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 $32-2v_y^2+8v_y({\sqrt 1-\frac{16-v_y^2}{4v_y^2}})=0$
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 $\tan\theta=\sqrt{3}$.