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Coefficient of resistution

  1. Dec 11, 2008 #1
    A smooth sphere A of mass 4m is moving on a smooth horizontal plane with speed u. It
    collides directly with a stationary smooth sphere B of mass 5m and with the same radius as A.

    The coefficient of restitution between A and B is 1/2

    (a) Show that after the collision the speed of B is 4 times greater than the speed of A.

    Sphere B subsequently hits a smooth vertical wall at right angles. After rebounding from the
    wall, B collides with A again and as a result of this collision, B comes to rest.

    Given that the coefficient of restitution between B and the wall is e,

    (b) find e.

    For part (b), I recieve an answer of 1/14. This is due to the fact that I have equated the coefficient of resisution of the two balls in the following way for the final collision:

    0.5 = v/(u+4eu)

    where v = velovity of ball A after the final collision, and u = the velocity of ball A before the collision (and hence the velocity of B is -4eu).

    This results in a result of e = 1/14 after solving simultaneously with a linear expression for the conservation of momentum.

    However, the markscheme has done the above as: 0.5 = (0-v)/(u+4eu) - note the presence of the negative sign. This results in a result of 0.5 for the second coefficient of resistution. Can someone please explain the logic behind this?

  2. jcsd
  3. Dec 13, 2008 #2
    Last edited by a moderator: May 3, 2017
  4. Dec 13, 2008 #3

    Doc Al

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    Staff: Mentor

    I didn't look at the problem in detail, but this piece seems to make sense. By definition of coefficient of restitution (e):
    V'a - V'b = e(Vb - Va)
    v - 0 = e(u - -4e'u) = 0.5(u + 4e'u)

    Where I use e' for the second COR.

    (Did I miss your point?)
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