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While solving a 2D-collision problem, I obtained the following 3 equations: $$0.866\,|\vec v_{Af}|+|\vec v_{Bf}| \cos \theta_{Bf}=6$$ $$0.5\,|\vec v_{Af}|+|\vec v_{Bf}| \sin \theta_{Bf}=0$$ $$|\vec v_{Af}|^2+|\vec v_{Bf}|^2=36$$ Where $$|\vec v_{Af}|=\text{Magnitude of final velocity of object A}$$ $$|\vec v_{Bf}|=\text{Magnitude of final velocity of object B}$$ $$\theta_{Bf}=\text{Angle made by the final velocity of object B with x-axis}$$ The first two equations are obtained by applying the law of conservation of momentum. The third equation is obtained by applying conservation of kinetic energy for elastic collision. Though there are 3 equations to find 3 unknowns, I am not able to solve them. If the masses are equal, I can use the relation ($\theta_{Af}-\theta_{Bf}=90^o$). Because, $\theta_{Af}$ is given in the problem. But I am confused as to how I would proceed when I get problems with two different masses. Please give me a method to solve the three equations without using the relation ($\theta_{Af}-\theta_{Bf}=90^o$)

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  • $\begingroup$ Eliminate va or vb with the first 2 eqns then sub into eqn3 and use quadratic solver. $\endgroup$ – PhysicsDave Mar 16 '19 at 16:48
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    $\begingroup$ You should use $\cos ^{2}\left( x\right) +\sin ^{2}\left( x\right) =1$ $\endgroup$ – Eli Mar 16 '19 at 22:26
  • $\begingroup$ The solution of 3 nonlinear equations, 3 unknowns, sounds like a question for Mathematics Stack Exchange. $\endgroup$ – JohnHoltz Mar 17 '19 at 17:16
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I could give you the following clue:

From your second equation: $$Ax + By \sin \theta = 0 $$ You know: $$ \sin \theta = -\frac{Ax}{By}$$ This implies: (form a triangle with opposite side and hypotenuse given) $$\cos \theta = \frac{\sqrt{B^2y^2 - A^2x^2}}{By}.$$ You can now use this, along your equation number one, to eliminate $\cos \theta$. Your problem now turns into a 2 equations with two variables. You can now solve this graphically for example.

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  • $\begingroup$ This method works $\endgroup$ – Nikhil Kumar Mar 17 '19 at 4:51
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equation (1)

$$a_1\,x+y\,\cos(\alpha)=c_1\tag 1$$

equation (2) $$a_2\,x+y\,sin(\alpha)=0\tag 2$$

equation (3)

$$x^2+y^2=c_2$$

from equation (1) you get: $$y\,\cos(\alpha)=c_1-a_1\,x$$

from equation (2) you get:

$$y\,\sin(\alpha)=-a_2\,x$$

$\Rightarrow$

$$y^2=(c_1-a_1\,x)^2+(a_2\,x)^2$$

$\Rightarrow\quad x_{1,2}=$ equation (3)

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    $\begingroup$ This method too works. I obtained the same results while using this method $\endgroup$ – Nikhil Kumar Mar 17 '19 at 4:52

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