block inelastically. The block is suspended to the ceiling by 2 strings that are 1.5m long. The height from the floor to the block is 1.25 m.

    I've done a and b (answers below), but the two strings are throwing me for a loop on the other ones.

    c. The maximum horizontal displacement of the block from its rest position following the collision.
    d. The maximum vertical displacement of the block following the collision.
    e. The angle between the ceiling and the strings when the block reaches the highest position.
    f. The tension in each string when the block swings back to its lowest position.

    I've figured out the speed of the block after the collision is 2.56 m/s through the momentum equations, and I've also figured out the kinetic energy of the bullet before the collision is 1227.4 J, while the block's kinetic energy after is 8.192 J.


    2. Relevant equations
    m1v1 +m2v2 = (m1 + m2)v
    Kbottom = Utop, or 1/2(m1 +m2)v^2 = (m1 +m2)gh
    The sine and cosine equations


    3. The attempt at a solution

    So, I know if this were one string, I would make a use the LOCOE above to find the height travelled and then make a triangle with the original given height as the hypotenuse and the h I got subtracted from it as the y. I could use inverse cosine to find the angle, and find the x to find the horizontal displacement. But what do I do differently when there are two strings?

    Edit: Also, I've read the FAQ and I'm sorry for putting urgent! Didn't meant to pressure anyone. Thank you for all the help I've received so far, you guys honestly don't know how much I appreciate it. =)
     
    Last edited: Dec 3, 2013
  2. jcsd
  3. Dec 3, 2013 #2
    Well how is the block attached by the 2 strings? Are the strings in the same plane or another plane to the bullet?
     
  4. Dec 3, 2013 #3
    Ah, sorry, I wasn't very clear on that! Here's how the whole situation looks:

    2yn48kh.png

    Does that help? I'm not a very good explainer.
     
  5. Dec 3, 2013 #4

    vela

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    Is energy still conserved when there are two strings? In other words, are there any non-conservative forces acting on the block?
     
  6. Dec 3, 2013 #5
    Well, the collision is inelastic, so some of the kinetic energy should be lost due to entropy anyways, right? The total energy should remain the same, but I guess some of the kinetic energy is converted to air/heat, etc.? I don't get how the number of strings would affect that... Well, if I visualize it, since there are two strings sharing the tension/work of carrying the box, the displacement of the box would be approximately halved. Maybe not that, but the speed and displacement would definitely be smaller. I just don't know how to translate that into matematical terms... Sorry for the physics illiteracy, I'm not a very math/science oriented person person hence why I didn't go anywhere near Physics C on my course selection.

    If I get this problem done, everyone who's helped gets a free puppy, circumstances withstanding.
     
  7. Dec 3, 2013 #6

    vela

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    Sorry, I wasn't clear enough. I was asking if energy is conserved after the bullet is embedded in the block. In solving the original problem you alluded to, you use conservation of momentum to find the speed of block-and-bullet combination after the collision, and then you use conservation of energy to figure out how high the combination goes. The reason you can do this is because the only force on the block that does work is gravity, which is a conservative force. The only other force on the block, the tension from the string, doesn't do any work. You should know why this is the case.

    Now you have two strings. How does this affect the analysis?
     
  8. Dec 3, 2013 #7
    Well, since we'd be using 1/2mv^2 = mgh and the mass of the strings is negligible... it wouldn't?
     
  9. Dec 3, 2013 #8

    vela

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    How do you know you can use conservation of energy?
     
  10. Dec 3, 2013 #9
    Oh, right, you'd have to use Conservation of Momentum for inelastic, right? Sorry, my bad. Wait, if we've already found the velocity, would it just turn into a projectile then?

    Edit: No, no, wait, sorry, that's when it flies off the string. I'm a little confused then as to what I'd do...
     
  11. Dec 3, 2013 #10

    haruspex

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    I don't think vela meant to imply that you cannot use conservation of energy for what happens after impact. The question was whether you can explain why it is valid.
     
  12. Dec 3, 2013 #11
    Well in my opinion, the situation is same as a single string if that is the case. You can try imagine it yourself - if the block is hit by a bullet, the only string pulling back the block is the first string. The other string is not doing anything since it is not taut at all.
     
  13. Dec 3, 2013 #12

    vela

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    If the second string didn't exert a force, the block would rotate, which it doesn't.
     
  14. Dec 3, 2013 #13

    haruspex

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    It doesn't say it doesn't rotate, but if we had to worry about that it would get quite complex. If the line of the nearer (to impact) string, extended down through the block, ever passes above the mass centre then it will exert a torque which the second string would be unable to counter. The second string would become slack and the block rotate. So whether the block rotates at some point will depend on the detailed geometry and the amplitude of the swing.
    But I take your point that it's a pair strings in order that it's feasible there is no rotation - it just omits to clarify that there is none.
    For so long as the line of the nearer string, extended down through the block, passes below the mass centre there will be tension in the second string.
     
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