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Abhisar

  • one year ago

A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after the collision will be, if coefficient of restitution is e

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  1. Abhisar
    • one year ago
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    @Michele_Laino

  2. Michele_Laino
    • one year ago
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    is your collision elastic?

  3. Michele_Laino
    • one year ago
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    or, how is defined the "coefficient of restitution"?

  4. Abhisar
    • one year ago
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    That's the only info given, for the sake of simplicity I tried solving it by considering the collision inelastic.

  5. Abhisar
    • one year ago
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    Oh well, the answer is \(\frac{1-e}{1+e}\)

  6. Abhisar
    • one year ago
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    That might help ....

  7. Michele_Laino
    • one year ago
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    if the collision is inelastic perfectly, then the final velocity of both spheres is equal to half of the initial velocity of the colliding sphere

  8. Abhisar
    • one year ago
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    So, you mean it can't be perfectly inelastic....Then how to find ratio of two final velocities in terms of e ?

  9. Michele_Laino
    • one year ago
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    no, sorry I'm not sure

  10. Abhisar
    • one year ago
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    It's ok, thnx for the effort though c:

  11. Michele_Laino
    • one year ago
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    let's suppose that our collision is perfectly inelastic, then we have this: |dw:1440942312488:dw|

  12. Michele_Laino
    • one year ago
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    using the conservation of momentum we can write this: \[\Large m{v_0} = 2mv\] and hence the final velocity is: \[\Large v = \frac{{{v_0}}}{2}\]

  13. Michele_Laino
    • one year ago
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    the two spheres are sticked together, after collision

  14. Abhisar
    • one year ago
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    Yeah i know that, but we need to find the ratio in terms of e. And I think I'll have to consider the collision elastic.

  15. Michele_Laino
    • one year ago
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    on the other hand, if the collision is elastic, then we have this: |dw:1440942582057:dw|

  16. Abhisar
    • one year ago
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    But in that case one of the final velocities will become 0, isn't it?

  17. Abhisar
    • one year ago
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    Yeah....

  18. Michele_Laino
    • one year ago
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    namely after collision the two spheres change their velocities

  19. Abhisar
    • one year ago
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    So basically, ratio will be 1:1 if the collision is inelastic...right?

  20. Abhisar
    • one year ago
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    And 0 if it's elastic?

  21. Michele_Laino
    • one year ago
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    the ratio is 1/2 if the collision is inelastic perfectly and 0 if that collision is elastic

  22. Abhisar
    • one year ago
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    So, the value of e must be in between 0-1

  23. Abhisar
    • one year ago
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    How, 1/2 , both the final velocities will be the same. Isn't it?

  24. Michele_Laino
    • one year ago
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    if collision is inelastic perfectly, then we can write this: \[\Large v = \frac{{{v_0}}}{2} \Rightarrow \frac{v}{{{v_0}}} = \frac{1}{2}\]

  25. Abhisar
    • one year ago
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    Umm..We need to find the ratio of velocities of both the ball after collision...

  26. Michele_Laino
    • one year ago
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    ok! then we have ratio=1

  27. Abhisar
    • one year ago
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    Thanks, I'll try to conclude to the given answer and meanwhile if something come up in your mind then please share c:

  28. Michele_Laino
    • one year ago
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    ok! :)

  29. IrishBoy123
    • one year ago
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    \[mu = mv_1 + mv_2 \implies u = v_1 + v_2\] \[e = \frac{v_2 - v_1}{u}\] solve for \(\large \frac{v_2 }{v_1} \)

  30. Michele_Laino
    • one year ago
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    By definition, from Wikipedia, we have that the coefficient of restitution is the ratio between the relative velocity after collision and the relative velocity before collision

  31. Michele_Laino
    • one year ago
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    here is the article which I read: https://en.wikipedia.org/wiki/Coefficient_of_restitution

  32. Michele_Laino
    • one year ago
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    let's suppose that part of the kinetic energy is lost due to collision, so we can write these equations: \[\Large \left\{ \begin{gathered} m{v_0} = m{u_1} + m{u_2} \hfill \\ \\ \frac{{mv_0^2}}{2} = \frac{{mu_1^2}}{2} + \frac{{mu_2^2}}{2} + \varepsilon \frac{{mv_0^2}}{2} \hfill \\ \end{gathered} \right.\] where \epsilon is such that: \[\Large 0 \leqslant \varepsilon \leqslant 1\] and \[\Large \varepsilon \frac{{mv_0^2}}{2}\] is the kinetic energy lost, due to collision, namely a fraction of the initial kinetic energy Developing those equation I get the subsequent final velocities: \[\Large {u_1} = {v_0}\frac{\varepsilon }{2},\quad {u_2} = {v_0}\left( {1 - \frac{\varepsilon }{2}} \right)\] therefore, we can write this: \[\Large \frac{{{u_2} - {u_1}}}{{{v_0}}} = 1 - \varepsilon \]

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