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Finding Angles after Perfectly Elastic Collisions

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data

    A 0.045kg Steel marble collides obliquely with an identical stationary marble, and continues at 55 degrees to its original direction. The collision is perfectly elastic. What is the angle between the direction taken of the target (previously static) ball and the original direction of the incident ball?

    So:

    m1: 0.045 kg
    v1: >0m/s
    p1: >0m/s

    m2: 0.045 kg
    v2: 0 m/s
    p2: 0 m/s

    p= m1v1
    p'=m1v1 + m2v2
    p = p'


    θ1': 55 degrees above or below previous.
    θ2': ?


    2. Relevant equations

    p = mv

    p = m1v1 + m2v2

    p' = m1v1' + m2v2'

    p = p'

    Sine Law

    Cosine Law


    3. The attempt at a solution

    Manipulating several formulas to find a result that required so few variables.

    No real luck there.

    Attempted to build a system where I added some variables, like v1' but that was unfruitful, given that the incident was feigned anyway.

    Any suggestions are certainly appreciated.
     
    Last edited: Jun 20, 2011
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  3. Jun 20, 2011 #2

    Hootenanny

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    There are two quantities that are conserved in an elastic collision: [kinetic] energy & momentum. Using this, can you write down two equations representing the two conservation laws.
     
  4. Jun 20, 2011 #3
    My instructor has neglected to teach me about energy laws yet (though it is in the next unit). Suggestions on sources to study?
     
  5. Jun 20, 2011 #4

    Hootenanny

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    Okay, perhaps a different tact is required. What do you know about the speed of the two marbles after the collision?

    A nice little primer on elastic collisions can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html
     
  6. Jun 20, 2011 #5
    I think I know very little. I could make some logical assumptions but I feel that that's not what you're trying to push me towards.

    Am I overlooking something obvious?
     
  7. Jun 20, 2011 #6

    Hootenanny

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    Okay, so we know that both energy and momentum are conserved. Now, I know you don't know a great deal about energy conservation, but I'm sure you know the basics. So, before the collision the total kinetic energy of the system (both masses) is [itex]T = \frac{1}{2}m{v^{(1)}}^2[/itex]. Without loss of generality, we can assume that the marble is initially moving in the positive x direction. Therefore, the initial momentum in the x-direction is [itex]p_x = mv^{(1)}[/itex]. Likewise, in the y-direction [itex]p_y = 0[/itex].

    Do you follow so far? Now, what can we say about the total kinetic energy and momenta immediately after the collision?
     
  8. Jun 20, 2011 #7
    I believe I'm with you.

    So, after the collision,

    [itex]p_x' = mv^{(x1')} + mv^{(x2')}[/itex]
    or
    [itex]p_x' = mv^{(1')}cos55 + mv^{(2')}cosθ[/itex]

    and

    [itex]p_y' = 0 = mv^{(y1')} + mv^{(y2')}[/itex]
    or
    [itex]p_y' = 0 = mv^{(1')}sin55 + mv^{(2')}sinθ[/itex]
    or
    [itex] -mv^{(1')}sin55 = mv^{(2')}sinθ[/itex]

    and

    [itex]T' = \frac{1}{2}m{v^{(1')}}^2 + \frac{1}{2}m{v^{(2')}}^2[/itex]

    A lot of these masses can be cancelled out.
    What's the next step with the kinetic energy equation?
     
    Last edited: Jun 20, 2011
  9. Jun 20, 2011 #8

    Hootenanny

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    With the kinetic energy, we have
    [tex]\frac{1}{2}m{v^{(1)}}^2 = \frac{1}{2}m{v^{(1')}}^2 + \frac{1}{2}m{v^{(2')}}^2\;,[/tex]
    or
    [tex]{v^{(1)}}^2 = {v^{(1')}}^2 + {v^{(2')}}^2\;.[/tex]
    Moreover, you can rewrite your equation for the x-component of momentum as
    [tex]v^{(1)} = v^{(1')}cos55 + v^{(2')}cosθ\;.[/tex]
    Can you use these two equations, together with your equation of the y-component of the momentum to find [itex]\theta[/itex]?
     
  10. Jun 20, 2011 #9
    I can rearrange the variables well, but I only find seem to end up with more complex derivatives of the original.

    Nothing seems to cancel. I fear I may be making rudimentary mistakes, given the hour and my lack of sleep.

    I don't want to give up now, and I'm not asking for the answer yet, but do you have any more suggestions?
     
  11. Jun 20, 2011 #10

    Hootenanny

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    Since we are only interested in the angles, you can also assume that [itex]v^{(1)}[/itex] is known, i.e. just set [itex]v^{(1)}=1[/itex].
     
  12. Jun 20, 2011 #11
    [itex]θ = cos^{-1}(({v^{(1)}-v^{(1')}cos55})/{v^{(2')}})[/itex]

    Is this right?
    I still can't find that cancel; I get extremely close but then exponent rules get in the way.

    My gut is telling me it's 55 degrees (which would be grand).
     
  13. Jun 20, 2011 #12
    i am starting from the first principle, although u must have learned a lot from the previous helping hands.
    Consider obvious points for an elastic collision:
    1) momentum is conserved (i.e, in both X and Y axes).
    Now, imagine the first ball was moving along negative X-axis toward origin, where it collided with the 2nd ball (stationary). Here only X-axis momentum was present before collision, so after collision the Y-axis components of momentum must cancel each other. After collision the first ball moves in an angle 55 deg, let us suppose, with positive X-axis, that is, , in +X/+Y quadrant of the co-ordinate system. It implies that the initially stationary ball would move in the +X/-Y plane because otherwise the Y-axis momentum would not cancel each other after collision and that cancellation is an obvious condition for conservation of momentum in Y-axis. This conclusion we can reach by conservation of momentum concept.

    Just draw the fig. and imagine the collision.

    2) kinetic energy must conserve. It implies,
    ½ mu2 = ½ mv12 + ½ mv22.
    Or, u square = v1 square + v2 square.

    It follows that v1 and v2 must be perpendicular to each other, if neither is zero. (imagine a right angled triangle obeying Pythagorian theorem having smaller sides v1 and v2 and with hypotenuse u. in this case v1 and v2 are mutually perpendicular).
    For head on collisions (not the case, we are presently dealing with) v1 becomes zero, so v2 = u. this is a special case, where the first ball stops after collision and the stationary ball moves with the velocity of the first ball possessed before collision.

    Now, if the 2nd ball makes an angle θ with positive X-axis in the +x/-Y plane, it follows that 55+ θ = 90 . the answer follows. this is a physical explanation of the proceedings, but you can find mathematically the same result also (i,e, 55+ θ = 90 and which u are trying, i have noticed, in the previous posts. it needs a bit of mathematical processes) with the three eqn.s of X-axis and Y-axis monentum conservation and K.E. conservation before and after collision.
    this is an interesting problem where we can conclude that when two equal masses collide elastically, one being stationary, and the collision is not head on, the angle between the velocities after collision will always be 90 deg, it is independent of the values of the individual masses (as long as they are equal) and the angle of deflection of the first mass.
     
    Last edited: Jun 20, 2011
  14. Jun 21, 2011 #13

    Hootenanny

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    You're on the right lines. Another good resource is here: http://www.scar.utoronto.ca/~pat/fun/NEWT3D/PDF/MOMENT3D.PDF [Broken]
     
    Last edited by a moderator: May 5, 2017
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