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Momentum and springs - A2 Physics

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A train of total mass 1.02x10^5 kg strikes a buffer that behaves like a spring of stiffness 320kN/m with an initial velocity of 0.15m/s
    Calculate the maximum compression of the spring.

    2. Relevant equations

    3. The attempt at a solution
    = 15300kgm/s

    Units of spring stiffness so assumed Stiffness = F/d

    so S=P/dt, v=d/t

    so, S=Pv/d^2

    d=sqrt (pv/s)
    final solution:

    sqrt ((1.53x10^4 * 0.15) / 320x10^3)

    = 0.0847m compression.

    Unsure of the real answer, and this method is totally invented. - Nor does it appear to take into account the changing force with respect to distance. - Is there a simpler method I have not taken into account? - This much assumption seems very off compared to the rest of the A2 syllabus

  2. jcsd
  3. Sep 18, 2009 #2


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    Gold Member

    The answer is correct. (Didn't check your method btw.)
  4. Sep 18, 2009 #3
    Hi Vanagib,

    i think you are right

    another approach:

    If the train is stopped by the spring, the hole kinetic energy of the train is saved in the spring, so

    [tex]T = V[/tex]​
    [tex]\frac{1}{2} m v^2 = \frac{1}{2} k x^2[/tex]​
    [tex]x_{max} = \sqrt{\frac{m}{k}} \cdot v = 0.0847 m[/tex]​

    with best regards!
  5. Sep 18, 2009 #4
    That's the correct answer.

    a simpler way to do this would to consider the Kinetic/Potential energy systems. The train has initial Kinetic energy found by .5mv^2, and when the train stops the buffer has all of that energy transformed into potential energy, which for a spring is .5kx^2. set the two equal, and simple algebra to solve for x.

    but the way you did it is fine too.

    edit: it seems as in typing this i've been beat to my explanation, and with Latex too!
  6. Sep 18, 2009 #5
    Many thanks all :)
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