(Note: this question has elements of physics in it but I am looking for a solution to the purely mathematical side of it.)

In trying to work out the resultant velocity and angle of an elastic collision, I have come across a pair of trigonometric simultaneous equations.

$$ m_1\cdot v_1 + m_2\cdot v_2 \cos \theta_i = (m_1 + m_2) v_f \cdot \cos \theta_f $$ $$ m_2\cdot v_2 \sin \theta_i = (m_1 + m_2) v_f \cdot \sin \theta_f $$

In these, there are two colliding bodies, 1 and 2, of which each has a mass $m_n$ and a velocity $v_n$. $\theta_i$ is the angle of incidence, $\theta_f$ the final angle, and $v_f$ the final total velocity.

Now, I can solve this partially: I can rearrange an equation for $v_f$, substitute it into the other, and come out with another equation:

$$ v_f = \frac{m_1\cdot m_2 + m_2\cdot v_2 \cos \theta_i}{(m_1 + m_2) \cos \theta_f} $$ $$ \theta_f = \sin^{-1}\left(\frac{m_2\cdot v_2 \sin \theta_i}{(m_1 + m_2) \frac{m_1\cdot m_2 + m_2\cdot v_2 \cos \theta_i}{(m_1 + m_2) \cos \theta_f}}\right) $$

Convoluted. Normally, this would not be too much of a problem, but I can't for the life of me figure out how to rearrange this so that

  • All $\theta_f$ terms are on the left; and
  • All $\theta_f$ terms combine to one term - i.e. no extra trigonometrical functions

How is it done?


If you multiply the first equation by $\sin \theta_i$ and the second equation by $\cos \theta_i$ and then subtract them, the terms on the left with $\theta_i$ will cancel out leaving a term with $\sin \theta_i$ and the terms on the right will have $\cos \theta_f \sin \theta_i -\sin \theta_f \cos \theta_i =\sin(\theta_i-\theta_f) $.

You then have $a \sin \theta_i =b \sin(\theta_i-\theta_f) $ for certain $a$ and $b$.

With this, you can get $\theta_f$ in terms of $\theta_i$ and the other parameters.

  • $\begingroup$ What does that multiplication get you? I can do the basic multiplication but I don't see where you got your terms from. $\endgroup$ – ArtOfCode Jun 1 '15 at 22:31
  • $\begingroup$ Interestingly, if one squares both of the above equations listed in the question and then adding, one can easily solve for $\theta_i$. $\endgroup$ – T.A.Tarbox Mar 15 '17 at 23:35
  • $\begingroup$ Make that an answer. $\endgroup$ – marty cohen Mar 16 '17 at 0:10

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