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Yahoo! Group Title

The Spring mass System moves with a Speed of 1m/sand has a Complete inelastic collision woth a mass resting..Find the maximum compression of spring after collison?

  • one year ago
  • one year ago

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  1. Yahoo! Group Title
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    |dw:1356870345732:dw|

    • one year ago
  2. sauravshakya Group Title
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    Is it 0.1732m

    • one year ago
  3. Yahoo! Group Title
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    I dont Know....any how Can u Show the Steps...:)...Some times It may Help .:)

    • one year ago
  4. sauravshakya Group Title
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    Conservation of energy

    • one year ago
  5. Yahoo! Group Title
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    Wat would be the K.E of Sys

    • one year ago
  6. Yahoo! Group Title
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    We have to Consider Effective Mass Right ?

    • one year ago
  7. sauravshakya Group Title
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    Initial K.E of the system is 3J

    • one year ago
  8. sauravshakya Group Title
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    maximum extension is either 0.1732m or 0.15m

    • one year ago
  9. sauravshakya Group Title
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    But I am not sure which equation is correct 1/2 kx^2 + 1/2 *6*(0.5)^2 = 1/2*6*1^2 1/2 kx^2 = 1/2 6*1^2

    • one year ago
  10. Yahoo! Group Title
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    We have to Find the Compression After Collision

    • one year ago
  11. sauravshakya Group Title
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    Now I think it is 1/2 kx^2 + 1/2 *6*(0.5)^2 = 1/2*6*1^2 which gives x=0.15 m

    • one year ago
  12. Yahoo! Group Title
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    1/2 *6*(0.5)^2 ?????

    • one year ago
  13. sauravshakya Group Title
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    I think |dw:1356871645704:dw|

    • one year ago
  14. sauravshakya Group Title
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    It will be something like that after collision

    • one year ago
  15. naveenbabbar Group Title
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    Use conservation of momentum And the compression will be maximum when the relative velocity between the blocks is zero

    • one year ago
  16. gleem Group Title
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    Also keep in mind that just after the collision the spring exerts a force on the two 3kg masses and speeds them up so their final velocity is > than 0.5 m/sec

    • one year ago
  17. sauravshakya Group Title
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    U mean velocity of blocks are same

    • one year ago
  18. sauravshakya Group Title
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    all*

    • one year ago
  19. sauravshakya Group Title
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    Now, that gives 1/2 *kx^2 + 1/2 *9* (2/3)^2 =1/2 *6*(1)^2 x=0.1 m

    • one year ago
  20. Yahoo! Group Title
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    @Vincent-Lyon.Fr

    • one year ago
  21. Vincent-Lyon.Fr Group Title
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    I find 5 cm, but I am looking for a simpler answer than the one I have written so far.

    • one year ago
  22. Yahoo! Group Title
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    can u Show..hw u Did ?

    • one year ago
  23. Vincent-Lyon.Fr Group Title
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    How do you usually solve these problems? Do you stay in the lab's frame of reference, or to you go to the 'CoM frame of reference'?

    • one year ago
  24. Yahoo! Group Title
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    Centre of mass

    • one year ago
  25. Vincent-Lyon.Fr Group Title
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    Ok, so CoM is moving at velocity \(2v_o/3\), right?

    • one year ago
  26. Vincent-Lyon.Fr Group Title
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    Wait! It is easier to consider the first collision in lab frame. After inelastic collision, new mass 2M will move at \(v_o/2\), right?

    • one year ago
  27. Vincent-Lyon.Fr Group Title
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    Now, new problem in CoM frame is:|dw:1356887265577:dw| Do you agree? Now, working out KE of system is easy. When spring is fully compressed, both masses will be at rest and KE will be zero. PE will lead you to compression of spring.

    • one year ago
  28. Yahoo! Group Title
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    v0 / 6 ???

    • one year ago
  29. Vincent-Lyon.Fr Group Title
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    Velocities in CoM frame are: \(v^*_{M}=v_o-v(CoM)=v_o-2v_o/3=+v_o/3\) \(v^*_{2M}=v_o/2-v(CoM)=v_o/2-2v_o/3=-v_o/6\) which means that: M-block is moving to the right with speed \(v_o/3\) 2M-block is moving to the left with speed \(v_o/6\)

    • one year ago
  30. shubhamsrg Group Title
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    energy is never conserved in case of in elastic collision. momentum is conserved, also, reduced mass = (3*3)/(3+3) = 3/2 so question simplifies to a mass 3/2 kg attached to a spring of spring constant 200N/m travels with a speed 1m/s colliding with 3 kg mass from consvn of momentum (3/2) (1) = (3/2 +3)v v = 1/3 now, loss in KE must have been converted to spring energy 1/2 (200)(x^2) = 1/2 (3/2) 1^2 - 1/2 (3/2 +3) (1/3^2) am getting 7cm approx.. and there is high chance i have made mistakes.. !

    • one year ago
  31. Vincent-Lyon.Fr Group Title
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    I do not understand how you determine the reduced mass. After the collision, there are 2 masses. One is M, the other is 2M. The reduced mass is 2M/3 = 2 kg.

    • one year ago
  32. Yahoo! Group Title
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    @shubhamsrg The options are 5 / sqrt2 cm 10 sqrt [ 5/2]cm 10sqrt5 cm 5 cm

    • one year ago
  33. shubhamsrg Group Title
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    i got 10/sqrt2 i.e. 5 sqrt 2 .. hmm..

    • one year ago
  34. Vincent-Lyon.Fr Group Title
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    If you follow the equations and the reasoning I wrote above, you will find 5 cm.

    • one year ago
  35. shubhamsrg Group Title
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    ohh..there is reduced mass also after the collision ! :O didnt notice that.. allow me to try again then.. reduced mass before = 3/2 reduced mass after= 2 so now, 3/2 = 2v v= 3/4 then 1/2 (200)x^2 = 1/2 (3/2) 1^2 - 1/2 (2)(3/4)^2 now getting 4.3 cms approx.. where am i going wrong.. ? :/

    • one year ago
  36. shubhamsrg Group Title
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    maybe conservation of energy isnt applicable right ? as, not necessary all change in Ke would have been converted to spring energy.. heat must have also be generated! hmm..

    • one year ago
  37. shubhamsrg Group Title
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    been*

    • one year ago
  38. shubhamsrg Group Title
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    how about this approach : if we apply COLM only on right block in the spring mass system : 3(1) = (3+3)v v = 1/2 (to the right i.e. original direction) |dw:1356965408871:dw| now, applying consn of energy , 1/2 (3)(1^2) + 1/2 (6) (1/2^2) = 0 + 0 + 1/2 (200) x^2 now 15 cms! :(

    • one year ago
  39. sauravshakya Group Title
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    I got 17.32 cm , 15 cm and 10 cm

    • one year ago
  40. Vincent-Lyon.Fr Group Title
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    I do not know where you are going wrong, because I do not understand what laws you are following or in what frame of reference you are working. If you understand my drawing above and what I wrote: "Now, working out KE of system is easy. When spring is fully compressed, both masses will be at rest and KE will be zero. PE will lead you to compression of spring." You will get: \(KE\,^*=\Large\frac{1}{2}\normalsize M \Large(\frac{v_o}{3})^2 \normalsize + \Large\frac{1}{2}\normalsize (2M) \Large(\frac{v_o}{6})^2 \normalsize = \Large\frac{1}{12}\normalsize M v_o^2\) Then \(PE=\Large\frac{1}{12}\normalsize M v_o^2=\Large\frac{1}{2}\normalsize k \;\Delta l ^2\) will lead to \(\Delta l\) = 5 cm

    • one year ago
  41. shubhamsrg Group Title
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    maybe,,here is where i went wrong, both blocks during max compression will have some common velocity, 3(1) + 6(1/2) = 3v + 6v 2/3 =v so consn of energy will modify like this : 1/2(3)(1^2) + 1/2(6) (1/4) = 1/2(3+6)(4/9)+ 1/2(200) x^2 ohh! i got 5 cms! \m/ !

    • one year ago
  42. shubhamsrg Group Title
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    am sorry i didnt look at your method yet,, i wanted to do this by some other way,,and voila ! to others and mostly to my surprise,,i was able to! :D

    • one year ago
  43. shubhamsrg Group Title
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    i'll try to explain my approach now : first let us consider the inelastic collision between the 2 blocks, applying COLM , 3(1) = (3+3)v so v=1/2 now we are reduced to the question |dw:1356966700681:dw| and we need to find max compression in this case

    • one year ago
  44. shubhamsrg Group Title
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    at max compression, both blocks will acquire a common velocity, since no external force is acting, momentum will be conserved. 3(1) + 6(1/2) = 3v + 6v v = 2/3 so, initial energy was 1/2 (3) (1^2) + 1/2(6) (1/2^2) final energy = 1/2(3)(2/3)^2 + 1/2(6) (3/2)^2 + 1/2 (200)x^2 equate both and find x

    • one year ago
  45. Vincent-Lyon.Fr Group Title
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    This is correct :) You are staying in lab-frame, so final KE is not zero. I was working in CoM-frame, so final KE* is zero. Both methods lead to 5 cm compression.

    • one year ago
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