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A question on impulsive tensions

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Two identical blocks, A and b, each of mass 2kg, are fastened to the ends of a rope 2m long. Block A rests on a smooth table, 1m from the edge. The rope passes over a pulley at the edge of the table and block B is held at the height of the tabletop, as shown. B is is then released.

    *For visual purposes: the diagram can be found here:
    http://books.google.com.my/books?id...B, each of mass 2kg, are fastened to"&f=false

    Page 113 Exercise 5B Question 2


    2. Relevant equations
    Impulsive tension: mv-mu
    g=10ms^-2
    3. The attempt at a solution
    mgh = 1/2mu^2 -> this is used to find the velocity of block B just before the top becomes taut.
    u^2 = 4g
    u = √2g
    Take upward and rightward directions as negative.
    Impulse is denoted as I
    For block B: -I = 2v - 2√2g
    For block A: -I=-2v

    -2v-2v=-2√2g
    4v=2√2g
    v=√2g/2

    Take the v as the initial velocity for both blocks A and B (initial velocity of Block A which is moving towards the edge of the table and initial velocity of Block B moving 1m downwards)

    u=√2g/2
    s=1m
    v=0
    t=?
    s=0.5(u+v)t
    t=0.894

    However, the answer given in the book is 0.782s

    Please include detailed explanations along with your solution, thanks!! :D
     
  2. jcsd
  3. Jul 10, 2012 #2

    tiny-tim

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    hi jiayingsim123! :smile:

    (try using the X2 button just above the Reply box :wink:)

    i'm confused :confused:

    the string goes taut immediately

    the initial speed is zero

    call the speed v, and the angle θ, and use conservation of energy and one other equation :wink:
     
  4. Jul 11, 2012 #3

    NascentOxygen

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    That link shows nothing useful in my current old IE7, but from the wording and OP's approach, I'd infer that block B falls 1m before the rope goes taut.

    I can't imagine what angle θ might be? :confused:
     
  5. Jul 11, 2012 #4

    tiny-tim

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    Hi NascentOxygen! :smile:

    i can't see the figure either :redface:,

    but I'm assuming it's the standard case of the string starting horizontal and stretched :wink:
     
  6. Jul 11, 2012 #5

    NascentOxygen

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    I agree, but warn that you are courting trouble if you don't type this as √(2g)
    Or you could have said the KE of the moving block becomes the KE of the pair.
    Can you explain why you say v=0? (As I explained, I can't see the diagram at your link.)
     
  7. Jul 11, 2012 #6
    http://img266.imageshack.us/img266/650/dropb.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Jul 11, 2012 #7

    NascentOxygen

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    Fair enough. Poster had better describe the setup then.

    Ah, thanks azizlwl :smile:
     
    Last edited: Jul 11, 2012
  9. Jul 15, 2012 #8
    Hi everyone, sorry for not replying sooner, I was a tad busy these few days.
    The answer to that question is 0.782s, but I can never seem to get that answer.
    After further thought, I think v is not 0 as the block doesn't stop after going down by 1m. I'm thinking along the lines of mgh=0.5mv^2 to get the final speed. But I'm really not too sure! Thanks azizlwl for posting the question! :)
     
  10. Jul 16, 2012 #9

    tiny-tim

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    yes, thanks azizlwl! :smile:
    looks ok so far …

    now use s = ut + 1/2 at2 twice, once for the first metre (a = g), and once for the second metre (a = g/2) :wink:
     
  11. Jul 16, 2012 #10
    Hi tiny-tim, thanks. But what should be the value of u? And if you were to do it, would you get 0.782? Sorry for my persistent asking of questions but what is the rationale behind doing what you suggested? Thanks in advance! :)
     
  12. Jul 16, 2012 #11

    tiny-tim

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    for the first metre, u = 0

    for the second metre, u = (√2)g/2
    i've no idea

    you tell us! :smile:
     
  13. Jul 16, 2012 #12
    Hi tiny-tim, I tried doing what you suggested, but I got 0.774 instead of 0.782, which is actually really close.
    This is what I did:
    Take g=10m/s^2

    For the first metre:
    s=ut+0.5at^2
    1=0.5gt^2
    t=sqrt1/5
    = 0.447

    For the second metro
    s=ut+0.5at^2
    1= √(2g)/2t + 0.5(g/2)t^2
    t=0.327

    Total time elapsed=0.774

    Sorry tiny-tim, but could you explain why you suggested this? I cannot really see the rationale, I'm sorry! :( Thanks in advance!
     
  14. Jul 16, 2012 #13

    pcm

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    i get 0.782 if i take g=9.8 m/s^2.
     
  15. Jul 16, 2012 #14

    TSny

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    Since I feel a little grumpy today :grumpy: , I'm going to complain about how the problem is stated. Although it may seem natural to assume that both masses come to the same speed immediately after each receives the impulse from the rope, that's an assumption that should have been stated in the problem. It's similar to having a 2-body collision problem and just assuming that it's completely inelastic.

    It's interesting to rework the problem assuming an elastic (but still very stiff) rope where no mechanical energy is lost in the "collision".

    Of course you could have a case anywhere in between the completely inelastic and elastic cases.

    (Now I feel a little better :smile:)
     
  16. Jul 16, 2012 #15

    tiny-tim

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    hi jiayingsim123! :smile:

    because for the 1st metre, the mass m is falling on its own, with acceleration g

    and for the 2nd metre, the total mass 2m is falling, with external force only mg, so the acceleration is g/2 :wink:
    no, the energy lost in the collision isn't relevant, conservation of momentum determines the speeds

    (though i agree that if the rope is stretchy, that will add slightly to the time taken)
     
  17. Jul 16, 2012 #16

    TSny

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    Hi tiny-tim. If no mechanical energy is lost in the collision, what would the speed of each mass be right after the collision?
     
  18. Jul 16, 2012 #17

    tiny-tim

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    it is lost

    i'm only saying it isn't relevant

    the motion is determined by conservation of momentum and by the constraint v1 = v2 (ie the collision is completely inelastic) :wink:
     
  19. Jul 16, 2012 #18

    TSny

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    But aren't you adding the constraint v1 = v2 as an assumption that is not stated in the problem?

    Maybe this constraint is implied in the wording of the question, but I don't see where. (Hope I'm not being too cantankerous in my grouchy mood today.)
     
  20. Jul 16, 2012 #19

    tiny-tim

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    i don't think so …

    the rope has a given length, so we assume it's a fixed length​
     
  21. Jul 16, 2012 #20

    TSny

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    Well, I don't see how assuming a fixed length makes much difference. When two steel balls collide, they don't deform noticeably during the collision, yet the collision might be close to elastic. Likewise, I think I can imagine a rope that doesn't stretch noticeably but still doesn't absorb much mechanical energy (sort of like an extremely stiff bungee cord).

    Anyway, I'll let it go.
     
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