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Block move down the curved hill and hit attached block spring with diagram

  1. May 2, 2008 #1
    A 100grams block is moving at 2m/s down the curved hill. The block slides along the smooth surface and collides elastically with the 400gram block. The 400gram block is initially at rest and is attached to an ideal spring with spring constant of 500N/m
    a)Find the maximum distance the spring compresses
    b)Find the maximum height the 100gram block reaches after the collision.

    I added a diagram....... help me T.T
     

    Attached Files:

  2. jcsd
  3. May 2, 2008 #2
    oh~ height is 3.0m.......
     
  4. May 3, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi cecico! Welcome to PF! :smile:
    I don't understand … it will be accelerating … at which point is its speed 2m/s? :confused:
     
  5. May 3, 2008 #4
    Someone else asked this exact same question on another thread, and that poster made the same mistake you did-- you have not attempted the problem. You must show that you have worked on the problem before asking for help.
     
  6. May 3, 2008 #5
    Not someone else.

    To start you off. I would ask my self what theory(s) I would like to apply to solve this problem. What theories have you learned that involve finding velocities and displacements?

    Work-Energy theorem and Conservation of Energy come to mind. Which would be easier in this case?

    And this too. Where is its velocity 2 m/s ?
     
    Last edited: May 3, 2008
  7. May 3, 2008 #6
    Oh you're right SaladSamurai! That poster must be impatient.
     
  8. May 3, 2008 #7
    SaladSamurai your right........ I have to use Work-Energy theorem and Conservation of Energy and the speed is 2m/s at point A and the surface is smooth.......
     
  9. May 3, 2008 #8

    tiny-tim

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    Hi cecico! :smile:

    ok … then what is the speed of the 100grams block just before it hits the other block? :smile:
     
  10. May 3, 2008 #9
    according to my calculation I got 7.925m/s.........after this i have no idea.....
     
  11. May 3, 2008 #10

    tiny-tim

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    That's right. :smile:

    Now pretend that there's no spring, and work out the speeds of the two blocks immediately after the collision. :smile:
     
  12. May 3, 2008 #11
    immediately after the collision I got 1.585m/s.... is this right??????????
     
  13. May 3, 2008 #12

    tiny-tim

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    uuh? you should get two speeds … one for each block! :rolleyes:

    (and if you show your working, that'll save me time checking! :smile:)
     
  14. May 3, 2008 #13
    ok......now i really don't get it..... what equation do i have use??
    momentum equation???? Mv+Mv=Mv+Mv but theres two unknown.....
    how do i calculate this????????? -0-
     
  15. May 3, 2008 #14

    tiny-tim

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    Hi cecico! :smile:
    DON'T PANIC!

    (i can tell you're panicking because the number of question marks is increasing faster than exponentially!!)

    In collisions, momentum is always conserved.

    And you're told that this collision is elastic, and that means that energy is also conserved.

    Two equations … two unknowns … physicist's heaven! :smile:
     
  16. May 3, 2008 #15
    alright for 100gram I got -4.755m/s and for 400gram I got 3.17m/s......it seems wrong..T.T
     
  17. May 3, 2008 #16

    tiny-tim

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    Well, let's assume it's right.

    You now have an initial velocity for the 400gram block.

    So … what is the maximum distance the spring compresses? :smile:
     
  18. May 3, 2008 #17
    I got 0.08966m........
     
  19. May 3, 2008 #18

    tiny-tim

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    Well, that's the answer to part a) then! :smile:

    And the maximum height that the 100gram block goes to?
     
  20. May 3, 2008 #19
    I used 1/2mv2=mgh that I got 1.154m.........
     
  21. May 3, 2008 #20

    tiny-tim

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    Hi cecico! :smile:

    Well, that's the whole question then, isn't it?

    Or is something still worrying you about it? :smile:

    ( … if aok now, click on "Thread Tools" to mark the thread [SOLVED] … :smile: )
     
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