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Maximum angle for 2D collision

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass m1, with initial velocity u, collides elastically with mass m2, which is initially at rest. After collision, m1 deflects by angle θ. Find the maximum value of θ. The answer given is θmax=acos(sqrt(1-(m1/m2)^2)). Does this mean that the maximum angle cannot exist if m1>m2?

    2. Relevant equations
    Let m2 deflect by angle α from the initial direction of m1, and v1, v2 be the velocity of m1, m2 respectively after collision. Then
    1.(m1)(u)= (m1)(v1)(cosθ)+(m2)(v2)(cosα)
    2. (m1)(v1)(sinθ)=(m2)(v2)(sinα)
    3. (m1)(u)^2=(m1)(v1)^2 + (m2)(v2)^2
    3. The attempt at a solution
    I got θmax =acos(sqrt(1-(m2/m1)^2)) instead.
    By 1. and 2.
    [(m1)(u)-(m1)(v1)(cosθ)]^2 +[(m1)(v1)(sinθ)]^2 = [(m2)(v2)(cosα)]^2 +[(m1)(v1)(sinα)]^2
    Hence
    (m1u)^2 +(m1v1)^2 - 2(m1)^2(v1u cosθ)^2 = (m2v2)^2
    By 3.
    (m2v2)^2 = (m2m1)(u^2-v1^2)
    Thus
    m1(u^2) +m1(v1^2) - 2m1(v1u cosθ)^2 = (m2)(u^2-v1^2)
    Dividing by v1^2 , we get:
    (m1-m2)(u/v1)^2 -(2m1cosθ)(u/v1)+(m1+m2)=0
    Hence the equation (m1-m2)x^2-(2m1cosθ)x+(m1+m2)=0 has a solution. Since it has a solution, we must have b^2-4ac≥0. Thus:
    (2m1cosθ)^2 ≥4(m1-m2)(m2+m1)
    And hence
    cosθ≥sqrt(1-(m2/m1)^2). Since the maximum θ of means the minimum of cosθ, θmax =acos(sqrt(1-(m2/m1)^2)).
    What went wrong?
     
  2. jcsd
  3. Aug 3, 2015 #2
    I picture would help a lot, as well as distinguishing between vectors and scalars in your notation.
     
  4. Aug 3, 2015 #3

    jbriggs444

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    A moment's thought shows that the given answer cannot be right. By inspection, if m1 < m2 then any angle at all up to 180 degrees is attainable. Like bouncing a ping pong ball from a basketball.

    In addition, both the given solution and your solution could be dramatically simplified. Start by taking the cosine of both sides.

    cos θmax = sqrt(1-(m1/m2)2)

    Square both sides

    cos2 θmax = 1 - (m1/m2)2

    Apply a simple trig identity

    sin2 θmax = m1/m22

    Take the square root

    sin θmax = m1/m2
     
  5. Aug 3, 2015 #4

    haruspex

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    I suspect that the original problem either stated or assumed m1>m2. It is then clear that m1 and m2 are swapped in the given answer.
    Luca, I agree with your answer for the case where m1>m2, but you may still be puzzled by the result it gives when m1<=m2. Consider the inequality you had just before "And hence". What do you get with m1<m2 in there?
     
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