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Block 1 collides with block two which is attached to a spring. How far do they go?

  1. Oct 26, 2011 #1
    Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

    Answer 0.305 m



    Relavant Equations
    Fsp = -kx
    ______________________________

    How do I solve for this? I'm given k but what is Fsp? Is there any other way to solve for Fsp?

    After scratching my head for a good while I tried something out of pure desperation, despite my gut feeling that it was wrong. I said,

    m1v1^2 + m2v2^2 = m1u1^2 + m2u2^2
    m1v1^2 = m2u2^2
    u2^2 = (m1v1^2)/m2
    u2 = sqrt((m1v1^2)/m2)
    u2 = sqrt((1.7(3.6)^2)/1.10) = 4.4753

    Being unsure about the relationship I said u2 = Fsp. Therefore,
    Fsp = -kx
    4.4753 = -144x
    4.4753/-144 = x
    -0.03107 = x

    Now, I am way off with this attempt but I'm not quite sure how I'm suppose to approach this problem. Can somebody help explain to me what, and most importantly why, I have to do to solve this problem? I'm having difficulty with this section so baby steps is appreciated :).

    Thank you for taking the time to review my question.
     
  2. jcsd
  3. Oct 26, 2011 #2

    PeterO

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    Homework Helper

    Re: Block 1 collides with block two which is attached to a spring. How far do they go

    I think you need to use energy stored in the spring.

    When those two masses collide, conservation of momentum will enable you to calculate the speed with which the blocks move after collision. From that velocity you can calculate the Kinetic energy of the blocks.
    When they stop, that energy will be transferred to the spring..
     
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