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Elastic Angular Momentum problem

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A billiard ball strikes an identical billiard ball initially at rest and is deflected 45 degrees from its original position. Show that if the collision is elastic, the other ball must move at 90 degrees to the first and with the same speed.


    2. Relevant equations

    Momentum:
    mv = m1v1 + m2v2

    Kinetic energy = 1/2m1v12 =1/2m1v12 + m2v22


    3. The attempt at a solution

    I think I've come up with three equations so far in breaking the problem up but I have no idea if I'm the right track or not. Any input is much appreciated.

    1. Pfx=Pix
    vxcos 45° + m2vxcos θ = v0

    2. Pfy=Piy

    vysin 45° - vysin θ = 0
    vysin 45°=vysin θ

    3. 1/2v02 = v12+v22
     
  2. jcsd
  3. Nov 5, 2012 #2

    Simon Bridge

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    Some observations:
    If ##v_x## referrs to the x-component of velocity ##v## then why have you got it paired with trig functions?

    If the balls are identical, then there is no need to index mass like that ##m_2## and that mass does not belong in the relation anyway.

    If the initial momentum is ##m\vec{v}=mv\hat{\imath}## ...
    And the final momentum is ##m\vec{v}_1+m\vec{v}_2## then you want the components in terms of ##v_1## and ##v_2##.

    I also think you should be doing angles in radiens (get used to it).
    Note: ##\sin(\pi/4)=\cos(\pi/4)=1/\sqrt{2}##
     
  4. Nov 5, 2012 #3
    Thank you for your input, I'll have to back and reevaluate the equations I found. I set it up the way I did with the hope of eliminating and plugging unknowns into equations. I honestly am confused though on to best go about solving this problem...=/
     
  5. Nov 6, 2012 #4

    Simon Bridge

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    All conservation problems are worked pretty much the same way - you write a heading "before" and sketch the situation. Then write the equations that describe what you drew.
    Do the same for "after". Then state the conservation laws and apply them.

    In this case:
    before (one mass incoming along the x axis)
    ##p_x=mv##
    ##p_y=0##
    ##K=\frac{1}{2}mv^2##

    after (the incoming mass is deflected by θ1=π/4 from the x axis.[*])
    ##p_x=v_1\cos(\theta_1)+v_2\cos(\theta_2)##
    ##p_y=\cdots##
    ##K=\cdots##

    conservation of momentum[\b]
    ##p_{x,before} = p_{x,after}##
    ##p_{y,before}\cdots##

    conservation of kinetic energy (elastic collision)
    ##K_{before}\cdots##

    ------------------------

    [*] the trig depends on the geometry.
    Note: if the first (incoming) ball is deflected by π/4, and the other ball makes an angle of π/2 to the first - then what angle does it make to the x-axis?
     
  6. Nov 6, 2012 #5

    haruspex

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    Do you mean
    v1cos 45° + v2cos θ = v0
    where v1 and v2 are the speeds of the balls after collision?
     
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