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Yahoo!

  • 2 years ago

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?

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  1. Yahoo!
    • 2 years ago
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    |dw:1356870345732:dw|

  2. sauravshakya
    • 2 years ago
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    Is it 0.1732m

  3. Yahoo!
    • 2 years ago
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    I dont Know....any how Can u Show the Steps...:)...Some times It may Help .:)

  4. sauravshakya
    • 2 years ago
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    Conservation of energy

  5. Yahoo!
    • 2 years ago
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    Wat would be the K.E of Sys

  6. Yahoo!
    • 2 years ago
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    We have to Consider Effective Mass Right ?

  7. sauravshakya
    • 2 years ago
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    Initial K.E of the system is 3J

  8. sauravshakya
    • 2 years ago
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    maximum extension is either 0.1732m or 0.15m

  9. sauravshakya
    • 2 years ago
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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

  10. Yahoo!
    • 2 years ago
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    We have to Find the Compression After Collision

  11. sauravshakya
    • 2 years ago
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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

  12. Yahoo!
    • 2 years ago
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    1/2 *6*(0.5)^2 ?????

  13. sauravshakya
    • 2 years ago
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    I think |dw:1356871645704:dw|

  14. sauravshakya
    • 2 years ago
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    It will be something like that after collision

  15. naveenbabbar
    • 2 years ago
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    Use conservation of momentum And the compression will be maximum when the relative velocity between the blocks is zero

  16. gleem
    • 2 years ago
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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

  17. sauravshakya
    • 2 years ago
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    U mean velocity of blocks are same

  18. sauravshakya
    • 2 years ago
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    all*

  19. sauravshakya
    • 2 years ago
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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

  20. Yahoo!
    • 2 years ago
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    @Vincent-Lyon.Fr

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

  22. Yahoo!
    • 2 years ago
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    can u Show..hw u Did ?

  23. Vincent-Lyon.Fr
    • 2 years ago
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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'?

  24. Yahoo!
    • 2 years ago
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    Centre of mass

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

  26. Vincent-Lyon.Fr
    • 2 years ago
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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?

  27. Vincent-Lyon.Fr
    • 2 years ago
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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.

  28. Yahoo!
    • 2 years ago
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    v0 / 6 ???

  29. Vincent-Lyon.Fr
    • 2 years ago
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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 \(v_o/6\)

  30. shubhamsrg
    • 2 years ago
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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.. !

  31. Vincent-Lyon.Fr
    • 2 years ago
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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.

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

  33. shubhamsrg
    • 2 years ago
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    i got 10/sqrt2 i.e. 5 sqrt 2 .. hmm..

  34. Vincent-Lyon.Fr
    • 2 years ago
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    If you follow the equations and the reasoning I wrote above, you will find 5 cm.

  35. shubhamsrg
    • 2 years ago
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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.. ? :/

  36. shubhamsrg
    • 2 years ago
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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..

  37. shubhamsrg
    • 2 years ago
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    been*

  38. shubhamsrg
    • 2 years ago
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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! :(

  39. sauravshakya
    • 2 years ago
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    I got 17.32 cm , 15 cm and 10 cm

  40. Vincent-Lyon.Fr
    • 2 years ago
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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

  41. shubhamsrg
    • 2 years ago
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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/ !

  42. shubhamsrg
    • 2 years ago
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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

  43. shubhamsrg
    • 2 years ago
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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

  44. shubhamsrg
    • 2 years ago
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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

  45. Vincent-Lyon.Fr
    • 2 years ago
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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.

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