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Inelastic collision (finding e, don't understand concept)

  1. Mar 5, 2013 #1
    I have problems with a(ii). (please view the first image.)

    Now I have found that w = (13-8e)u/7 and v = (20e+13)u/7

    What I did is like this:

    (Using NEL)

    v = 4ue
    Where the LHS is actually v-0 because I assume Q = 0.

    (20e+13)u/7 = 4ue

    e = 33/28 >1 ∴ e cannot exist.

    1. Is this way correct?

    2. My teacher used another way. (please view second image.)

    I don't understand the reason behind his workings.

    Actually he like to use the formula like this :
    V1- V2 = -e(U1-U2) ,even when V2 is faster than V1. Because he said in his brain, 2 always comes after 1, so he doesn't want to use V2-V1. I don't follow him, I follow the book, using speed of separation = fast - slow. But I still don't understand why he used V-W = -e(4u) to assume Q is brought to rest. (or did he even make that assumption at all?)

    My interpretation is like this: Since he uses "slower-faster = negative e (speed of approach)"
    Then in his workings, W should be assumed faster than V. And in the situation, both P and Q are moving to the left after collision, which makes their velocities negative. So if you want to assume Q is brought to rest, then W = 0. Since 0 is larger than negative, then his assumption is valid.

    This is just simply a hypothesis. I only have a vague picture of the concept but I cannot fully grasp it. Someone please explain it to me.

    3. The second part of a(ii), determining the least possible value of w. I just substituted e=1 into v-w = 4ue and got the answer. But after thinking about it, I suddenly don't understand why it should be e=1 which is substituted inside. I tried e=0.9, and the resulting w is smaller than the answer when e=1. So why is the least value obtained when e=1?
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2013 #2

    haruspex

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    You went a bit astray in there. You should have got e = 13/8. But your method is fine.
    I'm afraid your teacher has blundered. He has the sign wrong on the RHS. His method should have given e = e.
    That's fine. In this case, that gives v-w = -e(-u - 3u), so v - w = 4eu.
    w = (13-8e)u/7, so as e increases w decreases. Therefore the smallest w goes with the largest e, and the largest possible e is 1.
     
  4. Mar 7, 2013 #3
    Why is it that when e=e, there is no solution? Isn't e equal to e? (err....)

    I thought the right hand side should be -e (speed of approach) ? Why is it (-u-3u)? The particles are moving in opposite directions, which makes their speeds add up to become the total speed, no?
     
  5. Mar 7, 2013 #4

    haruspex

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    Arriving at e = e does not mean there is no solution. What it means is that the method your teacher used was ineffective. This can happen quite often when manipulating equations. If you use the same equation twice you can find everything cancels out and you end up with 0 = 0.
    You must measure all the velocities in the same direction. The given u and v are in opposite directions, so you have to flip the sign of one of them.
     
  6. Mar 10, 2013 #5
    This (-u-4u) thing only applies when using my teacher's version right? If I use W-V instead of V-W for speed of separation, then it would be just (u+4u)?
     
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