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Coefficient of Restitution & Collisions

  1. May 17, 2008 #1
    Hi there,

    I have been struggling with this problem for a while. It's an A-Level Mechanics problem.

    A smooth groove in the form of a circle of radius a is carved out of a horizontal table. Two small spheres, A and B, lie at rest in the groove at opposite ends of a diameter. At time t = 0, the sphere A is projected along the groove and the first collision occurs at time = T. Given that e is the coefficient of restitution between the spheres, find the velocities of A and B after the first collision. Hence, or otherwise, show that the second collision takes place at time

    t = T(2+e)/e.

    I have included my solution of the first part in Solution1.txt and can confirm the answers. It's the last part that I have been unable to do.

    After collision 1, B is travelling faster than A and I am assuming they are both travelling in the same direction (I have tried assuming A rebounded but still cant get the answer!)

    So distance travelled by A before the second collision will be

    (Va is the velocity after the first collision, Sa is the distance travelled by A before 2nd impact.)

    Sa = t x Va

    The distance travelled by B before impact will be the circumference + Sa

    Sb = 2πa + Sa

    where Sb also equals

    Sb = t x Vb

    I've substituted the velocities for A and B but still come up with

    t = 2T/e

    Can somebody help?

    Attached Files:

  2. jcsd
  3. May 17, 2008 #2


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    Hi wanchosen! :smile:

    But 2T/e is right … add the original T, and you get T(2+e)/e. :rolleyes:

    btw, you're making this needlessly complicated …

    Once you've obtained Vb = V(1+e)/2, Va = (1-e)/2,

    all you need to know is Vb - Va, which is eV.

    It's only the relative velocity that matters! :smile:
  4. May 18, 2008 #3
    Hi tiny-tim,

    Thank you so much for that. That makes sense now! It was much easier working with the relative velocity! I completely forgot about adding the original time T to travel to the first collision! Thanks again.
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