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You have two balls of equal masses: Ball 1 and Ball 2.

You are given the initial vector velocities (i.e velocity represented as a vector, with x and y components) as well as the initial x and y positions of the two balls.

How would you find the final x and y velocities of the two balls?

Basically, for my program I need an angle-free representation, meaning the equation doesn't require the angles of the balls. Here are two equations that I found (sorry, i'm not allowed to post images!):

https://wikimedia.org/api/rest_v1/media/math/render/svg/14d5feb68844edae9e31c9cb4a2197ee922e409c

https://wikimedia.org/api/rest_v1/media/math/render/svg/f5b3b7479d07075c0c5ce3f30db29f25c8a0449c

The first equation does give you the final x-velocity and y-velocity, but it isn't angle-free. The second equation is angle-free, but doesn't give the x and y components of the velocity.

Is there any equation that could achieve this? (Being angle free and resulting in a vector)

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  • $\begingroup$ The first image definitely is in terms of the final velocity with zero mention of any angles and is in fact the one you want (it arrives from conservation of momentum and kinetic energy). The first image also does indeed result in two vectors $\endgroup$ – Triatticus Apr 12 '17 at 7:19
  • $\begingroup$ Triatticus, you might want to review the two pictures again. You're welcome! $\endgroup$ – Rapid Readers Apr 17 '17 at 5:54
  • $\begingroup$ I have, and the first one is entirely independent of angle and is completely in terms of components, I don't know what more you want. The first set of equations is literally what you asked for. $\endgroup$ – Triatticus Apr 17 '17 at 7:46
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I will explain why your first picture is exactly what you need for your program, it takes a very slight bit of algebra and definitions to write it out, for instance, given two vectors $\vec{a} = (a_1,a_2,a_3,\dots, a_n) $ and $\vec{b} = (b_1,b_2,b_3,\dots,b_n)$, then: $$\langle \vec{a}, \vec{b} \rangle \equiv \sum_i a_i b_i = a_1b_1 + a_2b_2 + \dots a_nb_n $$ When the field of scalars used is $\mathbb{R}$ this product is bilinear in its arguments, that is $$\langle a + c, b \rangle = \langle a, b \rangle + \langle c, b \rangle$$ and $$\langle a , b + c \rangle = \langle a, b \rangle + \langle a, c \rangle$$ hence $$\langle a + c, b+d \rangle = \langle a, b \rangle + \langle a, d \rangle + \langle c, b \rangle + \langle c, d \rangle$$ Additionally $$||\vec{a}|| \equiv \sqrt{\vec{a} \cdot \vec{a}} \rightarrow ||\vec{a}||^2 = \vec{a} \cdot \vec{a} = \langle \vec{a}, \vec{a} \rangle $$ The first picture contains such objects, lets start with the first one ($M=m_1 + m_2$): \begin{eqnarray*} \vec{v}_1' &=& \vec{v}_1 - \frac{2m_2}{M}\frac{\langle\vec{v}_1 - \vec{v}_2, \vec{x}_1 - \vec{x}_2 \rangle}{||\vec{x}_2 - \vec{x}_1 ||^2}(\vec{x}_2 - \vec{x}_1) = \vec{v}_1 - \frac{2m_2}{M}\frac{\langle\vec{v}_1 - \vec{v}_2, \vec{x}_1 - \vec{x}_2 \rangle}{\langle \vec{x}_2 - \vec{x}_1,\vec{x}_2 - \vec{x}_1 \rangle}(\vec{x}_2 - \vec{x}_1) \\ &=& \vec{v}_1 - \frac{2m_2}{M}\frac{(\vec{v}_1\cdot\vec{x}_1 - \vec{v}_1 \cdot \vec{x}_2 - \vec{v}_2 \cdot \vec{x}_1 + \vec{v}_2\cdot \vec{x}_2)}{(\vec{x}_2\cdot\vec{x}_2 - \vec{x}_2 \cdot \vec{x}_1 - \vec{x}_1 \cdot \vec{x}_2 + \vec{x}_1\cdot\vec{x}_1)}(\vec{x}_2 - \vec{x}_1) \\ &=& \vec{v}_1 - \frac{2m_2}{M}\frac{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}(\vec{x}_2 - \vec{x}_1) \end{eqnarray*} Those large pieces are all in terms of the given initial components of the velocities and positions where I have used subscripts to denote the first and second vectors x and y components. This single equation is also two equations in one for the x and y components of $\vec{v}_1'$. $$v_{1x}' = v_{1x} - \frac{2m_2}{M}\frac{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}(x_{2x} - x_{1x}) $$ And $$v_{1y}' = v_{1y} - \frac{2m_2}{M}\frac{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}{(v_{1x}x_{1x} + v_{1y}x_{1y} - v_{1x}x_{2x} - v_{1y}x_{2y} - v_{2x}x_{1x} - v_{2y}x_{1y} + v_{2x}x_{2x} + v_{2y}x_{y})}(x_{2y} - x_{1y}) $$ As can be seen this is entirely without any mention at all of the angle of impact, and is solely in terms of the initial components. A similar argument can be applied to the second equation to obtain the four equations for the components of the final velocity.

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  • $\begingroup$ Yep, this was very helpful for my program (although a tintsy bit too complicated and annoying to write!). Thank you very much, Triatticus. Sorry for undermining you in the reply! $\endgroup$ – Rapid Readers Apr 20 '17 at 5:41
  • $\begingroup$ It is indeed a very cumbersome thing to have to write, no problem $\endgroup$ – Triatticus Apr 20 '17 at 15:11
  • $\begingroup$ Is there any difference between the numerator and denominator of the x and y component equations? They look equal to me, which would mean that they can be simplified to one, right? $\endgroup$ – Cullub Oct 4 '18 at 21:12
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I am a intermediate student and have seen such questions . I believe you will not be getting an "ANGLE FREE x-y component separated" dish on your platter.

I believe you should notice that $$\vec {v'_1}=\vec v_1 - \frac{2m_2}{m_1+m_2} \frac{<\vec v_1 - \vec v_2, \vec x_1-\vec x_2>}{||\vec x_1-\vec x_2||}(\vec x_1-\vec x_2)$$ and $$\vec {v'_2}=\vec v_2 - \frac{2m_1}{m_1+m_2} \frac{<\vec v_2 - \vec v_1, \vec x_2-\vec x_1>}{||\vec x_2-\vec x_1||}(\vec x_2-\vec x_1)$$ are your best bet. U must be having vector components of these quantities ($\vec v_1 ,\vec v_2$ etc.) And using the i,j,k orthogonal components must do your job.

Suggestions are always welcome! Cheers

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