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Newton's Law of Restitution

  1. Oct 27, 2011 #1

    FeDeX_LaTeX

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    Gold Member

    1. The problem statement, all variables and given/known data
    "A small smooth sphere of mass 3 kg moving on a smooth horizontal plane with a speed of 8 m/s collides directly with a sphere of mass 12 kg which is at rest. Given that the spheres move in opposite directions after the collision, obtain the inequality satisfied by e."

    2. Relevant equations
    e = v/u


    3. The attempt at a solution

    I am sure I have the method right but I am just getting the wrong sign in my answer.

    Textbook answer is e > 0.25. I'm getting e < 0.25...

    Diagram:

    ---> = positive direction

    8 m/s..............0 m/s
    ->
    (3 kg).............(12 kg)
    <-.....................->
    A m/s..............B m/s

    By the conservation of momentum:

    24 = 12B - 3A
    ⇒8 = 4B - A

    e = (speed of separation)/(speed of approach)

    Speed of approach is 8 m/s.
    Speed of separation is A + B.

    ⇒ e = (A + B)/8
    ⇒ 8e = A + B

    So we have:

    4B - A = 8
    A + B = 8e

    Adding both equations gets us:

    5B = 8(1 + e)

    so e = (5B - 8)/8

    B > 0 since moving in positive direction. So e > -1.
    A < 0 since moving in negative direction.

    Since 8e = A + B and A < 0, we can say:

    8e - B < 0

    We say that e = (5B - 8)/8, so re-arranging in terms of B, we have B = (8e + 8)/5.

    So:

    8e - (8e + 8)/5 < 0

    ⇒ (32e - 8)/5 < 0

    ⇒ 32e < 8

    ∴ e < 0.25.

    But textbook's answer is e > 0.25. Why?

    Thanks.
     
  2. jcsd
  3. Oct 28, 2011 #2

    FeDeX_LaTeX

    User Avatar
    Gold Member

    Solved it, never mind.
     
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