How do I find the two solutions of this system of equations?

So I have the equations

1. $$m_1 v_1 + m_2 v_2 = m_1 v_1^\prime + m_2 v_2^\prime \tag{1}$$
2. $$\frac{m_1v_1^2}{2} + \frac{m_2v_2^2}{2} = \frac{m_1{v_1^2}^\prime}{2} + \frac{m_2{v_2^2}^\prime}{2} \tag{2}$$

and need to find $v_1^\prime$ and $v_2^\prime$.

I first I solved for $v_1^\prime$ in eq. 1, which is

$$v_1^\prime = \frac{m_2v_2-m_2v_2^\prime+v_1m_1}{m_1} \tag{$1^*$}$$

Then I solved eq. 2 for $v_2^\prime$, which is

$$v_2^\prime = \sqrt{\frac{m_1v_1^2-m_1{v_1^2}^\prime+v_2^2m_2}{m_2}} \tag{$2^*$}$$

Then I plugged $v_2^\prime$ into $v_1^\prime$, to solve for $v_1^\prime$ without the $v_2^\prime$.

And I got stuck at this point:

$$v_1^\prime = \frac{m_2v_2+v_1m_1-\sqrt{m_1m_2v_1^2-m_1m_2{v_1^2}^\prime+v_2^2m_2^2}}{m_1} \tag{3}$$

Now, I know both solutions from a CAS program:

$${v_{1_1}}^\prime = v_1$$ $${v_{1_2}}^\prime = \frac{m_1v_1-m_2v_1+2m_2v_2}{m_1+m_2}$$

But, when I square both sides of the equation, I only get the first solution, so there must be a $\pm$ somewhere after squaring (that's what I think). And the $\pm$ should be infront of the square root term after squaring, but it does not give the right answer. At least I think you should square, because how would you isolate $v_1^\prime$ then?

If one were to put the $\pm$ infront of the square root after squaring, you would get:

$${v_1^2}^\prime = \frac{m_2^2v_2^2+v_1^2m_1^2\mp(m_1m_2v_1^2-m_1m_2{v_1^2}^\prime+v_2^2m_2^2)}{m_1^2} \tag{4}$$

For the first the case where the square is negative, you get the first right answer, but for the second (I did not do this manually), you get

$$v_{1_{2,3}}^\prime = \pm\sqrt{\frac{(m_1^2 v_1^2 + m_1 m_2 v_1^2 + 2 m_2^2 v_2^2)}{(m_1^2 + m_1 m_2)}}, \tag{5}$$

which would imply $m_1m_2 = m_2^2$, which can not be the case, because they are independent of each other. There would also be 3 answers, which is also wrong.

After that it should be easy to plug in the $v_1^\prime$ solution into $v_2^\prime$ from the second equation.

So, what am I doing wrong here?

Also, is it just me, or does eq. 3 look awfully similar to the quadratic equation?

Solution 1:

Assume $v_1\ne v_1'. m_1,m_2\ne 0$. From 1, $$m_1(v_1-v_1')=m_2(v_2'-v_2). \qquad (1)$$ From 2, $$m_1(v_1-v_1')(v_1+v_1')=m_2(v_2'-v_2)(v_2'+v_2). \qquad (2)$$

Hence $$v_1+v_1'=v_2+v_2'. \qquad (3)$$

Adding $m_1\cdot (3)$ to $(1)$, we get

$$2m_1v_1=(m_2+m_1)v_2'-(m_2-m_1)v_2 \qquad (4) $$

Together (3), (4), gives a general solution.

Edit. Since by the OP only $v_1', v_2'$ are unknowns the system (3)+(4) has only one solution. Plus the solution $v_1'=v_1, v_2'=v_2$, we get two solutions.