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Spring fired marble distance needed

  1. Nov 19, 2005 #1
    OK, this one comes out of the mechanical conservation of energy section.

    A spring loaded cannon is compressed 1.1cm and shot off a the edge of a horizontal table. The marble lands 1.93m from the table. How far would you have to compress the spring to make the marble land 2.20m from the table?

    Ok, so I tried doing this with potential energy of a spring = kinetic energy of marble. Now, I can find the inital velocity with v2^2 = v1^2 + 2ax [a=-9.8; v2=0; x=1.93] and get that the 1.1cm compression gave an inital velocity of 6.15m/s. Same method, for a 2.20m distance it would need to give an inital velocity of 6.57m/s. So, Uspring = Ke. Since both have a factor of 1/2: kx^2 = mv^2. Now, I don't know m or k. So, I just wrote k = m[va]^2/[xa]^2, and used numbers for v and x (6.15 and 0.11), yielding k = 312500m . Then I wrote the same thing for the second firing, substituting the "solved" value of k, well, for k. So, m*v^2 = 312500*m*x^2. Now, v^2 = 312500*x^2. 6.57^2/312500 = x^2. x = sqrt(6.57^2/312500) which is .01175 = 1.18cm.

    Now, the book says the answer is 1.25 cm. Just to see, I tried this:
    [xa]/[xb] = [da]/[db] where da=1.93m, db=2.20m, and xa=0.11m.
    Solving for xb, the compression of the spring in the second case, I got 1.25cm.

    Now, why did that work? I was kind of hoping, since it was in the conservation of energy section, that I could use that but apparently I did something wrong in there. I also tried having mv^2=kx^2 and just dividing the two, leaving [va]^2/[vb]^2 = [xa]^2/[xb]^2 but that gave the 1.18cm answer again!
     
  2. jcsd
  3. Nov 19, 2005 #2

    Astronuc

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    Staff: Mentor

    v2^2 = v1^2 + 2ax is not correct in this situation, because the this implies a deceleration to 0 velocity in the x-direction, which is not necessarily the case, and the acceleration (-9.8 m/s2) applies in the vertical, not horizontal, direction.

    The question is, what is happening?

    For both cases, the marble starts at the same vertical height. The marble starts accelerating toward the surface below, and y = h = 1/2 gt2 from which one find t (the same t).

    In the X direction, the distance traveled Xa = v at and Xb = v bt in the same t.

    Now, certainly 1/2 kx2 = 1/2 mv2, or simply kx2 = mv2, and this can be rewritten as,

    x2 = K v2, where K = m/k.

    From above, one has X = v*t or v = X/t (assuming v is constant).

    The equation for relating spring displacement x, with distance of marble travel, X, becomes,

    x2 = K (X/t)2 or

    x = [itex]\frac{\sqrt{K}}{t}[/itex] X, which is the second relationship you used.
     
  4. Nov 19, 2005 #3
    oh, now i feel stupid trying to stop the x-direction motion. thanks for your help.
     
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