Using "symmetry" to deduce final velocities of two colliding particles

[brotherbobby]

TL;DR

A textbook says this : "... if a particle of mass travels at a velocity and undergoes an elastic collision with another initially stationary particle of mass , the expression for the final velocities of the particles should not be symmetric."

I am struggling to understand this. Additionally, the statement above makes just as little sense when I carry out the mathematics. I write below the details, realising that space is brief here.

Text excerpt : For context, I copy and paste the paragraph on "symmetry" from the text, underlining the relevant lines in red. I hope it's readable.

Problem situation : This concerns the last three lines from the text excerpt. They say that :

"... if a particle of mass $m_1$ travels at a velocity $u$ and undergoes an elastic collision with another initially stationary particle of mass $m_2$, the expression for the final velocities of the particles should not be symmetric."

Doubt(s) : I have several.

1. What does it even mean for the final velocities to be symmetric ? Does it mean that if we replace $m_1\rightarrow m_2$ in the expression for the velocity for $m_1$, we will obtain the expression for the velocity for $m_2$?
2. Why should the expression(s) for the final velocities be not symmetric?

Attempt : I attempt the problem in two parts, tackling the symmetric case first.

(The symmetry argument) : The particle of mass $m_1$ moving with velocity $u$ to the right relative to particle of mass $m_2$ is the same as the latter moving with velocity $-u$ to the left relative to the first. But the second particle has a different mass $m_2$. The two situations are not symmetric before collision, hence the final velocities will also not be following collision.


(The mathematics) : From the picture drawn for the problem, momentum conservation yields $\small{m_1 u=m_1v_1+m_2v_2\Rightarrow v_2=\dfrac{m_1}{m_2}(u-v_1)\quad\color{blue}{\dots 1}}$. From energy conservation : $\small{m_1 u^2=m_1v_1^2+m_2v_2^2=m_1v_1^2+\dfrac{m_1^2}{m_2^2}(u-v_1)^2\Rightarrow u^2=v_1^2+\dfrac{m_1}{m_2}(u-v_1)^2\Rightarrow \left(1+\dfrac{m_1}{m_2}\right)v_1^2-\dfrac{2m_1u}{m_2}v_1-u^2\left(1-\dfrac{m_1}{m_2}\right)=0}$, which can be solved to give :
$\small{v_1=\dfrac{\dfrac{2m_1u}{m_2}\pm\sqrt{\dfrac{4m_1^2u^2}{m_2^2}+4u^2\left( 1- \dfrac{m_1^2}{m_2^2}u^2\right)}}{2\left(1+\dfrac{m_1}{m_2}\right)}=\dfrac{\dfrac{m_1}{m_2}u\pm u}{1+\dfrac{m_1}{m_2}}=u\;,\; \dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}u}$.

So the final velocity of the first particle can be $\boxed{(v_1=)\,u}$. Using $\small{\color{blue}{1}, \boxed{v_2=0}}$, the final velocity of the second particle. If I am correct, then indeed no symmetry exists in the final velocities.

Or if the final velocity of the first particle $\small{\boxed{(v_1=)\,\dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}u}}\Rightarrow \boxed{v_2}=\dfrac{m_1}{m_2}u\left(1-\dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}\right)=\dfrac{m_1}{m_2}u\dfrac{2}{1+\dfrac{m_1}{m_2}}=\boxed{\dfrac{m_1}{m_2}\dfrac{2u}{1+\dfrac{m_1}{m_2}}=v_2}$, which is the final velocity of the second particle, again showing no symmetry exists.

Request : Am I correct in my arguments above?

 

[A.T.]

A textbook says this : "... the expression for the final velocities of the particles should not be symmetric."
... the statement above makes just as little sense when I carry out the mathematics.
[mathematics]
... indeed no symmetry exists in the final velocities.
... again showing no symmetry exists.

I don't understand. First you say the book statement makes little sense according to your math. Then you show the math and conclude from it the same as the book statement.

 

[robphy]

.... which can be solved to give :
$\small{v_1=\dfrac{\dfrac{2m_1u}{m_2}\pm\sqrt{\dfrac{4m_1^2u^2}{m_2^2}+4u^2\left( 1- \dfrac{m_1^2}{m_2^2}u^2\right)}}{2\left(1+\dfrac{m_1}{m_2}\right)}=\dfrac{\dfrac{m_1}{m_2}u\pm u}{1+\dfrac{m_1}{m_2}}=u\;,\; \dfrac{\dfrac{m_1}{m_2}-1}{\dfrac{m_1}{m_2}+1}u}$.

You can avoid the quadratic in the general case by writing the conservation equations is gain-loss form:
$\Delta K_1 =-\Delta K_2$
$\Delta p_1 =-\Delta p_2$

Since $\Delta K_1$ is a difference of squares,
$\Delta K_1/\Delta p_1$ is relatively simple to work with.
($\Delta K_1/\Delta p_1 = \Delta K_2/\Delta p_2$ is a linear equation that doesn't depend on the masses.
Together with $\Delta p_1 =-\Delta p_2$, you now have a system of linear equations.)

 

[brotherbobby]

I don't understand. First you say the book statement makes little sense according to your math. Then you show the math and conclude from it the same as the book statement.

I don't understand what symmetry really means. I am going by what I suspect.

 

[haruspex]

I don't understand what symmetry really means. I am going by what I suspect.

The Atwood example makes it clear what is intended. In that case, we should not expect the value of interest - the tension - to depend on which mass is $m_1$ and which is $m_2$. Having obtained an expression for the tension, you can do a sanity check by seeing that swapping the masses in the expression produces one that is algebraically the same.
On the other hand, if you are asked for the final velocity of a particular ball in the collision example (the one originally at rest, say), swapping the masses will not conserve the result.

 

[A.T.]

The Atwood example makes it clear what is intended.

It's not a very clear, because in the Atwood example they compute a single quantity $x$ that is not associated with either mass. So symmetry of the expression $E_{ij}$ means:

$E_{12} = x = E_{21}$

But then they talk about computing two quantities $x_1$ and $x_2$ that are associated with each mass. And in this case symmetry can mean two different things:

a) Symmetry of the expression (just like in the Atwood example):

$E_{12} = x_1= x_2 = E_{21}$

b) Symmetry of the equations (what @brotherbobby mentions):

$E_{12} = x_1\neq x_2 = E_{21}$

 

[haruspex]

It's not a very clear, because in the Atwood example they compute a single quantity $x$ that is not associated with either mass. So symmetry of the expression $E_{ij}$ means:

$E_{12} = x = E_{21}$

But then they talk about computing two quantities $x_1$ and $x_2$ that are associated with each mass. And in this case symmetry can mean two different things:

a) Symmetry of the expression (just like in the Atwood example):

$E_{12} = x_1= x_2 = E_{21}$

b) Symmetry of the equations (what @brotherbobby mentions):

$x_1 = E_{12}$
$x_2 = E_{21}$
but possibly:
$x_1 \neq x_2$

I agree the collision example is ugly.
First, it needs to be specified, symmetry wrt what? Let’s say it is the interchange of $m_1$ and $m_2$. And exactly what is the result to be considered? Is it the ordered pair of final velocities or the unordered pair?

I think it is hard to come up with an asymmetric case that isn’t fairly trivially so. The useful part of the text is the tension case, illustrating verifying that an answer that should be symmetric is.