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Finding the coefficient of restitution

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A body is fired from point P and strikes at Q inside a smooth circular wall as shown in figure. It rebounds to point S (diametrically opposite to P). The coefficient of restitution will be:

    (Ans: ##\tan^2\alpha##)


    2. Relevant equations



    3. The attempt at a solution
    Let ##v## (along PQ) be the velocity before collision and ##v'## (along QS) be the velocity after collision.

    The coefficient of restitution (e) is defined as:
    $$e=\frac{\text{Relative speed after collision}}{\text{Relative speed before collision}}$$

    The numerator is ##v'\sin\alpha## and denominator is ##v\cos\alpha##. (The relative speed is measured along normal at the point of collision)

    Hence,
    $$e=\frac{v'\sin\alpha}{v\cos\alpha}=\frac{v'}{v}\tan\alpha$$

    From conservation of linear momentum along PS:
    $$mv\cos\alpha=mv'\sin\alpha \Rightarrow \frac{v'}{v}=\cot\alpha$$
    Hence,
    $$e=1$$
    :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2014 #2

    TSny

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    Why would this be true? Notice that the problem says that the surface of the wall is smooth.
     
  4. Apr 12, 2014 #3
    Yes but I don't see how it has got anything to do with conservation of momentum. :confused:
     
  5. Apr 12, 2014 #4

    TSny

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    "Smooth" means that the wall will not exert any tangential component of force on the body.
     
  6. Apr 12, 2014 #5
    Why are we talking about the tangential force now? Had it been present, does it mean that momentum would be conserved? :confused:
     
  7. Apr 12, 2014 #6

    TSny

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    If there is no force on a body in a certain direction, what does that mean about the component of momentum of the body in that direction?
     
  8. Apr 12, 2014 #7
    The momentum is conserved in that direction.

    Do you mean I should conserve momentum in radial direction?
     
  9. Apr 12, 2014 #8

    TSny

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    Yes. So, smooth means that the tangential component of momentum will be conserved. That is, the component of momentum parallel to the wall at the point of collision will be conserved.

    Note that PS is not the tangential direction.

    No, the tangential component (parallel to the wall) is conserved.
     
  10. Apr 12, 2014 #9
    Thanks TSny! :smile:

    This time I get, ##v\sin\alpha=v'\cos\alpha## which gives the right answer.
     
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