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Math Help - Trigonometry in Newtonian Dynamics HARD

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    Trigonometry in Newtonian Dynamics HARD

    A particle of mass m\sb{1} is travelling along the x-axis with velocity (v\sb{1},0,0). It collides elastically with a stationary particle of mass m\sb{2}. After the collision, the particle of mass m\sb{1} is travelling with speed v\sb{1}' at angle a to the x-axis, and the particle of mass m\sb{2} is travelling with speed v\sb{2}' at angle b to the x-axis.

    The conservation of momentum laws for this collision are:
    (i) m\sb{1}v\sb{1}'sin(a)+m\sb{2}v\sb{2}'sin(b)=0
    (ii) m\sb{1}v\sb{1}'cos(a)+m\sb{2}v\sb{2}'cos(b)=m\sb{1  }v\sb{1}

    The conservation of energy equation is:
    (iii) m\sb{1}(v\sb{1}')\sp{2}+m\sb{2}(v\sb{2}')\sp{2}=m\  sb{1}(v\sb{1})\sp{2}

    I have to show that these 3 equations imply:
    sin\sp{2}(a+b)=sin\sp{2}(b)+ksin\sp{2}(a), where k=\frac{m\sb{1}}{m\sb{2}}
    Last edited by TweedyTL; August 26th 2009 at 12:53 PM.
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    Quote Originally Posted by TweedyTL View Post
    A particle of mass m\sb{1} is travelling along the x-axis with velocity (v\sb{1},0,0). It collides elastically with a stationary particle of mass m\sb{2}. After the collision, the particle of mass m\sb{1} is travelling with speed v\sb{1}' at angle a to the x-axis, and the particle of mass m\sb{2} is travelling with speed v\sb{2}' at angle b to the x-axis.

    The conservation of momentum laws for this collision are:
    (i) m\sb{1}v\sb{1}'\sin(a)+m\sb{2}v\sb{2}'\sin(b)=0
    (ii) m\sb{1}v\sb{1}'\cos(a)+m\sb{2}v\sb{2}'\cos(b)={\co  lor{red}m_1}v\sb{1}

    The conservation of energy equation is:
    (iii) m\sb{1}(v\sb{1}')\sp{2}+m\sb{2}(v\sb{2}')\sp{2}=m\  sb{1}(v\sb{1})\sp{2}

    I have to show that these 3 equations imply:
    \sin\sp{2}(a+b)=\sin\sp{2}(b)+k\sin\sp{2}(a), where k=\frac{m\sb{1}}{m\sb{2}}
    First step is to solve equations (i) and (ii) for v_1' and v_2'. Multiply (i) by \cos a, multiply (ii) by \sin a, and subtract. That gives m_2v_2'\sin(a-b)=m_1v_1\sin a (using the trig formula \sin a\cos b-\cos a\sin b = \sin(a-b)). Therefore v_2' = \frac{kv_1\sin a}{\sin(a-b)}. A similar procedure (multiplying (i) by \cos b and (ii) by \sin b) gives v_1' = -\frac{v_1\sin b}{\sin(a-b)}.

    Now square both of those expressions, and substitute those formulas for v_1'^{\,2} and v_2'^{\,2} into (iii). You'll find that after a bit of simplification and cancellation, it reduces to \sin^2b+k\sin^2a = \sin^2(a-b). Note: \sin^2(a-b), not \sin^2(a+b) as (wrongly) stated in the question. Note also the missing m_1 in equation (ii).
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