# Finding 3 unknowns in 2D-collision problem [closed]

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 ($$\theta_{Af}-\theta_{Bf}=90^o$$)

## closed as off-topic by StephenG, Ruslan, Jon Custer, GiorgioP, Kyle KanosMar 17 at 11:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – StephenG, Jon Custer, GiorgioP, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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.

• This method works – Nikhil Kumar Mar 17 at 4:51

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)

• This method too works. I obtained the same results while using this method – Nikhil Kumar Mar 17 at 4:52