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The system Shown ( the spring is without tension) travels towards the wall and has an elastic collision . The maximum Compression of the spring is ?

  • 2 years ago
  • 2 years ago

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  1. Yahoo! Group Title
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    |dw:1353427046355:dw|

    • 2 years ago
  2. Yahoo! Group Title
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    @experimentX @siddhantsharan

    • 2 years ago
  3. experimentX Group Title
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    1/2 mv^2 = 1/2 kx^2

    • 2 years ago
  4. Vincent-Lyon.Fr Group Title
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    Why did you close this question?

    • 2 years ago
  5. Yahoo! Group Title
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    @Vincent-Lyon.Fr

    • 2 years ago
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    @shubhamsrg

    • 2 years ago
  7. shubhamsrg Group Title
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    for maximum compression..whole kinetic energy is transferred to spring energy...that helps ?

    • 2 years ago
  8. shubhamsrg Group Title
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    and in such case,,effective mass = m1 m2 / (m1 + m2) am i helping ?

    • 2 years ago
  9. shubhamsrg Group Title
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    i mean are you getting ?

    • 2 years ago
  10. Yahoo! Group Title
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    wat is effective mass

    • 2 years ago
  11. shubhamsrg Group Title
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    2 masses are connected to a spring on 2 ends..say m1 and m2.. then the effective mass is m1 m2/ (m1 + m2)

    • 2 years ago
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    ok..)

    • 2 years ago
  13. Yahoo! Group Title
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    Wait....I mean....I did nt get...wat is effective mass shuld nt it be m1 + m2

    • 2 years ago
  14. shubhamsrg Group Title
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    effective mass or can be also called reduced mass of the system.. never heard of it ?

    • 2 years ago
  15. Yahoo! Group Title
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    No..)

    • 2 years ago
  16. Yahoo! Group Title
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    So.... 1/2 k x^2 = 1/2 effective mass * v^2

    • 2 years ago
  17. Yahoo! Group Title
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    right?

    • 2 years ago
  18. shubhamsrg Group Title
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    seems right..

    • 2 years ago
  19. shubhamsrg Group Title
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    well reduced mass is no big deal..has a simple derivation.. you wanna know ?

    • 2 years ago
  20. shubhamsrg Group Title
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    there isnt much need for the derivation for you to know since you're JEE aspirant..but still i can tell if you want..

    • 2 years ago
  21. Yahoo! Group Title
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    Yes..i am interested in it

    • 2 years ago
  22. shubhamsrg Group Title
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    kk..newton's 3rd law states that F1 = -F2 right ? both taken in vector form.. F1 will be one acting on 1st body,,F2 on second..

    • 2 years ago
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    ok...

    • 2 years ago
  24. shubhamsrg Group Title
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    all expressions are in vector form below : m1 a1 = - m2 a2 =>a1 = (-m2/m1) a2 now relative acceleration between the bodies a = a2 - a1 = a2 (1 + m2/m1) = m2 a2 ( m1 + m2)/(m1 m2) following ?

    • 2 years ago
  25. shubhamsrg Group Title
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    ??

    • 2 years ago
  26. shubhamsrg Group Title
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    tell me if not getting anything ? we are not done yet with the thing..

    • 2 years ago
  27. Yahoo! Group Title
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    Yup....i can follow u..)

    • 2 years ago
  28. Yahoo! Group Title
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    Tell me a Scenario where we will use Effective Mass

    • 2 years ago
  29. shubhamsrg Group Title
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    right here!!

    • 2 years ago
  30. Yahoo! Group Title
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    Another..Eg

    • 2 years ago
  31. shubhamsrg Group Title
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    now..m1m2 / (m1+ m2) = M (reduced or effective mass) so we have F2/M = a or F2 = Ma so what do you conclude ?

    • 2 years ago
  32. shubhamsrg Group Title
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    see,,any 2 body system can be converted to 1 body system with this effective mass... force acting between them will also act between this one body system..

    • 2 years ago
  33. Yahoo! Group Title
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    Nw..i get it...)...thxx....)

    • 2 years ago
  34. shubhamsrg Group Title
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    you sure?

    • 2 years ago
  35. Yahoo! Group Title
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    Yes.

    • 2 years ago
  36. shubhamsrg Group Title
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    hmm,,whats the conclusion you can draw then?

    • 2 years ago
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