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Thread: Trigonometry in Newtonian Dynamics HARD

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    Trigonometry in Newtonian Dynamics HARD

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

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

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

    I have to show that these 3 equations imply:
    $\displaystyle sin\sp{2}(a+b)=sin\sp{2}(b)+ksin\sp{2}(a)$, where $\displaystyle k=\frac{m\sb{1}}{m\sb{2}}$
    Last edited by TweedyTL; Aug 26th 2009 at 11:53 AM.
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    Quote Originally Posted by TweedyTL View Post
    A particle of mass $\displaystyle m\sb{1}$ is travelling along the x-axis with velocity $\displaystyle (v\sb{1},0,0)$. It collides elastically with a stationary particle of mass $\displaystyle m\sb{2}$. After the collision, the particle of mass $\displaystyle m\sb{1}$ is travelling with speed $\displaystyle v\sb{1}'$ at angle $\displaystyle a$ to the x-axis, and the particle of mass $\displaystyle m\sb{2}$ is travelling with speed $\displaystyle v\sb{2}'$ at angle $\displaystyle b$ to the x-axis.

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

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

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

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