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Collision of ball with inclined plane

  1. Oct 12, 2013 #1
    Hi friends,
    I have an issue in solving a Collision of ball with inclined plane.
    Please Help me in solving this.
    Thank you all in advance.

    The problem is as:

    https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1379765_1432382863655486_1628611844_n.jpg


    Attempt:

    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-ash3/q77/s720x720/1381857_1432380430322396_543886441_n.jpg

    So friends when I place vale of tanθ, the answer arises e = 1 But according to question the answer of the problem is tan2θ, which is Option (A) . Please try to help me in this.
    I will appreciate the help.
     
  2. jcsd
  3. Oct 13, 2013 #2

    haruspex

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    Yes, you will get tan θ if e=1, but e may be < 1. You have not made full use of the the information that the rebound is horizontal. Think about a different direction.
     
  4. Oct 14, 2013 #3
    My friend haruspex,

    I thought about the horizontal direction also. But the directions in which I am concentrating are along the line of impact and perpendicular to line of imapact.

    If I conserve momentum along the line of impact,

    mv cosθ = -mv' sinθ
    that gives, v'/v = cotθ.
    Putting this vale in equation (1), it gives, e = -1

    Conserving momentum along the direction perpendicular to line of impact,

    v sinθ = v' cosθ
    it gives, v'/v = tanθ.

    eq. (1) gives now the correct answer, e = tan2θ

    But this is wrong way.
    I can not conserve momentum along perpendicular to line of impact. ƩFnet ≠ 0 along this.

    By the way tanks for reverting back.
     
    Last edited: Oct 15, 2013
  5. Oct 15, 2013 #4

    haruspex

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    As long as you use two directions it should be fine. You only used the normal to the plane, so try thinking about parallel to the plane.
    Momentum is clearly not conserved in that line. On impact, what direction is the force?
     
  6. Oct 15, 2013 #5
    Along parallel the force could be like mg cosθ !!!
     
  7. Oct 15, 2013 #6

    haruspex

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    No, gravity doesn't matter during the impact. I may have misled you by asking about force instead of impulse. The impact is taken as occurring in a very short time, and impulse is force*time. This means gravity supplies very little impulse during impact, so ignore it. Just think about the impulse from the plane on the particle. What direction is that in?
     
  8. Oct 15, 2013 #7
    Perpendicular to the Plane on the ball and on the floor separately. i.e. Impulse will be the change in momentum of the ball the direction perpendicular to the floor plane.
     
  9. Oct 15, 2013 #8

    haruspex

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    Ok, so what can you say about momentum of the ball parallel to the plane?
     
  10. Oct 16, 2013 #9

    For the ball-

    I think , the force on the ball by the plane is only on the perpendicular direction hence in this direction momentum will change. But

    In the direction parallel to the plane there is no force, hence momentum of the ball will not change in that direction. Hence

    v sinθ = v' cosθ

    i.e. v'/v = tanθ
    and from eqn (1),

    e = tan2θ

    Which is the required answer.

    I really appreciate your help man. Thanks a lot.
     
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