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Spring Constant Problem

  1. Oct 31, 2003 #1
    I could really use some help. I've been looking at this problem for a week now and am getting nowhere. If I could just find a little more info(like the spring constant, or spring mass) I can solve the problems. Here it is:

    A launcher uses a spring assemble to launch balls in the air. It has two settings, one where the assembly is pushed back 5 cm and the second pushed back 10cm. On the second setting (10cm) a 10g ball goes 1.6m straight up. On the same setting a 100g ball goes 1.25m up. On the first setting, the 10g ball goes 0.65m up.

    1. Why doesn’t the 10g ball go 10x as high as the 100g ball?
    2. When the spring is on the second setting why doesn’t the ball rise 4x as high as the first setting?
    3. What is the spring constant?
    4. What is the effective mass of the spring assembly?
    5. How much is the spring compressed when the ball leaves the launcher?
    6. How high would a 60g ball go up if it were launched at the first setting?
    7. What’s the speed of the 100g ball as it leave the launcher straight up, on the second setting?
    8. If the 100g ball were launched horizontally from the second setting, what would its speed be?


    The spring must be compressed after the balls leave, but how do I find how much? Why would the horizontal speed be different from the vertical speed right after launching (questions 7 and 8)?
     
  2. jcsd
  3. Nov 1, 2003 #2
    Please show us your work
     
  4. Nov 2, 2003 #3
    Well i tried to solve for K just what u wanted, and i did. i will put down my calculations.. look over it just to make sure that what i did was correct or not.
    m = 0.01g
    d = 1.6m

    to find K, u use the equation, Ee = 0.5kx^2, u need Ee.

    Ee = Ek
    Ek = 0.5mv^2

    u have the mass, u need the velocity to find the energy. Meaning, its back to kinematics

    V2^2 = V1^2 + 2ad
    0 = V1^2 + 2(-9.81)1.6m
    31.392 = v1^2
    5.6 m/s = V1
    u have the velocity, now u can go back to the Ek equation and slove for the amount of energy.
    Ek = 0.5 * 0.01Kg * 5.6m/s
    Ek = 0.028J
    Ee = Ek
    0.028J = 0.5*k*(0.1^2)m
    k = 5.6
    (NOTE: this is all for the second setting. you can use the same method inorder to find K for the first setting)
    good luck. :)
     
  5. Nov 2, 2003 #4
    Wouldn't this be true if the spring were relaxed once it fire the ball, and that the spring's mass was very small and we could ignore its kinetic energy. When you run the same calculations for the 100g ball you get a totally different constant which tells me two things.
    1. The mass of the spring is substatial in the interaction.
    2. The spring is not relaxed once the ball is fire.

    Both of these statements are questions that I must answers as well

    I know how to find the ball's speed once launched which I should be able to use as the spring and the balls final speed just before launch. But that still leaves me with 3 unknows in my CoE equation.
    0=.5MbVfb^2+ .5MsVfs^2+ Mbgzf + Msgzf + .5ks(rf-ro)^2 - .5ks(ri-ro)^2
    unknown: Ms, Ks and (rf-ro).

    Maybe I'm out in left field but this is how I think I should be looking at it. Only problem is anytime I get anywhere I end up with negative values or loop back around to the Ms=Mb.

    I'm really lost. I might just have to solve the couple of questions I know and that's it.
     
  6. Nov 2, 2003 #5
    well, im not really sure, u might have a point there.

    I have physics tomorrow afternoon, so i'll print this out and ask the teacher tomorrow. I really want to know the answer of this :D

    i'll let u know tomorrow if anything comes up.

    ( by the way, is this grd12 physics or what?)
     
  7. Nov 2, 2003 #6
    No, but someone moved it to this board.
     
  8. Nov 2, 2003 #7
    Well, anyways, i'll tell u what ever i get tomorrow.
     
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