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Blocks and a spring

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Two blocks of mass 3 kg and 6kg are placed on a smooth horizontal surface.They are connected by a light spring of force constant k=200N/m. Initially the spring is unstretched. The block of mass 3 kg is imparted a velocity of 1m/s and the block of mass 6 kg is imparted 2 m/s (in opposite directions). Find the maximum extension of the spring.
    The answer is 30 cm.

    2. Relevant equations


    3. The attempt at a solution
    I thought we can equate the sum of the Kinetic energies of the two blocks to 1/2kx2. But will there be 2 extensions x1 and x2 for both the blocks? In that case how will I find both? IS there a relation between them?
     
  2. jcsd
  3. Dec 8, 2015 #2

    BvU

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    Well, what do you get if you follow your own suggestion ? :smile:
     
  4. Dec 8, 2015 #3
    Well I get this equation with x1 and x2 squared. I have no idea how to proceed after that.
     
  5. Dec 8, 2015 #4

    BvU

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    I meant
     
  6. Dec 8, 2015 #5

    BvU

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    Ah, I see: you worry about x1 and x2 instead of about "the compression" :smile: ?
    If you compress on one end with x1 and then on the other end with x2, do you need ##\sum kx^2_i## or ##k(\sum x_i)^2## ?
     
  7. Dec 8, 2015 #6
    The second one? So the total extension must be put in the 1/2kx2 equation?
    And i think I should put my question a little more clearer. They are being pulled outward when the respective velocities are imparted. So not compression.
    Though it makes no difference in the method I guess.
    I did the calculation only now. I get 36 cm. aaaaargh.
     
  8. Dec 8, 2015 #7
    You missed, that the kinetic energy is not zero, when the spring is stretched to its maximal value.

    Try to explain what happens in the scenario with both blocks and when the spring should be stretched to its maximum (i.e. the condition for maximal elongation).
     
  9. Dec 8, 2015 #8
    oh. I think the blocks will move with a common velocity after a while and that can be found using conservation of momentum. But I don't know if that will be when they are stretched to maximum.
    Please help.
     
  10. Dec 8, 2015 #9
    You are on the right path. That's why I suggested to try to imagine what happens in the scenario. I start:

    1) Both blocks with different masses move with different initial velocities in different directions - the spring has its initial length and starts to be stretched
    2) At a certain point the 3kg-mass will come to rest and change the direction of movement, whereas the 6kg-mass will still move in the same direction - the spring will be stretched further as the velocity of the 6kg-mass is larger than the velocity of the 3kg-mass (but from now both move in the same direction)
    3) ...

    Your turn
     
  11. Dec 8, 2015 #10

    haruspex

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    When they are moving with the same velocity, what can you say about the rate of change of the distance between them?
    If you want to find the maximum of something that varies continuously, how do you use the rate of change?
     
  12. Dec 8, 2015 #11
    um....ok so the 6 kg mass will keep stretching the spring until at one point, they will move with the same velocity and a particular elongation which is the maximum?
    Please forgive me if I am making no sense.
     
  13. Dec 8, 2015 #12
    Correct (and just for clarification: That's only for one particular moment)!

    However, now you are able to solve the problem:

    You can find the (equal) velocity of both masses with the conservation of momentum and then the elongation of the spring with the conservation of energy.
     
  14. Dec 8, 2015 #13

    BvU

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    You're doing fine.
    There is no friction, so the oscillations will go on. But from time to time the blocks are not moving with respect to one another. Even then, the conservation of momentum helps you find their speed. But you have to think carefully what momentum is conserved and what momentum is not conserved (because the spring will keep exchanging spring energy for kinetic energy). Imagine the system covered with a box; with what speed would you need to have the box move to keep up ?
     
  15. Dec 8, 2015 #14
    in reply to both BvU and stockzahn
    ok first
    I get the combined velocity as 1m/s. I put it in the equation 1/2(6+3) 12=1/2 *200*x2.Is that right? because now i'm getting 21cm
    second
    Oh.......So it will move with the same velocity at one point and then the 3 kg block moves to the larger? And this will go on?
    Wow. Hard to visualize all of this.
     
  16. Dec 8, 2015 #15
    Why should the kinetic energy be equal to the potential energy in the spring? You have to compare the energies of the initial situation (elongation of the spring is zero → only kinetic energy) and the all energies at the moment of maximal elongation (kinetic energy and potential energy).
     
  17. Dec 8, 2015 #16
    Oh got it.30 cm.
    Got a little confused, thought the kinetic energy at that point is converted into potential energy but in reality they both exist separately?
     
  18. Dec 8, 2015 #17
    Well done!

    In every momet the sum of all energies must be the same. During the oscillation there are moments, when the elongation of the spring is zero, so the total amont of energy must be stored in the movement of the masses. The rest of the time the total energy consists partly of varying kinetic energy (of the blocks) and varying potential energy (of the spring).
     
  19. Dec 8, 2015 #18
    Thank you all for the help. Really appreciate it. There's a lot of difference between just getting an answer and really thinking and understanding the depth behind it.You all help us achieve the latter. I salute you all. :smile:.
     
  20. Dec 8, 2015 #19
    oh oh oh. Just one more question. There will be one time when both the blocks are pulling in opposite directions and it'll reach some elongation till the rest happens. How will I find out if that elongation is not the maximum one and the one that happens later is?
    I suppose that's what I did the very first time. Equating the 1/2*3*1+1/2*6*22=1/2kx2.
    Is there any way of knowing in advance or must I just calculate both ?
     
  21. Dec 8, 2015 #20

    BvU

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    My estimate is that later on you will learn that there are two degrees of freedom in the system described.

    It's easier to imagine/visualize if you take two identical masses: One degree of freedom for the motion of the center of mass; since there are no external forces on the whole (Newton 3), the center of mass moves uniformly. The other degree for compression/expansion of the spring. For unequal masses it becomes a bit more involved, but there still are these same two independent degrees of freedom.
     
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