I have obtained a set of three nonlinear algebraic equations resulting from conservation laws for momentum and kinetic energy. $$ \begin{align} m_1 V_1 \cos(\theta) + m_2 V_2 \cos(\alpha) & = m_1 V_0 \tag{1}\\ m_1 V_1 \sin(\theta) - m_2 V_2 \sin(\alpha) & = 0 \tag{2} \\ m_1 V_1^2 + m_2 V_2^2 &= m_1 V_0^2 \tag{3} \end{align} $$
with the substituion $r = m_2/m_1$, they're simplified into
$$ \begin{align} V_1 \cos(\theta) + r V_2 \cos(\alpha) & = V_0 \tag{4} \\ V_1 \sin(\theta) - r V_2 \sin(\alpha) & = 0 \tag{5} \\ V_1^2 + r V_2^2 &= V_0^2 \tag{6} \\ \end{align} $$
$m_1, m_2$ are the colliding masses, $V_0$ is the velocity of $m_1$ before the collision, and whereas $\theta , \alpha$ are the scattering angles and $V_1,V_2$ are the speeds of $m_1,m_2$ after collision respectively. The unknowns are $V_1$, $V_2$, and $\alpha$ (displayed as $\cos(\alpha)$), all remaning parameters are given.
My attempt to get an analytic solution is as follows:
- square Eqs. 4 & 5
- then multiply squared Eq.4 with $\sin(\theta) \sin(\alpha)$, and squared Eq.5 with $\cos(\theta) \cos(\alpha)$
- add resulting equations to get rid of cross multiplied terms.
- Finally replace $V_0^2$ at the resulting RHS, with LHS of the Eq.6
This way, I get an equation (or equality?) between $V_1^2$ and $V_2^2$, after which I do equate their coefficients to get a constraint for th unknown angle $\alpha$ :
$$ \cos(\alpha) \sin(\alpha) = \cos(\theta) \sin(\theta) / r \tag{7} $$
Eq.7 apparently eliminates the unknown angle $\alpha$, and then I would use Eqs 4 & 5, which become two linear equations in two remaning unknown $V_1$ and $V_2$, by substituting the found value $\alpha$ from Eq.7.
After this solution procedure, the result satisfies Eqs. 4 & 5, but not Eq.6? What's wrong in this solution procedure?
I've used a symbolic package to get an alternate answer, and it seems that the angle $\alpha$ I found using Eq.7 is different from the one returned from the symbolic package...
Any help in either explaining what I have done wrong above, or how to solve this set of nonlinear equations will be appreciated.
Hint.
As
$$ \left(V_0-V_1\cos\theta\right)^2+\left(V_1\sin\theta\right)^2=r^2V_2^2 $$
or
$$ V_0^2-2V_0V_1\cos\theta+V_1^2=r^2V_2^2 $$
solving
$$ \cases{ V_0^2-2V_0V_1\cos\theta+V_1^2=r^2V_2^2\\ V_1^2+rV_2^2=V_0^2 } $$
for $V_1,V_2$ and after that, for $\alpha$