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Billiard Balls-Elastic Collision

  1. May 1, 2009 #1
    Billiard Balls--Elastic Collision

    I'm going through my old homework problems to prepare for the AP exam, and I got this right back in December, but now I have no idea how to solve it

    1. The problem statement, all variables and given/known data

    A cue ball traveling at 7.0 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30° with its original direction of travel. Find:
    a) the angle between the velocity vectors of the two balls after the collision
    b) the speed of each ball after the collision


    2. Relevant equations

    KE = 1/2mv2
    p = mv

    3. The attempt at a solution

    The collision is elastic, so both kinetic energy and momentum are conserved, so

    1/2mv02 = 1/2mv12 + 1/2 mv22
    v02 = v12 + v22

    and

    mv0 = mv1 + mv2
    v0 = v1 + v2

    The x component of the cue ball's final velocity is v1cos30
    The x component of the target ball's final velocity is v2cosΘ

    this gives me two equations, and three unknowns:

    49 = (v1cos30)2 + (v2cosΘ)2

    7 = v1cos30 + v2cosΘ

    Am I missing something that will tell me the angle of the velocity of the second ball? Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 1, 2009 #2
    Re: Billiard Balls--Elastic Collision

    For KE, you use the total magnitude of the velocity and don't break it up into x components. Also, the third equation you might be looking for is the y-component of momentum. Three equations and three unknowns is solvable.
     
  4. May 1, 2009 #3
    Re: Billiard Balls--Elastic Collision

    Okay, so then my equations should be:

    49 = v12 + v22
    7 = v1cos30 + v2cosΘ

    and, since i feel like the y components of momentum ought to cancel out for some reason,

    0 = v1sin30 - v2sinΘ

    The algebra is getting pretty thorny...
     
  5. May 1, 2009 #4
    Re: Billiard Balls--Elastic Collision

    The y-components of the momentum are correct, since there is no initial y-momentum. The equations do look correct, although it is pretty late and I might be forgetting something, but lets hope not. Now you need to wade through some messy algebra, and definitely check your final answers when you are done.
     
  6. May 2, 2009 #5

    Andrew Mason

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    Re: Billiard Balls--Elastic Collision

    Ok.

    This is correct only if these are vectors.

    [tex]\vec v_0 = \vec v_1 + \vec v_2[/tex]

    Draw the vector triangle made by these velocity vectors.

    Since

    [tex]v_0^2 = v_1^2 + v_2^2[/tex]

    what kind of triangle is it?

    AM
     
  7. May 2, 2009 #6
    Re: Billiard Balls--Elastic Collision


    i think the answer you wanted me to get for the triangle question is that it's a right triangle, probably because I'm supposed to recognize the Pythagorean theorem? Didn't see that, maybe next time.

    Anyway, I slogged through the algebra and got the correct answer. I'll remember to look for that in the future though.

    Thanks!
     
  8. May 2, 2009 #7

    Andrew Mason

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    Re: Billiard Balls--Elastic Collision

    In an elastic collision between two objects of equal mass at an oblique angle, the angle between the directions after the collision is always 90 degrees. That makes it very easy to solve this problem.

    You can observe this all the time in curling and billiards (except that in billiards the cue ball spin can dramatically change how the cue ball moves after collision).

    AM
     
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