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Motion Problem with two moving masses

  1. Sep 9, 2013 #1
    A block of mass m slides down a ramp. At the bottom, it strikes a block of mass M, which is at rest on a horizontal surface. If the collision is elastic and friction can be ignored, determine the upper limit on mass m if it is to rebound from M, slide up the ramp, stop, slide back down the ramp and collide with M again.

    So I have found that:

    m(Vi)^2 = m(Vf)^2 + M(W)^2

    And that m can only catch M if the velocity Vf is greater than W so the limit becomes Vf=W.

    Thanks
     
  2. jcsd
  3. Sep 9, 2013 #2

    ehild

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    The block M is in rest initially, so the other block certainly catches it.

    You need to give the limit for m compared to M, so the block of mass m travels backwards after collision. There is one more conservation law you need to take into account.

    ehild
     
  4. Sep 9, 2013 #3
    Is it momentum? If so then mVi = mVf + Mw
     
  5. Sep 9, 2013 #4

    ehild

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    Yes. Now find the relation between Vi and Vf, using both conservation laws.What is the condition of m so that it can catch M when slides down the ramp second time, that is |Vf|≥W?

    ehild
     
    Last edited: Sep 10, 2013
  6. Sep 10, 2013 #5
    Do I do this simultaneously? By substituting one of the equations into the other?
     
  7. Sep 10, 2013 #6

    ehild

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    Yes, do it.

    ehild
     
  8. Sep 10, 2013 #7
    What variable should I be eliminating?
     
  9. Sep 10, 2013 #8

    ehild

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    You have the equations
    m(Vi2-Vf2)=MW2
    and
    m(Vi-Vf)=MW

    Does not it look easier to eliminate W ? Square the second equation and divide the first one with it.

    ehild
     
  10. Sep 10, 2013 #9
    So from that I get:

    m=M(Vi^2 - Vf^2)/(Vi - Vf)^2

    Where do I progress from here?
     
  11. Sep 10, 2013 #10

    ehild

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    Factorize Vi^2-vf^2 and simplify.
     
  12. Sep 10, 2013 #11
    So now I get that m=M(Vi + Vf)/(Vi - Vf). So this means that m is smaller than M by a factor of (Vi + Vf)/(Vi - Vf).

    Is this the limit?
     
  13. Sep 10, 2013 #12

    ehild

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    No, you need the limit in terms of M.

    Isolate Vf. It must be negative in order that m return back to the slope. What does it mean for m compared to M?

    m goes up on the slope and then returns back. Without friction, it reaches the bottom with the same speed. It can catch M if that speed |Vf| exceeds the speed of M .

    So you have to find bot Vf and W in terms of m/M and Vi, and find m/M from the condition |Vf|≥W.

    ehild
     
  14. Sep 10, 2013 #13
    So if I rearrange that equation in terms of M I find that M=m(Vi - Vf)/(Vi + Vf).
     
  15. Sep 10, 2013 #14
    Ah I see that m is Larger than M
     
  16. Sep 10, 2013 #15
    So I find that w/Vf= 2m/(m-M) is that correct?
     
  17. Sep 10, 2013 #16

    ehild

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    No, just the opposite. m must go up to the slope so it has to move backwards after the collision, That means negative Vf. Vf<0.

    ehild
     
  18. Sep 10, 2013 #17

    ehild

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    That is al right. Now you have to ensure that the block of mass m catches the larger block.


    ehild
     
  19. Sep 10, 2013 #18
    Is this where Vf≥w? as vf≥w, then w/vf must be less than or equal to one. So then 1≥2m/(M-m).
     
  20. Sep 10, 2013 #19

    ehild

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    The magnitude of Vf is greater than W. Yes, 1≥2m/(M-m). What does it mean for m?


    ehild
     
  21. Sep 10, 2013 #20
    That M must be greater than or equal to 3m if m is to catch up with M and collide a second time?
     
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