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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.

Fat32
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  • seems dimensionally incorrect – user619894 Dec 15 '22 at 20:07
  • it's ok now, thanks – user619894 Dec 15 '22 at 20:15
  • With the help of MATHEMATICA Solve[{v1 Cos[t] + r v2 Cos[a] == v0, v1 Sin[t] - r v2 Sin[a] == 0, v1^2 + r v2^2 == v0^2}, {v1, v2, a}] /. {C[1] -> 0} // FullSimplify – Cesareo Dec 15 '22 at 20:37
  • @Cesareo Hi, thank you! But I was asking for a solution by hand, I've already used MATLAB to get a symbolic answer but I wonder (want to learn) why my own solution procedure yields a wrong solution? – Fat32 Dec 15 '22 at 20:43

1 Answers1

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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 $\alpha$

Cesareo
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  • Thank you, that's very nice! This shows not only how to solve it, but also how to do it the better way than I was trying. Indeed I've already figured out my mistake too: I have, shamefully, regarded the equation $a V_1^2 + bV_2^2 = c V_1^2 + d V_2^2$ to imply $a=c$, and $b=d$ to reach my constraint on $\alpha$ in Eq.7. What a shame ! :-) Instead, in the corrected form, from Eqs 1 & 2, I define $V_1$ and $V_2$ in terms of $\alpha$, and then plug them into Eq.3 to get the correct relation to solve for $\alpha$, then insert $\alpha$ into Eqs. 1 & 2 for a linear system solution of $V_1$ and $V_2$. – Fat32 Dec 15 '22 at 23:44