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I have been trying to solve a problem involving conservation of momentum, in the end I end up with 3 equations which I feel cannot be solved. I would really appreciate if someone could show me how to manipulate these as nothing I seem to do really helps.

  • $v_x^2 + v_y^2 = 16$
  • $v_x\sin{30°} = v_y \sin{\theta}$
  • $v_x \cos{30°} + v_y \cos{\theta} = 4$

The base of the problem is that $v_x$ and $v_y$ are the speeds of two objects. One object moving at angle $\theta$ the other at ${30°}$.

I feel like it may not be solvable due to the fact the first equation does not have a reference to $\theta$, but everything up to this point is correct.

Working through using $v_x = {\sqrt 16-v_y^2}$

Gets to the equation $32-2v_y^2+8v_y({\sqrt 1-\frac{16-v_y^2}{4v_y^2}})=0$

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  • $\begingroup$ ok edited question to use velocity symbols to facilitate understanding $\endgroup$
    – Nikos M.
    Sep 9 '15 at 16:40
  • $\begingroup$ one approach is to use 1st eq and express $v_x$ wrt $v_y$ and then use that in the rest equations to solve for $v_x$, $\theta$ but it becomes non-linear system, except if use knowledge of the physical systems (and its symmetries) to simplify problem (at least in some range) $\endgroup$
    – Nikos M.
    Sep 9 '15 at 16:44
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Hint :

I rewrite the equation system

$$x^2+y^2=16$$

$$xsin(30°)=ysin(u)$$

$$xcos(30°)+ycos(u)=4$$

This gives $$(ysin(u))^2+(ycos(u))^2=y^2=\frac{x^2}{4} +(4-xcos(30°))^2$$

Insert this term in the equation $x^2+y^2=16$

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  • $\begingroup$ sorry but i dont see how you would then insert that into the first equation $\endgroup$
    – Cheezy
    Sep 9 '15 at 16:54
  • $\begingroup$ The term for $y^2$ (only depending on $x$) can be inserted in $x^2+y^2=16$. $\endgroup$
    – Peter
    Sep 9 '15 at 16:55
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If the angles are expressed in degrees we have $$\sin 30^{\circ}=\frac{1}{2}\qquad\text{and}\qquad\cos 30^{\circ}=\frac{\sqrt 3}{2}$$ Then in the second equation we have $$v_x=2v_y\sin \theta$$ So, plugging this into the third equation we get $$(2v_y\sin \theta)\frac{\sqrt 3}{2}+v_y\cos\theta=4$$ Then$$v_y=\frac{4}{\sqrt{3}\sin \theta+\cos\theta}\qquad\text{and}\qquad v_x=\frac{8\sin \theta}{\sqrt{3}\sin\theta+\cos\theta}$$ So, first equation reduces to $$\frac{1+4\sin^2\theta}{(\sqrt{3}\sin \theta+\cos \theta)^2}=1$$ Then, by dividing by $\cos^2 \theta$ both terms in the quotient \begin{align} \frac{\sec^2\theta+4\tan^2\theta}{(\sqrt{3}\tan \theta+1)^2}&=1\\[5pt] \frac{5\tan^2 \theta+1}{(\sqrt{3}\tan \theta+1)^2}&=1\\[5pt] 5\tan^2\theta+1&=3\tan^2\theta+2\sqrt{3}\tan \theta+1\\[5pt] 2\tan^2\theta-2\sqrt{3}\tan\theta&=0 \end{align} Then, $\tan \theta =0$ or $\tan\theta=\sqrt{3}$.

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