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Coefficient of restitution

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, small object A (mass: M) collides with small object B (mass: m), which is initially at rest, on top of a horizontal stand fixed to a horizontal floor. Both A and B proceed to shoot horizontally off the stand and fall to the floor. Horizontal distance D from the edge of the stand to the point where A lands is 1/2 of the horizontal distance d to the point where B lands. Friction between the two objects and the stand is negligible.

    https://scontent-kul1-1.xx.fbcdn.net/hphotos-xfp1/v/t34.0-12/12767193_1257518357595899_654751098_n.jpg?oh=5ee0294835905cc57ce2d8b8b2eed855&oe=56CC2F3E

    What is the coefficient of restitution between A and B?

    2. Relevant equations

    a899eacd21a1ea3a57908b2beefff863.png

    3. The attempt at a solution
    If I simply substitute M1U1 + M2U2 = M1V1 + M2V2 and the equation above, I will come out with (M - m)/ 2M but that's wrong since I didn't put in the distance factor..

    I'm not sure how to bring in the distance factor into the restitution formula. Should I use the v^2 = u^2 + 2as formula?

    Also, is it that both A and B are in rest when they land on the floor? If that's the case, won't the final velocity of the object be 0?
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Feb 20, 2016 #2

    ehild

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    What do you think which object falls further?
    You have a collision and a horizontal projection following it.

    No, you do not know what happens after the objects reach the floor. They can rebound, slide further, but you are interested in the velocity just before impact.
    Push a ball from a table, and see what happens when it lands on the floor.
     
    Last edited by a moderator: May 8, 2017
  4. Feb 20, 2016 #3
    well i failed to understand - where you are wrong and what is correct according to you ?
    try to explain !
    do you think you are correctly describing the physical situation! check by doing a simple exercise of throwing a ball horizontally from a height h?
     
  5. Feb 21, 2016 #4
    Thanks for the replies!

    Below is my attempt according to the two formulas.

    M1U1 + M2U2 = M1V1 + M2V2
    Since U2 = 0, M1U1 = M1V1 + M2V2

    U1 = (M1V1 + M2V2) /M1
    V1 = (M1U1 - M2V2) / M1
    V2 = (M1U1 - M1V1) / M2

    Using the formula a899eacd21a1ea3a57908b2beefff863.png
    Cr = [ (M1U1 - M1V1) / M2) - (M1U1 - M2V2) / M1] / [(M1V1 + M2V2) / M1]
    = {[(M1M1U1 - M1M1V1) / M1M2 ] - [M1M2U1 - M2M2V2 / M1M2]} x M1 / (M1V1 + M2V2)
    = { [M1M1U1 - M1M1V1 - M1M2U1+ M2M2V2] / M1M2 } x (M1V1 + M2V2)
    = [ M1M1U1 - M1M1V1 - M1M2U1 + M2M2V2 ] / (M1M2V1 + M2M2V2)

    *edited
    sorry I'm kinda messed up with this..

    My bad, after this I realize that this should be the right answer for my attempt.
    But still, I doubt that this is the wrong answer. As this formula doesn't require the horizontal distance to come out with the answer... (and I guess this is a bit over complicated?)

    I see, this is my misunderstanding..

    Object B falls further?
     
    Last edited: Feb 21, 2016
  6. Feb 21, 2016 #5

    ehild

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    It is not an answer. You need to give CR with known quantities, the masses of the objects M and m and the ratio of the horizontal distances they reach. You do not know V1 and V2. It has no sense to make the original formula more complicated by still having the same unknowns.

    Yes, what does it mean for its velocity after the collision?
     
  7. Feb 21, 2016 #6
    I think its velocity after the collision will be faster.
     
  8. Feb 21, 2016 #7

    ehild

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    How much faster if it reaches twice the distance then A?
     
  9. Feb 21, 2016 #8
    The velocity will also be 2 times faster than A, so the ratio of both distance and velocity will be 1:2. Am I right?
    So the next step would be substitute the ratio into the formula?
     
  10. Feb 21, 2016 #9

    ehild

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    yes, into both formula. First find VA in terms of the masses and UA, then substitute both VA and VB=2VA into the formula for CR.
     
  11. Feb 21, 2016 #10
    Let V2 = 2V1

    M1U1 = M1V1 + M2V2
    M1U1 = M1V1 + 2M2V1
    V1(M1+2M2) = M1U1
    V1 = M1U1 / (M1 + 2M2)

    CR = (2V1 - V1) / U1
    CR = V1 / U1

    Sub V1 = M1U1 / (M1 + 2M2)
    CR = V1 = M1U1 / (M1 + 2M2) / U1
    CR = V1 = M1U1 / (M1 + 2M2) x (1 / U1)
    CR = M1 / (M1+2M2)

    This is the correct answer, thanks a lot! :D
     
  12. Feb 21, 2016 #11

    ehild

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    You are welcome :smile:
     
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